QUESTION
SOLUTION
1.
Taking laplace inverse,
And by
Code for C(t)—For t<0;
t=-2.*pi:1:0
C2=-6.*exp(-5.*t);
plot(t,C2);
xlabel(‘t’);
ylabel(‘C’);
And—for t>=0;
t=0:1:2.*pi
C1=1-6.*exp(-5.*t);
plot(t,C2);
xlabel(‘t’);
ylabel(‘C’);
2.By s=jω in C(s),we have,
C(j ω)= -5jexp(-jωτ)/( ω(jω+5)), where τ=1,
By plotting, 20 log|G(jω)| Vs log(ω) ,critical frequency and thus sampling interval ω=2πf,
f=1/T can be found.
Code For Bode Plot—
w=0:1:200;
y=20.*log(abs(-5.*(w-i).*(5-i.*w))./(w.*(25-w.^2)));
w1=log(w);
plot(w1,y);
xlabel(‘logw’);
ylabel(’20logC’);
3.By Z[exp(-τs)G(s)]=Z[5exp(-2τs)/(s+5)],
By exp(-τs)=1-τs,
And by s/(s+5)=1-5/(s+5),
Z[exp(-τs)G(s)]=Z[5/(s+5) – 10τs/(s+5)],
The discrete time System transfer function is={(10τ-1)5z/(z-exp(-5t)}-10τδ(t).
Note—4th and 5th questions can not be solved.
JG83
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