Timeline for Why is the output of an LTI system the convolution of the input funtion and the impulse response?
Current License: CC BY-SA 3.0
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| Jun 22, 2011 at 13:14 | comment | added | PFiver | All right. I was too locked in to use $f$ as the $f$unction letter. Also, our prof has been using $\omega$, the circular frequency, which equals to $2\pi f$, mostly. -- I am aware of the advantages of doing things in the frequency domain but I'd nevertheless like to see the mess I'm avoiding, just once, to have an better understanding of what I'm doing | |
| Jun 22, 2011 at 1:29 | comment | added | Steve | The input-output operator is just a fancy name for the function that maps the input to the output. The use of $f$, which stands for frequency, is fairly standard in this area. One very important reason to work frequency domain is important is that it is the analogue of diagonalizing an operator. The convolution is fairly messy to compute in time but in the frequency domain, it is just a multiplication operator. Also, in many applications, it is the frequency domain that is important. For example, an ideal lowpass filter would be specified by $H(f)=1$ for $|f|\leq f_0$ and $0$ elsewhere. | |
| Jun 22, 2011 at 0:17 | comment | added | PFiver | Thanks a lot. The answer is not that useful to me because I am not familiar with the term "input-output operator". I find $f$ a confusing choice for the name of a constant here. Furthermore I'd prefer to look at the problem without changing to the frequency domain. As to the "system function": How would it look like? | |
| Jun 21, 2011 at 23:11 | history | answered | Steve | CC BY-SA 3.0 |