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c Solutions Manual! to accompany System Dynamics, First Edition by William J. Palm III University of Rhode Island
Solutions to Problems in Chapter One
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1.1 W = mg = 3(32.2) = 96.6 lb. 1.2 m = W / g = 100 / 9.81 = 10.19 kg. W = 100(0.2248) = 22.48 lb. m = 10.19(0.06852) = 0.698 slug. 1.3 d = (50 + 5 / 12)(0.3048) = 15.37 m. 1.4 n = 1 / [60(1.341 × 10−3 )] = 12.43, or approximately 12 bulbs. 1.5 5(70 − 32) / 9 = 21.1◦ C 1.6
= 3000(2π ) / 60 = 314.16 rad/sec. Period P = 2π / = 60 / 3000 − 1 / 50 sec.
1.7 = 5 rad/sec. Period P = 2π / 0.796 Hz.
= 2π / 5 = 1.257 sec. Frequency f = 1 / P = 5 / 2π =
1.8
y − x = 5.66
x + y = 10.30
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1.9
Ae−1/τ 4.912 = = e2/tau 3.293 Ae−3/τ 2 4.912 = ln 3.293 2 =5 = ln(4.912 / 3.293) A = 4.912e1/τ = 6
1.10 y = [0, 7] and t = [0, 4(3)] = [0, 12] 1.11 y = [0, 6 + 12] = [0, 18]. Dominant time constant is 5. Thus t = [0, 4(5)] = [0, 20]. 1.12 10 sin 3t cos 2 + 10 cos 3t sin 2 = B sin 3t + C cos 3t
B = 10 cos 2 = −4.161
C = 10 sin 2 = 9.093
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1.13
A sin 6t cos A cos Thus
+ A cos 6t sin = −6.062 < 0
is in the third quadrant. = tan
−1
!
3.5 6.062
A=
#
"
= −6.062 sin 6t − 3.5 cos 6t
A sin
= −3.5 < 0
= 0.524 + π = 3.665 rad
(6.062)2 + (3.5)2 = 7
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1.14
y˙ = 0.5(7) cos(7t + 4) = 3.5 cos(7t + 4) y¨ = −0.5(7)2 sin(7t + 4) = −24.5 sin(7t + 4) Velocity amplitude is 3.5 m/s. Acceleration amplitude is 24.5 m/s2 .
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1.15
y (t) = A sin(3t + ) = A sin 3t cos
Thus
A sin
= 0.05 > 0
A cos
= 0.08 > 0
+ A cos 3t sin
is in the first quadrant. = tan
−1
!
0.05 0.08
"
= 0.559 rad
Thus
y (t) = 0.094 sin(3t + 0.559) The amplitude of the displacement is
A= For the velocity,
#
(0.05)2 + (0.08)2 = 0.094 ft
y˙ (t) = 3(0.094) cos(3t + 0.559) = 0.282 cos(3t + 0.559) The amplitude of the velocity is 0.282 ft/sec. For the acceleration,
y¨(t) = −3(0.282) sin(3t + 0.559) = 0.846 sin(3t + 0.559) The amplitude of the acceleration is 0.846 ft/sec2 .
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1.16 y˙ = 5(0.07) sin(5t + ). Thus y˙ (0) = 0.35 sin = sin
−1
!
0.041 0.35
"
= 0.041, and
= 0.117 or π − 0.117 = 3.02 rad
Without more information we cannot determine the quadrant of .
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1.17
y (0) = A sin y˙ (0) = 5A cos Thus
= 0.048 > 0 = 0.062 > 0
is in the first quadrant. −1
= tan
A=
$
%
0.048 = 1.318 rad (0.062 / 5)
#
(0.048)2 + (0.062 / 5)2 = 0.0496 m
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1.18 The radian frequency is 2π (10) = 20π . The displacement is y = A sin 20π t. The velocity and acceleration are y˙ = 20π A cos 20π t
Given the acceleration amplitude
y¨ = −(20π )2 A sin 20π t (20π )2 A = 0.7g
we solve for the displacement amplitude A as
A=
0.7g = 5.71 × 10−3 ft (20π )2
Thus the amplitude of the velocity is 20π A = 0.359 ft/sec.
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1.19 A = 0.15 mm. The acceleration is
x¨ = (2π f )2 A cos 2π f t Given the acceleration amplitude, (2π f )2 A = 0.6g we can solve for the frequency.
f=
&
0.6g = 0.997 Hz 4π 2 A
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1.20 y certainly lies within the range [−6, 6] and y disappears after about four time constants, 4(3) = 12.
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1.21
y¨ = −
A y˙ = − e−t/τ sin( t + ) + A e−t/τ cos( t + ) A 2
e−t/τ sin( t + )−
A
e−t/τ cos( t + )−
A
e−t/τ cos( t + )− A
2 −t/τ
e
sin( t + )
which reduces to
y¨ = A
!
1 2
−
2
"
e−t/τ sin( t + ) −
2A
e−t/τ cos( t + )
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1.22 Physical considerations require the model to pass through the origin, so we seek a model of the form f = kx. A plot of the data shows that a good line drawn by eye is given by f = 0.2x. So we estimate k to be 0.2 lb/in. Skipping ahead to Section 1.6, we can solve this problem using the least squares method, based on equation (1.6.3). The script file is x = f = num den k =
[4.7,7.2,10.6,12.9]4.7; [0,0.47,1.15,1.64]; = sum(x.*f); = sum(x.^2); num/den
The result is k = 0.1977.
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1.23 The script file is x = [0:0.01:1]; subplot(2,2,1) plot(x,sin(x),x,x),xlabel(#x (radians)#),ylabel(#x and sin(x)# ), . . . gtext(# x# ),gtext(#sin(x)# ) subplot(2,2,2) plot(x,sin(x)x),xlabel(#x (radians)# ),ylabel(#Error: sin(x)  x# ) subplot(2,2,3) plot(x,100*(sin(x)x). /sin(x)),xlabel(#x (radians)# ), . . . ylabel(#Percent Error# ),grid The plots are shown in the figure. 0
1
x and sin(x)
0.6
Error: sin(x) − x
x
0.8
sin(x)
0.4 0.2 0 0
0.5 x (radians)
1
0.5 x (radians)
1
−0.05 −0.1 −0.15 −0.2 0
0.5 x (radians)
1
Percent Error
0 −5 −10 −15 −20 0
Figure : for Problem 1.23. From the third plot we can see that the approximation sin x ≈ x is accurate to within 5% if x ≤ 0.5 radians.
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1.24 For
For
near π / 4,
near 3π / 4,
!
"!
π − 4
!
"!
3π − 4
π π f ( ) ≈ sin + cos 4 4
3π 3π + cos f ( ) ≈ sin 4 4
" "
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1.25 For
For
near π / 3,
near 2π / 3,
!
"!
π − 3
!
"!
2π − 3
π π f ( ) ≈ cos − sin 3 3
2π 2π − sin f ( ) ≈ cos 3 3
" "
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1.26 For h near 25,
f (h) ≈
√ 1 1 25 + √ (h − 25) = 5 + (h − 25) 10 2 25
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1.27 For r near 5, For r near 10,
f (r ) ≈ 52 + 2(5)(r − 5) = 25 + 10(r − 5) f (r ) ≈ 102 + 2(10)(r − 10) = 100 + 20(r − 10)
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1.28 For h near 16,
f (h) ≈
√ 1 1 16 + √ (h − 16) = 4 + (h − 16) 8 2 16
f (h) ≥ 0 if h > −16.
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1.29 Construct a straight line the passes through √ the two endpoints at p = 0 and p = 900. At p = 0, f (0) = 0. At p = 900, f (900) = 0.002 900 = 0.06. This straight line is
f (p) =
0.06 1 p= p 900 15, 000
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1.30 (a) The data is described approximately by the linear function y = 54x − 1360. The precise values given by the least squares method are y = 53.5x −1354.5 (see Problem 1.48a). (b) Only the loglog plot of the data gives something close to a straight line, so the data is best described by a power function y = bxm where the approximate values are m = −0.98 and b = 3600. The precise values given by the least squares method are y = 3582.1x−0.9764 (see Problem 1.48b). (c) Both the loglog and semilog plot (with the y axis logarithmic) give something close to a straight line, but the semilog plot gives the straightest line, so the data is best described by a exponential function y = b(10)mx where the approximate values are m = −0.007 and b = 2.1 × 105 . The precise values given by the least squares method are y = 2.0622 × 105 (10)−0.0067x (see Problem 1.48c).
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1.31 With this problem, it is best to scale the data by letting x = y ear − 1990, to avoid raising large numbers like 1990 to a power. Both the loglog and semilog plot (with the y axis logarithmic) give something close to a straight line, but the semilog plot gives the straightest line, so the data is best described by a exponential function y = b(10)mx . The approximate values are m = 0.035 and b = 9.98. Set y = 20 to determine how long it will take for the population to increase from 10 to 20 million. This gives 20 = 9.98(10)0.03x . Solve it for x: x = (log(20) − log(9.98)) / 0.035. The answer is 8.63 years, which corresponds to 8.63 years after 1990. More precise values are given by the least squares method (see Problem 1.49).
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1.32 (a) If C (t) / C (0) = 0.5 when t = 500 years, then 0.5 = e−5500b , which gives b = − ln(0.5) / 5500 = 1.2603 × 10−4 . (b) Solve for t to obtain t = − ln[C (t) / C (0)] / b using C (t) / C (0) = 0.9 and b = 1.2603 × 10−4 . The answer is 836 years. Thus the organism died 836 years ago. (c) Using b = 1.1(1.2603 × 10−4 ) in t = − ln(0.9) / b gives 760 years. Using b = 0.9(1.2603 × 10−4 ) in t = − ln(0.9) / b gives 928 years.
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1.33 Only the semilog plot of the data gives something close to a straight line, so the data is best described by an exponential function y = b(10)mx where y is the temperature in degrees C and x is the time in seconds. The approximate values are m = −3.67 and b = 356. The alternate exponential form is y = be(m ln 10)x = 356e−8.451x . The time constant is 1 / 8.451 = 0.1183 s. The precise values given by the least squares method are y = 356.0199(10)−3.6709x (see Problem 1.51).
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1.34 Only the semilog plot of the data gives something close to a straight line, so the data is best described by an exponential function y = b(10)mx where y is the bearing life thousands of hours and x is the temperature in degrees F. The approximate values are m = −0.007 and b = 142. The bearing life at 150 ◦ F is estimated to be y = 142(10)−0.007(150) = 12.66, or 12,600 hours. The alternate exponential form is y = be(m ln 10)x = 142e−0.0161x . The time constant is 1 / 0.0161 = 62.1 or 6.21 × 104 hr. The precise values given by the least squares method are y = 141.8603(10)−0.0070x (see Problem 1.52).
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1.35 Only the semilog plot of the data gives something close to a straight line, so the data is best described by an exponential function y = b(10)mx where y is the voltage and x is the time in seconds. The first data point does not lie close to the straight line on the semilog plot, but a measurement error of ±1 volt would account for the discrepancy. The approximate values are m = −0.43 and b = 96. The alternate exponential form is y = be(m ln 10)x = 96e−0.99x . The time constant is 1 / 0.99 = 1.01 s. The precise values given by the least squares method are y = 95.8063(10)−0.4333x (see Problem 1.53).
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1.36 A semilog plot generated by the following script file shows that the exponential function T − 70 = bemt fits the data well. t = [0:300:3000]; temp = [207,182,167,155,143,135,128,123,118,114,109]; DT = temp70; semi logy(t ,DT, t ,DT,’o’) Fitting a line by eye gives the approximate values m = −4 × 10−4 and b = 125. The −4 corresponding function is T (t) = 70 + 125e−4×10 t . The precise values given by the least squares method are m = −4.0317 × 10−4 and b = 125.1276 (see Problem 1.54).
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1.37 In the first printing of the text, the time data was reversed (the largest height should have the smallest time). This misprint also occurs in the data table on page 28. Plots of the data on a loglog plot and rectilinear scales both give something close to a straight line, so we try both functions. (Note that the flow should be 0 when the height is 0, so we do not consider the exponential function and we must force the linear function to pass through the origin by setting b = 0.) The three lowest heights give the same time, so we discard the heights of 1 and 2 cm. The power function fitted by eye in terms of the height h is approximately f = 4h0.9 . Note that the exponent is not close to 0.5, as it is for orifice flow. This is because the flow through the outlet is pipe flow. For the linear function f = mh, the best fit by eye is approximately f = 3.2h. Using the least squares method gives more precise results: f = 4.1595h0.8745 and f = 3.2028h (see Problem 1.55)
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1.38 In the first printing of the text, the time data was reversed (the largest height should have the smallest time). This misprint also occurs in the data table on page 28. Plots of the data on a loglog plot and rectilinear scales both give something close to a straight line, so we try both functions. (Note that the flow should be 0 when the height is 0, so we do not consider the exponential function and we must force the linear function to pass through the origin by setting b = 0.) The variable x is the height and the variable y is the flow rate. The three lowest heights give the same time, so we discard the heights of 1 and 2 cm. The power function fitted by eye in terms of the height h is approximately f = 4h0.9 . Note that the exponent is not close to 0.5, as it is for orifice flow. This is because the flow through the outlet is pipe flow. For the linear function f = mh, the best fit by eye is approximately f = 3.7h. Using the least squares method gives more precise results: f = 4.1796h0.9381 and f = 3.6735h (see Problem 1.56).
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1.39 Fitting a straight line by eye gives the approximate values m = 15 and b = 7. The precise values given by the least squares method are m = 15.0750 and b = 7.1500 (see Problem 1.57).
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1.40 Plot the data on a loglog plot. We must delete the first data point to avoid taking the logarithm of 0. The power function fitted by eye is approximately y = 7x3 . The precise values given by the least squares method are m = 2.9448 and b = 7.4053 (see Problem 1.58).
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1.41 Plot the data on a semilog plot. The exponential function fitted by eye is approximately y = 6e3x . The precise values given by the least squares method are y = 6.4224e2.8984x (see Problem 1.59).
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1.42 Fitting a straight line by eye through the origin gives the approximate values m = 17 and b = 0. The precise value given by the least squares method is m = 16.6071 (see Problem 1.60).
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1.43 Plot the data on a loglog plot. We must delete the first data point to avoid taking the logarithm of 0. The power function fitted by eye is approximately y = 7x3 . The precise value given by the least squares method is b = 7.4793 (see Problem 1.61).
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1.44 a) The equation for b is obtained as follows.
J=
n ' i=1
(bemxi − yi )2
n n ' ' ∂J = 2b e2mxi − 2 yi emxi = 0 ∂b i=1 i=1
( mx ye b = ( 2mx
e
b) Plot the data on a semilog plot. The exponential function fitted by eye is approximately y = 6e3x . The precise values given by the least squares method are y = 5.8449e3x (see Problem 1.62).
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1.45 The integral form of the sum of squares to fit the function y = 5x2 is
J=
)
0
4*
2
+2
5x − mx − b
dx =
)
4 0
,

25x4 − 10mx3 + (m2 − 10b)x2 + 2mbx + b2 dx
This evaluates to
J = 5120 − 640m + (m2 − 10b)
64 + 16mb + 4b2 3
Thus
128 ∂J = −640 + m + 16b = 0 ∂m 3 ∂J 640 =− + 16m + 8b = 0 ∂b 3 These give m = 20 and b = −13.333. Thus the line is y = 20x − 13.333.
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1.46 a) The integral form of the sum of squares to fit the function y = Ax2 + B x is
J= or
J=
)
L 0
)
0
L*
+2
Ax2 + B x − mx − b
dx
,

A2 x4 + (2AB − 2mA)x3 + (−2bA + B 2 + m2 − 2mB )x2 + (2mb − 2bB )x + b2 dx
This evaluates to
J = A2 Thus
L4 L2 L3 L5 + 2A(B − m) + (−2bA + B 2 + m2 − 2mB ) + 2b(m − B ) + b2 L 5 4 3 2 2 AL 4 2 ∂J =− + mL3 − B L3 + bL2 = 0 ∂m 2 3 3 2 ∂J = − AL3 + (m − B )L2 + 2bL = 0 ∂b 3
These give 4Lm + 6b = 4B L + 3AL2 3Lm + 6b = 2AL2 + 3B L which can be solved for m and b, given values of A, B , and L. b) With L = 2, A = 3, and B = 5, we obtain m = 11 and b = −2. The fitted straight line is y = 11x − 2.
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1.47 a) The integral form of the sum of squares to fit the function y = B eM x is
J= or
J=
)
0
This evaluates to
)
0
L*
+2
B e M x − mx − b
dx
L,

B 2 e2M x − 2mB xeM x − 2bB eM x + m2 x2 + 2bmx + b2 dx = + 2mB L + B 2 * 2M L 2mB * M L eM L + e −1 − e −1 2 2M M M 3 + 2bB * L + bmL2 + b2 L + 1 − eM L − m2 M 3
J =
Thus
+ 2mL3 2B L M L 2B * M L ∂J =− e + bL2 = 0 + 2 e −1 + ∂m M M 3 2B M L 2B ∂J =− e + mL2 + 2bL = 0 + ∂b M M
These give
+ 2B L M L 2B * M L 2L3 m + L2 b = e − 2 e −1 3 M M 2B M L 2B e L 2 m + 2L b = − M M which can be solved for m and b, given values of M , B , and L. b) With L = 1, M = −5, and B = 15, we obtain m = −10.973 and b = 8.466. The fitted straight line is y = −10.973x + 8.466.
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1.48 (a) The script file is x y p p
= [25:5:45]; = [5, 260, 480, 745, 1100]; = polyfi t(x,y,1) = 1.0e+003 * 0.0535 1.3545
The function is y = 53.5x − 1354.5.
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1.48 (b) Only the loglog plot of the data gives something close to a straight line, so the data is best described by a power function y = bxm . The script file to find the coe cients m and b is x y p m b
= = = = =
[2.5:0.5:6,7:10]; [1500,1220,1050,915,810,745,690,620,520,480,410,390]; polyfi t(log10(x), log10(y),1); p(1) 10^p(2)
The results are m = −0.9764 and b = 3582.1. Thus the power function is y = 3582.1x−0.9764 .
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1.48 (c) Both the loglog and semilog plot (with the y axis logarithmic) give something close to a straight line, but the semilog plot gives the straightest line, so the data is best described by a exponential function y = b(10)mx . The script file to find the coe cients m and b is x y p m b m
= [550:50:750]; =[41.2,18.62,8.62,3.92,1.86]; = polyfi t(x, log10(y),1); = p(1) = 10^p(2) = 0.0067 b = 2.0622e+005 This gives the results m = −0.0067 and b = 2.0622 × 105 . Thus the exponential function is y = 2.0622 × 105 (10)−0.0067x . The results for the power function are obtained from p = polyfi t(log10(x), log10(y),1); m = p(1) b = 10^p(2) This gives the results m = −9.9949 and b = 1.0601 × 1029 . Thus the power function is y = 1.0601 × 1029 x−9.9949 . A plot of the data and the two functions shows that both functions describe the data well, but the exponential curve passes closer to the first data point than the power curve. In addition, we should be careful about using this power function because its coe cient b has a very large value. This large coe cient means that any predictions of y values made with this function will be very prone to error unless we use very precise values of x. Thus the exponential function is the best choice to describe this data.
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1.49 With this problem, it is best to scale the data by letting x = y ear − 1990, to avoid raising large numbers like 1990 to a power. Both the loglog and semilog plot (with the y axis logarithmic) give something close to a straight line, but the semilog plot gives the straightest line, so the data is best described by a exponential function y = b(10)mx . The script file to find the coe cients m and b is year = [1990:1995]; x = x = year1990; pop=[10,10.8,11.7,12.7,13.8,14.9]; p=polyfi t(x, log10(pop),1) p = 0.0349 0.9992 m=p(1); b=10^p(2) This gives the results m = 0.0349 and b = 9.9817. Thus the exponential function is y = 9.9817(10)0.0349x . Set y = 20 to determine how long it will take for the population to increase from 10 to 20 million. This gives 20 = 9.9817(10)0.0349x . Solve it for x: x = (log(20)−log(9.9817)) / 0.0349 = 8.6483 years, which corresponds to 8.6483 years after 1990.
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1.50 (a) If C (t) / C (0) = 0.5 when t = 500 years, then 0.5 = e−5500b , which gives b = − ln(0.5) / 5500. In Matlab this calculation is ( b = log(0.5)/5500 The answer is b = 1.2603 × 10−4 . (b) Solve for t to obtain t = − ln[C (t) / C (0)] / b using C (t) / C (0) = 0.9 and b = 1.2603 × 10−4 . In MATLAB this calculation is ( t = log(0.9)/b The answer is 836.0170 years. Thus the organism died 836 years ago. (c) Using b = 1.1(1.2603 × 10−4 ) in t = − ln(0.9) / b gives 760 years. 0.9(1.2603 × 10−4 ) in t = − ln(0.9) / b gives 928 years.
Using b =
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1.51 Only the semilog plot of the data gives something close to a straight line, so the data is best described by an exponential function y = b(10)mx . The script file to find the coe cients m and b is t ime = [0:0.1:0.6]; temp = [300,150,75,35,12,5,2]; p=polyfi t(t ime, log10(temp),1) m=p(1) b=10^p(2) This gives the results: m = −3.6709 and b = 356.0199. Thus the exponential function is y = 356.0199(10)−3.6709x , where y is the temperature in degrees C and x is the time in seconds. The alternate exponential form is y = be(m ln 10)x = 356.0199e−8.4526x . The time constant is 1 / 8.4526 = 0.1183 s.
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1.52 Only the semilog plot of the data gives something close to a straight line, so the data is best described by an exponential function y = b(10)mx . The script file to find the coe cients m and b is temp = [100:20:220]; l ife = [28,21,15,11,8,6,4]; p = polyfi t(temp, log10(l ife),1); m = p(1) b = 10^p(2) This gives the results: m = −0.0070 and b = 141.8603. Thus the exponential function is y = 141.8603(10)−0.0070x , where y is the bearing life thousands of hours and x is the temperature in degrees F. The bearing life at 150 ◦ F is estimated to be y = 141.8603(10)−0.0070(150) = 12.6433, or 12,643 hours. The alternate exponential form is y = be(m ln 10)x = 141.8603e−0.0161x . The time constant is 1 / 0.0161 = 62.0421 or 6.20421 × 104 hr.
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1.53 Only the semilog plot of the data gives something close to a straight line, so the data is best described by an exponential function y = b(10)mx . The first data point does not lie close to the straight line on the semilog plot, but a measurement error of ±1 volt would account for the discrepancy. The script file to find the coe cients m and b is t ime = [0:0.5:4]; vol tage = [100,62,38,21,13,7,4,2,3]; p = polyfi t(t ime, log10(vol tage),1); m=p(1) b = 10^p(2) This gives the results: m = −0.4333 and b = 95.8063. Thus the exponential function is y = 95.8063(10)−0.4333x , where y is the voltage and x is the time in seconds. The alternate exponential form is y = be(m ln 10)x = 95.8063e−0.9977x . The time constant is 1 / 0.9977 = 1.002 s.
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1.54 The semilog plot generated by the following script file shows that the exponential function T − 70 = bemt fits the data well. t = [0:300:3000]; temp = [207,182,167,155,143,135,128,123,118,114,109]; DT = temp70; semi logy(t ,DT, t ,DT,’o’) p = polyfi t(t , log(DT),1) m = p(1) b = exp(p(2)) The results are m = −4.0317 × 10−4 and b = 125.1276. The time constant is 1 / 4.0317 × 10−4 = 2.4803 × 103 s, or 0.689 hr.
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1.55 In the first printing of the text, the time data was reversed (the largest height should have the smallest time). This misprint also occurs in the data table on page 28. The correct data are shown in the following script file. Plots of the data on a loglog plot and rectilinear scales both give something close to a straight line, so we try all both functions. (Note that the flow should be 0 when the height is 0, so we do not consider the exponential function and we must force the linear function to pass through the origin by setting b = 0.) The variable x is the height and the variable y is the flow rate. The three lowest heights give the same time, so we discard the heights of 1 and 2 cm. For the power function, we use (1.6.1) and (1.6.2) in terms of the variables x and Y = log y . t = [7,8,9,10,11,13,15,17,23]; x = [11:1:3]; y = 250. / t ; X = log10(x); Y = log10(y); a1 = sum(X.^2); a2 = sum(X); a3 = sum(Y.*X); a4 = sum(Y); n = length(x); A = [a1, a2; a2, n]; C = [a3; a4]; solut ion = A\C; m = solut ion(1) b = 10^solut ion(2) J = sum((b*x.^my).^2) S = sum((b*x.^mmean(y)).^2) r2 = 1  J/S The results are m = 0.8745, b = 4.1595, J = 5.3644, S = 493.5634, and r 2 = 0.9891. So the fitted function in terms of the height h is f = 4.1595h0.8745 . Note that the exponent is not close to 0.5, as it is for orifice flow. This is because the flow through the outlet is pipe flow. For the linear function f = mx, we use (1.6.3). m = sum(x.*y)/sum(x.^2) J = sum((m*xy).^2) S = sum((m*xmean(y)).^2) r2 = 1  J/S The results are m = 3.2028, J = 8.0247, S = 615.9936, and r 2 = 0.9870. The fitted function in terms of the height h is f = 3.2028h. Using only r 2 as the criterion, it is impossible to decide whether the linear or the power function is the best model. 148 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
1.56 In the first printing of the text, the time data was reversed (the largest height should have the smallest time). This misprint also occurs in the data table on page 28. The correct data are shown in the following script file. Plots of the data on a loglog plot and rectilinear scales both give something close to a straight line, so we try all both functions. (Note that the flow should be 0 when the height is 0, so we do not consider the exponential function and we must force the linear function to pass through the origin by setting b = 0.) The variable x is the height and the variable y is the flow rate. The three lowest heights give the same time, so we discard the heights of 1 and 2 cm. For the power function, we use (1.6.1) and (1.6.2)in terms of the variables x and Y = log y . t = [6, 7, 8, 9, 9, 11, 13, 17, 21]; x = [11:1:3]; y = 250. / t ; X = log10(x); Y = log10(y); a1 = sum(X.^2); a2 = sum(X); a3 = sum(Y.*X); a4 = sum(Y); n = length(x); A = [a1, a2; a2, n]; C = [a3; a4]; solut ion = A\C; m = solut ion(1) b = 10^solut ion(2) J = sum((b*x.^my).^2) S = sum((b*x.^mmean(y)).^2) r2 = 1  J/S The results are m = 0.9381, b = 4.1796, J = 13.5531, S = 729.3505, and r 2 = 0.9814. So the fitted function in terms of the height h is f = 4.1796h0.9381 . For the linear function f = mx, we use (1.6.3). t = [6, 7, 8, 9, 9, 11, 13, 17, 21]; x = [11:1:3]; y = 250. / t ; m = sum(x.*y)/sum(x.^2) J = sum((m*xy).^2) S = sum((m*xmean(y)).^2) r2 = 1  J/S
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The results are m = 3.6735, J = 14.7546, S = 809.8919, and r 2 = 0.9818. The fitted function in terms of the height h is f = 3.6735h. Using only r 2 as the criterion, it is impossible to decide whether the linear or the power function is the best model.
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1.57 The script file is x = [0:2:6]; y = [4.5, 39, 72, 94]; a1 = sum(x.^2); a2 = sum(x); a3 = sum(y.*x); a4 = sum(y); n = length(x); A = [a1, a2; a2, n]; B = [a3; a4]; solut ion = A\B; m = solut ion(1) b = solut ion(2) J = sum((m*x+by).^2) S = sum((m*x+bmean(y)).^2) r2 = 1  J/S The results are m = 15.0750, b = 7.1500, J = 43.5750, S = 4.5451 × 103 , r 2 = 0.9904.
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1.58 Following the procedure shown in Example 1.6.1, we fit X = log x and Y = log y to a linear function. We must delete the first data point to avoid taking the logarithm of 0. The script file is x = [1:4]; y = [8, 50, 178, 490]; X = log10(x); Y = log10(y); a1 = sum(X.^2); a2 = sum(X); a3 = sum(Y.*X); a4 = sum(Y); n = length(x); A = [a1, a2; a2, n]; C = [a3; a4]; solut ion = A\C; m = solut ion(1) b = 10^solut ion(2) J = sum((b*x.^my).^2) S = sum((b*x.^mmean(y)).^2) r2 = 1  J/S The results are m = 2.9448, b = 7.4053, J = 2.7494×103 , S = 1.1218×105 , and r 2 = 0.9755.
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1.59 We fit x and Y = log y to the linear function Y = mx + B , where B = log b. The script file is x = [0:0.4:1.2]; y = [6.3, 22, 60, 215]; Y = log10(y); a1 = sum(x.^2); a2 = sum(x); a3 = sum(Y.*x); a4 = sum(Y); n = length(x); A = [a1, a2; a2, n]; C = [a3; a4]; solut ion = A\C; m = solut ion(1)/ log10(exp(1)) b = 10^solut ion(2) J = sum((b*exp(m*x)y).^2) S = sum((b*exp(m*x)mean(y)).^2) r2 = 1  J/S The results are m = 2.8984, b = 6.4244, J = 77.4488, S = 2.5496 × 104 , and r 2 = 0.9970. The fitted function is y = 6.4244e2.8984x .
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1.60 The equation for m, obtained from (1.6.3), is (
The script file is
xi yi m= ( 2 xi
x = [0:2:6]; y = [4.5, 39, 72, 94]; m = sum(x.*y)/sum(x.^2) J = sum((m*xy).^2) S = sum((m*xmean(y)).^2) r2 = 1  J/S The results are m = 16.6071, J = 116.6071, S = 5.5420 × 103 , and r 2 = 0.9790.
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1.61 The equation for b is obtained from equation (1) in Example 1.6.3 with m = 3. ( 3 x y b= ( 6
x
x = [0:4]; y = [1, 8, 50, 178, 490]; b = sum((x.^3).*y)/sum(x.^6) J = sum((b*x.^3y).^2) S = sum((b*x.^3mean(y)).^2) r2 = 1  J/S
The results are b = 7.4793, J = 799.4139, S = 1.6176 × 105 , and r 2 = 0.9951.
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1.62 a) The equation for b is obtained as follows.
J=
n ' i=1
(bemxi − yi )2
n n ' ' ∂J = 2b e2mxi − 2 yi emxi = 0 ∂b i=1 i=1
( mx ye b = ( 2mx
e
b) The script file is
x = [0:0.4:1.2]; y = [6.3, 22, 60, 215]; b = sum(y.*exp(3*x))/sum(exp(6*x)) J = sum((b*exp(3*x)y).^2) S = sum((b*exp(3*x)mean(y)).^2) r2 = 1  J/S The results are b = 5.8449, J = 27.7380, S = 2.7279 × 104 , and r 2 = 0.9990.
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c Solutions Manual! to accompany System Dynamics, First Edition by William J. Palm III University of Rhode Island
Solutions to Problems in Chapter Two
c Solutions Manual Copyright 2004 by McGrawHill Companies, Inc. Permission required ! for use, reproduction, or display.
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2.1 mv˙ = mg. Thus v(t) = gt = 32.2t
x(t) =
1 2 gt = 16.1t2 2
Thus t=
!
h(t) = 20 − x(t) = 20 − 16.1t2
20 − h(t) 16.1
For h = 10, t = 0.788 sec. For h = 0, t = 1.115 sec.
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2.2 t= x=
60/5280 3600 = 0.455 sec 90
1 2 gt = 16.1(0.455)2 = 3.326 ft 2
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2.3 Summing forces in the direction parallel to the plane gives mv˙ = f1 − mg sin φ − µmg cos φ Substituting the given values, 10v˙ = f1 − 98.1 sin 25◦ − 0.3(98.1) cos 25◦ = f1 − 68.132 Thus v˙ > 0 if f1 > 68.132. If f1 = 100 the block will continue to move up the plane. If f1 = 50, v˙ = −1.813 and the speed is given by v(t) = −1.813t + v(0) = −1.813t + 2 Thus v(0) = 0 at t = 1.103 s.
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2.4 Let d be the vertical distance dropped by the time the mass leaves the surface. See the following figure. Then L d= tan θ The speed v0 of the mass when it leaves the surface is found from conservation of energy: KE =
1 mgL mv02 = P E = 2 tan θ
Thus v0 =
!
2gL tan θ
(1)
The horizontal and vertical velocity components are vx = v0x sin θ v0y = v0 cos θ
(2)
Establish a coordinate system at the point where the mass leaves the surface, with x positive to the right and y positive down. In the vertical direction : y¨ = g and 1 y = gt2 + v0y t 2 The mass hits the ground when y = H, and timetohit tH is found from 1 2 gt + 2v0y tH = H 2 H or tH =
−2v0y ±
"
2 + 8gH 4v0y
2g
(3)
There will be one positive solution and one negative solution. Take the positive solution. In the horizontal direction: m¨ x = 0 and x = (v0 sin θ) t and D = (v0 sin θ) tH
(4)
The solution is given by (1) through (4).
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Figure : for Problem 4
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2.5 Let the origin of the launch point be at x = y = 0. Assuming the projectile is launched with a speed v0 at an angle θ from the horizontal, Newton’s law in the x and y directions gives x ¨=0 y¨ = −g vx = v0 cos θ
x = (v0 cos θ)t
vy = v0 sin θ − gt g y = − t2 + (v0 sin θ)t 2
These can be manipulated to show that vy2 = (v0 − sin θ)2 − 2gy which gives v0 sin θ =
"
vy2 − 2gy
Using this relation, the above can also be manipulated to show that vy = − Solve for x. x= where y is computed from
gx " 2 + vy − 2gy vx
vx " 2 vx vy vy − 2gy − g g y = R sin φ
(1) (2)
The desired distance D can be computed from D = x + R cos φ where x is computed from (1).
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2.6 a) From Newton’s law,
¨ = T − mg mh h˙ = h=
So the time to reach 10,000 ft is
#
#
T − mg m t=
!
$
T −g t m $
2
t2 = 10, 000 ft 2
10, 000m T − mg
Using the minimum thrust value, T = 11, 000 lb, and m = 100 slug, we obtain t = 16.03 sec. b) If T = 12, 000 ft and t = 16.03 sec, the maximum height will be h=
#
12, 000 − 100(32.2) 100
$
(16.03)2 = 11, 281 ft 2
c) Using t = 16.03 sec for the burn time, the fuel mass burned will be 0.5(16.03) = 8.015 slug. Thus the smallest possible launch mass is 100 − 8.015 = 91.985 slug, and the height reached will not be greater than h=
#
12, 000 − 91.985(32.2) 91.985
$
(16.03)2 = 12, 624 ft 2
So the maximum height will be in the range 11, 281 ≤ h ≤ 12, 624 ft.
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2.7 IG = 2mr 2 /5. Apply the parallel axis theorem. 2 IO = IG + mR2 = mr 2 + mR2 5
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2.8 a) Let point O be the pivot point and G be the center of mass. Let L be the distance from O to G. Treat the pendulum as being composed of three masses: 1) m1 , the rod mass above point O, whose center of mass is 1 ft above point O; 2) m2 , the rod mass below point O, whose center of mass is 1.5 ft below point O, and 3) m3 , the mass of the 10 lb block. Then, summing moments about G gives m1 g(L + 1) − m2 g(1.5 − L) − m3 g(3.5 − L) = 0 where
(1)
2 6 m1 g = 3 = lb 5 5 9 3 m2 g = 3 = lb 5 5 m3 g = 10 lb
From equation (1): 9 6 (L + 1) − (1.5 − L) − 10(3.5 − L) = 0 5 5 which gives L = 2.85 ft. b) Summing moments about the pivot point O gives IO θ¨ = −mgL sin θ where m is the total mass. From the parallelaxis theorem, treating the rod as a slender rod, we obtain IO =
1 12
# $
3 (5)2 + g
# $
3 (0.5)2 + g
#
$
10 (3.5)2 = 4.022 slug − ft2 g
and mgL = 13(2.85) = 37.05 ftlb. Thus the equation of motion is 4.022θ¨ = −37.05 sin θ or
θ¨ + 9.2118 sin θ = 0
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2.9 a) IA = mL22 + mC L21 IA θ¨ = −mC L1 g sin θ + mgL2 cos(β − θ)
b) Set θ¨ = 0 and solve for mg.
mg =
mC L1 g sin θ L2 cos(β − θ)
c) Substitute the given values to obtain mg =
5(0.2)g sin 20◦ = 2.315g = 22.713 N 0.15 cos 10◦
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2.10 The easiest way is to use the energyequivalence method. Let ωB the speed of pulley B (positive counterclockwise) and ωC the speed of pulley C (positive clockwise). The kinetic energy of the entire system is 1 1 1 1 1 2 2 2 2 2 KE = IB ωB + IC ω C + mL vC + mC vC = me vA 2 2 2 2 2 where me is the mass of the equivalent translational system. The potential energy is P E = (mL + mC )g(−sC ) = Fe sA where Fe is the equivalent gravity force. Thus Fe = −
mL + mC g 2
Using the kinematic relations: 1 sC = − sA 2
1 vC = − vA 2
ωB =
1 vA RB
ωC =
1 vC RC
the kinetic energy expression becomes 1 KE = 2
%
IC mL + mC IB + + 2 2 4 RB 4RC
Thus me =
&
2 vA =
1 2 me vA 2
IB IC mL + mC + + 2 2 RB 4RC 4
The model is me v˙ A = FA −
mL + mC g 2
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2.11 a) Let T be the tension in the cable attached to mass m2 . See the following figure. Then the cable force pulling up on m1 is T /2 because of the pulleys. Note also that because of the pulleys, x = 2y. Summing forces acting on m2 parallel to the plane, we obtain m2 y¨ = T − m2 g sin θ
(1)
Summing the vertical forces acting on m1 , we obtain
Since x = 2y, this becomes
1 m1 x ¨ = m1 g − T 2
(2)
1 2m1 y¨ = m1 g − T 2
(3)
T = 2m1 g − 4m1 y¨
(4)
Solve for T : Substitute this into (1) and collect the y¨ terms to obtain y = 2m1 g − m2 g sin θ (4m1 + m2 )¨
(5)
The mass m1 will lift m2 if y¨ > 0; that is, if 2m1 − m2 sin θ > 0 b) Follow the same procedure as in part (a) but include the friction force. Equation (1) becomes m2 y¨ = T − m2 g sin θ − µd m2 g cos θ (6) Equations (2) through (4) remain the same, but (5) becomes
y = 2m1 g − m2 g(µd cos θ + sin θ) (4m1 + m2 )¨
(7)
The mass m1 will lift m2 if y¨ > 0; that is, if 2m1 − m2 (µd cos θ + sin θ) > 0 For the case m1 = m2 /2, this becomes 1 − (µd cos θ + sin θ) > 0
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Figure : for Problem 11
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2.12 Refer to the following figure for definitions. This simplelooking system actually has complicated kinematics because the component of the cable tension T normal to the rod is a complicated function of θ, and because the acceleration x ¨ of m1 is a complicated function ˙ and θ. ¨ of θ, θ, Note that sin φ = cos θ and cos φ = sin θ. From the law of sines, H B = sin φ sin ψ
or
sin ψ =
H H sin φ = cos θ B B
(1)
From the law of cosines, B=
"
L2 + H 2 − 2LH cos φ =
Summing moments about point O,
'
L2 + H 2 − 2LH sin θ
(2)
T LH m2 g 1 L m2 L2 θ¨ = (T sin ψ) L − (m2 g cos θ) = cos θ − cos θ 12 2 B 2
(3)
Summing forces on m1 in the x direction: m1 x ¨ = m1 g − T ¨ because the cable is assumed Solving this for T , substituting into (3), and noting that x ¨=B inextensible, we obtain ( ) 1 m1 LH ¨ − m2 gL cos θ m2 L2 θ¨ = cos θ g − B 12 B 2
(4)
¨ To obtain the required expression for B ¨ as a function of θ and its derivaThis contains B. tives, differentiate (2) twice to obtain (
¨ = LH θ˙ 2 sin θ − θ¨ cos θ B +
)(
L2 + H 2 − 2LH sin θ
( )−3/2 LH ˙ θ cos θ −2LH θ˙ cos θ 2
(5)
)−1/2
The model consists of (2), (4), and (5).
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Figure : for Problem 12
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2.13 Let FR be the reaction force that acts on the contacting gear teeth. Then IG1 ω˙ 1 = T1 − r1 FR IG2 ω˙ 2 = T2 − r2 FR
If ω˙ 1 = ω˙ 2 = 0, these equation give T1 = r1 FR and T2 = r2 FR , and thus T1 = r1
T2 1 = T2 r2 N
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2.14 Let F be the contact force between the two gears. For gear 1, IG1 ω˙ 1 = T1 − r1 F For gear 2, If ω˙ 1 = ω˙ 2 = 0, or if IG1 = IG2 = 0,
IG2 ω˙ 1 = T2 − r2 F
T1 = r1 F which give T1 =
T 2 = r2 F r1 1 T2 = T 2 r2 N
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2.15 Let r2 be the radius of pulley 2. The equivalent inertia felt on shaft 1 is Ie = I1 +
1 1 1 I2 + 2 m2 r22 + 2 m3 r22 2 N N N
With N = 2, Ie = I1 + The equation of motion is Ie ω˙ 1 = T1 −
) 1( I2 + m2 r22 + m3 r22 4
m2 gr2 m3 gr2 gr2 + = T1 − (m3 − m2 ) N N 2
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2.16 The total kinetic energy is KE =
1 1 (Is + I) θ˙ 2 + mx˙ 2 2 2
KE =
) 1( Is + I + mR2 θ˙ 2 2
Substituting x = Rθ we obtain
Thus the equivalent inertia is
Ie = Is + I + mR2
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2.17 The total kinetic energy is KE =
1 1 1 (I1 + IS1 ) ω12 + (I2 + IS2 ) ω22 + mv 2 2 2 2
Substituting ω2 = r1 ω1 /r2 and v = r1 ω1 we obtain #
1 1 r1 ω 1 KE = (I1 + IS1 ) ω12 + (I2 + IS2 ) 2 2 r2 or
*
$2
1 + m (r1 ω1 )2 2
#
$2
+
#
$2
+ mr12
1 r1 KE = I1 + IS1 + (I2 + IS2 ) 2 r2
mr12
+
ω12
Thus the equivalent inertia is r1 Ie = I1 + IS1 + (I2 + IS2 ) r2
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2.18 a) With everything reflected to the load shaft (shaft 2), Newton’s law gives ω˙ 2 =
T2 + N T1 I2 + N 2 I1
b) To maximize ω˙ 2 , differentiate the above expression with respect to N , set the derivative equal to 0. This gives ∂ ω˙ 2 (I2 + N 2 I1 )T1 − 2I1 N (T2 + N T1 ) =0 = ∂N (I2 + N 2 I1 )2 This is true if the numerator is 0. Thus I1 T1 N 2 + 2I1 T2 N − I2 T1 = 0 The positive solution for N is T2 N =− + T1
! #
T2 T1
$2
+
I2 I1
This is the ratio that maximizes ω˙ 2 . (This can be confirmed to give a maximum rather than a minimum by showing that ∂ 2 ω˙ 2 /∂N 2 < 0). ' If the load torque2 T2 is 0, the optimal ratio for this case is denoted No and is: No = I2 /I1 , or I1 = I2 /No . This says that the ratio that maximizes the load acceleration is the ratio that makes the reflected load inertia (felt by the motor) equal to the motor’s inertia. This is the principle of inertia matching. If T2 = 0 and the actual ratio N differs from the optimal value No such that N = γNo , the efficiency E is actual ω˙ 2 2γ E= = max ω˙ 2 1 + γ2 Because E(γ) = E(1/γ), the efficiency when “overgearing” is the same as when “undergearing”. For example, if γ = 2 or γ = 1/2, E = 0.8, so the acceleration is 80% of the maximum possible.
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2.19 With I1 = I2 = I3 = 0, the total kinetic energy is 1 1 KE = I4 ω12 + I5 ω32 2 2 Substituting ω2 = 1.4ω3 and ω1 = 1.4ω2 = (1.4)2 ω3 = 1.96ω3 , and I4 = 0.02, I5 = 0.1, we obtain 1 KE = (0.177)ω32 2 and the equivalent inertia is Ie = 0.177 kg·m2 . The equation of motion is Ie ω˙ 3 = (1.4)2 T , or 0.177ω˙ 3 = 1.96T
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2.20 The total kinetic energy is KE =
1 1 1 (I4 + I1 ) ω12 + I2 ω22 + (I3 + I5 ) ω32 2 2 2
Substituting ω2 = 1.4ω3 and ω1 = 1.4ω2 = (1.4)2 ω3 = 1.96ω3 and the given values of the inertias, we obtain 1 KE = (0.203)ω32 2 and the equivalent inertia is Ie = 0.203 kg·m2 . The equation of motion is Ie ω˙ 3 = (1.4)2 T , or 0.203ω˙ 3 = 1.96T
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2.21 a)
ω4 ω3 ω2 73 ω4 3 = 2.1 = = ω1 ω3 ω2 ω1 65
b) The torque T1 felt on shaft 4 is T1 /2.1 and the equation of motion is I ω˙ 4 =
T1 2.1
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2.22 Using kinetic energy equivalence, 1 1 1 KE = mv 2 + Is ω 2 = Ie ω 2 2 2 2 The mass translates a distance x when the screw rotates by θ radians. When θ = 2π, x = L. ˙ ˙ we have Thus x = Lθ/2π and x˙ = v = Lθ/2π. Because ω = θ, KE =
#
$2
Ie =
mL2 + Is 4π 2
1 Lω m 2 2π
Solve for Ie to obtain
1 1 + Is ω 2 = Ie ω 2 2 2
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2.23 The expression for the kinetic energy is KE = KE of 2 rear wheels + KE of front wheel + KE of body # $ 1 1 1 = 2 Ir ωr2 + If ωf2 + mb v 2 2 2 2 But for the rear wheels, v = Rr ωr = 4ωr , and for the front wheel, v = Rf ωf = 2ωf . The inertias are calculated as follows. For the rear wheels: 1 1 500 2 4000 Ir = mr Rr2 = 4 = 2 2 g g For the front wheel: If =
1 1 800 2 1600 mf Rf2 = 2 = 2 2 g g
Also, mb = 9000/g. Thus KE =
4950 2 v g
and the equivalent mass is me = 2(4950)/g = 9000/g. The equation of motion is me v˙ = mg sin 10◦ , where m = 2mr + mf + mb = 2 Thus
500 800 9000 10, 800 + + = g g g g
9900 v˙ = 10, 800 sin 10◦ g
or v˙ = 0.1894. Thus v = 0.1894t ft/sec.
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2.24 In the first printing of the text, the first line of the problem refers to “mass m”. It should say “mass m1 ”. Using the equivalent mass approach, the equivalent mass referenced to the coordinate x is I me = m1 + m2 + 2 R where I is the inertia of the cylinder about its center. The force acting on me due to the weight of the cylinder is m1 g sin β. The force acting on me due to the weight of m2 is m2 g sin φ. See the following figure. The equation of motion of the equivalent system is me x ¨ = m1 g sin β − m2 g sin φ
Figure : for Problem 24
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2.25 Let y be the translational displacement of the cylinder to the right. Using the equivalent mass approach, the kinetic energy of the system is 1 1 1 KE = m2 y˙ 2 + I ω˙ 2 + m1 x˙ 2 2 2 2 where I = m2 R2 /2 is the inertia of the cylinder about its center. Because y˙ = 2x˙ and ω = ˙ y/R , we have ( ) 1 1 KE = m2 4x˙ 2 + 2 2
%
m2 R 2 2
&#
2x˙ R
$2
1 1 + m1 x˙ 2 = (6m2 + m1 ) x˙ 2 2 2
Thus the effective mass referenced to the coordinate x is me = 6m2 + m1 . The equation of motion is me x ¨ = m1 g or x = m1 g (6m2 + m1 )¨
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2.26 Summing moments about the pivot gives IO θ¨ = T − mgL sin θ where the effect of the motor torque is T = [2(1.5)]Tm = 3Tm and the inertia is IO = mL2 + Im [2(1.5)]2 + IG1 [2(1.5)]2 + IG2 (1.5)2 + IG3 (1.5)2 + IG4 = 1.936 kg · m2 Thus
1.936θ¨ = 3Tm − 29.43 sin θ
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2.27 a) The speed ratio of the sprocket drive at the driving shaft is the ratio of the sprocket diameters, which is 0.05/0.15 = 1/3. The equivalent inertia felt at the motor shaft is ,
Ie = I1 + I2 + IS1 + mc1 (0.05)2 +
,
IS2 + Id + 4Iw +
2mc2 rd2
 # 1 $2
+
10
mL rd2
 # 1 1 $2
3 10
where IS1 , IS2 , Id , and Iw are the inertias of the shafts 1 and 2, the drive shaft, and the drive wheels. The masses mc1 , mc2 , and mL are the masses of the sprocket chain, the drive chain, and the load. This expression evaluates to Ie = 0.0143 kg·m2 . The magnitude of the friction torque felt at the motor shaft is 54/[3(10)] = 1.8 N·m. The equation of motion is 0.0143ω˙ 1 = T1 − 1.8sgn(ω1 )
(1)
b) If T1 = 10, ω˙ 1 = 573.427 and ω1 = 573.427t rad/s. c) Solve (1) for T1 assuming ω1 > 0: T1 = 0.0143ω˙ 1 + 1.8 From the trapezoidal profile, 300
0 ≤ t ≤ 0.5 ω˙ 1 = 0 0.5 < t < 2.5 −300 2.5 ≤ t ≤ 3
Thus T1 =
300Ie + 1.8
1.8
1.8 − 300I e
0 ≤ t ≤ 0.5 0 ≤ t ≤ 0.5 6.09 0.5 < t < 2.5 = 1.8 0.5 < t < 2.5 2.5 ≤ t ≤ 3 −249 2.5 ≤ t ≤ 3
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2.28 a) The equation of motion for rotation is I ω˙ = Rft where I = mR2 /2 = 1.631 kg·m2 is the inertia of the cylinder about its center, and ft is the tangential force between the cylinder and the ground. The equation of motion for translation is I ω˙ m¨ x = f cos φ − ft = f cos φ − R Substituting x ¨ = Rω˙ we obtain (
)
mR2 + I ω ¨ = Rf cos φ
Noting that m = 800/9.81 = 81.549 kg, and substituting the given values, we obtain 16.636¨ ω = f cos φ b) Using the equivalent mass approach, the equivalent mass referenced to the coordinate x is I me = m + 2 R where I = mR2 /2 = 1.631 kg·m2 is the inertia of the cylinder about its center. The equation of motion is me x ¨ = f cos φ Substituting the given values, we obtain 83.18¨ x = f cos φ
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2.29 See the following figure for the coordinate definitions and the definition of the reaction force R. Let P be the point on the axle. Note that yP = 0. The coordinates of the mass center of the rod are xG = xP −
L sin θ 2
yG = −
L cos θ 2
Thus
L¨ L θ cos θ + θ˙ 2 sin θ 2 2 L L y¨G = θ¨ sin θ + θ˙ 2 cos θ 2 2 Let m be the mass of the rod. Summing forces in the x direction: ¨P − x ¨G = x
m¨ xG = f
#
$
L L m x ¨P − θ¨ cos θ + θ˙ 2 sin θ = f 2 2
or
(1)
Summing forces in the y direction: m¨ yG = R − mg
or
m
#
$
L¨ L θ sin θ + θ˙ 2 cos θ = R − mg 2 2
(2)
Summing moments about the mass center: L L IG θ¨ = f cos θ − R sin θ 2 2 Substituting for R from (2) and using the fact that IG = mL2 /12, we obtain mgL mL2 1 mL2 ˙ 2 fL mL2 θ¨ = cos θ − sin θ − sin2 θ θ¨ − θ sin θ cos θ 12 2 2 4 4
(3)
The model consists of (1) and (3) with m = 20 kg and L = 1.4 m.
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Figure : for Problem 29
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2.30 See the following figure for the coordinate definitions and the definition of the reaction forces Rx , Ry , N , and ft . Let P be the point on the axle. Note that yP = 0. The coordinates of the mass center of the rod are xG = xP −
L sin θ 2
yG = −
L cos θ 2
Thus
L¨ L θ cos θ + θ˙ 2 sin θ 2 2 L L y¨G = θ¨ sin θ + θ˙ 2 cos θ 2 2 Let m1 be the mass of the rod, m2 the mass of the wheel, and I the moment of inertia of the wheel about its center. Summing forces on the wheel in the x direction: x ¨G = x ¨P −
m2 x ¨P = f − Rx − ft
(1)
Summing forces on the wheel in the y direction: m2 y¨P = N − Ry − m2 g
(2)
Summing moments about the mass center of the wheel: Iω ¨ = Rft But x˙ P = Rω and x ¨P = Rω. ˙ Thus I x ¨ P = ft R2
(3)
Substitute this into (1) to obtain #
$
I m2 + 2 x ¨P = f − Rx R
(4)
Summing forces on the rod in the x direction: m1 x ¨ G = Rx
#
or
m1 x ¨P −
$
L¨ L θ cos θ + θ˙ 2 sin θ = Rx 2 2
(5)
Summing forces on the rod in the y direction: m1 y¨G = Ry − m1 g
or
m1
#
$
L¨ L θ sin θ + θ˙ 2 cos θ = Ry − m1 g 2 2
(6)
Summing moments about the mass center of the rod: L L IG θ¨ = Rx cos θ − Ry sin θ 2 2 235 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Substituting for Rx and Ry from (5) and (6), canceling terms using the identity sin2 θ + cos2 θ = 1, and using the fact that IG = m1 L2 /12, we obtain m1 L2 ¨ 1 m1 gL m1 L m1 L2 θ¨ = x ¨P cos θ − sin θ − θ 12 2 2 4 which can be rearranged as follows: m1 gL m1 L 1 m1 L2 θ¨ = x ¨P cos θ − sin θ 3 2 2
(7)
Substituting for Rx from (4) into (5) gives #
$ ) I m1 L ( ˙ 2 m1 + m2 + 2 x ¨P = F − θ sin θ − θ¨ cos θ R 2
(8)
The model consists of (7) and (8) with m1 = 20 kg, m2 = 3 kg, L = 1.4 m, R = 0.05 m, and I = m2 R2 /2.
Figure : for Problem 30
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2.31 Summing forces in the x direction: maGx = mg sin θ − F
(1)
Summing forces in the y direction: maGy = N − mg cos θ − F For no bounce, aGy = 0 and
N = mg cos θ
(2)
(3)
Summing moments about G: IG α = F r
(4)
From (1) and (4): maGx = mg sin θ − With slipping, aGx $= rα, but
F = µd N
IG α r
(5)
(6)
From (4) and (6): IG α = rµd N = rµd mg cos θ Thus, α=
(7)
rµd mg cos θ IG
From (1), aGx = g sin θ − µd g cos θ
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2.32 Follow the procedure of Example 2.4.2. Summing forces in the x direction: maGx = mg sin θ − F
(1)
Summing forces in the y direction: For no bounce, aGy = 0 and
maGy = N − mg cos θ − F N = mg cos θ
(2)
(3)
a) Summing moments about G: IG α = F r
(4)
From (1) and (4): maGx = mg sin θ −
For no slip,
IG α r
aGx = rα From (5) and (6): aGx =
(6)
mgr 2 sin θ mr 2 + IG
Since IG = mr 2 , aGx = and α=
g sin θ 2
(5)
(7)
(8)
aGx g sin θ = r 2r
b) Summing moments about P : MP = (mg sin θ)r = IG α + maGx r = IG Thus aGx =
aGx + maGx r r
mgr 2 sin θ g sin θ = mr 2 + IG 2
and
g sin θ aGx = r 2r c) The maximum friction force is Fmax = µs N = µs mg cos θ. From (4), (6), and (7), α=
F =
IG aGx IG mg sin θ = r2 mr 2 + IG
If Fmax > F there will be no slip; that is, if µs cos θ >
IG sin θ sin θ = 2 mr + IG 2
there will be no slip. 238 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
2.33 See the following figure. Summing forces in the x direction: maGx = µs NB − mg sin θ
(1)
Summing forces in the y direction: NA + NB = mg cos θ
(2)
Summing moments about G: NA LA − NB LB + µs NB H = 0
(3)
Solve (2) and (3): NA =
mg cos θ (µs H − LB ) 16, 677 cos θ (µs − 2.1) = µs H − LA − LB µs − 4.6
NB = −
mgLA cos θ 40, 025µs cos θ = µs H − LA − LB 4.6 − µs
The maximum acceleration is, from (1), aGx =
23.544µs cos θ − 9.81 sin θ 4.6 − µs
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Figure : for Problem 33
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2.34 a) Integrate the equation of motion as follows. v(t) =
2
v(t)
dv =
0
0
*
eat v(t) = 35 a 35 v(t) = a With a = 1 and g = 9.81,
2 t(
3#
#
1 t− a $
)
35teat − g dt $+t
0
− gt 4
1 at 1 t− e + − gt a a
,

v(t) = 35 (t − 1) et + 1 − 9.81t You can plot this to get an idea of the approximate value of t that makes v(t) = 300. It is approximately t = 2. To obtain a more precise value, create the following userdefined function file veldiff.m. function dv = veldiff(t) global a V dv = V  ((35/a)*((t1/a).*exp(a*t)+1/a)9.81*t); In the Command window, type %global a V %a = 1; V = 300; %T = fzero(# veldiff#,2) The answer returned is T = 2.0493 s.
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2.34 b) Create the following file vel2.m, using the time 2.05 s for a = 1 as the starting guess. global a V a = 1; T0 = fzero(# veldiff#,2.05); k = 0; for a = [1:0.01:5] k = k + 1; T(k) = fzero(#veldiff# ,T0); end plot([1:0.01:5],T) In the Command window, type %vel2 The plot is similar to that shown in Figure 2.5.1b, except that the maximum value is 2.05.
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2.35 Modify the program given in Example 2.5.2. a) % Set the values for initial speed, gravity, and angle. v0 = 40; g = 9.81; theta = 30*pi/180; % Compute the timetohit. t_hit = 2*v0*sin(theta)/g; % Compute the arrays containing time, height, and speed. t = [0:t_hit/100:t_hit]; y = v0*t*sin(theta)  0.5*g*t.^2; v = sqrt(v0^2  2*v0*g*sin(theta)*t + g^2*t.^2); % Determine when the height is no less than 15. u = find(y>=15); % Compute the corresponding times. t_1 = (u(1)1)*(t_hit/100) t_2 = u(length(u)1)*(t_hit/100) The results are 1.0194 and 3.0581 s. Thus the height is no less than 15 m for 1.0194 ≤ t ≤ 3.0581 s. b) % Set the values for initial speed, gravity, and angle. v0 = 40; g = 9.81; theta = 30*pi/180; % Compute the timetohit. t_hit = 2*v0*sin(theta)/g; % Compute the arrays containing time, height, and speed. t = [0:t_hit/100:t_hit]; y = v0*t*sin(theta)  0.5*g*t.^2; v = sqrt(v0^2  2*v0*g*sin(theta)*t + g^2*t.^2); % Determine when the height is no less than 15, % and the speed is no greater than 36. u = find(y>=15&v LI(RKT + A2 ) If the system is stable, the final value theorem can be applied. It gives θss = lim s s→0
LIs3
+ (RI +
Lc)s2
A Vi A = Vi 2 + (LKT + Rc)s + RKT + A s RKT + A2
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6.34 a) With vf = 0, if = 0, and IsΩ(s) = −cΩ(s) − TL (s) Thus
Ω(s) 1 =− TL (s) Is + c
b) I Thus
d2 θ dθ = K T if − c − T L 2 dt dt
$
%
Is2 + cs Θ(s) = KT If (s) − TL (s)
Also,
Vf (s) = (Lf s + Rf )If (s) and therefore
This gives
$
%
Is2 + cs Θ(s) = KT
Vf (s) − TL (s) (Lf s + Rf )
Θ(s) KT KT = = 2 2 Vf (s) (Lf s + Rf )(Is + cs) s[ILf s + (Rf I + Lf c)s + Rf c] and
Θ(s) 1 =− TL (s) s(Is + c)
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6.35 See the following figure. The equivalent inertia and damping at the motor shaft are Ie = Im +
IL N2
ce =
cL + cm N2
The transfer functions are Ωf (s) KT /N = Vf (s) (Lf s + Rf )(Ie s + ce ) Ωf (s) 1/N 2 = TL (s) Ie s + ce
Figure : for Problem 6.35
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6.36 Summing moments on I1 and I2 gives I1
d2 θ1 = T − kT (θ1 − θ2 ) = KT if − kT (θ1 − θ2 ) dt2 I2
d2 θ2 = kT (θ1 − θ2 ) − cθ˙2 dt2
For the field circuit,
dif + Rif dt Transforming these equations and solving for Θ2 (s) gives vf = Lf
Θ2 (s) =
s2 (L
kT K T 2 f s + Rf )[I1 I2 s + cI1 s + kT (I1 + I2 )]
The differential equation is I1 I2 Lf
d5 θ2 d4 θ2 d3 θ2 d2 θ2 +(cI1 I2 Lf +Rf ) 4 +[Lf kT (I1 +I2 )+Rf cI1 ] 3 +Rf kT (I1 +I2 ) 2 = kT KT vf 5 dt dt dt dt
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6.37 There are two circuits and one inertia in this system, and we must write an equation for each. For the generator circuit, Kirchhoff’s voltage law gives vf = Rf if + Lf
dif dt
For the motor circuit, Kirchhoff’s voltage law gives va = Ra ia + La
dia + Kb ω dt
where va = Kf if . For the inertia I, Newton’s law gives: I
dω = T − cω − TL dt
where T = KT ia . Substitute for va and T , and rearrange to obtain: Lf
dif + Rf if = vf dt
dω + cω = KT ia − TL dt dia La + R a ia = K f if − K b ω dt I
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6.38 Equations (6.6.5) and (6.6.6) give ia =
5 × 10−4 Va + 0.2TL 7.5 + 200TL = −4 2 (0.8)5 × 10 + (0.2) 40.4 ω=
0.2Va − 0.8TL 300 − 80TL = −2 4.04 × 10 4.04
(1)
(2)
Setting TL = 0 in (2) gives the noload speed ω = 300/4.04 = 74.26 rad/s or 709 rpm. From (1) we obtain the noload current: ia = 7.5/40.4 = 0.186 A. Setting ω = 0 in (2) gives the stall torque TL = 0.2(15)/0.8 = 3.75 N·m. From (1) we obtain the stall current: ia = [7.5 + 200(15/4)]/4.04 = 187.5 A.
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6.39 (a) The characteristic equation is La Is2 + (La c + Ra I)s + Ra c + Kb KT = 0 The roots are s=
−(La c + Ra I) ±
&
Ra2 I 2 − 2Ra ILa c + L2a c2 − 4La IKb KT 2La I
b) For c = 0 the roots are −219 and −47.5. The speed and the current will not oscillate. It will take about 4(1/47.5) = 0.08 second for the speed and the current to become constant. For c = 0.01 the roots are −196 ± 73.5i. The speed and current will oscillate with a frequency of 73.5 rad/s and a period of 0.085 s. It will take about 4(1/196) = 0.02 second for the speed and current to become constant. Because this time is less than the period, we will not see any oscillations in the plots. For c = 0.1 the roots are −1240 and −277. The speed and current will not oscillate. It will take about 4(1/277) = 0.014 second for them to become constant.
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6.40 Use the transfer functions given by (6.6.1) through (6.6.4). All four transfer functions have the same denominator, which is D(s) = 20 × 10−7 s2 + 4.02 × 10−4 s + 4 × 10−4 Current Response to Command Input Assuming va is a unit step input, we have from (6.6.1): 5 × 10−4 (s + 1) 250(s + 1) Ia (s) = = (1) 2 sD(s) s(s + 201s + 2020) The characteristic roots are s = −100.5 ± 100.5j. Thus the denominator can be expressed as (s + 100.5)2 + (100.5)2 . Expanding Ia (s) in a partial fraction expansion gives Ia (s) =
C1 100.5C2 (s + 100.5)C3 + + 2 2 s (s + 100.5) + (100.5) (s + 100.5)2 + (100.5)2
(2)
This gives a solution of the form: ia (t) = C1 + C2 e−100.5t sin 100.5t + C3 e−100.5t cos 100.5t Reducing (2) to a single fraction using the least common denominator s[(s + 100.5)2 + (100.5)2 ], we obtain Ia (s) =
C1 [(s + 100.5)2 + (100.5)2 ] + [100.5C2 + (s + 100.5)C3 ]s s[(s + 100.5)2 + (100.5)2 ]
(3)
Comparing the numerators of (1) and (3), we obtain C1 [(s + 100.5)2 + (100.5)2 ] + [100.5C2 + (s + 100.5)C3 ]s = 250(s + 1) or (C1 + C3 )s2 + (201C1 + 100.5C2 + 100.5C3 )s + 20, 200C1 = 250s + 250 Comparing like powers of s gives C1 = −C3 = 0.01238
C2 = 2.47518
Thus, after multiplying by 10, the magnitude of the step input, we have ia (t) = 0.1238 + 24.7518e−100.5t sin 100.5t − 0.1238e−100.5t cos 100.5t Speed Response to Command Input Assuming va is a unit step input, we have from (6.6.3): 0.2 105 Ω(s) = = (4) sD(s) s(s2 + 201s + 2020) Expanding Ω(s) in a partial fraction expansion gives Ω(s) =
C1 100.5C2 (s + 100.5)C3 + + 2 2 s (s + 100.5) + (100.5) (s + 100.5)2 + (100.5)2
(5)
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This gives a solution of the form: ω(t) = C1 + C2 e−100.5t sin 100.5t + C3 e−100.5t cos 100.5t Reducing (5) to a single fraction using the least common denominator s[(s + 100.5)2 + (100.5)2 ], we obtain Ω(s) =
C1 [(s + 100.5)2 + (100.5)2 ] + [100.5C2 + (s + 100.5)C3 ]s s[(s + 100.5)2 + (100.5)2 ]
(6)
Comparing the numerators of (5) and (6), we obtain C1 [(s + 100.5)2 + (100.5)2 ] + [100.5C2 + (s + 100.5)C3 ]s = 105 or (C1 + C3 )s2 + (201C1 + 100.5C2 + 100.5C3 )s + 20, 200C1 = 105 Comparing like powers of s gives C1 = −C3 = 4.9505
C2 = −C1 = −4.9505
(7)
Thus, after multiplying by 10, the magnitude of the step input, we have $
ω(t) = 49.505 1 − e−100.5t sin 100.5t − e−100.5t cos 100.5t
%
Current Response to Disturbance Input Comparing (6.6.2) with (6.6.3), and noting that Kb = KT , we see that Ia (s) Kb Ω(s) Ω(s) = = TL (s) KT Va (s) Va (s) Thus the response of ia to a unitstep disturbance TL will be the same as the response of ω to a unitstep voltage va . Thus we can use the coefficients given in (7) and multiply them by 0.2, the magnitude of TL , to obtain $
ıa (t) = 0.9901 1 − e−100.5t sin 100.5t − e−100.5t cos 100.5t
%
Speed Response to Disturbance Input To obtain the response of ω to TL , use (6.6.4). Assuming TL is a unit step input, we have from (6.6.4): Ω(s) =
4 × 10−3 s + 0.8 5 × 10−4 s + 0.1 = sD(s) s(s2 + 201s + 2020)
(8)
Expanding Ω(s) in a partial fraction expansion gives Ω(s) =
C1 100.5C2 (s + 100.5)C3 + + 2 2 s (s + 100.5) + (100.5) (s + 100.5)2 + (100.5)2
(9)
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This gives a solution of the form: ω(t) = C1 + C2 e−100.5t sin 100.5t + C3 e−100.5t cos 100.5t Reducing (9) to a single fraction using the least common denominator s[(s + 100.5)2 + (100.5)2 ], we obtain Ω(s) =
C1 [(s + 100.5)2 + (100.5)2 ] + [100.5C2 + (s + 100.5)C3 ]s s[(s + 100.5)2 + (100.5)2 ]
(10)
Comparing the numerators of (9) and (10), we obtain C1 [(s + 100.5)2 + (100.5)2 ] + [100.5C2 + (s + 100.5)C3 ]s = 5 × 10−4 s + 0.1 or (C1 + C3 )s2 + (201C1 + 100.5C2 + 100.5C3 )s + 20, 200C1 = 5 × 10−4 s + 0.1
Comparing like powers of s gives
C1 = −C3 = 4.9505 × 10−6
C2 = −4.187 × 10−7
Thus, after multiplying by 0.2, the magnitude of the step input, we have ω(t) = 9.901 × 10−7 − 8.37 × 10−8 e−100.5t sin 100.5t − 9.901 × 10−7 e−100.5t cos 100.5t
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6.41 Given Va = 20 and Istall = 25
ino
load
= 0.6
Thus Ra =
Va istall
ωno
=
load
=
2400 2π = 80π rad/s 60
20 = 0.8 Ω 25
From (6.6.6) and (6.6.6) with TL = 0 and Kb = KT , ino
load
= 0.6 =
ωno
load
= 80π =
20c 0.8c + KT2 20KT 0.8c + KT2
These two equations have the solution KT = Kb = 0.078 N·m/A and c = 1.85 × 10−4 N·m·s/rad.
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6.42 Refer to Table 6.6.1. θf = 3π/4 rad. Also, I = 0.215 t1 = 0.3
Td = 4.2 t2 = 1.7
tf = 2 R=4
KT = 0.3 = Kb Note that L is not needed. The calculated quantities are E = 35.7 J per cycle ωmax = 1.386 rad/s = 13.2 rpm Tmax = 5.19 N · m Trms = 4.24 N · m imax = 17.3 A irms = 14.1 A vmax = 69.6 V
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6.43 Refer to Table 6.6.1. θf = 11(2π) = 22π rad. Also, I = 0.05
Td = 3.6
tf = 3
t1 = 0.5
t2 = 2.5
R=3
KT = 0.4 = Kb Note that L is not needed. The calculated quantities are E = 872 J per cycle ωmax = 27.65 rad/s = 264 rpm Tmax = 6.36 N · m Trms = 3.94 N · m imax = 15.9 A irms = 9.85 A vmax = 58.8 V
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6.44 a) The transfer function is Y (s) s2 = T (s) = 2 Z(s) s + 18s + 100 With s = 120j we obtain ' ' ' ' −(120)2 ' ' M = T (120j) = ' ' = 0.9957 ' 100 − (120)2 + 18(120)j '
φ = # T (120j) = − tan−1
Thus the steady state response is
18(120) = −2.992 100 − (120)2
y(t) = 9.957 sin(120t − 2.992) b) The transfer function is Y (s) s2 = T (s) = 2 Z(s) s + 1800s + 106 With s = 120j we obtain ' ' ' ' −(120)2 ' ' M = T (120j) = ' 6 ' = 0.0144 ' 10 − (120)2 + 1800(120)j '
φ = # T (120j) = − tan−1
Thus the steady state response is
1800(120) = −0.2157 106 − (120)2
y(t) = 0.144 sin(120t − 0.2157) For case (a), the amplitude of the response (9.957) is almost identical to the amplitude of the displacement input (10). Thus the instrument functions as a vibrometer. For case (b), the amplitude of the response (0.144) is equal to 10−6 times the amplitude of the acceleration of the input, which is 10(120)2 = 0.144 × 106 . Thus the instrument functions as an accelerometer with a gain of 10−6 .
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6.45 Summing forces on the mass m gives m
d2 x dx +c + kx = fs + Kf i 2 dt dt
or (ms2 + cs + k)X(s) = Fs (s) + Kf I(s)
(1)
Since the applied voltage is zero, the circuit equation is L
di dx + Ri + Kb =0 dt dt
or (Ls + R)I(s) + Kb sX(s) = 0
(2)
Solving (1) for X(s) and substituting into (2), and rearranging, gives the transfer function. I(s) −Kb s = 3 2 Fs (s) mLs + (cL + mR)s + (kL + cR + Kb Kf )s + kR
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6.46 a) The roots of the secondorder model are s = −4333 ± 13 462j. b) The roots of the thirdorder model are s = −1643 ± 16 513j, which is the dominant pair, and s = −8715.
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6.47 The script file is KT = 0.2; Kb = 0.2; c = 5e4; Ra = 0.8; La = 4e3; I = 5e4; den = [La*I, Ra*I + c*La, c*Ra + Kb*KT]; % Speed transfer function sys1 = tf(KT, den); % Current transfer function sys2 = tf([I, c], den); [om, t1] = step(sys1); [ia, t2] = step(sys2); subplot(2,1,1) plot(t1, 10*om),xlabel($t (s)$ ),ylabel($\omega (rad/s)$ ) subplot(2,1,2) plot(t1, 10*ia),xlabel($t (s)$ ),ylabel($i_a (A)$ ) The plot is shown in the following figure. The peak current is approximately 8 A. The more accurate value of 8.06 A can be found with the step(sys2) function by rightclicking on the plot, selecting characteristics, then peak response, and multiplying the answer by 10. 60
! (rad/s)
50 40 30 20 10 0
0
0.01
0.02
0.03 t (s)
0.04
0.05
0.06
0
0.01
0.02
0.03 t (s)
0.04
0.05
0.06
10 8
ia (A)
6 4 2 0 −2
Figure : for Problem 6.47
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6.48 The script file is KT = 0.2; Kb = 0.2; c = 5e4; Ra = 0.8; La = 4e3; I = 5e4; den = [La*I, Ra*I + c*La, c*Ra + Kb*KT]; % Speed transfer function sys1 = tf(KT, den); % Current transfer function sys2 = tf([I, c], den); % Applied voltage t = [0:0.001:0.05]; va = 10*ones(size(t)); om = lsim(sys1, va, t); ia = lsim(sys2, va, t); subplot(2,1,1) plot(t, om),xlabel($t (s)$ ),ylabel($\omega (rad/s)$ ) subplot(2,1,2) plot(t, ia),xlabel($t (s)$ ),ylabel($i_a (A)$ ) The plot is shown in the following figure. The peak current is approximately 8 A. The more accurate value of 8.06 A can be found with the lsim(sys2, va, t) function by rightclicking on the plot, selecting characteristics, then peak response.
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60
! (rad/s)
50 40 30 20 10 0
0
0.005
0.01
0.015
0.02
0.025 t (s)
0.03
0.035
0.04
0.045
0.05
0
0.005
0.01
0.015
0.02
0.025 t (s)
0.03
0.035
0.04
0.045
0.05
10 8
ia (A)
6 4 2 0 −2
Figure : for Problem 6.48
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6.49 The resistance R was incorrectly given as 104 Ω. It should be 103 Ω. See Example 6.3.3 for the transfer functions. The script file is R = 1e+3; C = 2e6; L = 2e3; den = [L*R*C, L, R]; % Transfer function for v1: sys1 = tf(1, den); % Transfer function for v2: sys2 = tf([R*C, 0], den); % Response to unitstep voltage v1: t = [0:0.00001:0.025]; i31 = step(sys1, t); % Voltage v2: v2 = 4*sin(2*pi*60*t); % Response to voltage v2: i32 = lsim(sys2, v2, t); % Total response: i3 = 5*i31 + i32; plot(t, i3),xlabel($t (s)$ ),ylabel($i_3 (A)$ ) The plot is shown in the following figure. The steadystate response is a sine wave.
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0.016
0.014
0.012
3
i (A)
0.01
0.008
0.006
0.004
0.002
0
0
0.005
0.01
0.015
0.02
0.025
t (s)
Figure : for Problem 6.49
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6.50 See equations (6.6.1) through (6.6.4) for the transfer functions, where I and c must be replaced with the equivalent inertia and equivalent damping: Ie = I + 4 × 10−4 /N 2 and ce = c + 1.8 × 10−3 /N 2 . The script file is KT = 0.2; Kb = 0.2; c = 3e4; Ra = 0.8; La = 4e3; I = 4e4; N = 3; % Equivalent inertia: Ie = I + 1e3/N^2; % Equivalent damping: ce = c + 1.8e3/N^2; den = [La*Ie, Ra*Ie + ce*La, ce*Ra + Kb*KT]; % Speed command transfer function sys1c = tf(KT, den); % Current command transfer function sys2c = tf([I, c], den); % Speed disturbance transfer function sys1d = tf([La, Ra], den); % Current disturbance transfer function sys2d = tf(Kb, den); % Unit Command response: [omc, t1c] = step(sys1c); [iac, t2c] = step(sys2c); % Unit Disturbance response: omd = step(sys1d, t1c); iad = step(sys2d, t2c); % Total response: om = 20*omc + 0.04*omd; ia = 20*iac + 0.04*iad; subplot(2,1,1) plot(t1c, om),xlabel($t (s)$ ),ylabel($\omega (rad/s)$) subplot(2,1,2) plot(t2c, ia),xlabel($t (s)$ ),ylabel($i_a (A)$ ) The plot is shown in the following figure. The peak current is 12.73 A, which can be found by displaying the values in the array ia.
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120
! (rad/s)
100 80 60 40 20 0
0
0.01
0.02
0.03 t (s)
0.04
0.05
0.06
0
0.01
0.02
0.03 t (s)
0.04
0.05
0.06
15
ia (A)
10
5
0
−5
Figure : for Problem 6.50
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6.51 The main script file is Problem6p51.m. % Program Problem6p51.m KT = 0.05; Kb = KT; c = 0; Ra = 0.8; La = 3e3; I = 8e5; trapezoid motortf current = lsim(currenttf, v, t); speed = lsim(speedtf, v, t); subplot(2,1,1) plot(t,current),xlabel($t (s)$ ),ylabel($Current (A)$ ) subplot(2,1,2) plot(t,speed),xlabel($t (s)$ ),ylabel($Speed (rad/s)$ ) performance
This program calls the following files. % trapezoid.m Trapezoidal voltage profile t1 = 0.5; t2 = 2; tfinal = 2.5; t3 = 4; max_v = 30; dt = t3/1000; t = [0:dt:t3]; for k = 1:1001 if t(k) τ1 τ2 This implies that τ2 > τ1 , which says R1 R2 C > R1 C R1 + R2 Canceling R1 C gives
R2 >1 R1 + R2
which is impossible to satisfy. Thus the circuit cannot be a lowpass filter.
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9.8 a)
b)
20 5 ! ! yss = ! 5= ! 20 = 34.92 2 2 2 (10ω) + 1 (4ω) + 1 [10(0.2)] + 1 [4(0.2]2 + 1 0.5 0.5 yss = ! 32 = ! 32 = 0.178 2 2 2 (100 − ω ) + (10ω) (100 − 52 )2 + [10(5)]2
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9.9 a)
4 jω(100 − ω 2 + 10ωj)
T (jω) = M (ω) = Thus M (9) = 0.0048.
4 ω (100 − ω 2 )2 + 100ω 2 !
φ(ω) = −! jω − ! [(100 − ω 2 ) + 10ωj] So φ(9) = π/2 − tan−1 (90/19) = 0.2081. Thus yss (t) = 6M (9) sin[9t + φ(9)] = 0.0288 sin(9t − 0.2081) b)
10 + ωj)
T (jω) =
−ω 2 (1
M (ω) =
10 √ ω2 1 + ω2
Thus M (2) = 1.118. φ(ω) = −! (−10/ω 2 ) − ! [(1 + ωj) So φ(2) = π − tan−1 (2) = 2.0344. Thus yss (t) = 3M (2) sin[2t + φ(2)] = 3.354 sin(2t + 2.0344) c) ωj (1 + 2ωj)(1 + 5ωj) ω √ M (ω) = √ 2 1 + 4ω 1 + 25ω 2 T (jω) =
Thus M (0.7) = 0.1118. φ(ω) = ! ωj − ! (1 + 2ωj) − ! (1 + 5ωj) So φ(0.7) = π/2 − tan−1 (1.4) − tan−1 (3.5) = −0.6722. Thus yss (t) = 5M (0.7) sin[0.7t + φ(0.7)] = 0.559 sin(0.7t − 0.6722)
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d) T (jω) =
−ω 2 (1 + 2ωj)(1 + 5ωj)
M (ω) = √
ω2 √
1 + 4ω 2 1 + 25ω 2
Thus M (0.7) = 0.0782. φ(ω) = ! (−ω 2 ) − ! (1 + 2ωj) − ! (1 + 5ωj) So φ(0.7) = −π/2 − tan−1 (1.4) − tan−1 (3.5) = −3.8138. Thus yss (t) = 5M (0.7) sin[0.7t + φ(0.7)] = 0.391 sin(0.7t − 3.8138)
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9.10 The resonant frequency and peak magnitude are given by $
ωr = ωn 1 − 2ζ 2
Mr =
√ Note that ωn = 2. Thus a) For ζ = 0.1, ωr = 1.4 and Mr = 5.025. b) For ζ = 0.3, ωr = 1.28 and Mr = 1.747.
1 2ζ 1 − ζ 2 !
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9.11 The maximum amplitude occurs at ω = ωr and is 22Mr , where Mr = The damping ratio is
1 2ζ 1 − ζ 2 !
c c ζ= √ = √ 2 50 10 2
Thus 22Mr =
22 =3 2ζ 1 − ζ 2
Solve for ζ by squaring each side to obtain
!
4ζ 4 − 4ζ 2 − 13.44 = 0
√ which has the positive root ζ 2 = 4.2, or ζ = 2.05. Thus c = 10 2(2.05) = 28.99.
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9.12 The transfer function is T (s) =
13s2
From (9.2.20)
1 + 2s + k
$
ωr = ωn 1 − 2ζ 2
where
ωn =
%
k 13
2 1 ζ= √ =√ 2 13k 13k We want ωr = 4. Thus we must solve the following for k. 4=
%
k 13
&
1−
2 1√ = 13k − 2 13k 13
This gives k = 208 lb/ft. This gives ζ = 0.0192. From (9.2.21), 1 Mr = ! = 26.295 2ζ 1 − ζ 2
Thus the amplitude of the steadystate response is 26.295(10) = 262.95 The phase shift of the response is (with ω = 4) φ = −! [(k − 13ω 2 ) + 2ωj] = −! [−8 + 8j] = −2.356 Thus the steadystate response is xss (t) = 262.95 sin(4t − 2.356)
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9.13 a) T (jω) = M (ω) =
7 ωj(58 − ω 2 + 6ωj)
7 ω (58 − ω 2 )2 + 36ω 2 !
Since M (0) = ∞, the resonant frequency is at ω = 0. b) 7 T (jω) = 2 (174 − 3ω + 18ωj)(58 − 2ω 2 + 8ωj)
7 ! M (ω) = ! (174 − 3ω 2 )2 + 324ω 2 (58 − 2ω 2 )2 + 64ω 2
A plot of M versus ω shows that M has a single peak near ω = 4.97, which is the resonant frequency. We might have expected two peaks since there are two quadratic factors in the denominator. These quadratic factors have resonant frequencies of 4.58 and 6.32. Because these frequencies are so close in value, their two peaks merge into a single peak at ω = 4.97.
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9.14 Applying the voltage law to each loop gives v1 − Ri1 − L
di3 =0 dt
'
1 di3 i2 dt − L =0 C dt From conservation of charge, i3 = i1 + i2 . Take the Laplace transform of each equation using zero initial conditions. Solve the first two for I1 (s) and I2 (s): V1 (s) − LsI3 (s) I1 (s) = R v2 −
I2 (s) = CsV2 (s) − LCs2 I3 (s)
Substitute these into the third equation to obtain
(RLCs2 + Ls + R)I3 (s) = V1 (s) + RCsV2 (s) The two transfer functions are I3 (s) 1 1 = = −5 2 2 V1 (s) RLCs + Ls + R 10 s + 0.1s + 100 I3 (s) RCs 10−4 s = = V2 (s) RLCs2 + Ls + R 10−5 s2 + 0.1s + 100 The roots are s = −1127 and s = −8873. Thus the corner frequencies are ω1 = 1127 rad/s and ω2 = 8873 rad/s. First consider the transfer function I3 (s)/V1 (s). At low frequencies the m curve approaches 20 log(105 /[(1127)(8873)] = −40 dB. The m curve breaks down at ω1 = 1127 and again at ω2 = 8873 rad/s. So the system acts like a low pass filter for the input voltage v1 , and it tends to filter out frequency components in v1 that are higher than ω1 = 1127 rad/s. The low frequency gain is M = 10m/20 = 10−40/20 = 0.01. Now consider the transfer function I3 (s)/V2 (s). At low frequencies the m curve approaches 20 log(0) = −∞. At ω = 10, "
m = 20 log √
10(10) √ 2 1127 + 102 88732 + 102
#
= −100 dB
The m curve has a slope of 20 dB/decade until ω1 = 1127, when the slope becomes approximately zero. The curve breaks downward at ω2 = 8873 rad/s, and the slope becomes −20 dB/decade for higher frequencies. So the system acts like a band pass filter for the input voltage v2 , and it passes frequency components in v2 that are between ω1 = 1127 rad/s and ω2 = 8873 rad/s.
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9.15 In the first printing of the text, Figure P9.15 was incorrect. The damper should be between the mass and the support. The displacement y should act on the spring. a) The equation of motion is m¨ x = k(y − x) − cx˙ or x ¨ + cx˙ + 600x = 600y The transfer function is
X(s) 600 = 2 Y (s) s + cs + 600 √ √ √ This has the same form as (9.2.14) with ωn = 600 = 10 6 and ζ = c/(20 6). So, provided that c is such that ζ ≤ 0.707, we can use (9.2.20) and (9.2.21) to obtain T (s) =
$ √ $ ωr = ωn 1 − 2ζ 2 = 10 6 1 − 2ζ 2 rad/s
Mr =
1 2ζ 1 − ζ 2 !
√ Since c = 20ζ 6, ζ will be no greater than 0.707 if 0 ≤ c ≤ 34.64 N·s/m. For 0 ≤ c ≤ 34.64, a plot of ωr versus c shows that the resonant frequency varies from 3.6 (for c = 34.64) to 24.5 rad/s (for c = 0). For 0 ≤ c ≤ 34.64, a plot of Mr versus c shows that Mr varies from ∞ (for c = 0) to 1 rad/s (for c = 34.64). For values of ζ > 0.707 (for c > 34.64), there is no resonant peak and thus no resonant frequency. b) The transfer function is T (s) =
X(s) k k/m ωn2 = = = Y (s) ms2 + cs + k s2 + (c/m)s + k/m s2 + 2ζωn s + ωn2
This has the same form as (9.2.14), so provided ζ ≤ 0.707, we can use (9.2.20) and (9.2.21) to compute ωr and Mr . For values of ζ > 0.707, there is no resonant peak and thus no resonant frequency.
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9.16 Applying the voltage law gives vs −
1 C
'
i dt − L
di − vo = 0 dt
where v0 = Ri. Use the latter equation to eliminate i: LC
d2 vo dvo dvs + RC + vo = RC 2 dt dt dt
For the given values, 5 × 10−8 The transfer function is
For R = 10,
d2 vo dvo dvs + 10−5 R + vo = 10−5 R dt2 dt dt Vo (s) 103 Rs = 2 Vs (s) 5s + 103 Rs + 108 Vo (s) 104 s = 2 Vs (s) 5s + 104 s + 108
The roots are s = −1000 ± 4359j. For R = 100, Vo (s) 105 s = 2 Vs (s) 5s + 105 s + 108 The roots are s = −100.05 and s = −199, 900. The plots are shown below. The peak is approximately the same for each. It is −0.000185 dB, or M = 10−0.000185/20 = 1 at 5170 rad/s. Thus an increase in the value of R by a factor of 100 does not significantly change the peak value or the peak frequency, but it does change the spread of the plot about the peak.
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0
−10
R = 1000
−20
Magnitude (dB)
−30
−40
R = 10 −50
−60
−70
−80
−90 0 10
1
10
2
10
3
10
4
10
5
10
6
10
7
10
Frequency (rad/sec)
Figure : Log magnitude plots for Problem 9.16
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9.17 The equations of motion are I1 ω˙ 1 = T1 − c1 (ω1 − ω2 ) I2 ω˙ 2 = c1 (ω1 − ω2 ) − c2 ω2
Applying the Laplace transform for zero initial conditions, and eliminating Ω1 (s), we obtain the transfer function: Ω2 (s) 100 = 2 T1 (s) s + 5s + 2 From this we obtain the magnitude ratio and the phase angle. 100 M (ω) = ! (2 − ω 2 )2 + (5ω)2
5ω 2 − ω2 Evaluation of M and φ for ω = 0, 1.5, and 2 gives the following results. φ(ω) = − tan−1
M (0) = 50 φ(0) = 0
M (1.5) = 13.3259
φ(1.5) = −1.604 rad
Thus the steady state response is
M (2) = 9.8058 φ(2) = −1.763 rad
ω2ss = 50(4) + 13.3259(2) sin(1.5t − 1.604) + 9.8058(0.9) sin(2t − 1.763) or ω2ss = 200 + 26.6518 sin(1.5t − 1.604) + 8.8252 sin(2t − 1.763)
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9.18 The forcing frequency is ω = 5.2 and the natural frequency is ωn = beat period is 2π 2π = = 31.4159 ω − ωn  5.2 − 5 The vibration period is
!
75/3 = 5. The
4π 4π = = 1.232 ω + ωn 10.2
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9.19 From (9.3.4) with fo = 0.2, k = 64, and ωn = 8, "
sin 8t x(t) = 0.0125 − t cos 8t 8
#
The plot is shown below. It takes approximately 8.2 sec for x(t) to become greater than 0.1 ft. 0.1
0.08
0.06
0.04
x(t) (ft)
0.02
0
−0.02
−0.04
−0.06
−0.08
−0.1
0
1
2
3
4
5 t (sec)
6
7
8
9
10
Figure : Plot for Problem 9.19
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9.20 From the given data
3400(2π) = 356 rad/s 60 √ c = 2ζ mk = 548 N · s/m
ω=
Also
Thus
Ft (s) cs + k = Fr (s) ms2 + cs + k √ k + cωj k 2 + c2 ω 2 Ft (ω) = Fr  = Fr  ! 2 k − mω + cωj (k − mω 2 )2 + c2 ω 2
For rotating unbalance
Fr  = mu 'ω 2 = 0.1(0.02)(356)2 = 253.5 N Substituting the values of k, c, m, and ω, we obtain Ft  = 253.5(0.1027) = 26 N
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9.21 For base motion,
√ X k 2 + c2 ω 2 =! Y (k − mω 2 )2 + c2 ω 2
Plotting this expression versus ω for the given values of m, c, and k, we find that the ratio X/Y is never less than 1. Therefore, X will never be greater than 2 mm as long as Y is never greater than 2 mm.
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9.22 The vibration frequency of the base motion as a function the vehicle velocity v is ω=
"
5280 20
#"
#
1 (2π)v = 0.4608v 3600
From the results of Example 8.7.1, D3 = (m1 s2 +c1 s+k1 )(m2 s2 +c1 s+k1 +k2 )−(c1 s+k1 )2 = m1 m2 s4 +(m1 k1 +m1 k2 +m2 k1 )s2 +k1 k2 where we have taken c1 = 0. Using the given values, we obtain 114.472s4 + 4.0838 × 105 s2 + 8 × 107 = 0 The roots are s = ±14.4j and s = ±57.96j. Thus the resonant frequencies are 14.4 and 57.96 rad/s, and the resonant speeds are v=
14.4 = 31.25 ft/sec 0.4608
and
v=
57.96 = 125.8 ft/sec 0.4608
or v = 21.3 mph and 85.8 mph. Note that the amplitude of the road surface variation, although given in the problem statement, was not needed.
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9.23 Using ρ = 2388 kg/m3 for the density of steel, the beam mass is calculated to be mb = 2388(2)(0.3)(0.03) = 42.98 kg The equivalent system mass is me = 50 + 0.23mb = 59.886 kg The stiffness of the beam is k=
Ewh3 2 × 1011 (0.3)(0.03)3 = = 1.6875 × 106 N/m 4L3 4(2)3
The forcing frequency is ω=
3400(2π) = 356 rad/s 60
and the amplitude of the unbalance force is Fr  = mu 'ω 2 = 0.02(356)2 = 2535 N Thus the displacement amplitude is, since c = 0,
or 0.428 mm.
Fr  Fr  X=! = = 4.29 × 10−4 m 2 2 k − me ω 2  (k − me ω )
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9.24 Using ρ = 2388 kg/m3 for the density of steel, the beam mass is calculated to be mb = 2388(2)wh = 4776wh where w is the beam width and h is the beam thickness. The equivalent system mass is me = 50 + 0.5mb = 50 + 2388wh The stiffness of the beam is k=
16Ewh3 = 4 × 1011 wh3 L3
The forcing frequency is ω=
3400(2π) = 356 rad/s 60
and the amplitude of the unbalance force is Fr  = mu 'ω 2 = 0.02(356)2 = 2535 N Thus the displacement amplitude is, since c = 0, Fr  Fr  X = 0.01 = ! = 2 2 k − me ω 2  (k − me ω )
(1)
This expression is a complicated function of w and h since both k and me are functions of w and h. So let us temporarily neglect the beam mass. Then me ≈ 50 kg and the previous expression becomes X = 0.01 =
2535 k − 6.3368 × 106 
This gives k = 6.5903 × 106 N/m. Thus wh3 =
6.5903 × 106 = 1.6 × 10−5 4 × 1011
There is an infinite number of solutions to this equation, but we should choose one that results in an equivalent beam mass that is small compared to 50 kg, because we neglected the beam mass in obtaining this solution. One solution is h = 0.04 m, which gives w = 0.25 m, both reasonable dimensions. These give an equivalent beam mass of 0.5(47.76) = 23.88 kg and a total equivalent mass of me = 73.88 kg. Using this value in equation (1) we obtain X=
2535 = 9.14 × 10−4 m 6.59 × 106 − 73.88(356)2 
which is much less than the maximum allowable value of 0.01 m. So our solution is valid. 927 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
9.25 We √ are given m = 1500 kg, k = 20 000 N/m, ζ = 0.04, and Y = 0.01 m. Thus c = 2ζ mk = 438.18. From (9.3.9), with s = jω, !
At resonance, from (9.2.20),
mω 2 k 2 + (cω)2 Ft  = Y  ! (k − mω 2 )2 + (cω)2 $
ω = ωr = ωn 1 − 2ζ 2 = 0.998
%
k = 3.6442 m
Thus Ft  = 2500 N
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9.26 The static deflection is δ = mg/k. Thus k = mg/δ = 200/0.003 = 66 667 N/m. Also, ω = 40 Hz = 251 rad/s ωn =
%
k = m
%
66 667 = 57.2 rad/s (200/9.81)
From (9.3.7) with c = 0, X k =! = 0.055 Y  (k − mω 2 )2
Thus 5.5% of the airframe motion is transmitted to the module.
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9.27 a) Neglect damping in the isolator, and determine its required stiffness k. From (9.3.7) with c = 0, X k k/m ωn2 ! ! =! = = = 0.1 Y  (k − mω 2 )2 (k/m − ω 2 )2 (ωn2 − ω 2 )2
which gives (ω/ωn )2 = 11. Thus ωn2 =
ω2 [3000(2π)/60]2 (314)2 = = 11 11 11
and
2 (314)2 = 556.7 lb/ft 32.2 11 √ b) Let r = ω/ωn . From part (a), ωn = 314/ 11, we have k = mωn2 =
r1 =
2500(2π)/60 √ = 2.77 314/ 11
r2 =
3500(2π)/60 √ = 3.87 314/ 11
and
From (9.3.7) with c = 0,
X 1 = Y  1 − r 2 
Thus the highest percentage of motion will be transmitted at the lowest r value, which is r1 , the value corresponding to 2500 rpm. For 2500 rpm, X 1 = = 0.15 Y  1 − r12  For 3500 rpm,
X 1 = = 0.07 Y  1 − r22 
Thus at most, 15% of the crane motion will be transmitted to the module.
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9.28 The transmitted force is given by (9.3.14) and (9.3.17): Ft  = mu 'ω If we let
!
k 2 + (cω)2 (k − mω 2 )2 + (cω)2
2!
r = ω/ωn then we can express the previous equation as Ft  = mu 'ω
2
%
1 + 4ζ 2 r 2 (1 − r 2 )2 + 4ζ 2 r 2
We are given that mu = 0.05/32.2 = 0.00155 slug, m = 50/32.2 = 1.55 slug. ' = 0.1.12 = 0.0083 ft, and ω = 1000(2π)/60 = 104.7 rad/sec. Thus r= Thus
ω 104.7 =! = 5.83 ωn 500/1.55
Ft  = (0.00155)(0.0083)(104.7)2
%
1 + 136ζ 2 = 0.141 1088 + 136ζ 2
%
1 + 136ζ 2 1088 + 136ζ 2
a) For ζ = 0.05, Ft = 0.0049 lb. b) For ζ = 0.7, Ft = 0.034 lb.
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9.29 The transmitted force is given by (9.3.14) and (9.3.17): Ft  = mu 'ω If we let
!
k 2 + (cω)2 (k − mω 2 )2 + (cω)2
2!
r = ω/ωn then if c = 0, we can express the previous equation as Ft  = mu 'ω 2
1 1 − r 2 
Here Ft = 15 and ωR = 200(2π)/60 = 20.9 rad/s. Thus r= Thus
ω 20.9 20.9 =! =! = 3.62 ωn k/m 2500/75 Ft  = mu '(20.9)2 (0.0826) = 15
Solve for mu ' to obtain mu ' = 0.416 kg·m.
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9.30 For a vibrometer we require that the natural frequency ωn be much less than the forcing frequency ω. Here ω = 2π(200) = 400π rad/s and ωn = So we require that
%
%
k 0.1
k ) 400π 0.1
or k ) 1.579 × 105 N/m
Suppose that k = 1.6×104 . Then the static deflection will be δst = mg/k = 6.1×10−5 m. The designer must verify that there will be enough space in the instrument to accommodate this deflection.
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9.31 In the first printing of the text, Figure P9.31 was incorrect. The damper should be between the mass and the support. The displacement y should act on the spring. The equation of motion is m¨ x = k(y − x) − cx˙ or
The transfer function is T (s) =
200¨ x + 2000x˙ + 2 × 104 x = 2 × 104 y X(s) 2 × 104 100 = = 2 2 4 Y (s) 200s + 2000s + 2 × 10 s + 10s + 100
This has the same form as (9.2.14) with ωn = 10 and ζ = 0.5. So we can use (9.2.20) and (9.2.21) to obtain $ ωr = ωn 1 − 2ζ 2 = 7.07 rad/s Mr =
1 = 1.155 2ζ 1 − ζ 2 !
For the bandwidth, equation (2) of Example 9.4.2 gives r=
&
1−
2ζ 2
$
± 2ζ 1 − ζ 2 = 1.167, 0.605j
Because there is only one positive, real solution, the bandwidth extend from ω = 0 to ω = 1.167ωn = 11.69 rad/s.
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9.32 The transfer function becomes T (s) =
10−6 s 3 × 10−4 s2 + 10−2 s + 1
From the frequency response plot shown below we can tell that the bandwidth is between 45 and 76 rad/s. You can obtain this plot in MATLAB by typing: sys = tf([1e6,0],[3e4,1e2,1]); bodemag(sys)
Bode Diagram −80
−85
−90
Magnitude (dB)
−95
−100
−105
−110
−115
−120
−125 0 10
1
2
10
10
3
10
Frequency (rad/sec)
Figure : Plot for Problem 9.32
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9.33 a) The transfer function for the circuit in part (a) of the figure was derived in Chapter 6. It is Vo (s) G = 2 2 2 Vs (s) R C s + 2RCs + 1 Its natural frequency is ωn = Its damping ratio is
&
1 R2 C 2
=
1 RC
2RC ζ= √ =1 2 R2 C 2
From Equation (1) in Example 9.4.2, with r = ω/ωn = RCω, 1 kM = ! (1 − r 2 )2 + 4r 2 √ which has a maximum of 1 at r = 0. Setting kM = 1/ 2 and solving for r gives r = 0.644. Thus the bandwidth corresponds to 0 ≤ r ≤ 0.644 or 0≤ω≤
0.644 RC
b) The transfer function for the circuit in part (b) of the figure was derived in Chapter 6. It is Vo (s) G = 2 2 2 Vs (s) R C s + 3RCs + 1 Its natural frequency is ωn = Its damping ratio is
&
1 R2 C 2
=
1 RC
3RC ζ= √ = 1.5 2 R2 C 2
From Equation (1) in Example 9.4.2, with r = ω/ωn = RCω, 1 kM = ! (1 − r 2 )2 + 9r 2 √ which has a maximum of 1 at r = 0. Setting kM = 1/ 2 and solving for r gives r = 0.374. Thus the bandwidth corresponds to 0 ≤ r ≤ 0.374 or 0≤ω≤
0.374 RC
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9.34 The circuit equation is LC v¨1 + RC v˙ 1 + v1 = vs or 10−7 v¨1 + 10−6 Rv˙ 1 + v1 = vs Its natural frequency and damping ratio are √ √ ωn = 107 = 103 10 10R 5R √ ζ= √ = 1000 10 2 107 The resonant frequency is $ √ ωr = ωn 1 − 2ζ 2 if ζ ≤ 1/ 2
Thus there is a resonant peak and a resonant frequency only if 5R 1 √ ≤√ 1000 10 2 which implies that
√ 1000 10 √ R≤ = 447 Ω 5 2
If there is a resonant peak, its value is Mr =
1 2ζ 1 − ζ 2 !
A plot of Mr versus R for 0 ≤ R ≤ 447 shows that Mr = 3.2026 for R = 100. As R is increased to 447, Mr decreases to Mr = 1. For R > 447, the maximum value of M occurs at ω = 0. The following plots show the variation of M versus r = ω/ωn , for R = 100, 250, 400, 500, 750, and 1000 Ω. The plots show that as R is increased, the system changes from a bandpass filter to a low pass filter. For R ≤ 447, increasing R decreases the bandwidth. For R > 447, increasing R decreases the bandwidth.
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3.5 3
100
2.5 M
2 1.5
250
1 400
0.5 0
0
0.2
0.4
0.6
0.8
1 r
1.2
1.4
1.6
1.8
2
1.2
1.4
1.6
1.8
2
1 0.8
500
750 M
0.6 0.4
1000
0.2 0
0
0.2
0.4
0.6
0.8
1 r
Figure : Plot for Problem 9.34
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9.35 The magnitude ratio is M (ω) = √ and the phase angle is
1 1 =√ 1 + R2 C 2 ω 2 1 + 36 × 10−8 ω 2
φ(ω) = − tan−1 RCω = − tan−1 6 × 10−4 ω The bandwidth of this circuit is 0 ≤ ω ≤ 1/RC = 1667 rad/s. Thus the cos 720πt term and the higher terms lie outside the bandwidth. Evaluation of M and φ for ω = 0, 240π and 480π gives the following results. M (0) = 1 φ(0) = 0
M (240π) = 0.9111
φ(240π) = −0.425 rad
Thus the steady state response is voss =
M (480π) = 0.7415 φ(480π) = −0.735 rad
20 40 40 − 0.9111 cos(240πt − 0.425) − 0.7415 cos(480πt − 0.735) π 3π 15π
or voss = 6.3662 − 3.8664 cos(240πt − 0.425) − 0.6294 cos(480πt − 0.735)
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9.36 The magnitude ratio is M (ω) = √
1 1 =√ 1 + R2 C 2 ω 2 1 + 10−6 ω 2
and the phase angle is φ(ω) = − tan−1 RCω = − tan−1 10−3 ω The bandwidth of this circuit is 0 ≤ ω ≤ 1/RC = 1000 rad/s. Thus the sin 360πt term and the higher terms lie outside the bandwidth. Evaluation of M and φ for ω = 0 and 120π gives the following results. M (0) = 1 φ(0) = 0 Thus the steady state response is
M (120π) = 0.9357 φ(120π) = −0.361 rad
voss = 5 + 0.9357
20 sin(120πt − 0.361) π
or voss = 5 + 5.9569 sin(120πt − 0.361)
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9.37 The equation of motion is m¨ x + cx˙ + kx = cy˙ + ky or x ¨ + 98x˙ + 4900x = 98y˙ + 4900y The transfer function is
The magnitude ratio is
and the phase angle is
X(s) 98s + 4900 = 2 Y (s) s + 98s + 4900 √ 49002 + 982 ω 2 M (ω) = ! (4900 − ω 2 )2 + 982 ω 2 φ(ω) = tan−1
98ω 98ω − tan−1 4900 4900 − ω 2
A plot √ of M versus ω shows that M has a peak of 1.2764 at ω = 55 rad/s. Noting that 1.2764/ 2 = 0.9026 and that M (0) = 1, we see that the lower bandwidth frequency is 0. The plot also shows that M (111) = 0.906 which is close to 0.9026. So the upper bandwidth frequency is 111 rad/s. The circuit is a low pass filter with the bandwidth of 0 ≤ ω ≤ 111 rad/s. So we see that only the terms up to and including the sin 30πt term lie within the bandwidth. Evaluation of M and φ for ω = 0, 10π, 20π, and 30π gives the following results. M (0) = 1 φ(0) = 0
M (10π) = 1.1623
φ(10π) = −0.1057 rad
Thus the steady state response is xss =
M (20π) = 1.263 φ(20π) = −0.5187 rad
M (30π) = 1.0395 φ(30π) = 2.2467 rad
1 1 1 1 −1.1623 sin(10πt−0.1057)−1.263 sin(20πt−0.5187)−1.0395 sin(30πt+2.2467) 20 10π 20π 30π
or xss = 0.05 − 0.037 sin(10πt − 0.1057) − 0.0201 sin(20πt − 0.5187) − 0.011 sin(30πt + 2.2467)
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9.38 a) The transfer function is T (s) =
1 0.1s + 1
The time constant is τ = 0.1 sec. and the bandwidth is 1/τ = 10 rad/sec. b) The magnitude ratio and phase angle are given by M (ω) = √
1 0.01ω 2 + 1
φ(ω) = − tan−1 (1 + 0.1ωj)
The only components of the input that lie within the bandwidth are sin 4t and sin 8t. Thus we evaluate M (ω) and φ(ω) at ω = 4 and ω = 8.
Thus, or
M (4) = 0.928
φ(4) = −0.381 rad
M (8) = 0.781
φ(8) = −0.675 rad
y(t) ≈ 0.5(0.928) sin(4t − 0.381) + 2(0.781) sin(8t − 0.675) y(t) ≈ 0.464 sin(4t − 0.381) + 1.562 sin(8t − 0.675)
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9.39 a) The transfer function is T (s) =
X(s) 0.2 1 = = 2 2 F (s) 0.05s + 0.4s + 5 0.01s + 0.08s + 1
From the frequency response plot we can √ determine that the bandwidth is from 0 to 12 rad/sec. (The peak is Mr = 0.27, and Mr / 2 = 0.19, which occurs at approximately 2 rad/sec). b) Only the first two terms in the expansion have frequencies within the bandwidth. Thus 1 f (t) ≈ −0.2(sin 3t + sin 9t) 3 and at steady state: x(t) ≈ 0.2(−0.2)M1 sin(3t + φ1 ) + 0.2(−0.2/3)M2 sin(9t + φ3 ) where M (3) = 1.063 M (9) = 1.343 Thus
φ(3) = −0.258 rad φ(9) = −1.313 rad
x(t) ≈ −0.0425 sin(3t − 0.258) − 0.0179 sin(9t − 1.313)
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9.40 First normalize the vo data by dividing by the amplitude of the input. Let v=
vo 20
Then compute m = 20 log v dB and log ω, and plot m versus log ω. The plot looks like a first order system plot (see Figure 9.1.6). The peak value of m is −11.25 dB, and m is approximately 3 dB below that value at ω = 0.8. Thus we estimate the time constant to be τ = 1/0/8 = 1.25 s. Thus we surmise that the form of the transfer function is Vo (s) K = Vs (s) 1.25s + 1 To estimate K, note that for low frequencies (ω 0 and KP > 0). For a unit step disturbance, the steady state error is found by setting Td (s) = 1/s and Θr (s) = 0 in equation (1), and using the final value theorem. ess
"
1 = lim sE(s) = lim s s→0 s→0 20s2 + (10 + KD )s + KP
#
1 1 = s KP
if the system is stable. Thus the steady state errors are the same for P and PD control. So they both satisfy specification #1, and specification #2 if KP ≥ 10. The difference between the two lies in their transient performance. P control: For P control, the characteristic equation is (set KD = 0 in equation (2)): 20s2 + 10s + KP = 0 Thus
10 ζ= √ 2 20KP 1037
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To satisfy specification #3 (ζ = 0.707), KP must be 2.5, which is smaller than the value required to satisfy specification #2. Thus P control cannot satisfy simultaneously both specification #2 and specification #3. PD control. For PD control, since ζ < 1 we can write the following formula for the time constant [the real part of the roots of equation (1) is −(10 + KD )/40]: (3) The formula for ζ is (4)
τ=
40 10 + KD
10 + KD ζ= √ = 0.707 2 20KP
A solution that exactly satisfies specification #3 is KP = 10. From equation (4) we find that KD = 10. From equation (3) we see that τ = 40/20 = 2. Other solutions are possible for values of KP greater than 10. Part (b). In summary 1. With P control, the requirements on damping ratio and disturbance error cannot be met simultaneously. 2. With PD control, one solution is KP = 10, KD = 10 Thus P control does not work. PD control works but has numerator dynamics that increases the overshoot.
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10.33 Because ζ = cos β, the value ζ = 0.707 corresponds to β = 45◦ , which is a 45◦ line on the complex plane. Thus ζ = 0.707 corresponds to a pair of roots whose real and imaginary parts have the same magnitude; that is, s = −a ± ja. The time constant of these roots is τ = 1/a. Thus the roots must be s = −1 ± j if τ = 1 and ζ = 0.707. These roots correspond to the polynomial equation (s + 1 − j)(s + 1 + j) = (s + 1)2 + 1 = s2 + 2s + 2 = 0 or 10s2 + 20s + 20 = 0 Compare this with the system’s characteristic equation obtained from the denominator of the transfer function: 10s2 + (3 + KD )s + KP = 0 Thus KP = 20 and 3 + KD = 20, or KD = 17.
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10.34 From the figure we obtain Θ(s) =
KP 1 Θr (s) − Td (s) 20s2 + (10 + K2 )s + KP 20s2 + (10 + K2 )s + KP
The steadystate unitstep response for the command input is θss =
KP =1 KP
which is perfect. The steadystate deviation caused by a unitstep disturbance is ∆θss = −
1 KP
Thus KP ≥ 10 is required to meet the specification. If ζ ≤ 1, 40 τ= = 0.1 10 + K2 which gives K2 = 390. Thus
10 + K2 200 ζ= √ =√ ≤1 2 20KP 20KP
if KP ≥ 2000. So an acceptable design is KP = 2000, K2 = 390, which gives ζ = 1, τ = 0.1, and ∆θss = −
1 2000
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10.35 (a) With the specific parameter and gain values used in Problem 10.33, the resulting command transfer function is Θ(s) 20 + 17s = 2 Θr (s) 10s + 20s + 20
(1)
A MATLAB file to compute the unitramp response is the following. systheta = tf([17,20],[10,20,20]); t = [0:0.001:4]; input = t; lsim(systheta,input,t) Even though the system is underdamped, we do not see oscillations in the response to a unit ramp command because the oscillation period is 2π = 6.28, which is greater than 4τ = 1. Thus the oscillations die out before one period has occurred. The steadystate error can be found using the final value theorem. From the block diagram in Figure 10.7.4, with the specific parameter and gain values used here, we obtain $
20 + 17s E(s) = Θr (s) − Θ(s) = Θr (s) 1 − 10s2 + 20s + 20
%
= Θr (s)
(
10s2 + 3s 10s2 + 20s + 20
)
With Θr (s) = 1/s2 , we have ess
1 = lim sE(s) = lim s 2 s→0 s→0 s
(
10s2 + 3s 10s2 + 20s + 20
)
=
3 20
From Figure 10.7.4 with the specific parameter and gain values used here and Θr (s) = 1/s2 , we obtain the following actuator equation: T (s) =
170s3 + 251s2 + 60s Θr (s) 10s2 + 20s + 20
Note that the MATLAB tf function cannot be used to compute this transfer function because the order of the numerator is greater than that of the denominator. We can, however, work around this difficulty as follows. With Θr (s) = 1/s2 , and canceling an s term from the numerator and denominator, we obtain 170s2 + 251s + 60 1 T (s) = 10s2 + 20s + 20 s This is equivalent to the unitstep response of the following transfer function T (s) 170s2 + 251s + 60 = Ωr (s) 10s2 + 20s + 20 A MATLAB file to compute the unitramp response is the following. 1041 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
systorque = tf([170,251,60],[10,20,20]); step(systorque) The torque approaches the constant value of 3 needed to counteract the damping torque 3θ˙ = 3(1) = 3. (b) Actual disturbance inputs are not always “clean” functions like steps, ramps, and sine waves. We often do not know the exact functional form of the disturbance. It is often a random function, such as the disturbance torque due to wind gusts of a rotating radar antenna. The analysis of random inputs is beyond the scope of this text. However, although frequency response plots strictly speaking describe only the steadystate response for periodic inputs, the plots are used to obtain a rough idea of the system’s transient response to fluctuating inputs that are not periodic. The disturbance transfer function for the system in question is Θ(s) −1 = Td (s) 10s2 + 20s + 20
(1)
The frequency response plot can be obtained with the following MATLAB file. sys=tf(1,[10, 20,20]); bodemag(sys) Rightclick on the plot, select “Characteristics”, then “Peak Response”, to see the peak value and the corresponding frequency. They are m = −26 dB at ω = 0 rad/s. The plot shows that the system responds more to slowly varying disturbances, and tends to reject disturbances whose frequencies lie above the frequency ω = 1.4 radians per unit time. For disturbance frequencies within the bandwidth of 0 ≤ ω ≤ 1.4, the system attenuates the input by approximately m = −26 db, or by a multiplicative factor of M = 10−26/20 = 0.05.
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10.36 a) Because this system is third order, we do not have formulas to use for the damping ratio and the time constant. In addition, we must interpret the specifications to apply to the dominant root or root pair, and must choose the secondary root somewhat arbitrarily. The values ζ = 0.707 and τ = 1 correspond to the dominant root pair s = −1 ± j. The third root must be less than −1 so that it will not be dominant. We arbitrarily select the third root to be s = −2. This choice can be investigated later if necessary. These three roots correspond to the polynomial equation *
+
(s + 2)(s + 1 − j)(s + 1 + j) = (s + 2) (s + 1)2 + 1 = s3 + 4s2 + 6s + 4 = 0 To compare this with the system’s characteristic equation, we multiply it by 10. 10s3 + 40s2 + 60s + 40 = 0 Compare this with the system’s characteristic equation: 10s3 + (3 + KD )s2 + KP s + KI Thus KP = 60, KI = 40, and 3 + KD = 40, or KD = 37. b) The resulting disturbance transfer function for this system is Θ(s) −s = 3 Td (s) 10s + 40s2 + 60s + 40 The frequency response plot can be obtained with the following MATLAB file. sys=tf([1,0],[10, 40,60,40]); bodemag(sys) Rightclick on the plot, select “Characteristics”, then “Peak Response”, to see the peak value and the corresponding frequency. They are m = −33.7 dB at ω = 1.14 rad/s. The plot shows that the system attenuates disturbance inputs by a factor of m = −33.7 dB or more. This corresponds to an amplitude reduction of M = 10−33.7/20 = 0.0207. The system rejects by an even greater amount any disturbances whose frequencies lie below or above the frequency ω = 1.14. Compare this performance with the PD control system of Problem 10.35, which does not have as great an attenuation and responds more to low frequency disturbances than high frequency ones.
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10.37 The requirements are (for unit step inputs) 1. the steadystate command error must be zero. 2. the magnitude of the steadystate disturbance error must be ≤ 0.1 3. the time constant must be 0.1 We can do P control and PD control at the same time by finding the error equation for PD control, and then setting KD = 0 to obtain the results for P control. From Figure 10.7.4 with I = 20 and c = 10, the error equation can be written as follows (after doing some algebra): (1) E(s) = Θr (s)−Θ(s) =
20s2 + 10s 1 Θr (s)+ Td (s) 20s2 + (10 + KD )s + KP 20s2 + (10 + KD )s + KP
The characteristic equation is obtained from the denominator: (2)
20s2 + (10 + KD )s + KP = 0
For a unit step command, the steady state error is found by setting Θr (s) = 1/s and Td (s) = 0 in equation (1), and using the final value theorem. ess
&
20s2 + 10s = lim sE(s) = lim s s→0 s→0 20s2 + (10 + KD )s + KP
'
1 =0 s
if the system is stable (that is, if 10 + KD > 0 and KP > 0). For a unit step disturbance, the steady state error is found by setting Td (s) = 1/s and Θr (s) = 0 in equation (1), and using the final value theorem. ess
"
1 = lim sE(s) = lim s s→0 s→0 20s2 + (10 + KD )s + KP
#
1 1 = s KP
if the system is stable. Thus the steady state errors are the same for P and PD control. So they both satisfy specification #1, and specification #2 if KP ≥ 10. The difference between the two lies in their transient performance. P control. For P control, we have 10 ζ= √ 2 20KP Thus, if KP ≥ 10, ζ ≤ 0.35. Because ζ < 1, the time constant gives τ=
40 =4 10
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Thus specification #3 is not satisfied. PD control. For PD control, specifications #1 and #2 are met if KP ≥ 10. There are two ways to meet specification #3: i) set ζ ≤ 1 or ii) set ζ > 1. Case (i). For case (i), (ζ ≤ 1), the time constant is given by τ=
40 = 0.1 10 + KD
if KD = 390. For this value, the damping ratio is found from equation (4) to be 10 + 390 400 ζ= √ =√ 2 20KP 20KP Thus ζ ≤ 1 only if KP ≥ 8000. Thus one solution is KP = 8000 and KD = 390. This gives a damping ratio ζ = 1. Case (ii). For case (ii), (ζ > 1), the dominant root must have a real part equal to −10 for the time constant to be 0.1. The other root must be placed to the left of s = −10. Thus there are an infinite number of solutions, depending on where the second root is placed. For example, if the second root is placed at s = −b, the characteristic equation will factor as 20(s + 10)(s + b) = 20s2 + (200 + 20b)s + 200b. Comparing this to the characteristic equation shows that KP = 200b and KD = 190 + 20b. Thus the larger b is, the greater the gains KP and KD . For example, using an arbitrary root separation factor of 10, one solution is b = 100, KP = 20, 000 and KD = 2190. Generally, placing the second root far to the left requires higher gains, which is not desirable. In this case, additional specifications are needed to arrive at a unique solution. Part (b). In summary, 1. Using P control, specifications #2 and #3 cannot be met simultaneously. 2. Using PD control, a solution with ζ ≤ 1 is: KP = 8000, KD = 390. A solution with ζ > 1 is: KP = 20, 000, KD = 2190. Other solutions are possible.
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10.38 From the following diagram with Td (s) = 0 we obtain Θ(s) =
1 {Kf Θr + (KP + KD s) [Θr (s) − Θ(s)]} s(Is + c)
Thus
Kf + KP + KD s Θ(s) = 2 Θr (s) Is + (c + KD )s + KP
The steadystate unitstep response for the command input is θss =
Kf + KP =1 KP
only if Kf = 0. The error equation is E(s) = Θr (s) − Θ(s) =
Is2 + cs − Kf Θr (s) Is2 + (c + KD )s + KP
For a unitramp command, ess = lim sE(s) = ∞ s→0
regardless of the value of Kf . So Kf does not improve the steadystate response for a step or a ramp input.
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Figure : for Problem 10.38
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10.39 a) Try Iaction: G(s) = KI /s. Thus C(s) KI = 2 R(s) 20s + 0.2s + KI and the error equation with a unitramp input is 1 E(s) = R(s) − C(s) = 2 s
(
)
20s2 + 0.2s 20s2 + 0.2s + KI √ Thus ess = 0.2/KI = 0.01 if KI = 20. This gives ζ = 0.2/(2 20KI ) = 0.005 and τ = 2(20)/0.2 = 200. b) Try PIaction: G(s) = KP + KI /s. Thus C(s) KP s + KI = 2 R(s) 20s + (0.2 + KP )s + KI and the error equation with a unitramp input is 1 E(s) = R(s) − C(s) = 2 s
(
20s2 + 0.2s 20s2 + (0.2 + KP )s + KI
)
Thus ess = 0.2/KI = 0.01 if KI = 20. This gives ζ=
0.2 + KP 0.2 + KP √ = =1 40 2 20KI
if KP = 38.8. This gives τ = 2(20)/(0.2 + KP ) = 1. c) Try PIDaction: G(s) = KP + KI /s + KD s. Thus C(s) K D s2 + K P s + K I = R(s) (20 + KD )s2 + (0.2 + KP )s + KI and the error equation with a unitramp input is 1 E(s) = R(s) − C(s) = 2 s
(
20s2 + 0.2s (20 + KD )s2 + (0.2 + KP )s + KI
)
Thus ess = 0.2/KI = 0.01 if KI = 20. The damping ratio is 0.2 + KP 0.2 + KP ζ= ! = ! =1 2 (20 + KD )KI 2 20(20 + KD )
Because ζ = 1, the expression for the time constant is τ=
2(20 + KD ) = 0.1 0.2 + KP
The solution is KP = 3.8 and KD = −19.8. 1048 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
10.40 The error equation is: (1)
E(s) =
s2 + s 1 Θr (s) + 2 Td (s) 2 s + (1 + KD )s + KP s + (1 + KD )s + KP
For a unit ramp command, the steady state error is found by setting Θr (s) = 1/s2 and Td (s) = 0 in equation (1), and using the final value theorem. ess
(
s2 + s = lim sE(s) = lim s 2 s→0 s→0 s + (1 + KD )s + KP
)
1 1 = 2 s KP
if the system is stable (that is, if 1 + KD > 0 and KP > 0). For a unit ramp disturbance, the steady state error is found by setting Td (s) = 1/s2 and Θr (s) = 0 in equation (1), and using the final value theorem. ess = lim sE(s) = lim s s→0
s→0
$
1 2 s + (1 + KD )s + KP
%
1 =∞ s2
if the system is stable.
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10.41 The system’s closedloop transfer function can be found from the block diagram. It is C(s) KP + KD s = 2 R(s) 2s + (2 + KD )s + KP Thus the characteristic equation is 2s2 + (2 + KD )s + KP = 0. Because ζ is specified to be less than 1, we can write the following formula for the time constant. τ=
2(2) =1 2 + KD
This gives KD = 2. The damping ratio is given by 2 + KD 2+2 2 ζ= √ = √ =√ = 0.9 2 2KP 2 2KP 2KP This gives KP = 2.469. If the command is a step input with a magnitude m, then R(s) = m/s, and the Final Value Theorem gives css
$
KP + KD s = lim sC(s) = lim s 2 s→0 s→0 2s + (2 + KD )s + KP
%
m =m s
Thus the steady state error is zero, so all three specifications are satisfied.
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10.42 a) The output equation is Θ(s) =
K D s2 + K P s + K I s Θr (s) − Td (s) 3 2 3 10s + (2 + KD )s + KP s + KI 10s + (2 + KD )s2 + KP s + KI
The characteristic equation 10s3 + (2 + KD )s2 + KP s + KI = 0
(1)
For case 1, the desired characteristic equation can be expressed as 10(s + 0.5)[(s + 5)2 + 25] = 10s3 + 105s2 + 550s + 250 = 0
(2)
Comparing coefficients in equations (1) and (2), we obtain KP = 550
KI = 250
KD = 103
For case 2, the desired characteristic equation can be expressed as 10(s + 0.5)(s + 1)(s + 2) = 10s3 + 35s2 + 35s + 10 = 0
(3)
Comparing coefficients in equations (1) and (3), we obtain KP = 35
KI = 10
KD = 33
b) The MATLAB file is KP1 = 550; KD1 = 103; KI1 = 250; systheta1 = tf([KD1,KP1,KI1],[10,2+KD1,KP1,KI1]); KP2 = 35; KD2 = 33; KI2 = 10; systheta2 = tf([KD2,KP2,KI2],[10,2+KD2,KP2,KI2]); step(systheta1,systheta2) Rightclick on the plot, select “Characteristics”, then “Peak Response” to determine the maximum percent overshoot and the peak time. For Case 1, the maximum percent overshoot is 22%; for Case 2, it is15%. For the gains given in Example 10.7.7, it is 10%. The separation factors are 10, 2, and 10, respectively. On the basis of this example we would conclude that a large separation factor is better, provide that the dominant root is complex. b) The disturbance transfer function is Θ(s) s =− 3 Td (s) 10s + (2 + KD )s2 + KP s + KI The MATLAB file, which is a continuation of the previous file, is 1051 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
sysdist1 = tf([1,0],[10,2+KD1,KP1,KI1]); sysdist2 = tf([1,0],[10,2+KD2,KP2,KI2]); bodemag(sysdist1,sysdist2) Rightclick on the plot, select “Characteristics”, then “Peak Response” to determine the peak response and the corresponding frequency. For Case 1, the maximum response is m = −54.2 dB at ω = 2.44; for Case 2, it is −30 dB at ω = 0.585. For the gains given in Example 10.7.7, it is m = −34.1 dB at ω = 0.7. The separation factors are 10, 2, and 10, respectively. On the basis of this example we would conclude that a large separation factor gives more attenuation, provided that the dominant root is real.
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10.43 With I action only, the transfer function is C(s) KI K = 2 R(s) τ s + s + KI K and With DI action,
and
1 ζ= √ 2 τ KI K C(s) K(KD s2 + KI ) = R(s) (τ + KKD )s2 + s + KI K 1 ζ= ! 2 (τ + KKD )KI K
If KD > 0 the derivative action makes ζ smaller, and thus the response is more oscillatory. So D action does not improve the response.
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10.44 The compensated system diagram is shown in the following figure.
Figure : for Problem 10.44 Command compensation cannot affect the characteristic roots or the disturbance response, so we set D(s) = 0 here. The error equation is E(s) =
Is2 + cs − Kf R(s) Is2 + cs + KP
For a unitstep input, the final value theorem gives ess = −Kf /KP as long as the system is stable. Thus ess = 0 only if there is no command compensation (Kf = 0)! For a unitramp input, the final value theorem gives ess = ∞. Thus the command compensation does not improve the ramp response. Therefore, command compensation for this system degrades the step response, and does not improve the ramp response.
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10.45 a) The closedloop transfer function is C(s) 3KP + 3KD s = 2 R(s) s + 3KD s + 3KP − 4 The damping ratio is
The time constant is
3KD ζ= √ = 0.707 2 3KP − 4 τ=
2 = 0.1 3KD
These two conditions give KP = 68 and KD = 20/3. b) The closedloop transfer function with the negative rate feedback gain K1 is C(s) 3KP = 2 R(s) s + 3K1 s + 3KP − 4 This denominator has the same form as the transfer function in part (a) with KD replaced by K1 . Thus KP = 68 and K1 = 20/3. c) For part (a), C(s) 20s + 204 = 2 R(s) s + 20s + 200 This has numerator dynamics and will overshoot more than the design in part (b), for which C(s) 204 = 2 R(s) s + 20s + 200
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10.46 Since the desired value of θ is 0, the error signal is e = θr − θ = 0 − θ = −θ. Trying ˙ Substituting this into the equation of PD control we use f = KP e + KD e˙ = −KP θ − KD θ. motion we obtain M Lθ¨ − (M + m)gθ = f = −KP θ − KD θ˙ The characteristic equation is
M Ls2 + KD s + KP − (M + m)g = 0 Using M = 40, m = 8, L = 20, and g = 32.2 this becomes 800s2 + KD s + KP − 1545.6 = 0 A 2% settling time of 10 seconds corresponds to 4τ = 10 or τ = 2.5 seconds. Because ζ < 1, we can use the following formula for the time constant. τ=
2(800) = 2.5 KD
This gives KD = 640 lb/rad/sec. The formula for the damping ratio is KD 640 ζ= ! = ! = 0.707 2 800(KP − 1545.6) 2 800(KP − 1545.6)
Solve this for KP to obtain KP = 1801.6 lb/rad.
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10.47 The closedloop transfer function is Θ(s) 2412 + 1200s = Θr (s) 1500τ s3 + 1500s2 + (1200 − 1932τ )s + 480 The characteristic equation is 1500τ s3 + 1500s2 + (1200 − 1932τ )s + 480 = 0 From the RouthHurwitz criterion, we can tell that the system is stable if 0 ≤ τ ≤ 0.498, so the case where τ = 1 is unstable. If τ = 0.1, the roots are s = −0.3418 ± 0.4761j
The time constant is 1/0.3418 = 2.926, and the damping ratio is 0.583. The specifications call for a dominant time constant of 2.5 and a damping ratio of 0.707.
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10.48 The closedloop transfer function is Θ(s) (2412 + 1200s)(τ s + 1) = Θr (s) 1500τ s3 + 1500s2 + (1200 − 1932τ )s + 480 The characteristic equation is 1500τ s3 + 1500s2 + (1200 − 1932τ )s + 480 = 0 From the RouthHurwitz criterion, we can tell that the system is stable if 0 ≤ τ ≤ 0.498, so the case where τ = 1 is unstable. If τ = 0.1, the roots are s = −0.3418 ± 0.4761j
The time constant is 1/0.3418 = 2.926, and the damping ratio is 0.583. The specifications call for a dominant time constant of 2.5 and a damping ratio of 0.707.
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10.49 In the first printing of the text, Figure P10.49 is incorrect. The feedback gains should be ce and ke , and the gain in the control algorithm block should be me , not m. a) From Figure P10.49, (ms2 +cs+k)X(s) = Fd (s)+(ce s+ke )X(s)+me
,
K D s2 + K P s + K I s2 Xr (s) + [Xr (s) − X(s)] s

If me = m, ce = c, and ke = k, this equation becomes (ms3 + mKD s2 + mKP s + mKI )X(s) = sFd (s) + (ms3 + mKD s2 + mKP s + mKI )Xr (s) Thus
X(s) =1 Xr (s)
which is perfect, and X(s) s = 3 2 Fd (s) ms + mKD s + mKP s + mKI For a step disturbance, Fd (s) = 1/s, Xss = 0. For a unitramp disturbance, Fd (s) = 1/s2 , Xss = 1/mKI . Since the characteristic equation is third order, there is no expression for the damping ratio or time constant. We can, however, obtain a dominant root pair that meets the requirements that ζ = 1 and τ = τd . This means that the dominant root pair must be two identical roots at s = −1/τd . Thus the third root will be real. Denote it by s = −b. Then the characteristic equation can be factored as $
1 ms +mKD s +mKP s+mKI = m 2 + τd 3
2
%2
&
$
2 (s+b) = m s + b + τd 3
%
s + 2
(
1 2b + τd2 τd
)
b s+ 2 τd
'
Comparing coefficients, we see that the gains must be chosen so that KP = b + KP =
2 τd
1 2b + τd τd2
KI =
b τd2
The value of b must be selected such that the root at s = −b is not the dominant root. This requires that b > 1/τd . b) In addition to requiring accurate estimates of m, c, and k, this scheme requires that the second derivative, x ¨r , of the command input must be computed in real time. This can be difficult to do for noisy inputs or inputs having discontinuities in xr or x˙ r , such as steps and ramps. 1059 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
10.50 a) From the diagram, ΩL (s) =
1 KT [Vm (s) − Kb Ωm (s)] N (Ie s + ce ) La s + Ra
and Ωm (s) = N ΩL (s). Thus ΩL (s) KT = 2 Vm (s) N Ie La s + N (ce La + Ra Ie )s + ce Ra N + N KT Kb With the given values, ΩL (s) 1.4 × 105 4.997 = = Vm (s) 4.368s2 + 894.4s + 2.8015 × 104 1.5592 × 10−4 s2 + 0.0319s + 1 Since the two roots are real, this can be expressed as ΩL (s) K = Vm (s) (τ1 s + 1)(τ2 s + 1) where K = 4.997, τ1 = 2.591 × 10−2 and τ2 = 6.02 × 10−3 . b) Using PI control with the above transfer function results in a thirdorder system, for which there are no convenient design formulas. Since τ2 ( τ1 , we will neglect τ2 and express the transfer function as ΩL (s) K = Vm (s) τ1 s + 1 With PI control, the closedloop transfer function is ΩL (s) K(KP s + KI ) = 2 Ωr (s) τ1 s + (1 + KKP )s + KKI Since we will set ζ < 1, we can express the closedloop time constant as τ= Thus KP = The damping ratio is
2τ1 = 0.05 1 + KKP
2τ1 − 0.05 = 7.284 × 10−3 0.05K
1 + KKP 20τ1 ζ= √ =√ 2 τ1 KKI τ1 KKI
Thus KI =
400τ12 ζ 2 τ1 K
Choosing ζ = 0.707, we obtain KI = 4.1481 1060 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
The resulting thirdorder transfer function is ΩL (s) 0.0364s + 20.73 = Ωr (s) 0.0001559s3 + 0.03193s2 + 1.036s + 20.73 The roots of the thirdorder system are s = −170.39 and s = −17.21 ± 22j. Thus the dominant time constant is 1/17.21 = 0.058, which is close to the desired value. The damping ratio of the dominant root pair is ζ = cos[tan−1 (22/17.21)] = 0.616, which is within the required range.
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10.51 The value from Problem 10.19 is KP = 3.6772. From Example 10.8.6, Tm (s) N KP KT (Ie s + ce ) = Ωr (s) D(s) and thus, since ia = TM /KT , Ia (s) N KP (Ie s + ce ) = Ωr (s) D(s) where
*
D(s) = N La Ie s2 + (Ra Ie + ce La )s + Ra ce + KT Kb + KP KT Note that the term N in D(s) cancels N in the numerator of Ia (s)/Ωr (s). The MATLAB program is the following.
+
KT = 0.2; Kb = 0.2; Ie = 1.775e3;ce = 1e3; Ra = 0.8;La = 4e3; KP = 3.672; num = KP*[Ie, ce]; den = [La*Ie, Ra*Ie+ce*La, Ra*ce+KT*Kb+KP*KT]; sys = tf(num, den); step(209.4*sys) Note that we can use the step function by multiplying the transfer function by magnitude of the step input. The plot is shown next. The maximum current is 389 A! This unattainable value can be reduced by using a modified step input that increases slowly (see Problem 10.53 for a discussion of this type of input).
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Step Response 400
300
Current ia (A)
200
100
0
−100
−200
0
0.01
0.02
0.03
0.04
0.05
0.06
Time (sec)
Figure : for Problem 10.51.
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10.52 The new transfer functions are Ω(s) KP s + KI = Ωr (s) 0.5s3 + 5.2s2 + (2 + KP )s + KI 0.5s2 + 5.2s + 2 T (s) KP s + KI = Ωr (s) 0.1s + 1 0.5s3 + 5.2s2 + (2 + KP )s + KI Ω(s) −s(0.1s + 1) = Td (s) 0.5s3 + 5.2s2 + (2 + KP )s + KI The three cases are: 1. KP = 18, KI = 40; 2. KP = 18, KI = 20; 3. KP = 108, KI = 200; a) The MATLAB program is KP = [18, 18, 108]; KI = [40, 20, 200]; % Command transfer function k = 1; sysa1 = tf([KP(k), KI(k)],[0.5, 5.2, 2 + KP(k), KI(k)]); k = 2; sysa2 = tf([KP(k), KI(k)],[0.5, 5.2, 2 + KP(k), KI(k)]); k = 3; sysa3 = tf([KP(k), KI(k)],[0.5, 5.2, 2 + KP(k), KI(k)]); % Actuator transfer function k = 1; numb1 = conv([KP(k), KI(k)],[0.5, 5.2, 2]); denb1 = conv([0.1, 1],[0.5, 5.2, 2 + KP(k), KI(k)]); sysb1 = tf(numb1, denb1) k = 2; numb2 = conv([KP(k), KI(k)],[0.5, 5.2, 2]); denb2 = conv([0.1, 1],[0.5, 5.2, 2 + KP(k), KI(k)]); sysb2 = tf(numb2, denb2) k = 3; numb3 = conv([KP(k), KI(k)],[0.5, 5.2, 2]); denb3 = conv([0.1, 1],[0.5, 5.2, 2 + KP(k), KI(k)]); sysb3 = tf(numb3, denb3) subplot(2,1,1), step(sysa1, sysa2, sysa3) subplot(2,1,2), step(sysb1, sysb2, sysb3) The characteristic roots are as follows: KP KI Intended ζ Intended Roots Actual Roots 18 40 0.707 −2 ± 2j (dominant) −2.2353 ± 2.9147j (ζ = 0.6086), −5.9294 18 20 1 −2, −2 (dominant) −1.5016, −4.4492 ± 2.6159j 108 200 1.74 −2, −20 (dominant) −1.9664, −4.2168 ± 13.6248j The plots are shown in the following figure. The peak actuator values are 16.2, 14.2, and 52. 1064 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Step Response 1.5
3
1
Speed (rad/sec)
2 1
0.5
0
0
0.5
1
1.5
1
1.5
Time (sec)
Step Response
Actuatot Response
60
3
40
20
1 2
0
−20
0
0.5 Time (sec)
Figure : for Problem 10.52a.
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b) The MATLAB program is KP = [18, 18, 108]; KI = [40, 20, 200]; % Disturbance transfer function k = 1; sysc1 = tf([0.1, 1, 0],[0.5, 5.2, 2 + KP(k), KI(k)]); k = 2; sysc2 = tf([0.1, 1, 0],[0.5, 5.2, 2 + KP(k), KI(k)]); k = 3; sysc3 = tf([0.1, 1, 0],[0.5, 5.2, 2 + KP(k), KI(k)]); bodemag(sysc1, sysc2, sysc3) The plots are shown in the following figure. The peak values are −23.3, −24.7, and −30.8 dB. c) The peak actuator values are less than in Example 10.6.4 because the actuator responds more slowly in this problem (it has a time constant of 0.1, whereas in the example the actuator responded instantaneously). The peak disturbance response values are slightly higher than in the example. Bode Diagram −20
1
−25
2 −30
3 −35
Magnitude (dB)
−40
−45
−50
−55
−60
−65
−70 −1 10
0
1
10
10
2
10
Frequency (rad/sec)
Figure : for Problem 10.52b.
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10.53 a) The transfer functions are Ω(s) KP s + KI = 2 Ωr (s) 5s + (2 + KP )s + KI T (s) (KP s + KI )(5s + 2) = 3 Ωr (s) 5s + (2 + KP )s + KI The three cases are: 1. KP = 18, KI = 40; 2. KP = 18, KI = 20; 3. KP = 108, KI = 200; a) The MATLAB program is KP = [18, 18, 108]; KI = [40, 20, 200]; % Command transfer function k = 1; sysa1 = tf([KP(k), KI(k)],[5, 2 + KP(k), KI(k)]); k = 2; sysa2 = tf([KP(k), KI(k)],[5, 2 + KP(k), KI(k)]); k = 3; sysa3 = tf([KP(k), KI(k)],[5, 2 + KP(k), KI(k)]); % Actuator transfer function k = 1; sysb1 = tf(conv([KP(k), KI(k)],[5, 2]),[5, 2 + KP(k), KI(k)]) k = 2; sysb2 = tf(conv([KP(k), KI(k)],[5, 2]),[5, 2 + KP(k), KI(k)]) k = 3; sysb3 = tf(conv([KP(k), KI(k)],[5, 2]),[5, 2 + KP(k), KI(k)]) tc = [0:.001:2]; rc = 1  exp(20*tc); subplot(2,1,1),lsim(sysa1,sysa2,sysa3,rc,tc) ta = [0:.001:0.8]; ra = 1  exp(20*ta); subplot(2,1,2),lsim(sysb1,sysb2,sysb3,ra,ta) The plots are shown in the following figure. The peak actuator values are 14.5, 13.4, and 40.1. The peak actuator values are less than in Example 10.6.4 because the command input does not act as fast as the step input used in the example. This gives the system more time to respond, and thus makes less demand on the actuator.
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Linear Simulation Results 1.4
1
1.2
Amplitude
1
3
Command
2
0.8 0.6 0.4 0.2 0
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
Time (sec)
Linear Simulation Results
Amplitude
40
3
30
20
1 10
0
2 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Time (sec)
Figure : for Problem 10.53.
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10.54 The new transfer functions are Ω(s) KI = 3 2 Ωr (s) 0.5s + 5.2s + (2 + K2 )s + KI T (s) KI (5s + 2) = Ωr (s) 0.5s3 + 5.2s2 + (2 + K2 )s + KI Ω(s) −s(0.1s + 1) = 3 Td (s) 0.5s + 5.2s2 + (2 + K2 )s + KI The three cases are: 1. K2 = 18, KI = 40, ζ = 0.707; 2. K2 = 18, KI = 20, ζ = 1; 3. K2 = 108, KI = 200, ζ = 1.74; a) The MATLAB program is K2 = [18, 18, 108]; KI = [40, 20, 200]; % Command transfer function k = 1; sysa1 = tf(KI(k),[0.5, 5.2, 2 + K2(k), k = 2; sysa2 = tf(KI(k),[0.5, 5.2, 2 + K2(k), k = 3; sysa3 = tf(KI(k),[0.5, 5.2, 2 + K2(k), % Actuator transfer function k = 1; sysb1 = tf(KI(k)*[5, 2],[0.5, 5.2, 2 + k = 2; sysb2 = tf(KI(k)*[5, 2],[0.5, 5.2, 2 + k = 2; sysb3 = tf(KI(k)*[5, 2],[0.5, 5.2, 2 + subplot(2,1,1), step(sysa1, sysa2, sysa3) subplot(2,1,2), step(sysb1, sysb2, sysb3)
KI(k)]); KI(k)]); KI(k)]); K2(k), KI(k)]); K2(k), KI(k)]); K2(k), KI(k)]);
The characteristic roots are as follows: K2 KI Intended ζ Intended Roots Actual Roots 18 40 0.707 −2 ± 2j (dominant) −2.2353 ± 2.9147j (ζ = 0.6086), −5.9294 18 20 1 −2, −2 (dominant) −1.5016, −4.4492 ± 2.6159j 108 200 1.74 −2, −20 (dominant) −1.9664, −4.2168 ± 13.6248j The plots are shown in the following figure. The peak actuator values are 8.64, 4.87, and 11.3.
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Step Response 1.4
" = 0.707
1.2
!(t)
1
" = 1.74
0.8
"=1
0.6 0.4 0.2 0
0
0.5
1
1.5
2
2.5
3
3.5
2
2.5
3
3.5
Time (sec)
Step Response 12
" = 1.74
10
" = 0.707 T(t)
8 6
"=1
4 2 0
0
0.5
1
1.5 Time (sec)
Figure : for Problem 10.54a.
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b) The MATLAB program is K2 = [18, 18, 108]; KI = [40, 20, 200]; % Disturbance transfer function k = 1; sysc1 = tf([0.1, 1, 0],[0.5, 5.2, 2 + K2(k), KI(k)]); k = 2; sysc2 = tf([0.1, 1, 0],[0.5, 5.2, 2 + K2(k), KI(k)]); k = 3; sysc3 = tf([0.1, 1, 0],[0.5, 5.2, 2 + K2(k), KI(k)]); bodemag(sysc1, sysc2, sysc3) The plots are shown in the following figure. The peak values are −23.3, −24.7, and −30.8 dB. c) The peak actuator values are greater than in Example 10.6.5. The peak disturbance response values are slightly higher than in the example. Bode Diagram −20
" = 0.707 −30
" = 1.74
"=1 −40
Magnitude (dB)
−50
−60
−70
−80
−90
−100
−110 −2 10
−1
10
0
1
10
10
2
10
Frequency (rad/sec)
Figure : for Problem 10.54b.
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10.55 The characteristic equation is 10s3 + (2 + KD )s2 + KP s + KI = 0 With KP = 55 and KD = 58, we have 10s3 + 60s2 + 55s + KI = 0 or 1+
KI 1 =0 3 10 s + 6s2 + 5.5s
The root locus gain is K = KI /10. In MATLAB type sys = tf(1, [1, 6, 5.5, 0]); rlocus(sys) The plot follows. In Example 10.7.7, KI = 25. This gives ζ = 0.707 and τ = 2. The plot shows that to reduce the error by increasing KI , the dominant time constant will become larger and the damping ratio will decrease, making the system slower and more oscillatory. If we decrease the error by half by doubling KI to 50, the plot shows that the new damping ratio will be approximately 0.4 and the new time constant will be approximately 1/0.435 = 2.3.
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Root Locus 10
8
6
4
Imaginary Axis
2
0
−2
−4
−6
−8
−10 −14
−12
−10
−8
−6
−4
−2
0
2
4
Real Axis
Figure : for Problem 10.55.
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10.56 The characteristic equation is s3 + KD s2 + (KP − 4)s + KI = 0 With KP = 604 and KD = 40, we have s3 + 40s2 + 600s + KI = 0 or 1 + KI
1 =0 s3 + 40s2 + 600s
The root locus gain is K = KI . In MATLAB type sys = tf(1, [1, 40, 600, 0]); rlocus(sys) The plot follows. In Example 10.8.3, KI = 4000. This gives ζ = 0.707 and τ = 0.1. The plot shows that to reduce the error by increasing KI , the dominant time constant will become larger and the damping ratio will decrease, making the system slower and more oscillatory. If we decrease the error by half by doubling KI to 8000, the plot shows that the new damping ratio will be approximately 0.3 and the new time constant will be approximately 1/5.68 = 0.18.
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Root Locus 40
30
20
Imaginary Axis
10
0
−10
−20
−30
−40 −60
−50
−40
−30
−20
−10
0
10
Real Axis
Figure : for Problem 10.56.
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10.57 From equations (2) and (3) in the example, ΩL (s) (0.002s + 0.6)/1.5 =− −6 2 TL (s) 5.407 × 10 s + 1.623 × 10−3 s + 2.8 × 10−3 + 0.04KP The program is given below. KP = 0.63; num = [0.002, 0.6]/1.5; den = [5.407e6, 1.623e3, 2.8e3 + 0.04*KP]; sys = tf(num, den); t1 = [0:0.001:0.05]; t2 = [0.051:0.001:0.3]; t = [t1, t2]; TL1 = 40*t1; TL2 = 2*ones(size(t2)); TL = [TL1, TL2]; lsim(sys, TL, t) The plot is shown below. The steadystate speed deviation is 28.4 rad/s, or 271 rpm, below the desired speed of 1000 rpm.
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Linear Simulation Results 5
Load Torque (N m)
0
−5
Amplitude
−10
Speed Deviation (rad/s) −15
−20
−25
−30
0
0.05
0.1
0.15
0.2
0.25
0.3
Time (sec)
Figure : for Problem 10.57.
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10.58 Let K= where Ie = I1 + and K = 12.7547. The transfer function is
Kpot Ka KT 6Ie
"
#
1 1 I2 + I3 = 0.0157 4 9
Θ(s) K(KP s + KD ) = Θr (s) Ls3 + Rs2 + KKD s + KKP a) For L = 0,
Θ(s) K(KP s + KD ) = 2 Θr (s) Rs + KKD s + KKP
To obtain τ = 0.5 and ζ = 1, τ = 0.5 = and
2R KKD
KKD ζ=1= √ 2 RKKP
These give
4R = 0.0941 K Parts (b), (c), and (d) are done with the following MATLAB program. KD = KP =
I1 = 0.01; I2 = 5e4; I3 = 0.2; Ka = 1; KT = 0.6; Kpot = 2;R = 0.3; Ie = I1+(1/4)*(I2+(1/9)*I3) K = Kpot*Ka*KT/(6*Ie) KD = 4*R/K KP = 4*R/K sys1 = tf([K*KD,K*KP],[R,K*KD,K*KP]) L = 0.015; sys2 = tf([K*KD,K*KP],[L,R,K*KD,K*KP]) step(sys1,sys2) The resulting plot is shown on the next page. With L = 0, the overshoot is 13.5% and the settling time is 2.7 s. With L = 0.015, the overshoot is 16.2% and the settling time is 2.59 s. So the response predicted by the second order model, which makes the calculation of the gains KP and KD much easier, is close to that of the third order system. This will not be true if L is much larger than 0.015. 1078 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Step Response 1.4
1.2
L = 0.015 L=0
1
Amplitude
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2
2.5
3
3.5
Time (sec)
Figure : for Problem 10.58.
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10.59 Since the acceleration time is 4 s, we set τ = 1 so that the controllers can follow the ramp input. The time constant expressions for the three controllers are: 1. τ = I/(c + KP ) = 10/(3 + KP ) for P control. 2. τ = I/(c + KP ) = 10/(3 + KP ) for PI control, if ζ ≤ 1. 3. τ = I/(c + K2 ) = 10/(3 + K2 ) for Modified I control, if ζ ≤ 1. Requiring, for example, a maximum steady state error of 10% of the slew speed means that ess ≤ 0.1. The slope of the input is 1/4, and the steady state errors for a ramp command of slope 1/4 is 1. ess = 0.25/(c + KP ) = 0.25/(3 + KP ) for P control. 2. ess = 0.25c/KI = (0.25)3/KI for PI control. 3. ess = 0.25(c + K2 )/KI = 0.25(3 + K2 )/KI for Modified I control. Combining these requirements and satisfying the requirement on τ exactly, we obtain the following gains: 1. KP = 7 for P control. 2. KP = 7 , KI = 7.5 for PI control. 3. K2 = 8 , KI = 110 for Modified I control. The following program is a modification of the programs command_tf.m, actuator_tf.m, and trap1.m given in Section 10.9. The program calls on the files named plot_command and plot_actuator listed in Table 10.9.3. KPa = 120; KPb = 7; KIb = 7.5; K2 = 8;KIc = 110; I = 10;c = 3; % Create the command transfer functions. sysa = tf(KPa,[I,c+KPa]); sysb = tf([KPb,KIb],[I,c+KPb,KIb]); sysc = tf(KIc,[I,c+K2,KIc]); % Create the actuator transfer functions sysaACT = tf(KPa*[I,c],[I,c+KPa]); sysbACT = tf(conv([KPb,KIb],[I,c]),[I,c+KPb,KIb]); syscACT = tf(KIc*[I,c],[I,c+K2,KIc]); % A specific trapezoidal profile t = [0:0.01:19]; for k = 1:length(t) 1080 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
if t(k) 0. For this plot, the farthest to the left the dominant root can lie is at the breakaway point. Thus the smallest possible dominant time constant is 1/0.4833 = 2.07.
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11.13 b) PD action will add a zero that will pull the locus to the left, stabilize the system, and allow a smaller time constant to be obtained, if the zero is placed properly. With PD action, the openloop transfer function becomes !
"
KP TD s + T1D KP (1 + TD s) = s(s + 1)(s + 8) s(s + 1)(s + 8)
The poles are still at s = 0, s = −1, and s = −8, and the zero is at s = −1/TD . The root locus parameter is K = KP TD . We want τ = 0.5. Thus the zero at s = −1/TD must be placed far enough to the left to pull the locus to the left of s = −2. Thus we choose TD = 0.5 to place the zero at s = −2. The resulting root locus is shown in the following figure. Next we choose ζ = 0.707, which along with τ = 0.5, specifies the point s = −2 ± 2j. From the plot we can see that the line corresponding to ζ = 0.707 passes through the locus, so we know that a solution exists. At s = −2 ± 2j, we find that K = 19.7. Thus KP = 19.7/TD = 2(19.7) = 39.4. The third root lies at s = −5 and thus is not dominant. Therefore, one solution is TD = 0.5, KP = 39.4. Other solutions are possible, depending on the choices made for TD and ζ. Root Locus 0.707 4
3
s = −2 + 2j
Imaginary Axis
2
1
0
−1
−2 0.707 −5
−4
−3
−2
−1
0
1
2
3
4
Real Axis
Figure : for Problem 11.13
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11.14 The root locus equation is 1 + KP
s + 10 =0 (s + 2)(s + 3)
The root locus plot is shown in the following figure. The line tangent to the circle gives the smallest ζ. This line has an angle of 49◦ . Thus the smallest ζ is ζ = cos 49◦ = 0.66. The 45◦ line corresponds to ζ = 0.707. This line intersects the circle at two points. The leftmost point gives the smallest time constant. This point, which can be found graphically or analytically, is s = −6.73 + 6.73j, which corresponds to #
(s + 2)(s + 3) ## KP = − = 8.46 # s + 10 s=−6.73+6.73j
The time constant is τ = 1/6.73 = 0.149. Root Locus 8
6
! = 0.66
! = 0.707 4
Imaginary Axis
2
0
−2
−4
−6
−8
−18
−16
−14
−12
−10
−8
−6
−4
−2
0
Real Axis
Figure : for Problem 11.14
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11.15 The characteristic equation is found from 1 + KP Gp (s) = 0 which becomes 1+2
5(s + 4) =0 (s + 3)(s + p)
or s2 + 13s + 40 + p(s + 3) = 0 Let K = p − 7. Then the characteristic equation becomes s2 + 20s + 61 + K(s + 3) = 0 where K ≥ 0. The poles are s = −3.76 and s = −16.2. The zero is s = −3. The root locus plot is sketched in the following figure. As p increases above 7, the dominant root moves from −3.76 to −3, and the other root moves from −16.2 to −∞. The roots are always real, so there will be no oscillations in the free response. The dominant time constant varies from τ = 1/3.76 = 0.266 to τ = 0.333 as p increases above p = 7. Root Locus
8
6
4
Imaginary Axis
2
0
−2
−4
−6
−8
−25
−20
−15
−10
−5
0
Real Axis
Figure : for Problem 11.15
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11.16 Express the controller transfer function as Gc (s) =
KP (s +
KI KP
)
s
=
KP (s + b) s
where b = KI /KP . The root locus equation is 1 + KP
s+b =0 s(s + 1)(s + 2)
The poles are s = 0, s = −1, and s = −2. The zero is at s = −b. The asymptotic angles are θ = ±90◦ . The asymptotes intersect at σ=
b−3 2
Sketches of the root locus plots for three cases are shown in the following figure. These three cases correspond to a) 0 < b < 1, using b = 0.5 b) 1 < b < 2, using b = 1.5, and c) b > 2, using b = 3. Case (a) does not satisfy the specification that the dominant root have a damping ratio of ζ = 0.707. Case (b) allows a smaller dominant time constant than Case (c). Root Locus 4
3
2
Imaginary Axis
1
0
−1
−2
−3
−4
−6
−5
−4
−3
−2
−1
0
1
2
3
4
Real Axis
Figure : for Problem 11.16, case (a)
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Root Locus 5
4
3
2
Imaginary Axis
1
0
−1
−2
−3
−4
−5
−6
−4
−2
0
2
4
Real Axis
Figure : for Problem 11.16, case (b)
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Root Locus 5
4
3
2
Imaginary Axis
1
0
−1
−2
−3
−4
−5 −8
−6
−4
−2
0
2
4
Real Axis
Figure : for Problem 11.16, case (c) The characteristic equation is s3 + 3s2 + (KP + 2)s + bKP = 0 Because we require that ζ = 0.707, the dominant root pair has the form s = −c ± cj. The characteristic equation can be factored as [(s + c)2 + c2 ](s − s3 ) = s3 + (2c − s3 )s2 + (2c2 − 2cs3 )s − 2c2 s3 = 0 where the third root is s3 and must be real. Comparing coefficients, we see that 2c − s3 = 3
2c2 − 2cs3 = KP + 2
− 2c2 s3 = bKP
Thus the third root is s3 = 2c − 3. In order for s = −c ± cj to be dominant, s3 < −c, and thus we must choose c < 1. To minimize the dominant time constant τ = 1/c, we should choose c as large as possible, subject to the restriction that c < 1. However, the closer c is to 1, the closer the secondary root s3 is to the dominant root. If s3 is too close to the dominant root, the response expected for ζ = 0.707 might not be obtained. Thus we should try a value for c and check the effects of the secondary root by analysis or simulation. For example, one solution is c = 0.9. This gives s3 = −1.2, b = 1.09, and KP = 1.78.
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11.17 The characteristic equation is 1 + Gc (s)Gp (s) = 0 which becomes
&
'
KI 4 1 + KP + + KD s =0 2 s 3s + 3
or
3s3 + 4KD s2 + (3 + 4KP )s + 4KI = 0
(1)
√ To achieve τ = 1 and ζ = 0.5, the desired dominant roots are s = −1 ± j 3. The third root is s = −b, where b has some arbitrary value such that b > 1 (so that s = −b will not be the dominant root). The polynomial corresponding to these three roots is [(s + 1)2 + 3](s + b) = 0 or s3 + (b + 2)s2 + (4 + 2b)s + 4b = 0 To compare this with equation (1), we must multiply by 3: 3s3 + 3(b + 2)s2 + 3(4 + 2b)s + 12b = 0
(2)
Comparing the coefficients of (1) and (2), we obtain 9 + 6b 4 KI = 3b 3(b + 2) KD = 4 The gains can be computed once a value for b > 1 has been selected. KP =
11.18 With a PD compensator, the closedloop transfer function is C(s) (KP + KD s)Gp (s) KP + KD s = = 2 R(s) 1 + (KP + KD s)Gp (s) 5s + KD s + KP − 6
The specifications require that
ζ = 0.707 Thus ωn =
+
ωn = 0.5
KP − 6 = 0.5 5
which gives KP = 7.25. Also,
which gives KD = 3.5355.
KD KD ζ= % = √ = 0.707 2 5(KP − 6) 2 6.25 1151
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√ 11.19 a) For K√= 4 the roots are s = −1 ± 3j. For ωn = 4 and ζ = 0.5, the roots must be s = −2 ± 2 3j. A simple gain adjustment will not give the desired roots. The angle deficiency is # # 4 # $ = $ 4 − $ s − $ (s + 2) = −210◦ s(s + 2) #s=−2+2√3j
Thus the compensator must add (210 − 180) = 30◦ to the system. This implies a lead compensator. Place the compensator’s pole and zero to the left of s = −2. Using the same procedure as in Example 11.2.2, with µ = 10, we obtain T = 0.0138 and T = 0.293. Choose T = 0.0138 because the second value would place the zero to the right of s = −2, and thus violate the assumed geometry. The compensator’s pole is s = −1/T = −72.6, and its zero is s = −1/aT = −7.26. The compensated openloop transfer function is s + 7.26 K s + 72.6 s(s + 2) √ The value of K required to place the root at s = −2 ± 2 3j is K = 155.6. Other solutions are possible, depending on the choice for µ. For C = 1µF , R1 R2 T = 0.0138 = 10−6 R1 + R2 Gc (s)G(s) =
µ=
R1 + R2 = 10 R2
These give R1 = 1.38 × 105 and R2 = 1.53 × 104 Ω.
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√ 11.19 b) A ramp error of 0.05 implies that Cv = 20. With K = 4, s = −1 ± j 3 and Cv = 2. To obtain Cv = 20 we need to increase the gain by a factor of 10, but without changing the root locations appreciably. Therefore we use a lag compensator. Because the PM and GM are not specified, we use a root locus design method. Step 1: With no compensation, K = 4 will place the roots at the desired locations. Thus KP 1 = 4. Step 2: To obtain Cv = 20, K must be increased from 4 to 40. Thus KP 2 = 40 and µ = KP 1 KP 2 = 0.1. Step 3: Choose T large. Note that the plant has a pole at s = −2. Select the compensator’s pole and zero to be well to the right of s = −2. So try s = −0.02 and s = −0.2. This gives T = 100. Step 4: The openloop compensated transfer function is 1 2 + 0.2 K 10 s + 0.02 s(s + 2) √ The precise value of the desired roots s = −1 ± j 3 probably will not lie exactly on the new root locus, but if T is chosen large enough, they should be close. The two specifications are ζ = 0.5 and ωn = 2, so we can look for a root that satisfies one specification exactly, and hope that the other is close to being satisfied. For T = 100, the root locus shows that s = −0.8989 + 1.56j gives ζ = 0.5, but ωn = 1.8, when K = 36. If ωn is close enough to 2, we can stop. Otherwise a larger value of T can be chosen. Note that the roots s = −0.8989 ± 1.56j are “dominant” only if we consider the pole at s = −0.2 to be canceled by the zero at s = −0.2. Gc (s)G(s) =
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11.20 Try a PID compensator. The transfer function is C(s) 5(KD s2 + KP s + KI ) = 3 R(s) s + 5KD s2 + 5KP s + 5KI The steadystate error for a step command input will be zero, as required. The characteristic equation is s3 + 5KD s2 + 5KP s + 5KI = 0 (1) To achieve τ = 1 and ζ = 0.45, the desired dominant roots are s = −1 ± j1.985. The third root is s = −b, where b has some arbitrary value such that b > 1 (so that s = −b will not be the dominant root). The polynomial corresponding to these three roots is [(s + 1)2 + (1.985)2 ](s + b) = 0 or s3 + (b + 2)s2 + (4.94 + 2b)s + 4.94b = 0
(2)
Comparing the coefficients of (1) and (2), we obtain KP =
4.94 + 2b 5
4.94b 5 b+2 KD = 5 The gains can be computed once a value for b > 1 has been selected. KI =
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11.21 In the first printing of the text, the last line of the problem statement says ”but increases Cv to 50”. It should say ”but increases Cv to 5”. desired valu Because the desired roots are obtainable with a simple gain selection, we need a lag compensator to achieve the desired Cv value. The lag compensator has the form Gc (s) = K Thus Cv =
s+z s+p
Kz 2p
To achieve Cv = 5 with K = 1, we need z/p = 10. Choosing z = 0.05 and p = 0.005 satisfies this requirement, and places the compensator’s pole and zero near the origin, far away from the desired root locations of s = −0.338 ± 0.562j. Using the root locus plot, we can determine the value of K required to place the closedloop roots near the desired location. This value is K = 1.024. Thus one solution is the compensator transfer function Gc (s) = 1.024
s + 0.05 s + 0.005
The resulting closedloop roots are s = −2.326, −0.055, and −0.312 ± 0.55j, which are close to the desired locations. The closedloop zero at s = 0.05 approximately cancels the closedloop pole at s = −0.055. The resulting Cv value is Cv = 5.12, which is close to the desired value.
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11.22 Note: In the first printing of the text, the transfer function was printed incorrectly. The correct transfer function is 1 Gp (s) = 2 s +s Also, the desired value of Cv was incorrectly given to be 5. It should be 10. √ The values ζ = 0.5 and ωn = 2 correspond to a dominant root location of s = −0.5±j 3. These roots can be obtained by setting K = 1. So the gain KP required to achieve the desired transient performance has already been established as KP 1 = K = 1. A lag compensator is indicated, because the steadystate error is too large. The second step is to determine the value of the parameter µ. For this system, the coefficient Cv is Cv = lim s s→0
KP = KP s(s + 1)
and KP 2 is the value of KP that gives Cv = 10. Thus, KP 2 = 10, and the parameter µ = KP /KP 2 = 1/10. The compensator’s pole and zero must be placed close to the imaginary axis, with the ratio of their distances being 1/10. Noting that the plant has a pole at s = −1, we select locations well to the right of this pole, say, at s = −0.01 and s = −0.1 for the pole and zero, respectively. This gives T = 100. The openloop transfer function of the compensated system is thus Gc (s)G(s)H(s) =
0.1Kc (s + 0.1) s(s + 1)(s + 0.01)
The root locus is shown in the following figure. For the desired damping ratio of 0.5, the locus essentially lies on the asymptote that passes through σ=
$
$
sp − sz 0 − 1 − 0.01 + 0.1 = = −0.455 P −Z 3−1
So the real part of the root is approximately −0.455, and the imaginary part will be 0.455 tan 60◦ = 0.788 to obtain ζ = 0.5. Thus the root is approximately s = −0.455 + 0.788j. The value of Kc required to obtain this root is found from # # # s(s + 1)(s + 0.01) # # # s + 0.1
0.1Kc = − ##
= 0.9128 s=−0.455+0.788j
or Kc = 9.128. This is only a tentative estimate because the root does not lie exactly on the locus. Using this value of Kc in the characteristic equation, we find that the actual roots are s = −0.4496 ± 0.788j and s = −0.1109. So our estimate turns out to be very close to the true value. The error coefficient is thus Cv = 9.128 and less than the desired value of 10. Also, the dominant roots differ somewhat from the desired locations at s = −0.5 + j0.866. If these differences are too large, the compensator’s pole and zero can be placed closer to the imaginary axis, say, at s = −0.01 and s = −0.001, respectively, with T = 1000. This will decrease the compensator’s influence on the locus near the desired root locations. 1156 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Root Locus 0.4
0.3
0.2
Imaginary Axis
0.1
0
−0.1
−0.2
−0.3
−0.4
−1
−0.9
−0.8
−0.7
−0.6
−0.5
−0.4
−0.3
−0.2
−0.1
0
Real Axis
Figure : for Problem 11.22
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11.23 Step 1: Adjust KP to meet Cv = 20. Cv = K/2 = 20. Thus KP = 40. Use KP = 40 in the following steps. Step 2: Check phase margin and gain margin with KP = 40, for G(s) =
20 s(0.5s + 1)
The phase margin is 17◦ and the gain margin is infinite. Thus we must add 40 − 17 = 23◦ to obtain a phase margin of 40◦ . This requires a lead compensator. Following the method of Example 11.3.3, we add a 5◦ safety factor and choose φm to be 23 + 5 = 28◦ . Then 1 + sin 28◦ µ= = 2.7698 1 − sin 28◦ √ Now find the frequency at which the uncompensated gain equals −20 log µ = −4.42 db. √ This occurs at about ω = 8.2 rad/sec. Thus choose ωm = 8.2 and ωm = 1/T µ = 8.2, which gives T = 0.073. Thus the compensator’s parameters are µ = 2.7698 and T = 0.073. The pole and zero are s = −1/T = −13.699 and s = −4.946. The openloop transfer function of the compensated system is Gc (s)G(s) =
s + 4.946 40 s + 13.699 s(s + 2)
As s → 0, the compensator’s gain is seen to introduce a gain factor of 4.946/13/699 = 0.361 = 1/µ. Thus, in order to preserve Cv = 20, we must increase KP by a factor of 1/0.361 = 2.7698. Therefore we set KP = 40(2.7698) = 110.79. The compensated openloop transfer function is Gc (s)G(s) = 110.79
s + 4.946 1 20(0.2022s + 1) = s + 13.699 s(s + 2) s(0.073s + 1)(0.5s + 1)
The frequency response plots show that the phase margin is 42◦ , the gain crossover frequency is ωg = 8, and the gain margin is infinite. Thus the specifications are satisfied. The closedloop transfer function is Gc (s)G(s) 20(0.2022s + 1) = 1 + Gc (s)G(s) 0.0365s3 + 0.573s2 + 5.044s + 20 The roots are s = −7.104 and s = −4.297 ± 7.659j, with a dominant time constant of 0.233, a dominant damping ratio of 0.489, and natural frequency of 8.78.
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11.24 In the first printing of the text, the transfer function was printed incorrectly. The correct transfer function is 2.5KP G(s) = s(s + 2)(0.25s + 2) With no compensation, Cv = 0.625, which is much less than the desired value of Cv = 80. Thus a lag compensator is needed to increase the gain by 80/0.625 = 128. Lead compensation is needed because it is not possible to place the roots at the desired location with a simple gain adjustment. Thus a simple lead or a simple lag compensator will not work. We therefore will design a laglead compensator. The angle deficiency is found from $
G(s)s=−2+2
√
3j
=$
#
# 10 # = 120 − 90 − 30 = −240◦ s(s + 2)(s + 8) #s=−2+2√3j
Thus, the angle deficiency is 180 − 240 = 60◦ . For the lead compensator G1 (s), $
G1 (s)s=−2+2√3j = $
#
1 + µT1 s ## = 60◦ 1 + T1 s #s=−2+2√3j
This is satisfied if the lead compensator has a pole at s = −53 and a zero at s = −3.65. Thus 1/µT1 = 3.65 and 1/T1 = 53. These give T1 = 0.0189 and µ = 14.5. For the lag compensator, the choice of T2 = 100 gives # #s + # # #s +
µ T2 1 T2
# # # # s + 0.145 ## # # # = 0.967 #=# # s + 0.01 #s=−2+2√3j
which we take to be close enough to 1 to indicate adequate polezero cancelation. The laglead compensator is thus given by Gc (s) = 128
s + 3.65 s + 0.145 s + 53 s + 0.01
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11.25 We try a compensator of the form Gc (s) = K
s+a s+b
The characteristic equation is found from 1 + Gc (s)Gp (s) = 0 which becomes 1+K
s+a 10 =0 2 s + b s (0.1s + 1)
or 0.1s4 + (1 + 0.1b)s3 + bs2 + 10Ks + 10Ka = 0
(1)
For τ = 1 and ζ = 0.5, the required dominant root pair is s = −1 ± 1.732j. So the characteristic equation must be factored as 0.1[(s + 1)2 + (1.732)2 ](s2 + cs + d) = 0 or 0.1(s2 +2s+4)(s2 +cs+d) = 0.1s4 +0.1(2+c)s3 +0.1(4+2c+d)s2 +0.1(4c+2d)s+0.4d = 0
(2)
Comparing the coefficients of equations (1) and (2), we obtain 1 + 0.1b = 0.1(2 + c)
(3)
b = 0.1(4 + 2c + d)
(4)
10K = 0.1(4c + 2d)
(5)
10Ka = 0.4d
(6)
We can choose c and solve (3) for b and (4) for d. The choice of c = 14 gives b = 6 and d = 28. So the secondary roots from s2 + cs + d = 0 are s = −2.417 and s = −11.58. Then solve (5) for K to obtain K = 1.12. Finally, solve (6) for a to obtain a = 1. The resulting compensator is s+1 Gc (s) = 1.12 s+6
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11.26 The root locus equation is 1+
K s+a =0 5 s2 (s + b)
The poles are s = 0, s = 0, and s = −b. The zero is s = −a. The asymptotic angles are θ = ±90◦ . The asymptotes intersect at σ=
a−b 2
For stability, the intersection point should be negative; thus, we should select b > a. For a 5% overshoot, the damping ratio of the dominant root must be ζ = 0.69, which corresponds to a complex root. For a time constant of τ = 0.5, the dominant roots must be s = −2 ± 2.096j to achieve ζ = 0.69. The corresponding factor is (s + 2)2 + (2.096)2 = s2 + 4s + 8.393. Thus the characteristic equation can be factored as (s2 + 4s + 8.393)(s − s3 ) = s3 + (4 − s3 )s2 + (8.393 − 4s3 )s − 8.393s3 = 0 The characteristic equation is s3 + bs2 +
K K s+a =0 5 5
Comparing coefficients, we see that s3 = 4 − b
K = 8.393 − 4s3 5
a=−
8.393s3 K/5
The third root must lie to the left of the dominant root, whose real part is −2. Therefore, b must be chosen so that b > 6. The other restriction is that b > a. One solution is b = 10, which gives s3 = −6, K = 161.965, and a = 1.55. If we try to cancel one of the poles at s = 0 by letting a → 0, the root locus equation becomes K 1 1+ =0 5 s(s + b) and the characteristic equation is s2 + bs +
K =0 5
The specifications that ζ = 0.69 and τ = 0.5 require that s = −2 ± 2.096j. This is achieved with b = 4 and K = 41.95.
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11.27 The openloop poles are s = 0 and −2 ± 3j. a) The root locus equation for the lead compensator is 1+
s+ s
1 µT + T1
s(s2
Kc =0 + 4s + 13)
where µ > 1 and the root locus parameter is K = Kc ≥ 0. b) The root locus equation for the lag compensator is 1+
s+ s
1 µT + T1
µKc =0 s(s2 + 4s + 13)
where µ < 1 and the root locus parameter is K = µKc ≥ 0. c) The root locus equation for the reversereaction compensator is 1+
s− s+
1 T1 1 T2
− TT12 Kc
s(s2 + 4s + 13)
=0
where the root locus parameter is K = −T1 Kc /T2 ≤ 0. The rootlocus plots are shown in the following figures. All three compensators can produce an unstable system if the gain Kc is too large. However, only the reversereaction compensator can pull the complexroot paths starting at s = −2±3j to the left. This enables a dominant root to be obtained that is not close to the imaginary axis. The optimum root locations are shown on the plot.
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Figure : For Problem 11.27
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11.28 a) The plot for K = 2 is shown in figure (a). It can be obtained with the following MATLAB file. K = 2; sys = tf(5*K,[1,6,5,0]) margin(sys) The results are that the gain margin is 9.54 dB; the phase crossover frequency is 2.24 rad/sec; the phase margin is 25.4◦ ; and the gain crossover frequency is 1.23 rad/sec. b) The plot for K = 20 is shown in figure (b). It can be obtained in a manner similar to that used for part (a). The results are that the gain margin is −10.5 dB (which means that the system is unstable); the phase crossover frequency is 2.24 rad/sec; the phase margin is −23.7◦ (another indication of instability); and the gain crossover frequency is 3.91 rad/sec. c) For K = 6 both the gain and phase margins are 0. Thus they both are a limiting factor for stability.
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11.29 The openloop transfer function is G(s) = Gc (s)Gp (s) =
2 + 19s 100s2 + s
The Bode plots are shown in following figure. They can be obtained with the MATLAB margin function. The gain margin is infinite because the phase curve is always above the −180◦ line. The phase margin is 66◦ . The system is stable. Bode Diagram Gm = Inf , Pm = 66.3 deg (at 0.212 rad/sec) 150
Magnitude (dB)
100
50
0
Phase (deg)
−50 −90
−120
−150 −4
10
−3
10
−2
10
−1
10
0
10
1
10
Frequency (rad/sec)
Figure : For Problem 11.29
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11.30 The openloop transfer function is G(s) = Gc (s)Gp (s) =
25(7s + 64) 175s + 1600 = 3 5s3 + 6s2 + 5s 5s + 6s2 + 5s
The Bode plots are shown in following figure. They can be obtained with the MATLAB margin function. Both the gain and phase margins are negative, so the system is unstable. Bode Diagram Gm = −47.3 dB (at 1.07 rad/sec) , Pm = −41.5 deg (at 7.46 rad/sec) 150
Magnitude (dB)
100
50
0
−50
−100 −90
Phase (deg)
−135
−180
−225
−270 −2
10
−1
10
0
1
10
10
2
10
3
10
Frequency (rad/sec)
Figure : For Problem 11.30
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11.31 The openloop transfer function is G(s) =
0.1KP (10) KP = (0.2s + 1)(s2 + s + 10) 0.2s3 + 1.2s2 + 3s + 10
The Bode plots, phase margins, and gain margins can be obtained with the MATLAB margin function. The results are as follows: a) KP = 1: Gain margin = 20.9 dB, Phase margin = 88.3◦ b) KP = 10: Gain margin = 0.916 dB, Phase margin = 70.4◦ c) KP = 100: Gain margin = −19.1 dB, Phase margin = −115◦ Cases (a) and (b) are stable, but Case (c) is unstable. This means that any initial roll angle will continue to increase.
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11.32 One solution is K = 0.376. 11.33 The answers are given in the following table.
Case (a) (b) (c)
Type No. 1 0 2
Cp ∞ 20 ∞
Cv 20 0 ∞
Table : for Problem 11.33 Ca Step Error Ramp Error 0 0 1/20 0 1/21 ∞ 7 0 0
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11.34 The openloop transfer function is G(s) = KP e−(D1 +D2 )s
1 1 = KP e−100s 100s + 1 100s + 1
The delay of 100 s lowers the phase curve by 100ω rad, or 100ω(180/π) degrees . A plot of m and φ for KP = 1 shows that instability is caused by the negative gain margin. The phase crossover frequency is approximately 0.02 rad/sec, and m at this frequency is approximately −7 dB. To achieve a positive gain margin, we must therefore increase KP to at least 107/20 = 2.239. For this value of KP both the phase margin and the gain margin are zero. The calculations for KP = 1 can be done in MATLAB as follows. sys = tf(1,[100,1]); w = [0.001:0.001:0.02]’; [mag, phase] = bode(sys,w); m = 20*log10(mag(:)); phasetotal = phase(:)100*w*(180/pi); semilogx(w,m,w,phasetotal),grid
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11.35 10.31 With D = 0, the openloop frequency response plot of 10/(0.1s + 1) has a gain crossover at ω ≈ 100, and a phase margin of 95◦ . For PM = 40◦ , the dead time D can reduce the phase curve at ω = 100 by no more than 95 − 40 = 55◦ . Because $
P (iω)e−iωD = $ P (iω) − ωD
we have 100D ≤ 55
π rad 180
or D ≤ 0.0096.
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11.36 In the first printing of the text, the required value for Cv was incorrectly stated as Cv = 0.2. The required value should be Cv = 5. One solution is the compensator transfer function Gc (s) =
s + 0.1 s + 0.01
This gives a gain margin of 14.3 dB and a phase margin of 41.6◦ .
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11.37 One solution is the compensator transfer function Gc (s) = 42
s + 4.4 s + 18
This gives an infinite gain margin and a phase margin of approximately 50◦ .
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11.38 The problem statement should read “at least 45◦ ”. One solution is the compensator transfer function Gc (s) = 20
6.66s + 1 66.6s + 1
This gives an infinite gain margin and a phase margin of approximately 57◦ .
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11.39 In the first printing of the text, the required value for Cv was incorrectly stated as Cv = 0.05. The required value should be Cv = 20. Also, in the second line after the equation, delete the phrase “the steadystate ramp error will be”. One solution is the compensator transfer function Gc (s) = 20
0.34s + 1 0.07s + 1
This gives an infinite gain margin and a phase margin of approximately 50◦ .
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11.40 In the first printing of the text, the required value for Cv was incorrectly stated as Cv = 0.02. The required value should be Cv = 50. One solution is the compensator transfer function Gc (s) = 50
(s + 1)(s + 0.16) (s + 0.01)(s + 16)
The closedloop transfer function is C(s) 50s2 + 58s + 8 = R(s) 0.2s5 + 4.402s4 + 20.24s3 + 66.2s2 + 58.16s + 8 There is polezero cancelation of the factor s + 1, so the transfer function becomes C(s) s + 0.16 = 250 R(s) (s + 17.185)(s + 0.168)(s2 + 3.656s + 13.846) The roots are s = −17.1837, s = −0.1681, and s = −1.8291 ± 3.2401j. If we take the roots s = −0.1681 to be canceled approximately by the compensator zero at s = −0.16, the root pair s = −1.8291 ± 3.2401j is the dominant root. It has a damping ratio of ζ = cos(tan−1 (3.2401/1.8291)) = 0.4916, which is approximately the desired value of 0.5.
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11.41 One solution is the compensator transfer function Gc (s) = 20
(s + 0.7)(s + 0.15) (s + 7)(s + 0.015)
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c Solutions Manual! to accompany System Dynamics, First Edition by William J. Palm III University of Rhode Island
Solutions to Problems in Chapter Twelve
c Solutions Manual Copyright 2004 by McGrawHill Companies, Inc. Permission required ! for use, reproduction, or display.
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12.1 The equation of motion is m¨ x + kx = ky where y(t) = Y sin 6πt = 4 × 10−3 sin 6πt. Also, ωn2 =
500 k = = 1000 rad/s m 0.5
From (12.1.9), with ζ = 0 and r 2 = (6π)2 /1000, ! !
X = Y !!
!
1 !! = 6.2 × 10−3 m 1 − r2 !
or 6.2 mm
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12.2 For a period of 20 feet and a vehicle speed of v (mph), the frequency ω is ω=
"
5280 20
#"
#
1 (2π) = 0.4608v 3600
rad/sec
Thus ω = 0.4608(20) = 9.216 rad/sec for v = 20 mph, and ω = 0.4608(50) = 23.04 rad/sec for v = 50 mph. For a car weighing lb, the quartercar mass is m = 500/32.2 slugs. Its natural $ 2000 $ frequency is ωn = k/m = 2000/(500/32.2) = 11.35 rad/sec. Its frequency ratio at 20 mph is r = ω/ωn = 9.216/11.35 = 0.812, and at 50 mph it is r = 23.04/11.35 = 2.03. Its damping ratio is 360 ζ= $ = 1.02 2 2000(500/32.2)
Now substitute these values of r and ζ and Y = 0.03 ft into the following expressions, obtained from (12.1.9) and (12.1.12). X=Y
%
1 + 4ζ 2 r 2 (1 − r 2 )2 + 4ζ 2 r 2
Ft = r 2 kX This gives the following table.
v (mph) 20 50
2000 lb Car (ζ = 1.02) r X (ft) Ft (lb) 0.812 0.034 45.12 2.03 0.025 203.4
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12.3 We are given m = 1500 kg, k = 20, 000 N/m, ζ = 0.04, and Y = 0.01 m. At resonance, $
ωr ωn 1 − 2ζ 2 r= = = 0.998 ωn ωn and Ft = r kY 2
%
4ζ 2 r 2 + 1 = 2500 N (1 − r 2 )2 + 4ζ 2 r 2
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12.4 The static deflection is δ = mg/k. Thus k = mg/δ = 200/0.003 = 66 667 N/m. Also, ω = 40 Hz = 215 rad/s r = 2
From (12.1.9) with c = 0,
"
ω ωn
ωn =
%
#2
"
=
k = m
251 57.2
%
#2
66 667 = 57.2 rad/s (200/9.81)
= 19.26
X 1 = = 0.055 Y 1 − r 2 
Thus 5.5% of the airframe motion is transmitted to the module.
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12.5 a) Neglect damping in the isolator, and determine its required stiffness k. From (12.1.9), 1 X = = 0.1 Y 1 − r 2 
which gives r 2 = 11. Thus
ωn2 =
ω2 [3000(2π)/60]2 (314)2 = = r2 11 11
and
2 (314)2 = 556.7lb/ft k = mωn2 = 32.2 11 √ b) r = ω/ωn and ωn = 314/ 11. Thus
From (12.1.9)
r1 =
2500(2π)/60 √ = 2.77 314/ 11
r2 =
3500(2π)/60 √ = 3.87 314/ 11 1 X = Y 1 − r 2 
Thus the highest percentage of motion will be transmitted at the lowest r value, which corresponds to 2500 rpm. For 2500 rpm, X 1 = = 0.15 Y 1 − r12  For 3500 rpm,
1 X = = 0.07 Y 1 − r22 
Thus at most, 15% of the crane motion will be transmitted to the module.
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12.6 We are given m = 5/32.2 slug and r = 2
From (12.1.9) with ζ = 0,
"
ω ωn
#2
=
(30(2π)/60)2 32.2k/5
1 X = = 0.1 Y 1 − r 2 
which gives r 2 = 11. Thus
(30(2π)/60)2 = 11 32.2k/5
Solve for k to obtain k = 0.139 lb/ft. The transmitted force is "
Ft = r
2X
Y
#
kY = 11(0.1)0.139(0.003) = 4.59 × 10−4 lb
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12.7 a) From Newton’s law, m¨ x = c(y˙ − x) ˙ − kx
or
mx˙ + cx˙ + kx = cy˙ T (s) =
X(s) cs = 2 Y (s) ms + cs + k
Thus T (jω) = This gives T (jω) =
cωm mk j
−mω 2
cωj/k cωj = + cωj + k 1 − r 2 + cωj/k
1 − r 2 + cωmj/mk
=
$
2ζωn ωj/ωn2 2ζrj = 2 2 1 − r + 2ζωn ωj/ωn 1 − r 2 + 2ζrj
where we have used the fact that ωn = k/m, r = ω/ωn , and c/m = 2ζωn . The magnitude is 2ζr X=$ Y 2 (1 − r )2 + (2ζr)2 b) Thus
2ζrk Ft = kX = $ Y (1 − r 2 )2 + (2ζr)2
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12.8 The displacement transmissibility with r = 2 is X = Y
%
4ζ 2 r 2 + 1 = (1 − r 2 )2 + 4ζ 2 r 2
%
1 + 16ζ 2 9 + 16ζ 2
This has a minimum of 1/3 when ζ = 0. This is the best choice for ζ if the velocity increases (r > 2). However, if the velocity decreases (so that r → 1, and the system approaches resonance), an nonzero value of ζ would be a better choice to limit the resonant response. With ζ = 0 and a 20% increase in r (r = 2.4), we have X/Y = 0.21. With ζ = 0 and a 20% decrease in r (r = 1.6), we have X/Y = 0.64.
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12.9 The relations are Ft = r 2 kX
X=Y where
%
1 + 4ζ 2 r 2 (1 − r 2 )2 + 4ζ 2 r 2
c ζ= √ 2 mk
r=
ω ωn
At 40 mph, the forcing frequency is ω=
"
#"
5280 20
#
1 2π(4) = 18.432 rad/sec 3600
and r = 18.432/ωn , where ωn =
%
k m
To minimize Ft , choose either 1. r small (with ζ near 1 and ωn large; this means k large and c large), or 2. r > 2 (this means ωn small). For example, with m = 800/32.2, choosing ζ = 1 and r = 0.5, implies that ωn = 2(18.432) = 36.864 k = m(36.864)2 = 33 763 lb/ft √ c = 2 mkζ = 1832 lb sec/ft This gives X =Y
%
and
4ζ 2 r 2 + 1 = 0.0589 ft (1 − r 2 )2 + 4ζ 2 r 2 Ft = r 2 kX = 497 lb
On the other hand, choosing ζ = 1 and r = 3 for example, we obtain k=
mωn2
800 = 32.2
c=2 This gives X =Y
&
%
"
18.432 3
#2
= 937.9 lb/ft
800 937.9 = 305 lb sec/ft 32.2
4ζ 2 r 2 + 1 = 0.0317 ft (1 − r 2 )2 + 4ζ 2 r 2
1210 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
and Ft = r 2 kX = 268 lb Note that the plot in Figure 12.1.3 is a plot of Ft /kY . Thus the curves in the plot must be interpreted with the value of k kept in mind. Even though the curve for r = 0.5 is lower that the curve for r = 3, for ζ = 1, the corresponding values of k are different, and thus the case r = 3 gives a lower value of Ft .
1211 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
12.10 For the lumpedmass equivalent system, we have the following spring constant and equivalent mass (after converting inches to feet): (4.32 × 109 )(0.333)(0.03125)3 Ewh3 = 4L3 4(0.5)3 = 8.78 × 104 lb/ft
k =
We compute the equivalent mass of the beam as follows. Using 15.2 slug/ft3 for the density of steel, and including 23% of the beam’s mass, we obtain me =
20 + 0.23(15.2)(0.333)(0.5)(0.03125) = 0.640 32.17
slug
The unbalanced mass is m = 1/32.17 = 0.0311 slug. The model for the system is ¨ + cx˙ + kx = f (t) = mu %ω 2 sin ωt me x This gives the transfer function 1 X(s) = 2 F (s) me s + cs + k 1/k me 2 c k s + ks + 1
T (s) = = Thus,
T (jω) = where 1/k = 1.139 × 10−5 and ωn = ratio is.
$
1−
'
1/k ω ωn
(2
+
2ζω ωn j
k/me = 370 rad/sec = 3537 rpm. The log magnitude +
ω2 m(ω) = 20 log(1/k) − 10 log 1 − 2 ωn
,2
+
"
2ζω ωn
#2
The motor speed of 1750 rpm gives a forcing frequency of ω = 183 rad/sec. Using ζ = 0.1 and ω/ωn = 183/370, we see that m = 2.36 − 98.9 = −96.5 db, which corresponds to a magnitude ratio of 1.5 × 10−5 . The amplitude of the forcing function is mu %ω 2 = 0.0311(0.01)(183)2 = 10.4 lb. Thus, the steadystate amplitude is 10.4(1.5 × 10−5 ) = 15.6×10−5 ft. On the downward oscillation, the total amplitude as measured from horizontal is the preceding value plus the static deflection, or 15.6 × 10−5 + 20/87800 = 3.84 × 10−4 ft. Because ω is not close to ωn in this problem, the preceding results are not very sensitive to the assumed value of ζ = 0.1. For example, the calculated amplitudes of vibration for ζ = 0.05 and 0.2 are close to the amplitude for ζ = 0.1 (the amplitudes are 15.7 × 10−5 ft and 15.2 × 10−5 ft, respectively). However, if we had used a motor with a speed of 3500 rpm = 366 rad/sec, this choice would put the forcing frequency very close to the 1212 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
natural frequency. In this region, the assumed value of ζ would be critical in the amplitude calculation. In practice, such a design would be avoided. That is, in vibration analysis, the most important quantity to know is the natural frequency ωn . If the damping is slight, the resonant frequency is near ωn . If ωn is designed so that it is not close to the forcing frequency, it is often not necessary to know the precise amount of damping.
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12.11 We are given that the natural frequency is ωn1 = 900 rpm. Thus, at 1750 rpm, r1 =
1750 = 1.94 900
If we decrease the stiffness k by 1/2, then the new natural frequency will be ωn2 =
%
1 900 k/2 = √ ωn1 = √ rpm m 2 2
Therfore, r2 =
1750 √ = 2.75 900/ 2
From the rotating unbalance equation (12.2.6) with ζ = 0, X=
mu % r 2 m 1 − r 2 
(1)
We are given that at r1 , X1 = 8 mm. We need to compute X2 . From (1), X2 = X1
"
r2 r1
#2
−2.7636 1 − r12 = 0.842 =2 2 −6.5625 1 − r2
Thus X2 = 0.842X1 = 6.73 mm
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12.12 We are given that k= Thus ωn1 =
%
500 = 24, 000 lb/ft 0.25/12
k = m
%
24, 000 = 39.31 rad/sec 500/32.2
From the rotating unbalance equation (12.2.6) with ζ = 0, mu % r 2 m 1 − r 2 
X= and thus
X2 = X1
where
"
r2 r1
#2
1 − r12 1 − r22
1750(2π)/60 = 4.66 39.31 After the block is added, the new mass is m2 = 4m1 and the new natural frequency is r1 =
ωn2 = Therefore, r2 = Thus,
%
1 k = ωn1 4m1 2
ω ω =2 = 2r1 ωn2 ωn1
X2 −20.7156 1 − r12 = 0.9651 = (2)2 =4 X1 −85.8624 1 − r22
and given that X1 = 0.1 in, we have
X2 = 0.9651(0.1) = 0.0965 in
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12.13 The transmitted force is given by Ft = mu %ω
2
%
1 + 4ζ 2 r 2 (1 − r 2 )2 + 4ζ 2 r 2
We are given that mu = 0.05/32.2 = 0.00155 slug, m = 50/32.2 = 1.55 slug. R = 0.1/12 = 0.0083 ft, and ω = 1000(2π)/60 = 104.7 rad/sec. Thus r= Thus
ω 104.7 =$ = 5.83 ωn 500/1.55
Ft = (0.00155)(0.0083)(104.7)2
%
1 + 136ζ 2 = 0.141 1088 + 136ζ 2
%
1 + 136ζ 2 1088 + 136ζ 2
a) For ζ = 0.05, Ft = 0.0049 lb. b) For ζ = 0.7, Ft = 0.034 lb.
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12.14 We have Ft = mu %ω 2 Tr where Ft = 15, ω = 200(2π)/60 = 20.9 rad/s, and because c = 0, Tr = We have r= Thus Tr = 0.0826 and
r2
1 −1
ω 20.9 20.9 =$ =$ = 3.62 ωn k/M 2500/75 Ft = mu %(20.9)2 (0.0826) = 15
Solve for mu % to obtain mu % = 0.416 kg m.
1217 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
12.15 We want Tr = 0.1. Neglecting damping, we have from (12.2.12) r2 = But r= √ Thus ωn = 314/ 11 = 94.7. But
1 + Tr = 11 Tr
314 √ 3000(2π)/60 = = 11 ωn ωn
ωn =
%
k = m
%
k 3
from which we obtain k = 3ωn2 = 3(94.7)2 = 26 904 N/m.
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12.16 Damping is assumed to be negligible. Thus for the vertical motion, Ft 1 = Tr = 2 F r −1
where r=
1750(2π)/60 ω ω = $ = 10.24 =$ ωn 4k/m 8000/25
Thus Tr = 0.0096. The equation of motion for rotation is
I θ¨ = −(4kRθ)D/2 $
Thus the natural frequency for rotation is ωn = 4kD 2 /4I = rotational motion, 1 Tt = Tr = 2 T r −1 where 1750(2π)/60 ω = 9.16 = r= ωn 20
$
80/0.2 = 20. For the
Thus Tr = 0.012.
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12.17 We are given m = 20/g slug, mu = 1/g slug, % = 0.01 ft, and ω = 3500(2π)/60 = 366.5 rad/sec. Thus the unbalance force amplitude is mu %ω = 2
"
#
1 (0.01)(366.5)2 = 41.719 lb 32.2
The beam stiffness is k=
Ewh3 4.32 × 109 (1/3)(3/96)3 = = 8.79 × 104 lb/ft 4L3 4(1/2)3
The beam mass is m = ρV = 15.2(1/3)(3/96)(1/2) = 0.079 slug. The first design equation for the absorber is r2 = 1, or r2 =
ω 366.5 =$ =1 ωn2 k2 /m2
This implies that k2 /m2 = (366.5)2 . The second design equation is X2 = 0.25/12 = 1/48 ft, where 1 41.719 X2 = F = k2 k2 Thus k2 = 41.719(48) = 2002.5 lb/ft. Substitute this value into the first design equation to obtain k2 2002.5 m2 = = 0.0149 slug 2 (366.5) (366.5)2
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12.18 We are given that X2 ≤ 0.08/12 ft, ω = 6000(2π)/60 = 628.3 rad/sec, and that the unbalance force amplitude is mu %ω 2 = 60 lb. Thus 60 60 = = 0.0955 2 ω 628.3
mu % =
The first design equation for the absorber is k2 = The second design equation is
F 60 = = 9000 lb/ft X2 0.08/12 %
Thus m2 =
k2 = ω = 628.3 m2
k2 = 0.023 slug (628.3)2
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12.19 We are given that ω = 200(2π)/60 = 20.94 rad/sec, the amplitude of the unbalance force is mu %ω 2 = 4 lb, and that X2 ≤ 1/12 ft. Assume that the table legs are rigid. The first design equation for the absorber is %
k2 = ω = 20.94 m2
The second design equation is k2 =
F 4 = = 48 lb/ft X2 1/12
m2 =
48 = 0.109 slug (20.94)2
Thus
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12.20 The machine mass is m1 = 8 kg. a) We are given that ωn = 2π(6) = 12π rad/sec, mu %ω 2 = 50 N, ω = 4(2π) = 8π rad/sec, and X2 ≤ 0.1 m. The design equation for the absorber is % k2 = ω = 8π m2 Thus k2 = 50/0.1 = 500 N/m, and m2 =
k2 500 = = 0.792 kg 2 (8π) (8π)2
b) We have that k1 = m1 ωn2 = (12π)2 m1 = 1421m1 = 11368 N/m. From (12.3.7), 1 X1 (jω) = T1 (jω) = F (jω) 11368 where
! ! ! ! 1 − r24 ! ! 0 !/ 2 4 ! ! b r2 − [1 + (1 + µ)b2 ] r22 + 1 !
2 ωn2 8π = = ωn1 12π 3 0.792 m2 = µ= = 6.336 m1 8 13 3.168 + = 1.84 1 + (1 + µ)b2 = 9 8 b=
The amplitude of F (jω) is mu %ω 2 , where r2 = ωωn2 = ω/8π. Thus 1 X1 = mu % 11368 or
! ! X1 1 !! 3 = mu % 11368 !! 4 9
The plot is shown in the following figure.
! ! ! ω 2 11 − r 2 2 ! ! ! 2 !4 4 ! ! r2 − 1.84r22 + 1 ! 9
'
ω2 1 −
ω4 4096π 4
ω2 64π 2
(
2
ω − 1.84 64π 2
! ! ! 4 !! +1 !
1223 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
1
0.9
0.8
0.6
0.5
1
u
X /m " (slug−1)
0.7
0.4
0.3
0.2
0.1
0
0
10
20
30
60 50 40 Forcing frequency ! (rad/s)
70
80
90
100
Figure : for Problem 12.20.
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12.21 The first printing of the text had incorrect values for some of the frequencies. The correct values are as follows. The operating speed range is from 1500 to 3000 rpm. Without an absorber, excessive vibration was observed at 2100 rpm. After the absorber was attached, resonance was observed at 1545 and 2850 rpm. An absorber tuned to 2100 rpm requires that %
k2 = m2
%
2100(2π) k1 = 219.9 rad/sec = m1 60
Because we are given that m2 = 5/32.2 = 0.155, we have that k2 = (219.9)2 m2 = 9747.8. Also, k1 = (219.9)2 m1 (1) The characteristic equation of the combined system is given by the denominator of (12.3.1), and is m1 m2 s4 + (m2 k1 + m2 k2 + m1 k2 )s2 + k1 k2 = 0 (1) One of the observed resonances of the combined system is 2850 rpm, or 2850(2π)/60 = 298.45 rad/sec. Thus substituting s = 298.45j and relation (1) into (2), along with the values m2 = 0.155 and k2 = 9747, we can determine the value of m1 , which is m1 = 0.4 slug. Thus gives k1 = 19, 344 lb/ft. With m1 and k1 determined, we now calculate the required values of m2 and k2 . Suppose we choose to put the resonances just outside the operating range, say at 1400 and 3100 rpm (146.6 and 324.6 rad/sec). Let λ = s2 , and let λ1 and λ2 denote the desired values of s2 . We can factor the characteristic equation (2) as follows: m1 m2 (λ − λ1 )(λ − λ2 ) = m1 m2 λ2 − m1 m2 (λ1 + λ2 )λ + m1 m2 λ1 λ2 = 0
(3)
Comparing the coefficients of (3) with (2), we see that
and
(m2 k1 + m2 k2 + m1 k2 = −m1 m2 (λ1 + λ2 ) m1 m2 λ1 λ2 = k1 k2
(4)
(5)
We can solve (5) for k2 as follows. k2 =
m 1 m 2 λ1 λ2 = Dm2 k1
where D=
(6)
m1 λ1 λ2 k1
Substitute (6) into (4) and solve for m2 . m2 = −
m1 (λ1 + λ2 ) + k1 + m1 D D
The desired values are λ1 = −(146.6)2 and λ2 = −(324.6)2 . These give the absorber values m2 = 0.2706 slug and k2 = 12, 672 lb/ft. 1225 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
12.22 a) From Newton’s law, m1 x ¨1 = f − kx1 + c(x˙ 2 − x˙ 1 ) m2 x ¨2 = −c(x˙ 2 − x˙ 1 )
Transform these equations with zero initial conditions to obtain (m1 s2 + cs + k)X1 (s) − csX2 (s) = F (s) −csX1 (s) + (m2 s2 + cs)X2 (s) = 0
The solutions obtained with Cramer’s rule are X1 (s) =
m2 s2 + cs F (s) D(s)
X2 (s) =
cs F (s) D(s)
where Cramer’s determinant is D(s) = s(m1 m2 s3 + c(m1 + m2 )s2 + km2 s + ck) Define the following parameters: µ=
m2 m1
ω12 =
c ζ= √ 2 m 1 k1
k1 m1
r=
ω ω1
Then D(jω) can be written as
or
! !
! !
D(jω) = !ωj[−m1 m2 ω 3 j − c(m1 + m2 )ω 2 + km2 ωj + kc]! 5
D(jω) = m21 ω14 r (2ζr)2 [1 − (1 + µ)r 2 ]2 + µ2 r 2 (1 − r 2 )2 $
The numerator of kX1 (jω)/F (jω) is  − m2 ω 2 + cωj = m1 rω12 4ζ 2 + µ2 r 2 . The numerator of X2 (jω)/F (jω) is cωj. Thus, $
and
kX(jω) 4ζ 2 + µ2 r 2 =$ F (jω) (2ζ)2 [1 − (1 + µ)r 2 ]2 + µ2 r 2 (1 − r 2 )2 1 X2 (jω) 2ζ = 2$ 2 F (jω) ω1 (2ζ) [1 − (1 + µ)r 2 ]2 + µ2 r 2 (1 − r 2 )2
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12.23 a) From Newton’s law, m1 x ¨1 = f − kx1 + k2 (x2 − x1 ) + c(x˙ 2 − x˙ 1 ) m2 x ¨2 = −k2 (x2 − x1 ) − c(x˙ 2 − x˙ 1 )
Transform these equations with zero initial conditions to obtain (m1 s2 + cs + k1 + k2 )X1 (s) − (cs + k2 )X2 (s) = F (s) −(cs + k2 )X1 (s) + (m2 s2 + cs + k2 )X2 (s) = 0
The solutions obtained with Cramer’s rule are X1 (s) =
m2 s2 + cs + k2 F (s) D(s)
X2 (s) =
cs + k2 F (s) D(s)
where Cramer’s determinant is D(s) = m1 m2 s4 + c(m1 + m2 )s3 + (m2 k1 + m2 k2 + m1 k2 )s2 + ck1 s + k1 k2 Define the following parameters: µ=
m2 m1
ω12 =
k1 m1
ω22 =
k2 m2
ω1 ω r= ω2 ω1 c k2 ζ= √ λ= k1 2 m 1 k1 α=
Then D(jω) can be written as
or
! !
! !
D(jω) = !m1 m2 ω 4 − (m2 k1 + m2 k2 + m1 k2 )ω 2 + k1 k2 + [ck1 ω − (m1 + m2 )cω 3 ]j ! ! !
! !
D(jω) = !m21 ω 4 [µr 4 − (1 + λ + µλ)r 2 + µα2 ] + [2ζr − 2(1 + µ)r 3 ]j ! !
!
The numerator of k1 X1 (jω)/F (jω) is k2 − m2 ω 2 + cωj = k1 ![λ − µr 2 + 2ζri]!. The numerator of k1 X2 (jω)/F (jω) is cωj + k2  = k1 (λ + 2ζrj). Thus, $
and
k1 X(jω) (λ − µr 2 )2 + (2ζr)2 =5 F (jω) [µr 4 − (1 + λ + µλ)r 2 + µα2 ]2 + (2ζr)2 [1 − (1 + µ)r 2 ]2 $
k1 X2 (jω) λ2 + (2ζr)2 =5 F (jω) [µr 4 − (1 + λ + µλ)r 2 + µα2 ]2 + (2ζr)2 [1 − (1 + µ)r 2 ]2
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12.24 The equations of motion are m1 x ¨1 = −k1 x1 − k2 (x1 − x2 ) m2 x ¨2 = k2 (x1 − x2 )
From these equations we can write the modal amplitude equations by substituting x1 = A1 est and x2 = A2 est , and using the given parameter values. The result is (10s2 + 30, 000)A1 − 20, 000A2 = 0 These give the solution
−20, 000A1 + (30, 000s2 + 20, 000)A2 = 0
s2 + 3000 A1 (1) 2000 The roots are found from Cramer’s determinant of the modal equations, which is A2 =
3s4 + 11, 000s2 + 2 × 106 = 0 The roots are s2 = −3475 and s2 = −192. Substitute s2 = −3475 into (1) to obtain A2 = −0.2375A1 . Substitute s2 = −192 into (1) to obtain A2 = 1.404A1 . √In the first mode, the masses oscillate in opposite directions with a radian frequency of 3475. The displacement amplitude of mass 2 is 0.2375 times that of mass 1. In √ the second mode, the masses oscillate in the same direction with a radian frequency of 192. The displacement amplitude of mass 2 is 1.404 times that of mass 1.
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12.25 The equations of motion are m1 L2 θ¨1 = −mg1 Lθ1 − kd(dθ1 − dθ2 ) m2 L2 θ¨2 = −m2 gLθ2 + kd(dθ1 − dθ2 )
From these equations we can write the modal amplitude equations by substituting θ1 = A1 est and θ2 = A2 est . The result is (m1 L2 s2 + m1 gL + kd2 )A1 − kd2 A2 = 0 −kd2 A1 + (m2 L2 s2 + m2 gL + kd2 )A2 = 0
Using the given parameter values and g = 9.81 m/s2 , these equations give the solution A2 = (3.13s2 + 7.13)A1
(1)
The roots are found from Cramer’s determinant of the modal equations, which is 2500s4 + 10810s2 + 11 586 = 0 The roots are s2 = −1.96 and s2 = −2.36. Substitute s2 = −1.96 into (1) to obtain A2 = 0.995A1 . Substitute s2 = −2.36 into (1) to obtain A2 = −0.257A1 . √ In the first mode, the masses oscillate in the same direction with a radian frequency of 1.96. The displacement amplitude of mass 2 is 0.995 times that of mass 1. In√the second mode, the masses oscillate in the opposite direction with a radian frequency of 2.36. The displacement amplitude of mass 2 is 0.257 times that of mass 1.
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12.26 The equations of motion are I1 θ¨1 = k2 (θ2 − θ1 ) − k1 θ1 I2 θ¨2 = −k2 (θ2 − θ1 )
From these equations we can write the modal amplitude equations by substituting θ1 = A1 est and θ2 = A2 est . The result is (I1 s2 + k1 + k2 )A1 − k2 A2 = 0 −k2 A1 + (I2 s2 + k2 )A2 = 0
Using the given parameter values, these equations give A2 =
I1 s2 + k1 + k2 s2 + 4 A1 θ1 = k2 3
(1)
The roots are found from Cramer’s determinant of the modal equations, which is 5s4 + 23s2 + 3 = 0 The roots are s2 = −0.134 and s2 = −4.47. Substitute s2 = −0.134 into (1) to obtain A2 = 1.29A1 . Substitute s2 = −4.47 into (1) to obtain A2 = −0.157A1 . √ In the first mode, the masses oscillate in the same direction with a radian frequency of 0.134. The displacement amplitude of mass 2 is 1.29 times that of mass 1. In√the second mode, the masses oscillate in the opposite direction with a radian frequency of 4.47. The displacement amplitude of mass 2 is 0.157 times that of mass 1.
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12.27 The equations of motion are m¨ x = −2k sin 45◦ x = −1.41kx m¨ y = −ky − 2k sin 45◦ x = −2.41kx
$
In the first mode, the mass oscillates in the x direction with a radian frequency of 1.41k/m. In $ the second mode, the mass oscillates in the y direction with a radian frequency of 2.41k/m.
1231 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
12.28 From (12.4.8): (750)(1350)s4
+
6
7
730[(1.95 × 104 )(1.5)2 + 2.3 × 104 (1.1)2 ] + 1350(4.25 × 104 ) s2
+ 1.95(2.3) × 108 (2.6) = 0 or
s4 + 111.329s2 + 1.166 × 104 = 0
This gives or
s2 = −99.43
and
s2 = −11.9
s = ±9.971j
and
s = ±3.45j
These correspond to frequencies of 1.587 Hz and 0.549 Hz.
x 3.95 × 103 5.411 k1 L1 − k2 L2 A1 = = = = 2 2 2 4 A2 θ ms + k1 + k2 730s + 4.25 × 10 s + 58.2192 For mode 1 (s2 = −99.43), x = −0.131 m ahead of the mass center θ For mode 2 (s2 = −11.9), x = 0.1168 m behind the mass center θ
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12.29 Referring to Figure 12.4.4, we are given that m1 g = 1000 lb, c1 = 0, and k2 = 1300 lb/in. The ride rate should be ke =
m1 g 1000 = = 102 ∆ 9.8
lb/in
Thus the suspension stiffness should be k1 =
ke k2 102(1300) = = 111 lb/in k2 − ke 1300 − 102
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12.30 Define the following (refer to Figure 12.4.2): k1 = rear quartercar suspension stiffness. k2 = front quartercar suspension stiffness. kr = total rear suspension stiffness = 2k1 . kf = total front suspension stiffness = 2k2 . ker = total rear ride rate (including suspension and tire stiffness). kef = total front ride rate (including suspension and tire stiffness). ke1 =quartercar rear ride rate. ke2 =quartercar front ride rate. kt = individual tire stiffness. Referring to the guidelines on page 851, and using equations (4) and (5) of Example 12.4.3 as approximations for the bounce and pitch dynamics, we have kef = 0.7ker 1 ωbounce = 2π 2π ωpitch 1 = 2π 2π
%
%
(1)
kf + kr ≤ 1.3 Hz m
(2)
kf L21 + kr L22 ≤ 1.3 Hz IG
ke1 =
(m/4)g ∆1
(4a)
ke2 =
(m/4)g ∆2
(4b)
(3)
From the above definitions, kef = 2ke2
ker = 2ke1
(5)
Because the tire stiffness is in parallel with the suspension stiffness, ke1 =
k1 kt k1 + kt
ke2 =
k2 kt k2 + kt
(6)
These relations must be satisfied by k1 , k2 , and kt . Using the given values m = 4800/32.2, IG = 1800, L1 = 3.5, and L2 = 2.5, these equations can be rearranged as follows: $ 0.02019 k1 + k2 ≤ 1.3 (7) $
0.0375 24.5k1 + 12.5k2 ≤ 1.3 k1 =
k2 =
1200 ∆ 1 kt kt − 1200 ∆1
1200 ∆ 2 kt kt − 1200 ∆2
(8)
(9) (10)
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∆1 = 0.7∆2
(11)
The procedure is to select suitable values for kt and ∆2 , solve (9) and (10) for k1 and k2 , and see if (7) and (8) are satisfied. Trying ∆2 = 9.8/12 ft, as suggested on page 851, and kt = 1200(12) = 14400 lb/ft, we obtain from (9) and (10) k1 = 2457 lb/ft and k2 = 1636 lb/ft. With these values, the lefthand sides of (7) and (8) are 1.29 and 1.07. Thus the requirements have been met.
1235 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.