Conv two continuous time functions

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DDD
DDD 2015-5-26
评论： zhitao Luo 2020-6-2
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

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采纳的回答

Thorsten
Thorsten 2015-5-26
编辑：Thorsten 2015-5-26
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;

更多回答（1 个）

Immanuel Manohar
Immanuel Manohar 2019-10-2
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 个评论
zhitao Luo
zhitao Luo 2020-6-2
Hello, Immanuel Manohar, I also encountered the same problem, is there any more detailed answer？

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