Conv two continuous time functions

8 vues (au cours des 30 derniers jours)
DDD
DDD le 26 Mai 2015
Commenté : zhitao Luo le 2 Juin 2020
given y(t) and x(t), it is asked to conv them. Note: x(t)=dirac(t-3)-dirac(t-5). The conv result should sum y(t-3)-y(t-5) but it gives me:
y=@(t) 1.0*(t>=0).*exp(-3*t);
x=@(t) 1.0*(t==3)-1.0*(t==5);
delta=0.0001;
tx=2:delta:6; %tx=(-200:300)*delta;
ty=-1:delta:1.5; % ty=(-100:300)*delta;
c=conv(y(ty),x(tx))*delta;
tc=(tx(1)+ty(1)):delta:(tx(end)+ty(end));
figure()
title('c')
subplot(3,1,1)
plot(tx,x(tx))
xlabel('n'); title('x(t)'); ylim([min(x(tx))-1,max(x(tx))+1]); grid on
subplot(3,1,2)
plot(ty,y(ty))
xlabel('n'); title('h(t)'); ylim([min(y(ty))-1,max(y(ty))+1]); grid on
subplot(3,1,3)
plot(tc,c);
xlabel('n'); title('x(t)*h(t)');ylim([min(c)-1,max(c)+1]); grid on
What can i do to solve the problem?
Thanks

Réponse acceptée

Thorsten
Thorsten le 26 Mai 2015
Modifié(e) : Thorsten le 26 Mai 2015
The y-axis is too large to show the data. You can rescale them by, e.g.,
axis([1 8 -delta delta])
or with your code, use
ylim([min(c),max(c)]);
or get rid of the *delta in
c=conv(y(ty),x(tx))*delta;

Plus de réponses (1)

Immanuel Manohar
Immanuel Manohar le 2 Oct 2019
Your dirac Delta is wrong... you're attempting continuous time convolution but you are using unit impulse instead of dirac delta for convolution. To get the correct answer, your dirac delta approximation should have the height of 1/delta.
  1 commentaire
zhitao Luo
zhitao Luo le 2 Juin 2020
Hello, Immanuel Manohar, I also encountered the same problem, is there any more detailed answer?

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