0.3 Sampling, up--sampling, down--sampling, and multi--rate  (Page 5/5)

 Page 5 / 5
$FT\left\{{⨿}_{T}\left(t\right)\right\}={\int }_{-\infty }^{\infty }\sum _{n}\delta \left(t-nT\right)\phantom{\rule{0.166667em}{0ex}}{e}^{-j\omega t}\phantom{\rule{0.166667em}{0ex}}dt=\sum _{n}\int \delta \left(t-nT\right)\phantom{\rule{0.166667em}{0ex}}{e}^{-j\omega t}\phantom{\rule{0.166667em}{0ex}}dt$
$=\sum _{n}{e}^{-j\omega nT}=\left\{\begin{array}{cc}\infty \hfill & \omega =2\pi /T\hfill \\ 0\hfill & \text{otherwise}\hfill \end{array}$
$=\frac{\mathrm{2\pi }}{T}{⨿}_{\mathrm{2\pi /T}}\left(\omega \right)$

The multiplicative constant is found from knowing the result for a single delta function.

These “shah functions" will be useful in sampling signals in both the continuous time and discrete time cases.

Up–sampling, signal stretching, and interpolation

In several situations we would like to increase the data rate of a signal or, to increase its length if it has finite length. This may be part of amulti rate system or part of an interpolation process. Consider the process of inserting $M-1$ zeros between each sample of a discrete time signal.

$y\left(n\right)=\left\{\begin{array}{cc}x\left(n/M\right)\hfill & M=0\left(\text{or}\phantom{\rule{5pt}{0ex}}n\phantom{\rule{5pt}{0ex}}=\mathrm{kM}\right)\hfill \\ 0\hfill & \text{otherwise}\hfill \end{array}$

For the finite length sequence case we calculate the DFT of the stretched or up–sampled sequence by

${C}_{s}\left(k\right)=\sum _{n=0}^{MN-1}y\left(n\right)\phantom{\rule{0.166667em}{0ex}}{W}_{MN}^{nk}$
${C}_{s}\left(k\right)=\sum _{n=0}^{MN-1}x\left(n/M\right)\phantom{\rule{0.166667em}{0ex}}{⨿}_{M}\left(n\right)\phantom{\rule{0.166667em}{0ex}}{W}_{MN}^{nk}$

where the length is now $NM$ and $k=0,1,\cdots ,NM-1$ . Changing the index variable $n=Mm$ gives:

${C}_{s}\left(k\right)=\sum _{m=0}^{N-1}x\left(m\right)\phantom{\rule{0.166667em}{0ex}}{W}_{N}^{mk}=C\left(k\right).$

which says the DFT of the stretched sequence is exactly the same as the DFT of the original sequence but over $M$ periods, each of length $N$ .

For up–sampling an infinitely long sequence, we calculate the DTFT of the modified sequence in Equation 34 from FIR Digital Filters as

${C}_{s}\left(\omega \right)=\sum _{n=-\infty }^{\infty }x\left(n/M\right)\phantom{\rule{0.166667em}{0ex}}{⨿}_{M}\left(n\right)\phantom{\rule{0.166667em}{0ex}}{e}^{-j\omega n}=\sum _{m}x\left(m\right)\phantom{\rule{0.166667em}{0ex}}{e}^{-j\omega Mm}$
$=C\left(M\omega \right)$

where $C\left(\omega \right)$ is the DTFT of $x\left(n\right)$ . Here again the transforms of the up–sampled signal is the same as the original signal except over $M$ periods. This shows up here as ${C}_{s}\left(\omega \right)$ being a compressed version of $M$ periods of $C\left(\omega \right)$ .

The z-transform of an up–sampled sequence is simply derived by:

$Y\left(z\right)=\sum _{n=-\infty }^{\infty }y\left(n\right)\phantom{\rule{0.166667em}{0ex}}{z}^{-n}=\sum _{n}x\left(n/M\right)\phantom{\rule{0.166667em}{0ex}}{⨿}_{M}\left(n\right)\phantom{\rule{0.166667em}{0ex}}{z}^{-n}=\sum _{m}x\left(m\right)\phantom{\rule{0.166667em}{0ex}}{z}^{-Mm}$
$=X\left({z}^{M}\right)$

which is consistent with a complex version of the DTFT in [link] .

Notice that in all of these cases, there is no loss of information or invertibility. In other words, there is no aliasing.

Down–sampling, subsampling, or decimation

In this section we consider the sampling problem where, unless there is sufficient redundancy, there will be a loss of information caused byremoving data in the time domain and aliasing in the frequency domain.

The sampling process or the down sampling process creates a new shorter or compressed signal by keeping every ${M}^{th}$ sample of the original sequence. This process is best seen as done in two steps. The first isto mask off the terms to be removed by setting $M-1$ terms to zero in each length- $M$ block (multiply $x\left(n\right)$ by ${⨿}_{M}\left(n\right)$ ), then that sequence is compressed or shortened by removing the $M-1$ zeroed terms.

We will now calculate the length $L=N/M$ DFT of a sequence that was obtained by sampling every $M$ terms of an original length- $N$ sequence $x\left(n\right)$ . We will use the orthogonal properties of the basis vectors of the DFT which says:

$\sum _{n=0}^{M-1}{e}^{-j2\pi nl/M}=\left\{\begin{array}{cc}M\hfill & \text{if n is an integer multiple of}M\hfill \\ 0\hfill & \text{otherwise.}\hfill \end{array}$

We now calculate the DFT of the down-sampled signal.

${C}_{d}\left(k\right)=\sum _{m=0}^{L-1}x\left(Mm\right){W}_{L}^{mk}$

where $N=LM$ and $k=\mathrm{0,1,...,L-1}$ . This is done by masking $x\left(n\right)$ .

${C}_{d}\left(k\right)=\sum _{n=0}^{N-1}x\left(n\right){x}_{M}\left(n\right){W}_{L}^{nk}$
$=\sum _{n=0}^{N-1}x\left(n\right)\left[\frac{1}{M}\sum _{l=0}^{M-1}{e}^{-j2\pi nl/M}\right]{e}^{-j2\pi nk/N}$
$=\frac{1}{M}\sum _{l=0}^{M-1}\sum _{n=0}^{N-1}x\left(n\right){e}^{j2\pi \left(k+Ll\right)n/N}$
$=\frac{1}{M}\sum _{l=0}^{M-1}C\left(k+Ll\right)$

The compression or removal of the masked terms is achieved in the frequency domain by using $k=\mathrm{0,1,...,L-1}$ This is a length-L DFT of the samples of $x\left(n\right)$ . Unless $C\left(k\right)$ is sufficiently bandlimited, this causes aliasing and $x\left(n\right)$ is not unrecoverable.

It is instructive to consider an alternative derivation of the above result. In this case we use the IDFT given by

$x\left(n\right)=\frac{1}{N}\sum _{k=0}^{N-1}C\left(k\right)\phantom{\rule{0.166667em}{0ex}}{W}_{N}^{-nk}.$

The sampled signal gives

$y\left(n\right)=x\left(Mn\right)=\frac{1}{N}\sum _{k=0}^{N-1}C\left(k\right)\phantom{\rule{0.166667em}{0ex}}{W}_{N}^{-Mnk}.$

for $n=0,1,\cdots ,L-1$ . This sum can be broken down by

$y\left(n\right)=\frac{1}{N}\sum _{k=0}^{L-1}\sum _{l=0}^{M-1}C\left(k+Ll\right)\phantom{\rule{0.166667em}{0ex}}{W}_{N}^{-Mn\left(k+Ll\right)}.$
$=\frac{1}{N}\sum _{k=0}^{L-1}\left[\sum _{l=0}^{M-1},C,\left(k+Ll\right)\right]{W}_{N}^{-Mnk}.$

From the term in the brackets, we have

${C}_{s}\left(k\right)=\sum _{l=0}^{M-1}C\left(k+Ll\right)$

as was obtained in [link] .

Now consider still another derivation using shah functions. Let

${x}_{s}\left(n\right)={⨿}_{M}\left(n\right)\phantom{\rule{0.166667em}{0ex}}x\left(n\right)$

From the convolution property of the DFT we have

${C}_{s}\left(k\right)=L\phantom{\rule{0.166667em}{0ex}}{⨿}_{L}\left(k\right)*C\left(k\right)$

therefore

${C}_{s}\left(k\right)=\sum _{l=0}^{M-1}C\left(k+Ll\right)$

which again is the same as in [link] .

We now turn to the down sampling of an infinitely long signal which will require use of the DTFT of the signals.

${C}_{s}\left(\omega \right)=\sum _{m=-\infty }^{\infty }x\left(Mm\right)\phantom{\rule{0.166667em}{0ex}}{e}^{-j\omega Mm}$
$=\sum _{n}x\left(n\right)\phantom{\rule{0.166667em}{0ex}}{⨿}_{M}\left(n\right)\phantom{\rule{0.166667em}{0ex}}{e}^{-j\omega n}$
$=\sum _{n}x\left(n\right)\left[\frac{1}{M},\sum _{l=0}^{M-1},{e}^{-j2\pi nl/M}\right]{e}^{-j\omega n}$
$=\frac{1}{M}\sum _{l=0}^{M-1}\sum _{n}x\left(n\right)\phantom{\rule{0.166667em}{0ex}}{e}^{-j\left(\omega -2\pi l/M\right)n}$
$=\frac{1}{M}\sum _{l=0}^{M-1}C\left(\omega -2\pi l/M\right)$

which shows the aliasing caused by the masking (sampling without compression). We now give the effects of compressing ${x}_{s}\left(n\right)$ which is a simple scaling of $\omega$ . This is the inverse of the stretching results in [link] .

${C}_{s}\left(\omega \right)=\frac{1}{M}\sum _{l=0}^{M-1}C\left(\omega /M-2\pi l/M\right).$

In order to see how the various properties of the DFT can be used, consider an alternate derivation which uses the IDTFT.

$x\left(n\right)=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }C\left(\omega \right)\phantom{\rule{0.166667em}{0ex}}{e}^{j\omega n}\phantom{\rule{0.166667em}{0ex}}d\omega$

which for the down–sampled signal becomes

$x\left(Mn\right)=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }C\left(\omega \right)\phantom{\rule{0.166667em}{0ex}}{e}^{j\omega Mn}\phantom{\rule{0.166667em}{0ex}}d\omega$

The integral broken into the sum of $M$ sections using a change of variables of $\omega =\left({\omega }_{1}+2\pi l\right)/M$ giving

$x\left(Mn\right)=\frac{1}{2\pi }\sum _{l=0}^{M-1}{\int }_{-\pi }^{\pi }C\left({\omega }_{1}/M+2\pi l/M\right)\phantom{\rule{0.166667em}{0ex}}{e}^{j\left({\omega }_{1}/M+2\pi l/M\right)Mn}\phantom{\rule{0.166667em}{0ex}}d{\omega }_{1}$

which shows the transform to be the same as given in Equation 9 from Chebyshev of Equal Ripple Error Approximation Filters .

Still another approach which uses the shah function can be given by

${x}_{s}\left(n\right)={⨿}_{M}\left(n\right)\phantom{\rule{0.166667em}{0ex}}x\left(n\right)$

which has as a DTFT

${C}_{s}\left(\omega \right)=\left(\frac{2\pi }{M}\right)\phantom{\rule{0.166667em}{0ex}}{⨿}_{2\pi /M}\left(\omega \right)*C\left(\omega \right)$
$=\frac{2\pi }{M}\sum _{l=0}^{M-1}C\left(\omega +2\pi l/M\right)$

which after compressing becomes

${C}_{s}=\frac{2\pi }{M}\sum _{l=0}^{M-1}C\left(\omega /M+2\pi l/M\right)$

Now we consider the effects of down–sampling on the z-transform of a signal.

$X\left(z\right)=\sum _{n=-\infty }^{\infty }x\left(n\right)\phantom{\rule{0.166667em}{0ex}}{z}^{-n}$

Applying this to the sampled signal gives

${X}_{s}\left(z\right)=\sum _{n}x\left(Mn\right)\phantom{\rule{0.166667em}{0ex}}{z}^{-Mn}=\sum _{n}x\left(n\right)\phantom{\rule{0.166667em}{0ex}}{⨿}_{M}\left(n\right)\phantom{\rule{0.166667em}{0ex}}{z}^{-n}$
$=\sum _{n}x\left(n\right)\phantom{\rule{0.166667em}{0ex}}\sum _{l=0}^{M-1}{e}^{j2\pi nl/M}\phantom{\rule{0.166667em}{0ex}}{z}^{-n}$
$=\sum _{l=0}^{M-1}\sum _{n}x\left(n\right){\left\{{e}^{j2\pi l/M},\phantom{\rule{0.166667em}{0ex}},z\right\}}^{-n}$
$=\sum _{l=0}^{M-1}X\left({e}^{-j2\pi l/M}\phantom{\rule{0.166667em}{0ex}}z\right)$

which becomes after compressing

$=\sum _{l=0}^{M-1}X\left({e}^{-j2\pi l/M}\phantom{\rule{0.166667em}{0ex}}{z}^{1/M}\right).$

This concludes our investigations of the effects of down–sampling a discrete–time signal and we discover much the same aliasing properties asin sampling a continuous–time signal. We also saw some of the mathematical steps used in the development.

More later

We will later develop relations of sampling to multirate systems, periodically time varying systems, and block processing. This should be avery effective formulation for teaching as well as research on these topics.

Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
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
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
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
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
hi
Loga
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
Berger describes sociologists as concerned with
what is hormones?
Wellington
Got questions? Join the online conversation and get instant answers!