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I have an ideal low pass filter having unity gain and cutoff frequency 100Hz. If I provide an input having single frequency component at 100Hz(E.g.- cos(200pi t), will the low pass filter block the input frequency?

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  • $\begingroup$ Why not plot the case (Octave is free: gnu.org/software/octave/index )? $\endgroup$
    – Juha P
    Dec 19, 2021 at 11:24
  • $\begingroup$ @Abhishek Your question sounds almost like a trick question. Can you tell us exactly (very precisely) what you mean by an "ideal lowpass filter having a cutoff frequency 100Hz"? $\endgroup$ Dec 19, 2021 at 11:57
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    $\begingroup$ @RichardLyons: I think the question boils down to "what's the value of a rectangular function at the exact step?" I think it's 0.5 by I haven't done the math on this yet $\endgroup$
    – Hilmar
    Dec 19, 2021 at 13:47
  • $\begingroup$ I thought this was easy to derive, but it's not so far; Got a lot of integrals resisting Lebesgue $\endgroup$ Dec 19, 2021 at 14:51
  • $\begingroup$ Take the equation for the frequency response of the low-pass filter and evaluate it at $f=100$. $\endgroup$
    – MBaz
    Dec 19, 2021 at 16:20

2 Answers 2

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This is mostly an academic question since both a sine wave and ideal low pass filter are infinite in time and hence can't exist in a real world application.

A quick numerical hack Matlab hack would indicate the answer is indeed 0.5, regardless of the phase of the sine wave, but that could also be an artifact of my finite simulation.

The correct way to determine this is to solve the time domain convolution integral $$ y(t) = \int_{-\infty}^{-\infty}\frac{\sin(\pi\tau)\sin(\pi(t-\tau))}{\pi \tau}d\tau = \\ \frac{1}{2}\int_{-\infty}^{-\infty}\frac{\cos(2\pi \tau -\pi t) - \cos(\pi t)}{\pi \tau}d\tau $$ which is as far as I got.

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  • $\begingroup$ Using the sinc function for the impulse response is equivalent to defining the frequency response value at the cut-off frequency to be $\frac12$. That's simply a property of the Fourier transform. So the numerical hack is correct (but it just confirms what we already know). $\endgroup$
    – Matt L.
    Dec 20, 2021 at 12:11
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There is no good answer to your question. What you're basically asking is

What is the value of a discontinuous function at its discontinuity?

It is up to you to define a value of the function at the discontinuity if you really need one. E.g., for the Heaviside step function there are three common definitions of its value at the jump discontinuity (i.e., at zero argument). Note, however, that in many practical cases it is irrelevant (and even nonsensical) to talk about a function value at a jump discontinuity.


As discussed in the comments and in Hilmar's answer, it is of course possible to compute the output of an ideal lowpass filter in the time domain. If $\omega_0$ denotes the frequency of the sinusoidal input signal, and if it also equals the cut-off frequency of the ideal lowpass filter, the output signal is given by the convolution integral

$$y(t)=\int_{-\infty}^{\infty}\sin[\omega_0(t-\tau)]\frac{\sin(\omega_0\tau)}{\pi \tau}d\tau=\frac12\sin(\omega_0t)\tag{1}$$

However, note that the impulse response of the ideal lowpass filter is obtained by the inverse Fourier transform of the ideal brickwall frequency response. Consequently, the frequency response value $\frac12$ at the cut-off frequency is simply implied by the properties of the Fourier transform, and the result $(1)$ shouldn't come as a surprise. It's a pure consequence of the way discontinuities are dealt with by the (inverse) Fourier transform.

Clearly, the Fourier transform of a sinc lowpass filter impulse response evaluated at the cut-off frequency equals $\frac12$, which is just another way of stating the result of the convolution $(1)$:

$$\int_{-\infty}^{\infty}\frac{\sin(\omega_0t)}{\pi t}e^{-j\omega_0t}dt=\frac12\tag{2}$$


Derivation of Equations $(1)$ and $(2)$:

Eq. $(1)$: $$\begin{align}\int_{-\infty}^{\infty}\sin[\omega_0(t-\tau)]\frac{\sin(\omega_0\tau)}{\pi \tau}d\tau&=\sin(\omega_0t)\int_{-\infty}^{\infty}\frac{\cos(\omega_0\tau)\sin(\omega_0\tau)}{\pi \tau}d\tau+\\&+\cos(\omega_0t)\underbrace{\int_{-\infty}^{\infty}\frac{\sin^2(\omega_0\tau)}{\pi \tau}d\tau}_{0\textrm{ (integrand odd!)}}\\&=\sin(\omega_0t)\frac12\underbrace{\int_{-\infty}^{\infty}\frac{\sin(2\omega_0\tau)}{\pi \tau}d\tau}_{1\textrm{ (DC value of ideal LP)}}\\&=\frac12\sin(\omega_0t)\end{align}$$

Eq. $(2)$: $$\begin{align}\int_{-\infty}^{\infty}\frac{\sin(\omega_0t)}{\pi t}e^{-j\omega_0t}dt&=\int_{-\infty}^{\infty}\frac{\sin(\omega_0t)}{\pi t}\cos(\omega_0t)dt\\&=\frac12\int_{-\infty}^{\infty}\frac{\sin(2\omega_0t)}{\pi t}dt\end{align}$$

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  • $\begingroup$ hm but there's a theorem that if you take a discontinuous function, apply Fourier (or alike) transform on it, and the inverse of that, you get a function where the value at the discontinuity is in the middle of the left- and right-side limit of the original function. Gibb's phenomenon is a result of exactly that. $\endgroup$ Dec 19, 2021 at 18:49
  • $\begingroup$ @MarcusMüller: That's right! The inversion formula is only valid if we assume the function value at jump discontinuities is the average of the two one-sided limits. But that's just one way of defining a function value at a discontinuity. I mean, why should we apply a Fourier transform and its inverse in the first place? The problem has nothing to do with the way we need to define function values for the inverse Fourier transform to hold true. $\endgroup$
    – Matt L.
    Dec 19, 2021 at 18:56
  • $\begingroup$ point is that when we wonder about what happens at "cutoff frequency", we're asking ourselves what the result of the point-wise multiplication of the signal's Fourier transform (i.e. the $\pm f_c$ Diracs) and the system's Fourier transform is, transformed back to time – and this consideration says it's multiplication with a factor $\frac12$ in freq domain, so we'll arrive at a sine of half the original amplitude in time. $\endgroup$ Dec 19, 2021 at 19:04
  • $\begingroup$ @MarcusMüller: That's a reasonable definition, but it remains a definition in my opinion. $\endgroup$
    – Matt L.
    Dec 19, 2021 at 19:08
  • $\begingroup$ Hm, I think (my proof on paper here isn't airtight, so I'd love to say "I know", but really can't) this is actually the answer to question "what happens when you convolve sin x with (sin x)/x", and I don't think that's really something that should be dependent on the definition of what the value of the rect function is – because that's just an "intuition thing" for the human who likes to thing in the frequency domain, but "the ideal brickwall filter with cutoff frequency $f_c$ convolved with the signal $\sin(f_ct)$"should be independent of that - the first is unambigously a scaled sinc with $\endgroup$ Dec 19, 2021 at 19:14

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