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Let $f_{k}$ be $\sigma_{k}$-bandlimited, $k=1,2,...,K$ and define $f:=f_{1}\cdot f_{2}\cdot...\cdot f_{K}$; then how should I choose $T>0$ such that we can recover $f$ from samples $f(nT)$?

According to the Shannon sampling theorem, we must choose $T$ such that $\frac{1}{T}>2\cdot B$, where $B$ is the bandlimit. Hence the question boils down to how we determine $B$.

Now, we have to choose $B$ such that $\hat{f}(\xi)=0$ for $|\xi|>B$. If we compute

$$\hat{f}(\xi)=\mathscr{F}(f_{1}\cdot f_{2}\cdot...\cdot f_{K})(\xi)=\hat{f}_{1}(\xi)\ast\hat{f}_{2}(\xi)\ast...\ast\hat{f}_{K}(\xi),$$

then this still seems inconclusive to me, so I'd appreciate some help in this regard.

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  • $\begingroup$ i wish people would leave the semantic $f$ to represent frequency. it would make our lives easier especially when expressing the Fourier Transform. $\endgroup$ – robert bristow-johnson Jan 28 '17 at 22:56
  • $\begingroup$ @robertbristow-johnson I do apologise. The only explanation I can offer is that I come from a mathematics background. $\endgroup$ – Jason Born Jan 28 '17 at 23:00
  • $\begingroup$ i know. but there are reasons electrical engineers have changed some of the notation, beginning with "$j$" for the imaginary unit. $\endgroup$ – robert bristow-johnson Jan 28 '17 at 23:01
  • $\begingroup$ @robertbristow-johnson I am not aware of the reasons, although I will look them up for sure! Incidentally, I have insisted on using $i$ as my imaginary unit while doing signal processing. $\endgroup$ – Jason Born Jan 28 '17 at 23:03
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    $\begingroup$ the answer to your question is: $$ \frac{1}{T} > 2 \sum\limits_{k=1}^{K} \sigma_k $$ $\endgroup$ – robert bristow-johnson Jan 28 '17 at 23:04
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The definition of bandlimited is that $\hat f_k$ has compact support of size $\sigma_k$. To make our argument easier, we'll also note that we can decompose every $\hat f_k$ into $\tilde f_k * \delta_k$ (a convolution with a dirac impulse, just a shift) so that $\tilde f_k$ is "0-aligned", ie. $\sup f_k = \sigma_k$ and $\inf f_k=0$.

For $f_1\cdot f_2$ follows directly that the support of $\tilde f_1*\tilde f_2$ has the property

$$\text{supp}(f_1*f_2) \subset [-\sigma_2;\sigma_1]$$

from the way we calculate the convolution:

$$(f_1*f_2)(x)= \int\limits_{-\infty}^\infty f_1(y)f_2(x-y)\,dy$$

Now, that support has "length" (to be more exact, an upper boundary for measure) $\sigma_1+\sigma_2$.

It's very intuitive to now apply induction to show that this applies to all $k$.

So, in the end, $\text{supp}\left({\Large*}\left(\tilde f_k\right)_{k=1}^{K}\right)\subset [0, \sum\limits_{k=1}^K \sigma_k]$.

Since the shifting around in frequency domain doesn't theoretically (measure of $\delta_k0\,\forall k$) nor practically matter (compare: undersampling), we can just sample sufficiently by observing the sum bandwidth.

Note that this has very practical effects, for example, when sampling signals that are reflected by moving targets.

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