Understanding index transformation in derivation of Fourier transform for sampling rate reduction Was going over some notes regarding deriving fourier transform equation for Sampling Rate Reduction. Reference to Notes from below link https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-341-discrete-time-signal-processing-fall-2005/lecture-notes/lec05.pdf or from Book Discrete-Time Signal Processing by Alan V. Oppenheim (2nd Edition), equation 4.75.
$$r = i + kM$$
I am lost as to how this is obtained. I understand that every $M-1$ samples are dropped from original sampling results; but still cannot understand how this expression for $r$ is derived.
Could someone help me understand this?
It's not derived, it's just chosen in a smart way such that the relationship between the decimated and the original sequences becomes obvious.
It's just a rearrangement of the terms of the sum. As a simple example, take an infinite sum of numbers $a_r$:
$$S=\sum_{r=-\infty}^{\infty}a_r\tag{1}$$
Under certain conditions that we don't need to bother with now we can rearrange the sum and write it like
$$\begin{align}S&amp;=\ldots+a_{-2M}+a_{-M}+a_0+a_M+a_{2M}+\ldots\\&amp;\ldots+a_{-2M+1}+a_{-M+1}+a_1+a_{M+1}+a_{2M+1}+\ldots\\&amp;\vdots\\&amp;\ldots+a_{-2M+(M-1)}+a_{-M+(M-1)}+a_{0+(M-1)}+a_{M+(M-1)}+a_{2M+(M-1)}+\ldots\tag{2}\end{align}$$
with some arbitrarily chosen integer $M$. So we just start with element $a_0$ and add every $M^{th}$ element, then we move to element $a_1$ and add again every $M^{th}$ element, etc. If we do this $M$ times, we've added all elements, just like in the originial sum $(1)$.
Eq. $(2)$ can be written as
$$S=\sum_{i=0}^{M-1}\sum_{k=-\infty}^{\infty}a_{i+kM}\tag{3}$$
which means that we expressed the index $r$ as $r=i+kM$.