Why is the fourier transform of impulse train a impulse train? Is there a intuitive reason behind it?
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$\begingroup$ Also take a look at this answer. $\endgroup$– Matt L.Commented Sep 8, 2016 at 19:44
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$\begingroup$ Or this current question. $\endgroup$– robert bristow-johnsonCommented Sep 9, 2016 at 1:57
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2 Answers
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Intuition can sometimes be misleading. But here are some ideas that might help one move towards creating a mental picture.
An infinitely long pure sinewave in the time domain (consisting of just one frequency FT or DFT basis function) will be a single impulse in the frequency domain.
Distort the sinewave a little, but leave the waveform perfectly periodic, and the impulse will be followed by an evenly spaced harmonic series. Usually the narrower and sharper the distortion (but keeping the waveform still perfectly periodic), the longer the harmonic series. What might be considered a limiting case of maximum distortion, the narrowest waveform with the sharpest edge will have the longest harmonic series. Or an infinitely long sine wave in the time domain maximally distorted into just an infinitely periodic impulse train, will produce a impulse followed by an infinitely long harmonic series, which looks a lot like another periodic impulse train.
Make it an even (cosine) function, and all the impulses in the FT will be real and thus symmetric around 0. Add a DC offset to the distorted sine wave to complete the pulse train at 0.
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$\begingroup$ Nice intuitive explanation. I pictured the distortion as making the sine wave valleys wider and flatter, and making the peaks narrower and taller, until the sine is converted to a train of impulses. This clearly is a sine wave with harmonic distortion. $\endgroup$– MBazCommented Sep 8, 2016 at 17:46
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3$\begingroup$ Another related example of this is the derivative of a square wave. Namely the Fourier transform/series of a square wave is an infinite series of impulses in the frequency domain, whose amplitude is proportional to the inverse of the frequency. Taking the derivative is the same as multiplying by $s$ in the frequency domain, so therefore the amplitudes of the impulses will stay constant. In the time domain the derivative of the square wave looks like periodically and equally spaced positive and negative impulses. $\endgroup$ Commented Sep 8, 2016 at 17:57
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From the theory of Fourier series we know that a periodic function has a discrete spectrum. From the duality of the Fourier transform it follows that in general periodicity in one domain implies discretization in the other domain and vice versa. Now, an impulse train is periodic and discrete, so its Fourier transform must be discrete (due to periodicity in time) and periodic (due to discretization in time), which means that an impulse train in one domain corresponds to an impulse train in the other domain.
