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Let's say we have $h_c(t)$ as a continuous-time signal with bandwidth $B$ and we would like to sample it. To be able to reconstruct it correctly, the sampling rate must be greater than $2B$. Now assume we have $h_{c1}(t)=h_c(t)\cdot h_c(t)$ and $h_{c2}(t)=h_c(t)\circledast h_c(t)$ what is Nyquist rate of these two signals? Is it $4B$ for both of them because each $h_{c}(t)$ needs $2B$?

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  • $\begingroup$ a discrete time signal $h[n]$ is by definition bandlimited. Do you mean $h[n]$ is the samples of the continuous time signal $h_c(t)$ which was bandlimited to B, and also the other signals are, therefore, $h_1[n] = h_{1c}(nT)$ where $h_{1c}(t) = h_c(t)h_c(t)$ (multiplication) and $h_2[n] = h_{2c}(nT)$ where $h_{2c}(t) = hc(t) \star hc(t)$ (convolution) $\endgroup$ – Fat32 Apr 15 '16 at 16:56
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    $\begingroup$ The Nyquist rate makes no sense once the signal is sampled. Can you please reforms late your question in terms of $h_c(t)$ as @Fat32 suggests? $\endgroup$ – Peter K. Apr 15 '16 at 17:53
  • $\begingroup$ i would say that the question is formed well enough to know what David is asking about. the answer is that since they all are discrete signals after sampling, both $h_1[n]$ and $h_2[n]$ have the same bandwidth which is at $\omega = \pi$ not $B$. "$B$" is the assumed bandwidth of the analog $h(t)$ before it was sampled to be $h[n]$. now if you want to talk about the bandwidths of $h_1(t) = h(t)\cdot h(t)$ vs. $h_2(t) = h(t) \circledast h(t)$, that is a different question and has meaning. if $h(t)$ is bandlimited to $B$, so also is $h_2(t)$, but $h_1(t)$ is bandlimited to $2B$. $\endgroup$ – robert bristow-johnson Apr 15 '16 at 18:38
  • $\begingroup$ @robertbristow-johnson, yes. in fact, my question is about re-sampling a discrete signal. $\endgroup$ – David Apr 15 '16 at 18:51
  • $\begingroup$ well, no. now your question is about the bandwidths of two different continuous-time signals. and, if you add the subscript $\cdot_c$ to my previous comment/answer, that answers your question. so the answer is that the "Nyquist rate" (which is twice the bandwidth) for $h_{c2}(t)$ need not be greater than $2B$, but for $h_{c1}(t)$, it must be $4B$. $\endgroup$ – robert bristow-johnson Apr 15 '16 at 18:52
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squaring a signal effectively doubles the bandwidth of the signal (and doubles the Nyquist rate, which is the lower limit of the sampling rate). convolving a signal with anything, even itself, does not in general increase the bandwidth.

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Convolution spreads the signal in the domain in which the convolution takes place. Convolution in the time domain spreads the signal in time. Convolution in the frequency domain spreads the signal in frequency. Multiplying the signal by itself in time domain is equivalent to convolution in the frequency domain - so the resulting signal will have twice the bandwidth.

For convolution in time-domain - this corresponds to multiplication in the frequency domain, so the resulting signal will have the essentially the same bandwidth. The bandwidth may change slightly. If you consider the -3 dB point of the original signal - the multiplication or convolution will change the position of the -3 dB points slightly in the final signal. How much the position of these points change, will depend of the definition of bandwidth you use and the exact signal you use.

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  • $\begingroup$ what is the role of this "-3 dB"? $\endgroup$ – David Apr 16 '16 at 1:25
  • $\begingroup$ I will edit my answer. The -3 dB points are usually convenient measures for the resolution (e,g. beamforming or spectral resolution). I was trying to use it as an example. For sampling purposes you tend to using something like the -60 dB points (or some significant attenuation) to limit the amount of aliasing. $\endgroup$ – David Apr 16 '16 at 12:33

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