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As we all know that the inverse Fourier transform of a frequency domain unit pulse/rectangular function (which looks like low pass filter) is Sinc function. So, I was under the impression that any low pass filter impulse response would look like Sinc function.

enter image description here

However, the RC low pass filters impulse response look like an exponentially decaying function.

enter image description here

Since both are low pass filters with the similar looking frequency domain transfer functions, why RC low pass filter response does not look like Sinc function?

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    \$\begingroup\$ No, sinc function in the time domain only applies to rectangular frequency response (and vice versa). A first order (RC) LPF does have a decaying exponential. And there are other characteristics : uniquely, a Gaussian LPF has a Gaussian time domain response... There used to be a time when math courses taught the commonest Laplace transforms like these. \$\endgroup\$
    – user16324
    Commented Jan 11, 2021 at 23:02
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    \$\begingroup\$ Is a rectangular pulse the same as an exponentially decaying waveform? Not only they are not, but one is symmetric, the other is not. According to your logic -- "Since both are low pass filters with the similar looking frequency domain transfer functions" -- a Bessel lowpass is the same as an elliptic/Cauer lowpass. The transfer functions are really not the same. \$\endgroup\$
    – a concerned citizen
    Commented Jan 11, 2021 at 23:02

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Here's the short answer: a first-order lowpass and a rectangular-in-frequency filter don't look the same in the frequency domain. So why should they be the same when you take the inverse Fourier transform?

enter image description here

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  • \$\begingroup\$ SINC pulses do not need to be square. e.g. Blackman, Welsh etc \$\endgroup\$
    – D.A.S.
    Commented Jan 13, 2021 at 16:16
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    \$\begingroup\$ There is exactly one sinc function, and its Fourier transform is square. There may be functions like the sinc function -- but their Fourier transforms are going to be way more square than \$1/(\tau s + 1)\$. So I think my point stands. \$\endgroup\$
    – TimWescott
    Commented Jan 13, 2021 at 19:19
  • \$\begingroup\$ You think only square windows make SINC functions.? fiiir.com \$\endgroup\$
    – D.A.S.
    Commented Jan 13, 2021 at 22:02
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    \$\begingroup\$ By definition \$\mathrm{sinc} x = \lim_{\chi \to x} \frac{\sin \pi \chi}{\pi \chi}\$. That page you reference is generating what it calls "windowed sinc". Not plain ol' sinc. \$\endgroup\$
    – TimWescott
    Commented Jan 14, 2021 at 1:04
  • \$\begingroup\$ So in theoretical terms, it must be a "boxcar" but in practical terms it doesn't. \$\endgroup\$
    – D.A.S.
    Commented Jan 14, 2021 at 1:22

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