Frequency response of an ideal low pass filter

Question

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?

Answer 1

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.

Answer 2

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.

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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 $(1)$:

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

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