Timeline for Efficient algorithm for computing area under convolution?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Nov 5, 2014 at 22:10 | comment | added | user541686 | I completely understand. In fact, that was the entire point of my question. If you want to make the extra assumption that the signals are zero for $n < 0$ then my question would be how would you compute the given summation starting at an arbitrary $n = n_0$, not necessarily at zero. | |
| Nov 5, 2014 at 21:37 | comment | added | Matt L. | @Mehrdad: So what you're saying is that the signals $x_i[n]$ may have non-zero values for $n<0$. This is kind of important because it changes quite a few things. But does this make sense? | |
| Nov 5, 2014 at 21:35 | comment | added | user541686 | Huh? Why wouldn't there be? | |
| Nov 5, 2014 at 21:27 | comment | added | Matt L. | @Mehrdad: Why would there be any elements for $n<0$? | |
| Nov 5, 2014 at 19:25 | comment | added | user541686 | Why are you going through all elements, though? The question is about elements at n >= 0, not all n. | |
| Nov 5, 2014 at 16:40 | comment | added | Matt L. | @Mehrdad: I added a little Matlab/Octave script for clarity. | |
| Nov 5, 2014 at 16:39 | history | edited | Matt L. | CC BY-SA 3.0 | added 463 characters in body |
| Nov 5, 2014 at 12:36 | comment | added | Matt L. | @Mehrdad: Well, the product is over all $N$ signals, and the sums simply go over all elements of the respective signals. | |
| Nov 5, 2014 at 12:20 | comment | added | user541686 | What are the bounds on your summation? | |
| Nov 5, 2014 at 12:14 | history | answered | Matt L. | CC BY-SA 3.0 |