I'm trying to follow the steps from Oppenheim for the derivation of the frequency representation of a signal which is to be down sampled by a factor of M.

It makes sense to me that:

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But then the book says to make a change of variables from $r$ to $i$ and $k$ where $i$ = $0, 1, ... M-1$, and $k$ = (-$\infty,\infty$). However, I don't really understand how this change of variables comes logically. After making this change of variables it is then easy to see how the downsampled frequency representation is just a sum of scaled and shifted representations of the original sampled sequence. But I would like to be able to make the bridge better between steps and how such a change of variable was concluded.


1 Answer 1


In the final result, you want to express the spectrum $X_d(e^{j\omega})$ in terms of $X(e^{j\omega})$, the spectrum of $x[n]=x_c(nT)$. Since $X(e^{j\omega})$ is already periodic, it must be possible to represent $X_d(e^{j\omega})$ as a sum of a finite number ($M$) of shifted versions of $X(e^{j\omega})$. This is why the original infinite sum is split up into a finite sum of infinite sums, the latter being shifted versions of $X(e^{j\omega})$.

I think you shouldn't worry if you think you wouldn't have come up with that change of variables yourself. What is important is that you understand what's going on.

Furthermore, I think it's instructive to understand the derivation of the expression for $X_d(e^{j\omega})$ without introducing an auxiliary continuous-time signal $x_c(t)$:

$$\begin{align}X_d(e^{j\omega})&=\sum_{n=-\infty}^{\infty}x[Mn]e^{-jn\omega}\\&=\sum_{n=kM}x[n]e^{-jn\omega /M}\end{align}\tag{1}$$

where in the second sum we only sum over indices $n$ that are integer multiples of $M$. If we introduce a sequence $d[n]$ which equals $1$ for $n=kM$ ($k\in\mathbb{Z}$), and zero otherwise, we can rewrite $(1)$ as

$$X_d(e^{j\omega})=\sum_{n=-\infty}^{\infty}x[n]d[n]e^{-jn\omega /M}\tag{2}$$

An expression for the sequence $d[n]$ satisfying our requirements is

$$d[n]=\frac{1}{M}\sum_{l=0}^{M-1}e^{j2\pi ln/M}\tag{3}$$

Inserting $(3)$ into $(2)$ gives the final result:

$$\begin{align}X_d(e^{j\omega})&=\frac{1}{M}\sum_{n=-\infty}^{\infty}x[n]\sum_{l=0}^{M-1}e^{j2\pi ln/M}e^{-jn\omega /M}\\ &=\frac{1}{M}\sum_{l=0}^{M-1}\sum_{n=-\infty}^{\infty}x[n]e^{-jn(\omega-2\pi l)/M}\\ &=\frac{1}{M}\sum_{l=0}^{M-1}X\left(e^{j(\omega-2\pi l)/M}\right)\tag{4}\end{align}$$


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