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In Gonzalez book Digital Image Processing, section 4.34 (third edition), he writes:

Unfortunately, except for some special cases mentioned blow, aliasing is always present in sampled signals because, even if the original sampled function is band-limited, infinite frequency components are introduced the moment we limit the duration of the function, which we always have to do in practice.

For example, suppose that we want to limit the duration of a band-limited function $f(t)$ (i.e. a function whose Fourier transform is non-zero only on a closed interval of range of frequencies), to an interval say $[0, T]$. We can do this by multiplying $f(t)$ by the function

$h(t)= 1 $ if $t \in [0,T]$, and is $0$ otherwise.

Then from the convolution theorem we know that the transform of this product $h(t)f(t)$ is the convolution of the transforms of the functions. Even if the transform of $f(t)$ is band-limited, convolving it with $F(h(t))=H(\mu)$ will yield a result with frequency components that are infinite.

This very last statement is what I am not sure about. If the Fourier transform of $f$ is band-limited, then outside of a closed interval, the transformed function will be $0$, and so I am not sure for which frequency components the convolution of the transforms will have infinite frequency. Any insights appreciated.

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Note that in order to obtain the Fourier transform of the windowed time domain signal $f(t)h(t)$, you need to convolve the Fourier transforms of $f(t)$ and $h(t)$. We know that the Fourier transform of $f(t)$ is zero outside some interval (because $f(t)$ is band-limited). However, since $h(t)$ is time-limited, i.e., it is zero outside some time interval, we know that its Fourier transform cannot be band-limited. For this specific example of a rectangular window $h(t)$, we know that the corresponding Fourier transform is a sinc function. Convolving any function with a sinc function results in a function that extends from $-\infty$ to $+\infty$. Consequently, the Fourier transform of the windowed signal $f(t)h(t)$ also extends from $-\infty$ to $+\infty$. Consequently, the signal $f(t)h(t)$ cannot be band-limited.

In sum, a band-limited signal cannot be time-limited, and a time-limited signal cannot be band-limited. Note, however, that you cannot conclude that a signal is band-limited if it is not time-limited and vice versa. There are signal that are neither time-limited nor band-limited.

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  • $\begingroup$ "Convolving any function with a sinc function results in a function that extends from −∞ to +∞". I think this is the heart of my conusion. Can you elaborate on this, because it seems to me that we would be integrating the FT of $f(t)$ weighted by a translation of the sync in some interval, and zero elsewhere? I don't see why there would be non=zero values in the entire range of $\mathbb R$. Thanks for the help! $\endgroup$ – IntegrateThis May 16 at 19:52
  • $\begingroup$ * non-zero, typo. $\endgroup$ – IntegrateThis May 16 at 19:57
  • $\begingroup$ @IntegrateThis: Convolving a rectangle with a sinc is basically just a moving average of the sinc function, hence extending from $-\infty$ to $\infty$. $\endgroup$ – Matt L. May 16 at 20:26

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