3
Multiplication in the frequency domain is equivalent to circular convolution in the time domain with a period of NFFT. If you don't zero pad them to at least length(x1)+length(x2)-1 samples, the IFFT result would be aliased in the time domain. That's why overlap-save method discards part of the result.
3
The image in a digital camera/sensor is represented as a matrix of discrete numbers, as you say.
The (spatial) digitization consists of an array of sensels each with some non-zero area that are sensitive to light, framed by non-sensitive area, possibly micro-lenses that increase the fill-rate (the effective light sensitive area), possibly optical lowpass ...
3
Both results are correct. Hence, the equality
$$\frac{1}{T}\sum_{k=-\infty}^{\infty}\frac{1}{1+ j \left(\omega - k\frac{2\pi }{T}\right)}=\sum_{n=0}^{\infty}e^{-nT(1+j\omega)}\tag{1}$$
holds true.
One way to prove this is to realize that the term on the left-hand side of Eq. $(1)$ is periodic with period $2\pi/T$. Consequently, we can express this term by ...
2
Either someone didn't stress enough that there are four basic flavors of the Fourier Transform, or they did introduce that for you but you've forgotten.
For a sampled-time signal of infinite extent in time, you should use the DTFT (not to be confused with the discrete Fourier transform, of which the FFT is the fast version).
The DTFT is defined as $$X(\omega)...
2
It's the spectrum of a discrete signal: sampling in time $\Leftrightarrow$ periodizing in frequency - explained in detail here. Overlap means there's aliasing, and we require a higher sampling rate. FFT returns one period of this spectrum since it contains all the information. (More precisely, your image shows DTFT, and DFT (which FFT implements) is a ...
Only top voted, non community-wiki answers of a minimum length are eligible