I found the answers to this question and this question to be extremely helpful in understanding the derivation of the downsampling or decimation property of the DTFT. Thank you! I am now struggling to come up with a good example that shows clearly the role of the frequency-shifted term(s).

To summarize, let $y[n]$ be a discrete-time signal formed by downsampling $x[n]$ by an integer factor $M$, $$ y[n] = x[Mn]. $$ Then the discrete-time Fourier transforms of $y[n]$ and $x[n]$ are related through $$ Y \left( e^{j \omega} \right) = \frac{1}{M} \sum_{k=0}^{M-1} X \left( e^{j (\omega - 2 \pi k)/M} \right). $$

The simplest case is for $M=2$, $$ Y \left( e^{j \omega} \right) = \frac{1}{2} \, X \left( e^{j \omega/2} \right) + \frac{1}{2} \, X \left( e^{j (\omega/2 - \pi)} \right) . $$

I want to develop an example that clearly shows the role of the second term. I have already derived the result for a causal exponential signal, but for that example the role of the second term is not at all obvious from a plot of the spectrum. Therefore, let $x[n]$ be a sinc signal, $$ x[n] = \frac{\sin(\omega_0 n)}{\pi n} \qquad \Rightarrow \qquad X \left( e^{j \omega} \right) = \begin{cases} 1, & |\omega| < \omega_0 \\ 0, & \omega_0 < |\omega| < \pi. \end{cases} $$ I am suppressing the periodic repetition of the rectangular spectrum here for simplicity.

Assume for the moment that $\omega_0 < \pi/4$. The computation of the DTFT of $y[n]$ by the downsampling property gives $$ \begin{array}{rclcl} Y_1 \left( e^{j \omega} \right) & = & \dfrac{1}{2} \, X \left( e^{j \omega/2} \right) & = & \begin{cases} \frac{1}{2}, & |\omega| < 2 \omega_0 \\ 0, & 2 \omega_0 < |\omega| < \pi. \end{cases} \\ \\ Y_2 \left( e^{j \omega} \right) & = & \dfrac{1}{2} \, X \left( e^{j (\omega - 2 \pi)/2} \right) & = & \begin{cases} \frac{1}{2}, & \pi - 2 \omega_0 < |\omega| < \pi\\ 0, & |\omega| < \pi - 2 \omega_0. \end{cases}\\ \\ Y \left( e^{j \omega} \right) & = & Y_1 \left( e^{j \omega} \right) + Y_2 \left( e^{j \omega} \right) & = & \begin{cases} \frac{1}{2}, & |\omega| < 2 \omega_0 \\ \frac{1}{2}, & \pi - 2 \omega_0 < |\omega| < \pi\\ 0, & 2 \omega_0 < |\omega| < \pi - 2 \omega_0. \end{cases}\\ \end{array} $$

This expression tells us that the DTFT of $y[n]$ consists of two rectangular spectra, one centered around $\omega = 0$, and the other around $\omega = \pm \pi$. This would seem to be exactly what I am looking for. However, if we return to the original expression for $y[n]$, $$ y[n] = x[2n] = \frac{\sin(2 \omega_0 n)}{2 \pi n} . $$

Therefore, the DTFT of $y[n]$ should be $$ Y \left( e^{j \omega} \right) = \begin{cases} \frac{1}{2}, & |\omega| < 2 \omega_0 \\ 0, & 2 \omega_0 < |\omega| < \pi, \end{cases} $$

which contains only the low-frequency rectangular component. How can I resolve this apparent contradiction?


1 Answer 1


You got the term $X \left( e^{j (\omega - 2 \pi)/2} \right)$ wrong. It is centered at $2\pi$ and it is non-zero in the interval $(2\pi-2\omega_0,2\pi+2\omega_0)$. So the results obtained in the frequency domain and in the time domain, respectively, are identical.

Note that the term $X \left( e^{j \omega/2} \right)$ is $4\pi$-periodic, so the term $X \left( e^{j (\omega - 2 \pi)/2} \right)$ makes sure that the spectrum of the downsampled signal is $2\pi$-periodic. Clearly, there is no aliasing as long as $\omega_0<\pi/2$ is satsified.

  • $\begingroup$ Oh dear. You're right of course, thank you for the help. Then in the general case $X \left( e^{j \omega/M} \right)$ is $2M \pi$-periodic, and the $M-1$ frequency-shifted terms are required to make $Y \left( e^{j \omega} \right)$ $2 \pi$-periodic. This in turn means that any example showing the effect of the shifted spectral copies would be the result of aliasing, i.e. that $X \left( e^{j \omega} \right)$ has some nonzero-value for $\omega > \pi/M$. That makes sense of course. $\endgroup$
    – Steve J.
    Jul 3, 2020 at 17:50
  • $\begingroup$ @SteveJ.: That's right! $\endgroup$
    – Matt L.
    Jul 3, 2020 at 17:55

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