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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?

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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

$$\begin{align}S&=\ldots+a_{-2M}+a_{-M}+a_0+a_M+a_{2M}+\ldots\\&\ldots+a_{-2M+1}+a_{-M+1}+a_1+a_{M+1}+a_{2M+1}+\ldots\\&\vdots\\&\ldots+a_{-2M+(M-1)}+a_{-M+(M-1)}+a_{0+(M-1)}+a_{M+(M-1)}+a_{2M+(M-1)}+\ldots\tag{2}\end{align}$$

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 $r=i+kM$.

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  • $\begingroup$ thanks for info. What troubles me is that the down sampling samples only contain a(M) + a(2M) + a(3M) ... so in context of summation i would write it as $S=\sum_{k=-\infty}^{\infty}a_{kM}\tag{1}$ $\endgroup$ – niil87 Jun 17 at 18:54
  • $\begingroup$ @niil87: But that sum is not over time domain samples, this is a sum of shifted frequency spectra, there's no downsampling involved in that sum. $\endgroup$ – Matt L. Jun 17 at 19:14
  • $\begingroup$ thank you for reply, makes sense! $\endgroup$ – niil87 Jun 19 at 9:02

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