# 0.7 Decimation and downsampling

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## Pure downsampling

The easiest case consists in reducing the sampling rate by simply dropping samples.This procedure is called Downsampling . However, we need to be careful on the effect of this procedure. We note immediately,that the new sampling rate ${f}_{d}={f}_{e}/M$ needs to be still above Nyquist, i.e., ${f}_{e}>M·2B$ in order to avoid aliasing.

• Downsampling Assume that ${f}_{e}>M·2B$ .
$\begin{array}{ccc}\hfill \left({y}_{0},{y}_{1},{y}_{2},...\right)\to & ↓M& \to \left({y}_{0},{y}_{M},{y}_{2M},...\right)=\left({z}_{0},{z}_{1},{z}_{2},...\right)\hfill \end{array}$

Effect in the frequency domain In continuous time, downsampling the signal $y\left(t\right)$ corresponds to passing to the signal $z\left(t\right)=y\left(Mt\right)$ . Indeed, sampling $z$ at the same frequency as $y$ we obtain the samples ${z}_{k}={y}_{kM}$ as we should. In summary, using the properties of the Fourier transform

$z\left(t\right)=y\left(Mt\right)\phantom{\rule{2.em}{0ex}}Z\left(f\right)=\frac{1}{M}Y\left(\frac{f}{M}\right)$

This tells us, that the spectral copies of ${Z}_{e}$ (the Fourier transform of the sampled signal ${z}_{e}\left(t\right)$ ) should have the same overall shapeand at the same distance ${f}_{e}$ as those of ${Y}_{e}$ , but they are stretched in frequency by a factor $M$ and squeezed in amplitude by a factor $M$ .

Indeed, we arrive at the same conclusion computing via the sampled signals, using [link] :

${Z}_{e}\left(f\right)=\sum _{k}\tau {z}_{k}{e}^{-j2\pi fk\tau }=\sum _{k}\tau {y}_{kM}{e}^{-j2\pi fk\tau }=\frac{1}{M}\sum _{k}\left(M\tau \right){y}_{kM}{e}^{-j2\pi \left(f/M\right)k\left(M\tau \right)}=\frac{1}{M}{Y}_{d}\left(\frac{f}{M}\right)$

Recall that ${Y}_{d}\left(f\right)$ , the Fourier transform of $y$ sampled at frequency ${f}_{d}={f}_{e}/M$ , looks like ${Y}_{e}$ except that its spectral copies lie at distance ${f}_{d}$ , thus $M$ times closer than those of ${Y}_{e}$ .

For the discrete Fourier Transform we should expect to see roughly the same behavior. Recall, though, that the relation between continuous and discrete Fourier transform is only approximative.

Aliasing : We give two examples.

First, downsampling to a sampling rate that is too low can lead to aliasing. Downsampling the image of [link] left leads to the same effect as visible in the same figure center.

Second, downsampling a simple signal such as the filter $b$ of [link] by $M=2$ results in a new filter ${b}^{\text{'}}$ with twice the cutoff frequency, i.e. ${f}_{c}^{\text{'}}=1/4=0.25$ , but with a value ${\stackrel{^}{b}}_{k}=1/2=1/M$ over the pass-band, thus a power spectral value $1/4=1/{M}^{2}$ over the pass-band (see [link] ). No aliasing occurs since the condition ${f}_{e}>M·2B$ is satisfied.

Power : We consider first the case of a $T$ -periodic signal $y$ . Then, $z\left(t\right)=y\left(Mt\right)$ has period $T/M$ . Substituting $s=tM$ with $ds=Mdt$ we get

$\begin{array}{ccc}\hfill {P}_{z}& =& \frac{1}{T/M}{\int }_{0}^{T/M}{|z\left(t\right)|}^{2}dt=\frac{M}{T}{\int }_{0}^{T/M}{|y\left(Mt\right)|}^{2}dt=\frac{1}{T}{\int }_{0}^{T}{|y\left(s\right)|}^{2}ds={P}_{y}\hfill \end{array}$

From the simple properties we know ${Z}_{k}{{|}_{f=\frac{k}{T/M}}={Y}_{k}|}_{f=\frac{k}{T}}$ (same complex amplitudes There is no contradiction between ${Z}_{k}={Y}_{k}$ and $Z\left(f\right)=\frac{1}{M}Y\left(\frac{f}{M}\right)$ . Both express that $z\left(t\right)=y\left(Mt\right)$ , both allow to conclude that power does not change under downsampling, and both imply thatthe DFT of $z$ is $M$ times smaller and corresponds to frequencies which are $M$ times further apart than those of $y$ . Indeed, using [link] we get

${\stackrel{^}{z}}_{k}{|}_{f=\frac{k}{T/M}}=\frac{K}{M}{Z}_{k}=\frac{1}{M}K{Y}_{k}=\frac{1}{M}{\stackrel{^}{y}}_{k}{|}_{f=\frac{k}{T}}$
and using [link] with $L$ equal to the correct period and number of samples ( $T/M$ and $K/M$ for $z\left(t\right)$ respectively $T$ and $K$ for $y\left(t\right)$ ) we get again
${\stackrel{^}{z}}_{k}{|}_{f=\frac{k}{T/M}}=\frac{K/M}{T/M}Z\left(\frac{k}{T/M}\right)=\frac{K}{T}\frac{1}{M}Y\left(\frac{1}{M}\frac{k}{T/M}\right)=\frac{1}{M}\frac{K}{T}Y\left(\frac{k}{T}\right)=\frac{1}{M}{\stackrel{^}{y}}_{k}{|}_{f=\frac{k}{T}}$
, but belonging to different frequencies). Using Parseval we see again that the power does not change.See properties of Fourier Series.

Note that the energy of a finite energy signal would change under downsampling: computing in time

$\begin{array}{ccc}\hfill {||z||}^{2}& =& {\int }_{-\infty }^{\infty }{|z\left(t\right)|}^{2}dt={\int }_{-\infty }^{\infty }{|y\left(Mt\right)|}^{2}dt=\frac{1}{M}{\int }_{-\infty }^{\infty }{|y\left(s\right)|}^{2}ds=\frac{1}{M}{||y||}^{2}\hfill \end{array}$

A simliar computation can be done in frequency. See Comment 6 .

Comment 6 Computing in frequency with $f=Mg$ and $df=Mdg$ :

$\begin{array}{ccc}\hfill {||Z||}^{2}& =& {\int }_{-\infty }^{\infty }{|Z\left(f\right)|}^{2}df={\int }_{-\infty }^{\infty }{|\frac{1}{M}Y\left(\frac{f}{M}\right)|}^{2}df=\frac{1}{M}{\int }_{-\infty }^{\infty }{|Y\left(g\right)|}^{2}dg=\frac{1}{M}{||Y||}^{2}\hfill \end{array}$

For a discrete signal we are naturally in the periodic case. From the above we should expect that power stays approximatively the same under downsampling provided that ${f}_{e}>M·2B$ . Also, we noted earlier that power should not dependon the sampling rate, as long as the samples faithfully represent the signal, and at least approximatively.

For the simple signal $b$ , the filter from [link] , we may verify this explicitly. Denote the downsampled filter by ${b}^{\text{'}}$ (see [link] for an illustration with $M=2$ ). Since no aliasing occurs during downsampling, the pass-band is now $M$ times longer, meaning that $M$ times more of the ${\stackrel{ˆ}{{b}^{\text{'}}}}_{k}$ are different from zero (this makes the power increase by $M$ ). Further, their power spectral values $|{\stackrel{ˆ}{{b}^{\text{'}}}}_{k}{|}^{2}$ are by ${M}^{2}$ smaller (this makes the power decrease by ${M}^{2}$ ). Finally, the sample length is now $M$ times shorter (this increases the power my $M$ ; recall [link] ).

All in all, power is not changed, at least approximatively.One finds ${P}_{{b}^{\text{'}}}=0.{003}^{\text{'}}829$ which has to be compared to the power ${P}_{b}=0.{003}^{\text{'}}891$ of the original filter.

To obtain a low-pass filter one would have to normalize ${b}^{\text{'}}$ to ${b}^{\text{'}\text{'}}=M·{b}^{\text{'}}$ .

## Decimation

Let us now drop the assumption ${f}_{e}>M·2B$ .

To resample a signal $x\left(t\right)$ at an $M$ times lower rate, a first attempt would be to discard all but every $M$ -th sample: ${z}_{k}={x}_{kM}=x\left(kM\tau \right)$ . This step is called downsampling as we have seen above. However, to avoid aliasing effects caused by downsampling below Nyquist ratea low-pass filtering at cutoff ${B}^{*}={f}_{e}/\left(2M\right)$ is required before downsamling. The filter used is called anti-aliasing filter . The new utilized bandwidth will be only ${B}^{*}$ .

The procedure of first applying an anti-aliasing filter and then downsampling is called decimation .

Agreeably, the filtering before downsampling destroys information about $x\left(t\right)$ . However, this loss occurs in a controllable manner: It removes high-frequencyinformation. For an audio signal, we loose the high pitch sound. For an image we loose sharpness of edges. This should be compared to an uncontrolled lossof quality when no anti-aliasing filter is applied (see [link] ).

In summary, Decimation by $M$ means to resample at $M$ times lower rate ${f}_{d}={f}_{e}/M$ and consists of two steps:

1. low-pass filtering the samples at cutoff frequency $\frac{1}{2M}$
$\begin{array}{ccc}\hfill \left({x}_{0},{x}_{1},{x}_{2},...\right)\to & \cap \frac{1}{2M}& \to \left({y}_{0},{y}_{1},{y}_{2},...\right)\hfill \end{array}$
2. Down-sampling (in French: decimation)
$\begin{array}{ccc}\hfill \left({y}_{0},{y}_{1},{y}_{2},...\right)\to & ↓M& \to \left({y}_{0},{y}_{M},{y}_{2M},...\right)=\left({z}_{0},{z}_{1},{z}_{2},...\right)\hfill \end{array}$

Low-pass filtering will reduce power, as high frequencies are attenuated. Down-sampling will leave the new power roughly the same.

The matlab commands are decimate and downsample .

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