Timeline for Amplitude in frequency domain not as expected with discontinues time domain signal
Current License: CC BY-SA 3.0
11 events
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| Mar 8, 2018 at 1:46 | comment | added | Cedron Dawg | I know it as Parseval's. I looked it up, Plancherel's seems to apply to the continuous case and Parseval's for the discrete case. Not all the bins will be zero, only the even numbered ones. The 5Hz bin because of cancellation, the other even bins have orthogonal products in each half. The odd ones will have values. The power is proportional to the square of the amplitude. BTW, I don't have MATLAB so I can't test your code. I am getting prompted to move this to a chat room. If you want to discuss it further I would prefer you email me at cedron at exede dot net. Thanks. | |
| Mar 8, 2018 at 1:36 | comment | added | MBaz | One question: if all bins are zero, doesn't that contradict Plancherel's theorem? (It is in fact this theorem which gives basis to my assertion that the DFT amplitude depends on a signal's power). | |
| Mar 8, 2018 at 1:22 | comment | added | Cedron Dawg | It is not leakage, it is cancellation. My point about even and odd bins should hold for this case too. (No I haven't done it). You can call the odd bin values leakage if you like, in a sense, except you have a whole number of signal cycles and a whole plus a half number of basis cycles. | |
| Mar 8, 2018 at 1:17 | comment | added | MBaz | @CedronDawg BTW, I do see your point: the correlation of the two halves with the DFT complex exponential can be thought of as being of the form $X+(-X)=0$. I'm still thinking about it. | |
| Mar 8, 2018 at 0:52 | comment | added | MBaz |
@CedronDawg Try finding the complete DFT. You're getting 0 in that exact bin because of spectral leakage (my guess). This Matlab code produces this plot: fs=20; t=0:1/fs:10; s=(t<5).*cos(2*pi*5*t)+(t>=5).*cos(2*pi*5*t+pi); plot(abs(fft(s))).
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| Mar 8, 2018 at 0:16 | comment | added | Cedron Dawg | Well I hadn't, but I did. See the followup in my previous answer. | |
| Mar 7, 2018 at 23:10 | comment | added | MBaz | @CedronDawg Have you actually tried it? | |
| Mar 7, 2018 at 19:55 | comment | added | Cedron Dawg | Suppose instead that he added another 5Hz signal, one half cycle out of phase, to the second half. The DFT result would indicate zero 5Hz energy. | |
| Mar 7, 2018 at 17:46 | comment | added | MBaz | @CedronDawg He's adding energy at 10 Hz, but the average power at 5 Hz is being halved. | |
| Mar 7, 2018 at 16:34 | comment | added | Cedron Dawg | He is not adding another 5-second interval with zero energy, he is adding a different tone. | |
| Mar 7, 2018 at 16:04 | history | answered | MBaz | CC BY-SA 3.0 |