# Ideal low pass filter with cut-off B does not affect a band-limited baseband signal with maximum frequency B

Let $$x(t)$$ be a band-limited signal with spectrum (fourier transform) lying between -B and +B. Such a signal is not warped by an ideal low-pass filter with cut-off frequency equal or higher than +B.

So, let's consider an ideal (vertical edges) low-pass filter whose spectrum is $$H(f) = \mathrm{rect}(f/2B)$$. It's shape in time domain (i.e. pulse response) will be $$h(t) = 2B\cdot \mathrm{sinc}(2Bt)$$.

So, the first statement tells us that the signal $$x(t)$$ is not warped by the convolution with $$h(t)$$, i.e.:

$$x(t) = \int_{-\infty}^{+\infty} x(\theta)\cdot 2B\cdot \mathrm{sinc}(\frac{t-\theta}{2B})\,\,d\theta$$

Well, I understand that the low pass filter does not modify the signal spectrum and hence do not modify the time domain signal. But I wanted to figure it out in the time domain.

The previous convolution means that for each time instant $$\theta$$, the signal value $$x(\theta)$$ is multiplied by the $$\mathrm{sinc}$$ function whose peak value (considering also the 2B term) is one. The superposition of all these weighted sinc gives the original signal $$x(t)$$. I can't understand and visualize how this is possible. The sinc tails will overlap together. I'd understand it if the sinc would have just a peak equal to 1 and zero for all the other instants. I think I need an image to understand such an equation.

• Given that the whole notion of the frequency domain comes from the Fourier transform, I'm not sure how valid it is to insist that you don't go through the Fourier transform to find a solution. I'd suggest choosing a couple of known, easy, band-limited signals ($x(t) = \mathrm{sinc \theta}$ and $x(t) = \mathrm{sinc^2 \frac{\theta}{2}}$ come to mind), work out the convolutions, and see if that doesn't help with your intuitions. Jan 28, 2022 at 21:06