# DTFT and a Downsampled Sinc Function

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?

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.
• 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. – Steve J. Jul 3 at 17:50