experiment5

Eps d exponential respsonse kkk0110 f 2kkk

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Unformatted text preview: jU)'¥£ ‡ " "’’’   ¥ ©§  SUPPLEMENT: (5.67) 27 CHAPTER 5 MATLAB Solution Program % Experiment 5 % % PART 1 % z=1/2; p=[0 -1/2 1/3 -1/4]; K=1; [num,den]=zp2tf(z,p,K); % (a) Impulse respsonse k=0:1:20; h=dimpulse(num,den,k); % figure (1) plot(k,h,’*’); grid; xlabel(’Discrete time’); ylabel(’Impulse response’) print -deps figure5_1.eps % (b) Step respsonse ystep=dstep(num,den,k); % figure (2) plot(k,ystep,’*’); grid; xlabel(’Discrete time’); ylabel(’Step response’) print -deps figure5_2.eps % (c) Sinusiodal response kk=0:1:50; f=sin(2*kk); y=dlsim(num,den,f); % figure (3) plot(kk,y,’o’); grid; xlabel(’Discrete time’); ylabel(’Sinusiodal response’) print -deps figure5_3.eps % (d) Exponential respsonse kkk=0:1:10; f=(-2).ˆkkk; y=dlsim(num,den,f); % figure (4) plot(kkk,y,’*’); grid; xlabel(’Discrete time’); ylabel(’Exponential response’) print -deps figure5_4.eps % % PART 2 % % (a) Tranfer function factored form num=[1 -3 5 2 0]; den=[2 0 5 -3 1 -2 3 -1]; [zero,p,K]=tf2zp(num,den); % H(z)=K*((z-zero(1))*(z-zero(2))*...*(z-zero(4)))/((z-p(1))*...*(z-p(7))) % (b) Tranfer function partial fraction form and the impulse response num_z=[1 —3 5 2]; % H(z) divided by z [R,p,res]=residue(num,den); % H(z)/z=R(1)/(z-p(1))+R(2)/(z-p(2))+...+R(7)/(z-p(7)) + res(z) p p1=abs(p(1)) p3=abs(p(3)) p5=abs(p(5)) anglep1=angle(p(1)) anglep3=angle(p(3)) anglep5=angle(p(5)) R r1=abs(R(1)) r3=abs(R(3)) r5=abs(R(5)) r7=abs(R(7)) phi1=angle(R(1)) phi3=angle(R(3)) phi5=angle(R(5)) 28 LABORATORY EXPERIMENTS for LINEAR DYNAMIC SYSTEMS AND SIGNALS by ZORAN GAJIC h=2*r1*(abs(p(1)).ˆk).*cos(angle(p(1))*k+phi1)... +2*r3*(abs(p(3)).ˆk).*cos(...
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This note was uploaded on 11/25/2013 for the course EE 130 taught by Professor Ee130 during the Fall '07 term at Berkeley.

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