Conv two continuous time functions
15 visualizaciones (últimos 30 días)
Mostrar comentarios más antiguos
DDD
el 26 de Mayo de 2015
Comentada: zhitao Luo
el 2 de Jun. de 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:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/150558/image.jpeg)
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
0 comentarios
Respuesta aceptada
Más respuestas (1)
Immanuel Manohar
el 2 de Oct. de 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 comentario
zhitao Luo
el 2 de Jun. de 2020
Ver también
Categorías
Más información sobre Signal Processing Toolbox en Help Center y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!