Convolution - Signals and Systems

  • Thread starter Air
  • Start date
  • #1
Air
202
0
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?
 

Answers and Replies

  • #2
6
0
Convolution for a CT signal is defined as
[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!
 

Related Threads on Convolution - Signals and Systems

  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
5K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
4
Views
17K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
1K
Top