0

I've been asked to sketch why

$$y(n)=\sum_{k=0}^{\infty}x(k)h(n-k) = \sum_{k=0}^{\infty}h(k)x(n-k)$$

are equivalent expressions for convolution. I could explain this pretty easily with a proof, but can't think of a concise way to illustrate this. Does anyone have any ideas?

samp1920
  • 139
  • If you can prove it, what is left to illustrate? Are you looking for confirmation on the proof? Show it! – rschwieb Sep 15 '15 at 23:47
  • I'm not looking for confirmation of the proof. Just would like to show a graphical representation. Not for any reason other than being asked by my professor. Would convolving say x(n) = [1 1 2 0 0] with the impulse response be an ok example to use? – samp1920 Sep 15 '15 at 23:51
  • I have no idea what that means. Do you know the relationship of convolution with multiplication of power series? – rschwieb Sep 16 '15 at 00:26

0 Answers0