# what is nyquist rate of $h(t)\cdot h(t)$ and $h(t)\circledast h(t)$

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 $2B$?

• 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) – Fat32 Apr 15 '16 at 16:56
• 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? – Peter K. Apr 15 '16 at 17:53
• 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 $2B$. – robert bristow-johnson Apr 15 '16 at 18:38
• @robertbristow-johnson, yes. in fact, my question is about re-sampling a discrete signal. – David Apr 15 '16 at 18:51
• 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 $4B$. – robert bristow-johnson Apr 15 '16 at 18:52