Basic question of convolution

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JjjO
JjjO el 17 de Abr. de 2021
Comentada: JjjO el 17 de Abr. de 2021
Hi. I solved a problem using conv .
'y[n] = x[n] * h[n] when x[n] = h[n] = [1,3,2,4]'.
Like this:
x=[1,3,2,4];
h=[1,3,2,4];
y=conv(x, h)
stem(y,'r','LineWidth', 2);
But, when the equation is y[n] = x[n-1] * h[n+1], how can I make n-1 and n+1??

Respuesta aceptada

Matt J
Matt J el 17 de Abr. de 2021
Editada: Matt J el 17 de Abr. de 2021
But, when the equation is y[n] = x[n-1] * h[n+1], how can I make n-1 and n+1??
There is no need to in this case. because of the shift-invariance of convolution, the negative shift in x will be cancelled out by the positive shift in h.
  3 comentarios
Matt J
Matt J el 17 de Abr. de 2021
That is what I mean.
JjjO
JjjO el 17 de Abr. de 2021
Thank you :)

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AB WAHEED LONE
AB WAHEED LONE el 17 de Abr. de 2021
Editada: AB WAHEED LONE el 17 de Abr. de 2021
Look Convolution for discrete case is defined as y[n]=Σ(x[k]*h[n-k])
so y[n-1]=Σ(x[k]*h[n-1-k]) and y[n+1]=Σ(x[k]*h[n+1-k])

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