# ch10 - T E N Frequency Response Techniques SOLUTION TO CASE...

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T E N Frequency Response Techniques SOLUTION TO CASE STUDY CHALLENGE Antenna Control: Stability Design and Transient Performance First find the forward transfer function, G(s). Pot: K 1 = 10 π = 3.18 Preamp: K Power amp: G 1 (s) = 100 s(s+100) Motor and load: J = 0.05 + 5 ( 1 5 ) 2 = 0.25 ; D = 0.01 + 3 ( 1 5 ) 2 = 0.13; K t Ra = 1 5 ; K b = 1. Therefore, G m (s) = θ m (s) E a (s) = K t R a J s(s+ 1 J (D + K t K b R a )) = 0.8 s(s+1.32) . Gears: K 2 = 50 250 = 1 5 Therefore, G(s) = K 1 KG 1 (s)G m (s)K 2 = 50.88K s(s+1.32)(s+100) Plotting the Bode plots for K = 1,

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Solution to Case Study Challenge 409 a. Phase is 180 o at ω = 11.5 rad/s. At this frequency the gain is - 48.41 dB, or K = 263.36. Therefore, for stability, 0 < K < 263.36. b. If K = 3, the magnitude curve will be 9.54 dB higher and go through zero dB at ω = 0.94 rad/s. At this frequency, the phase response is -125.99 o . Thus, the phase margin is 180 o - 125.99 o = 54.01 o . Using Eq. (10.73), ζ = 0.528. Eq. (4.38) yields %OS = 14.18%. c. Program: numga=50.88; denga=poly([0 -1.32 -100]); 'Ga(s)' Ga=tf(numga,denga); Gazpk=zpk(Ga) '(a)' bode(Ga) title('Bode Plot at Gain of 50.88') pause [Gm,Pm,Wcp,Wcg]=margin(Ga); 'Gain for Stability' Gm pause '(b)' numgb=50.88*3; dengb=denga; 'Gb(s)' Gb=tf(numgb,dengb); Gbzpk=zpk(Gb) bode(Gb) title('Bode Plot at Gain of 3*50.88') [Gm,Pm,Wcp,Wcg]=margin(Gb); 'Phase Margin' Pm for z=0:.01:1 Pme=atan(2*z/(sqrt(-2*z^2+sqrt(1+4*z^4))))*(180/pi); if Pm-Pme<=0; break end end z percent=exp(-z*pi/sqrt(1-z^2))*100 Computer response: ans = Ga(s)
410 Chapter 10: Frequency Response Methods Zero/pole/gain: 50.88 ------------------ s (s+100) (s+1.32) ans = (a) ans = Gain for Stability Gm = 262.8585 ans = (b) ans = Gb(s) Zero/pole/gain: 152.64 ------------------ s (s+100) (s+1.32) ans = Phase Margin Pm = 53.9644 z = 0.5300 percent = 14.0366

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Answers to Review Questions 411 ANSWERS TO REVIEW QUESTIONS 1. a. Transfer functions can be modeled easily from physical data; b. Steady-state error requirements can be considered easily along with the design for transient response; c. Settles ambiguities when sketching root locus; (d) Valuable tool for analysis and design of nonlinear systems. 2. A sinusoidal input is applied to a system. The sinusoidal output's magnitude and phase angle is measured in the steady-state. The ratio of the output magnitude divided by the input magnitude is the magnitude response at the applied frequency. The difference between the output phase angle and the input phase angle is
412 Chapter 10: Frequency Response Methods the phase response at the applied frequency. If the magnitude and phase response are plotted over a range of different frequencies, the result would be the frequency response for the system. 3. Separate magnitude and phase curves; polar plot 4. If the transfer function of the system is G(s), let s=j ω . The resulting complex number's magnitude is the magnitude response, while the resulting complex number's angle is the phase response. 5. Bode plots are asymptotic approximations to the frequency response displayed as separate magnitude and phase plots, where the magnitude and frequency are plotted in dB. 6. Negative 6 dB/octave which is the same as 20 dB/decade 7. Negative 24 dB/octave or 80 dB/decade 8. Negative 12 dB/octave or 40 dB/decade 9. Zero degrees until 0.2; a negative slope of 45 o /decade from a frequency of 0.2 until 20; a constant -90 o phase from a frequency of 20 until 10.

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## This note was uploaded on 04/04/2008 for the course MECH COntrol Sy taught by Professor Khurshid during the Spring '08 term at Michigan State University.

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ch10 - T E N Frequency Response Techniques SOLUTION TO CASE...

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