Timeline for Intuition behind impulse response terms in convolution
Current License: CC BY-SA 3.0
4 events
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| Feb 9, 2016 at 9:04 | comment | added | Derek Elkins left SE | If instead of $z$ we used $D$ in the polynomials above, where $(Dx)[n] = x[n-1]$ then the polynomial would be a weighted sum of these delays. Multiplying the polynomials would, as you say, convolve the coefficients. The convolution formula says the weight at some point in time is the product of the weights of the delays that, combined, would shift us to this point in time. | |
| Feb 9, 2016 at 8:58 | comment | added | Derek Elkins left SE | Another way to write the convolution of $x$ and $h$ is: $\sum_{i+j=n} x[i]h[j]$ | |
| Feb 9, 2016 at 8:25 | comment | added | Ryan | The coefficients would be the discrete convolution of $x[n]$ and $h[n]$ (we did this as an exercise in class). I don't quite see what you're getting at though. | |
| Feb 9, 2016 at 8:16 | history | answered | Derek Elkins left SE | CC BY-SA 3.0 |