Timeline for Autocorrelation of convolution integral
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 24, 2014 at 2:46 | answer | added | Dilip Sarwate | timeline score: 2 | |
| Jun 23, 2014 at 20:43 | comment | added | tgoossens | @whuber sorry. I've corrected it (g had to be h) h is the impulse response of a linear system. | |
| Jun 23, 2014 at 20:42 | history | edited | tgoossens | CC BY-SA 3.0 |
g must be h
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| Jun 23, 2014 at 19:58 | comment | added | whuber♦ | I'm not sure--it depends on what you really mean by the "$h$" notation. I'm struggling to make sense of it, but maybe I'm overlooking something that's obvious to you. | |
| Jun 23, 2014 at 19:40 | comment | added | tgoossens | @whuber Thank you for clearing this up. So that means that the presented solution is nonsense? | |
| Jun 23, 2014 at 18:35 | comment | added | whuber♦ | This question is difficult because the notation doesn't make sense. Presumably, $h=g$. (If not, what is $h$?) Convolution is an operation on functions whereas expressions like $h(\tau)$ and $h(-\tau)$ are numbers. Thus, for instance, it makes sense to write expressions like $$(h * r_X)(t)=\int_\mathbb{R}h(t-u)r_X(u)du$$ but this is not equal to "$h(t)*r_X(t)$," which has no meaning. Because of this confusion, after a good start to the solution things begin to go awry at the second step when "$*$" first appears. If you straighten out this notation you might find the problem is simple. | |
| Jun 23, 2014 at 18:15 | history | edited | tgoossens | CC BY-SA 3.0 |
deleted 2 characters in body; edited title
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| Jun 23, 2014 at 17:25 | history | asked | tgoossens | CC BY-SA 3.0 |