Convolution - Signals and Systems

[Air]

I will make this my discussion thread. I have many questions to ask which I will post here. Please keep checking. All help will be appreciated.

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?

 

[bhupala]

Convolution for a CT signal is defined as
$$y(t) = \int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau$$
and for DT it is defined as
$$y[n] = \sum_{k=-\infty}^{\infty}x[k]h[n-k]$$

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!

 

[oswaler]

The convolution integral is valid for both discrete and continuous signals. However, the variables used may differ. For a discrete signal, the convolution integral is typically expressed as a summation, using the variable 'n'. For a continuous signal, the convolution integral is expressed as an integral, using the variable 't'. The tau and integral will be the same in both cases, as they represent the range of integration for the convolution operation.