# 0.7 Decimation and downsampling

 Page 1 / 1

## 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 .

find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!