# 4.1 Fundamentals of multirate signal processing

 Page 1 / 1
Here we present some background material on multirate signal processing that is necessary to understand the filterbank processing used in sub-band coding. In particular, we describe modulation, upsampling, and downsampling in several domains: the time-domain, z-domain, and DTFT domain. In addition, we describe the aliasing phenomenon.

The presence of upsamplers and downsamplers in the diagram of Figure 2 from "Introduction and Motivation" implies that a basic knowledge of multirate signal processing is indispensible to an understanding of sub-band analysis/synthesis.This section provides the required background.

• Modulation: [link] illustrates modulation using a complex exponential of frequency ω o . In the time domain,
$\begin{array}{|c|}\hline y\left(n\right)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}x\left(n\right){e}^{j{\omega }_{o}n}\\ \hline\end{array}.$
In the z -domain,
$\begin{array}{c}\hfill Y\left(z\right)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\sum _{n}y\left(n\right){z}^{-n}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\sum _{n}\left(x,\left(n\right),{e}^{j{\omega }_{o}n}\right){z}^{-n}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\sum _{n}x\left(n\right){\left({e}^{-j{\omega }_{o}},z\right)}^{-n}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\begin{array}{|c|}\hline X\left({e}^{-j{\omega }_{o}},z\right)\\ \hline\end{array}.\end{array}$
We can evaluate the result of modulation in the frequency domain by substituting $z={e}^{j\omega }$ . This yields
$\begin{array}{c}\hfill Y\left(\omega \right)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\sum _{n}y\left(n\right){e}^{-j\omega n}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\begin{array}{|c|}\hline X\left(\omega -{\omega }_{o}\right)\\ \hline\end{array}.\end{array}$
Note that $X\left(\omega -{\omega }_{o}\right)$ represents a shift of $X\left(\omega \right)$ up by ω o radians, as in [link] .
• Upsampling: [link] illustrates upsampling by factor N . In words, upsampling means the insertion of $N\phantom{\rule{-0.166667em}{0ex}}-\phantom{\rule{-0.166667em}{0ex}}1$ zeros between every sample of the input process.Formally, upsampling can be expressed in the time domain as
$\begin{array}{c}\hfill \begin{array}{|c|}\hline y\left(n\right)=\left\{\begin{array}{cc}x\left(n/N\right)\hfill & \phantom{\rule{4pt}{0ex}}\text{when}n=mN\phantom{\rule{4pt}{0ex}}\text{for}m\in \mathbb{Z}\hfill \\ 0\hfill & \text{else.}\hfill \end{array}\right)\\ \hline\end{array}\end{array}$
In the z -domain, upsampling causes
$\begin{array}{c}\hfill Y\left(z\right)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\sum _{n}y\left(n\right){z}^{-n}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\sum _{m}x\left(m\right){z}^{-mN}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\begin{array}{|c|}\hline X\left({z}^{N}\right)\\ \hline\end{array},\end{array}$
and in the frequency domain,
$\begin{array}{c}\hfill Y\left(\omega \right)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\sum _{n}y\left(n\right){e}^{-j\omega n}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\begin{array}{|c|}\hline X\left(N,\omega \right)\\ \hline\end{array}.\end{array}$
As shown in [link] , upsampling shrinks $X\left(\omega \right)$ by a factor of N along the ω axis.
• Downsampling: [link] illustrates downsampling by factor N . In words, the process of downsampling keeps every ${N}^{th}$ sample and discards the rest.Formally, downsampling can be written as
$\begin{array}{|c|}\hline y\left(m\right)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}x\left(mN\right).\\ \hline\end{array}$
In the z domain,
$\begin{array}{c}\hfill Y\left(z\right)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\sum _{m}y\left(m\right){z}^{-m}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\sum _{m}x\left(mN\right){z}^{-m}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\sum _{n}\stackrel{˜}{x}\left(n\right){z}^{-n/N},\end{array}$
where
$\stackrel{˜}{x}\left(n\right)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\left\{\begin{array}{cc}x\left(n\right)\hfill & \text{when}\phantom{\rule{4.pt}{0ex}}n=mN\phantom{\rule{4.pt}{0ex}}\text{for}\phantom{\rule{4.pt}{0ex}}m\in \mathbb{Z}\hfill \\ 0\hfill & \text{else}.\hfill \end{array}\right)$
The neat trick
$\begin{array}{c}\hfill \frac{1}{N}\sum _{p=0}^{N-1}{e}^{j\frac{2\pi }{N}np}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\left\{\begin{array}{cc}1\hfill & \text{when}\phantom{\rule{4.pt}{0ex}}n=mN\phantom{\rule{4.pt}{0ex}}\text{for}\phantom{\rule{4.pt}{0ex}}m\in \mathbb{Z}\hfill \\ 0\hfill & \text{else}\hfill \end{array}\right)\end{array}$
(which is not difficult to prove) allows us to rewrite $\stackrel{˜}{x}\left(n\right)$ in terms of $x\left(n\right)$ :
$\begin{array}{ccc}\hfill Y\left(z\right)& =& \sum _{n}x\left(n\right)\left(\frac{1}{N},\sum _{p=0}^{N-1},{e}^{j\frac{2\pi }{N}np}\right){z}^{-n/N}\hfill \\ & =& \frac{1}{N}\sum _{p=0}^{N-1}\sum _{n}x\left(n\right){\left({e}^{-j\frac{2\pi }{N}p},{z}^{1/N}\right)}^{-n}\hfill \\ & =& \begin{array}{|c|}\hline \frac{1}{N}\sum _{p=0}^{N-1}X\left({e}^{-j\frac{2\pi }{N}p},{z}^{1/N}\right)\\ \hline\end{array}.\hfill \end{array}$
Translating to the frequency domain,
$\begin{array}{c}\hfill \begin{array}{|c|}\hline Y\left(\omega \right)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{1}{N}\sum _{p=0}^{N-1}X\left(\frac{\omega -2\pi p}{N}\right)\\ \hline\end{array}.\end{array}$
As shown in [link] , downsampling expands each $2\pi$ -periodic repetition of $X\left(\omega \right)$ by a factor of N along the ω axis. Note the spectral overlap due to downsampling, called “aliasing.”
• Downsample-Upsample Cascade: Downsampling followed by upsampling (of equal factor N ) is illustrated by [link] . This structure is useful in understanding analysis/synthesis filterbanksthat lie at the heart of sub-band coding schemes. This operation is equivalent to zeroing all but the $m{N}^{th}$ samples in the input sequence, i.e.,
$\begin{array}{|c|}\hline y\left(n\right)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\left\{\begin{array}{cc}x\left(n\right)\hfill & \text{when}n=mN\phantom{\rule{4.pt}{0ex}}\text{for}m\in \mathbb{Z}\hfill \\ 0\hfill & \text{else}.\hfill \end{array}\right)\\ \hline\end{array}$
$\begin{array}{ccc}\hfill Y\left(z\right)& =& \sum _{n}y\left(n\right){z}^{-n}\hfill \\ & =& \sum _{n}x\left(n\right)\left(\frac{1}{N},\sum _{p=0}^{N-1},{e}^{j\frac{2\pi }{N}np}\right){z}^{-n}\hfill \\ & =& \frac{1}{N}\sum _{p=0}^{N-1}\sum _{n}x\left(n\right){\left({e}^{-j\frac{2\pi }{N}p},z\right)}^{-n}\hfill \\ & =& \begin{array}{|c|}\hline \frac{1}{N}\sum _{p=0}^{N-1}X\left({e}^{-j\frac{2\pi }{N}p},z\right)\\ \hline\end{array},\hfill \end{array}$
which implies
$\begin{array}{c}\hfill \begin{array}{|c|}\hline Y\left(\omega \right)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}\frac{1}{N}\sum _{p=0}^{N-1}X\left(\omega ,-,\frac{2\pi p}{N}\right)\\ \hline\end{array}.\end{array}$
The downsampler-upsampler cascade causes the appearance of $2\pi /N$ -periodic copies of the baseband spectrum of $X\left(\omega \right)$ . As illustrated in [link] , aliasing may result.

how can chip be made from sand
are nano particles real
yeah
Joseph
Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master?
no can't
Lohitha
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
has a lot of application modern world
Kamaluddeen
yes
narayan
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
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!