- #1

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My first question is: For discrete signal, we use variable 'n' and for continuous signal, we use variable 't'. But is the convolution integral valid for both. E.g. the only difference would be 'n' and 't'. Tau and integral will be the same?

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- Thread starter Air
- Start date

- #1

- 202

- 0

My first question is: For discrete signal, we use variable 'n' and for continuous signal, we use variable 't'. But is the convolution integral valid for both. E.g. the only difference would be 'n' and 't'. Tau and integral will be the same?

- #2

- 6

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[tex]

y(t) = \int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau

[/tex]

and for DT it is defined as

[tex]

y[n] = \sum_{k=-\infty}^{\infty}x[k]h[n-k]

[/tex]

Thus for DT signal we do not have integral but summation.

Just to distinguish for DT case we call it convolution summation and for CT case we call it integral. But the operation is same!

Bhupala!

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