ch13 - T H I R T E E N Digital Control Systems SOLUTIONS TO...

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T H I R T E E N Digital Control Systems SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Transient Design via Gain a. From the answer to the antenna control challenge in Chapter 5, the equivalent forward transfer function found by neglecting the dynamics of the power amplifier, replacing the pots with unity gain, and including the integration in the sample-and-hold is G e (s) = 0.16K s 2 (s + 1.32) But, Thus, G e (z) = 0.16K z - 1 z G z , or, G e (z) = 7.659x10 -4 K (z+0.95696) (z-1) (z-0.87634) b. Draw the root locus and overlay it over the ζ = 0.5 (i.e. 16.3% overshoot) curve. .
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Solutions to Case Studies Challenges 581 We find that the root locus crosses at approximately 0.93 ± j0.11 with 7.659x10 -4 K = 8.63x10 -3 . Hence, K = 11.268. c. K v = 1 T lim z 1 ( z 1) G e ( z ) = (7.659x10 4 K)(1.95696) 0.12366 = 0.1366; e ( ) = 1 K v = 7.321 d. Program: T=0.1; %Input sampling time numf=0.16; %Numerator of F(s) denf=[1 1.32 0 0]; %Denominator of F(s) 'F(s)' %Display label F=tf(numf,denf) %Display F(s) numc=conv([1 0],numf); %Differentiate F(s) to compensate %for c2dm which assumes series zoh denc=denf; %Denominator of continuous system %same as denominator of F(s) C=tf(numc,denc); %Form continuous system, C(s) C=minreal(C,1e-10); %Cancel common poles and zeros D=c2d(C,T,'zoh'); %Convert to z assuming zoh 'F(z)' D=minreal(D,1e-10) %Cancel common poles and zeros and display rlocus(D) pos=(16.3); z=-log(pos/100)/sqrt(pi^2+[log(pos/100)]^2); zgrid(z,0) title(['Root Locus with ' , num2str(pos), ' Percent Overshoot Line']) [K,p]=rlocfind(D) %Allows input by selecting point on %graphic Computer response: ans = F(s) Transfer function: 0.16 -------------- s^3 + 1.32 s^2 ans = F(z) Transfer function: 0.0007659 z + 0.000733 ---------------------- z^2 - 1.876 z + 0.8763 Sampling time: 0.1 Select a point in the graphics window selected_point = 9.2969e-001 +1.0219e-001i K = 9.8808e+000
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582 Chapter 13: Digital Control Systems p = 9.3439e-001 +1.0250e-001i 9.3439e-001 -1.0250e-001i Antenna Control: Digital Cascade Compensator Design a. Let the compensator be KG c (s) and the plant be G p ( s ) = 0.16 s ( s + 1.32) . For 10% overshoot and a peak time of 1 second, ζ = 0.591 and ω n = 3.895, which places the dominant poles at –2.303 ± j3.142. If we place the compensator zero at –1.32 to cancel the plant’s pole, then the following geometry results. X X -p c -2.303 j3.142 s-plane Hence, p c = 4.606. Thus, G c ( s ) = K ( s + 1.32) ( s + 4.606) and G c ( s ) G p ( s ) = 0.16 K s ( s + 4.606) . Using the product of pole lengths to find the gain, 0.16K = (3.896) 2 , or K = 94.87. Hence,
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Solutions to Case Studies Challenges 583 G c ( s ) = 94.87( s + 1.32) ( s + 4.606) . Using a sampling interval of 0.01 s, the Tustin transformation of G c (s) is G c ( z ) = 93.35( z 0.9869) ( z 0.955) = 93.35 z 92.12 z 0.955 . b. Cross multiplying , ( z - 0.955) X(z) = (93.35 z – 92.12) E(z) Solving for the highest power of z operating on X(z), zX(z) = (93.35 z – 92.12) E(z) + 0.955 X(z) Solving for X(z) , X(z) = (93.35 – 92.12 z -1 ) E(z) + 0.955 z -1 X(z) Implementing this equation as a flowchart yields the following diagram c. Program: 's-plane lead design for Challenge - Lead Comp' clf %Clear graph on screen. 'Uncompensated System' %Display label. numg=0.16; %Generate numerator of G(s). deng=poly([0 -1.32]); %Generate denominator of G(s). 'G(s)' %Display label.
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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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ch13 - T H I R T E E N Digital Control Systems SOLUTIONS TO...

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