Conv two continuous time functions

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

 Respuesta aceptada

Thorsten
Thorsten el 26 de Mayo de 2015
Editada: Thorsten el 26 de Mayo de 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;

Más respuestas (1)

Immanuel Manohar
Immanuel Manohar el 2 de Oct. de 2019

0 votos

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 comentario

zhitao Luo
zhitao Luo el 2 de Jun. de 2020
Hello, Immanuel Manohar, I also encountered the same problem, is there any more detailed answer?

Iniciar sesión para comentar.

Preguntada:

DDD
el 26 de Mayo de 2015

Comentada:

el 2 de Jun. de 2020

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