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Problem 1: The impulse response ht) for a particular LTI system is shown below hit) Be5e (4 cos(3t)+ 6 sin(3t) e. u(t) 1. Plo

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Answer #1

matlab code:

t=0:0.001:10;
f=3*exp(-3*t)+5*exp(-2*t)+exp(-t).*(4*cos(3*t)+6*sin(3*t))+exp(-4*t);
figure
plot(t,f),grid on
r=[3;5;2;2;-3*i;3*i;1];
p=[-3;-2;-1+3*i;-1-3*i;-1+3*i;-1-3*i;-4];
k=[];
[b a]=residue(r,p,k)
sys=tf(b,a)
figure
pzmap(sys)
p=pole(sys)   
z=zero(sys)
figure
bode(sys)
margin(sys)

output:


b =

13 142 834 3352 8536 13392 10824


a =

1 13 86 384 1180 2516 3560 2400


sys =

13 s^6 + 142 s^5 + 834 s^4 + 3352 s^3 + 8536 s^2 + 13392 s + 10824
---------------------------------------------------------------------
s^7 + 13 s^6 + 86 s^5 + 384 s^4 + 1180 s^3 + 2516 s^2 + 3560 s + 2400

Continuous-time transfer function.


p =

-1.0000 + 3.0000i
-1.0000 - 3.0000i
-1.0000 + 3.0000i
-1.0000 - 3.0000i
-4.0000 + 0.0000i
-3.0000 + 0.0000i
-2.0000 + 0.0000i


z =

-0.9147 + 3.7651i
-0.9147 - 3.7651i
-3.8609 + 0.0000i
-2.6326 + 0.0000i
-1.3001 + 1.9407i
-1.3001 - 1.9407i

impulse responce:

14 12 10 0 10

p-z map:

Pole-Zero Map 0 -0.5 1.5 -2 2.5 3.5 Real Axis (seconds)bode plot:

Bode Diagram Gm Inf, Pm 99.1 deg (at 12.8 radis) 20 15 10 -5 10 15 -20 45 0. -90 101 10° 101 102 Frequency (rad/s)

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