Timeline for Complex impulse response functions?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jun 6, 2012 at 11:31 | comment | added | Tormod | Wave is the input. Signal is the output (calculated by FFT of spectrum). The two series called IRF are calculated by convolution. I expect "Signal IRF" to be equal to Signal, just calculated in a different way | |
| Jun 5, 2012 at 15:01 | comment | added | Jim Clay | @Tormod It would be easier to understand and help if you could explain clearly what Signal, Signal IRF, Wave, Wave IRF, IRF 1/2/3, RAO, and RAO IRF 1/2 are. I don't necessarily need to know what the physical context- just things like what is the input, what is the filter, and what is the output. | |
| Jun 5, 2012 at 14:36 | comment | added | Tormod | The dashed lines are the output from convolution with the complex IRF and output = output.real + output.imag | |
| Jun 5, 2012 at 14:33 | comment | added | Tormod | Hi, this did change the results, but not quite as I had hoped. Can ther be differet reasons why the amplitude is less than what I expect? !new results | |
| Jun 5, 2012 at 14:32 | comment | added | Jim Clay | @JasonR I saw that that portion of the answer was not very clear, so I decided to get rid of it. | |
| Jun 5, 2012 at 14:28 | history | edited | Jim Clay | CC BY-SA 3.0 |
added 7 characters in body
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| Jun 5, 2012 at 14:15 | comment | added | Jason R | Convolution with a complex filter is not equivalent to convolving the real and imaginary parts separately. For a complex filter, convolution is specified as an inner product of the filter coefficients (its impulse response) and the input signal. There are cross-terms in that inner product where the real part of the impulse response multiplies the imaginary part of the input signal, and vice versa. If you perform two separate real convolutions, you don't get those terms. | |
| Jun 5, 2012 at 14:08 | history | answered | Jim Clay | CC BY-SA 3.0 |
