1
$\begingroup$

I know that if $f_\mathrm{m}$ is the "Nyquist frequency" (max frequency) and $f_\mathrm{s}$ sampling rate then $f_\mathrm{s}>2f_\mathrm{m}$.

  • Am I correct so far?

I have a signal $x(t)$ with max frequency $f_1$ and $h(t)$ with $f_2$ and we define $y(t)= h(t)*x(t)$ ($*$ for convolution) and we need to find the sampling frequency/Nyquist frequency of this function.

So the Nyquist frequency of $x(t)$ is $f_{\mathrm{s}_{x}} >2f_1$ and of $h(t)$ is $f_{\mathrm{s}_{h}} >2f_2$.

Now I saw that someone wrote that using the convolution theorem we get $Y(f)=H(f)X(f)$, so there must be that $f_\mathrm{s} \leq \min\{2f_1,2f_2 \}$ stating that this is an upper bound because frequency may cancel each other.

  • Why is that true?

I must mention that he wrote $Nq(x)$ instead of $f_{\mathrm{s}_{x}}$ (I just understood that he meant the same), also it's not supposed to be $f_\mathrm{s} \geq \min\{2f_1,2f_2 \}$ ?

I'm also not sure about is that $X(f)=0$ if $|f|>\frac{Nq(x)}{2}$.

  • Why is that?
Improve this question
$\endgroup$
13
  • $\begingroup$ "someone wrote": Nope, not arguing with some unknown source about something that we both suspect is wrong. Cite, or find a better source. $\endgroup$
    – Marcus Müller
    Feb 8, 2019 at 18:21
  • $\begingroup$ @MarcusMüller if he is wrong can you answer me what is the correct answer? Thanks $\endgroup$
    – user2323232
    Feb 8, 2019 at 18:54
  • 1
    $\begingroup$ you also need to fix your symbols a little bit. i would change "$x$" to "$t$". then change "$f(x)$" to "$x(t)$" and "$g(x)$" to "$y(t)$". then change "$F$" and "$G$" to "$X$" and "$Y$". then use "$\omega$" only for angular frequency and use "$f$" for ordinary frequency for the arguments of $F(\cdot)$ and and $G(\cdot)$ and $H(\cdot)$. $\endgroup$
    – robert bristow-johnson
    Feb 8, 2019 at 20:30
  • 1
    $\begingroup$ i don't want to think about the question until i am comfortable about the nomenclature. $\endgroup$
    – robert bristow-johnson
    Feb 8, 2019 at 20:34
  • 1
    $\begingroup$ i didn't fix it yet, but Nyquist frequency (as ordinary, not angular frequency) is simply half of the sample rate, $f_\mathrm{s}/2$. it is not the sample rate in any decent modern textbook or lit. that abuse of notation needs to be corrected. $\endgroup$
    – robert bristow-johnson
    Feb 8, 2019 at 20:50

2 Answers 2

Reset to default
1
$\begingroup$

Since nobody answered my I will answer using what I read on the internet, "Nyquist frequency" is the max frequency we can get using given sampling rate. the $f_s$ of convolution between 2 function is indeed the min of their max frequency, but I'm not sure yet about the part which stated that this is an upper bound because frequency may cancel each other. Since nobody here could answer it , but myself I'm marking this as an answer until someone will be able to answer it, thanks.

Improve this answer
$\endgroup$
0
$\begingroup$

I'll give it a try.

I know that if fm is the "Nyquist frequency" (max frequency) and fs sampling rate then fs>2fm.

Am I correct so far?

No.

The Nyquist frequency is simply defined as half the same rate. If your sample rate is 48kHz, your Nyquist frequency is 24kHz no matter what and how you sample anything.

You may have signals that have frequency content above the Nyquist frequency. If that's the case, you will get aliasing when you sample. However the Nyquist frequency is completely independent from the frequency content of the signal. The frequency content only determines whether you get aliasing or not.

Since you already started with a wrong assumption, the rest of your question as written doesn't make a lot of sense.

I think you are just confusing the terms "bandwitdh" and "Nyquist frequency" and what you are asking is "I convolve a signal of bandwidth $f_1$ with a signal of bandwidth $f_2$: what's the bandwidth of the result?". If that's your question, please re-phrase as such. That answer would be $min(f_1,f_2)$

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thank you , but it’s not supposed to be the min? $\endgroup$
    – user2323232
    Feb 9, 2019 at 13:46
  • $\begingroup$ Yes, of course. Corrected $\endgroup$
    – Hilmar
    Feb 9, 2019 at 15:09
  • $\begingroup$ also you didn’t referred to one of my question , the claim that it is less equal to it, since frequency may cancel each other.. $\endgroup$
    – user2323232
    Feb 9, 2019 at 16:28
  • $\begingroup$ Well, it's more complicated than that. For example, if you convolve two sine waves of different frequencies, the result will be zero. If your question is really about bandwidth, then please rephrase or ask a new one and so it is clear what specifically you want to know. $\endgroup$
    – Hilmar
    Feb 10, 2019 at 14:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.