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Let $X(\omega)$ be the Discrete Time Fourier Transform (DTFT) of $x[n]$, I want to construct $X(\omega/2)$.

Precisely, I use FFT function to compute the samples of $X(\omega)$ in one period, say $[0, 2\pi]$. I want Construct samples of the $Y(\omega)=X(\omega/2)$ in that period.

My work:

Since dividing frequency by 2 stretch the signal, the values in $[0, \pi]$ of $X(\omega)$ cover all the $[0, 2\pi]$ space of $Y(\omega)$. On the other hand, the values in $[\pi, 2\pi]$ of $X(\omega)$ cover all the $[0, 2\pi]$ space of $Y(\omega)$ too.

Let X be a vector with size of n that contains the samples of $X(\omega)$, so to construct vector Y that contains the samples of $Y(\omega)$ I use the following Matlab code

Y = (kron(x[1:n/2], [1, 1]) + kron(x[n/2+1:n], [1, 1]))/2

Is my work correct?
Thanks.

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    $\begingroup$ Are we talking in the continuous world or discrete? In the continuous as you wrote scaling on one domain is scaling by the inverse on the other domain. In the discrete world you can have it even simpler. Will that be enough? Namely for $ N $ samples signal you get $ N $ Samples on $ \left[ 0, 2 \pi \right] $ hence only $ \frac{N}{2} $ on the range $ \left[ 0, \pi \right] $. If you want $ N $ samples on the $ \left[ 0, \pi \right] $ range there is easy way to do it. $\endgroup$ – Royi Feb 5 '17 at 10:41
  • $\begingroup$ From your question, despite your notation $x(t)$, I gather that you're talking about discrete time. In that case, stretching $X(\omega)$ by a factor $2$ does not result in a valid DTFT because the stretched function has period $4\pi$ instead of $2\pi$. You have to reformulate your question to take that into account and do something about it, or conclude that what you're trying to do does not make much sense. $\endgroup$ – Matt L. Feb 6 '17 at 9:43
  • $\begingroup$ It is best if you don't travel between continuous and discrete times and rewrite your question (exactly what you want) in discrete domain to make it more understandable. $\endgroup$ – Qasim Chaudhari Feb 13 '17 at 4:42

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