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Fourier transform and spectrum of the signal

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Member level 4
Sep 28, 2010
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I know how to find the Fourier series of signal and how to take Fourier transforms but one point behind the logic of all these things remain unresolved in my head. Why do we get the frequency spectrum of a signal when we take the Fourier transform of the time-domain signal? I mean, as a mathematical tool I nearly understand the Fourier transform, but I cannot connect it to the signal domain. Why does the Fourier transform have this physical meaning for electrical signals?

Thanks a lot,


As far as I understand this topic, I'll try to help:

1. Every waveform (periodic function) can be expresed by the sum of some (probably infinite many) different complex sinusoids.

2. With the Discrete-Fourier-Transform we 'decompose' each waveform to its coefficients of some (finite, its an aproximation) complex sinusoids.

So for example, if we analyze a signal we can find that it 'contains' too much or to less of a specific frequency (also see about harmonics). And that its not aparrent with other ways, eg if we just see the waveform in an oscilloscope. You can also check about FFT witch is an algorithm to compute the DFT, many oscilloscopes have built-in real-time FFTs. The freq-domain results of the FFTs are showing the frequency 'componets' of the waveform.

The Fourier analysis allows to represent signals as sinusoidal waves. That is the sum of sines and cosines. The advantage of this is that you have now simpler waves than the original signal.
Furthermore the sinusoidal waves are guaranteed to produce a sinusoidal output (for linear circuits). Only the change in phase and amplitude are to be accounted.
Then the analysis of a circuit for sine waves can be done by the simpler method of the impedance abstraction. Instead of using differential equations that are required for complex time varying signals.

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