We are given that $f(x,y)$ is highest frequency is $\omega$ what will be the frequency sample rate if we want to restore the function of the form $g(x,y)=f^2(x,y)$

Would it be correct to say that because $\sin^2x=1-\cos 2x$, we will have to sample it at $4\pi$ ?

  • $\begingroup$ $f(x,y)$ is a function of two variables. If you are referring to "signal functions" then this is likely to be $y_1 = f(t), y_2 = g(t)$. On whether your reasoning is correct, I can tell you that it is down the right track (although, that is not exactly $\sin^2(x)$) but what if the input signal is not a simple sinusoid? $\endgroup$
    – A_A
    Commented Jan 23, 2019 at 18:27
  • $\begingroup$ @A_A it is an image $\endgroup$
    – gbox
    Commented Jan 23, 2019 at 19:07
  • 1
    $\begingroup$ Nice twist. What are your thoughts about $f(x,y)$ being a squared "image"? What is the role of spatial resolution in this case? $\endgroup$
    – A_A
    Commented Jan 23, 2019 at 19:31
  • $\begingroup$ @A_A Sorry, I do not understand $f(x,y)$ is let say an $N$ by $N$ matrix of 256 gray values. So it is a sampling $N^2$ times, I think $\endgroup$
    – gbox
    Commented Jan 23, 2019 at 20:03

1 Answer 1


By the Convolution Theorem multiplication in Time / Spatial domain is equivalent of Convolution in the Frequency Domain.

The sampling rate (In its classic interpretation) is proportional to the support of a function in the frequency domain.

So if a function has a certain support in frequency, what would be its support after convolution with itself? Indeed it will be doubled in each dimension.


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