# ch10 - T E N Sinusoidal Tools SOLUTION TO CASE STUDY...

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Unformatted text preview: T E N Sinusoidal Tools SOLUTION TO CASE STUDY CHALLENGE Antenna Control: Stability Design and Transient Performance First find the forward transfer function, G s . Pot: 1 10 3.18 K Preamp: K Power amp: 1 100 100 G s s s Motor and load: 2 2 1 1 1 0.05 5 0.25; 0.01 3 0.13; ; 1 5 5 5 t b a K J D K R . Therefore, 0.8 1.32 1 t m a m a t b a K s R J G s E s s s K K s s D J R . Gears: 2 50 1 250 5 K Therefore, 1 1 2 50.88 1.32 100 m K G s K KG s G s K s s s 10-2 Chapter 10: Sinusoidal Tools Plotting the Bode plots for 1 K , a. Phase is 180 at 11.5rad /s . At this frequency the gain is 48.41dB , or 263.36 K . Therefore, for stability, 263.36 K . b. If 3 K , the magnitude curve will be 9.54 dB higher and go through zero dB at 0.94rad/s . At this frequency, the phase response is 125.99 . Thus, the phase margin is 180 125.99 54.01 . Using Eq. (10.73), 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') Solutions to Problems 10-3 [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) 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 = 10-4 Chapter 10: Sinusoidal Tools Phase Margin Pm = 53.9644 z = 0.5300 percent = 14.0366 Solutions to Problems...
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ch10 - T E N Sinusoidal Tools SOLUTION TO CASE STUDY...

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