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.
