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For the square wave signal, why does a narrower square wave correspond to more spread in the frequency domain?

The "bell curve" is described by the Gaussian function, $e^{- k x^2}$, and is its own Fourier transform when $k = \pi$: $\mathcal F(e^{-\pi x^2}) = e^{-\pi x^2}$. Fourier transforms are ...
user121330's user avatar
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For the square wave signal, why does a narrower square wave correspond to more spread in the frequency domain?

For that kind of question ('how to get it'), I'd suggest regarding Fourier transform as a way of determining a level of similarity (or correlation) between the original signal and reference periodic ...
yury10578's user avatar
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5 votes

For the square wave signal, why does a narrower square wave correspond to more spread in the frequency domain?

Think of what happens when you play back a recording at 2x speed: each frequency of the sound doubles. When you make the pulse narrower, that's equivalent to speeding it up in the time domain. ...
jpa's user avatar
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For the square wave signal, why does a narrower square wave correspond to more spread in the frequency domain?

One way to view this concept is from energy conservation. With proper scaling, the Fourier transform is energy conserving. In order to use up its energy in a short window, the signal has to expel its ...
Baddioes's user avatar
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