$$FT\left\{{\u2a3f}_{T}\left(t\right)\right\}={\int}_{-\infty}^{\infty}\sum _{n}\delta (t-nT)\phantom{\rule{0.166667em}{0ex}}{e}^{-j\omega t}\phantom{\rule{0.166667em}{0ex}}dt=\sum _{n}\int \delta (t-nT)\phantom{\rule{0.166667em}{0ex}}{e}^{-j\omega t}\phantom{\rule{0.166667em}{0ex}}dt$$
$$=\sum _{n}{e}^{-j\omega nT}=\{\begin{array}{cc}\infty \hfill & \omega =2\pi /T\hfill \\ 0\hfill & \text{otherwise}\hfill \end{array}$$
$=\frac{\mathrm{2\pi}}{T}{\u2a3f}_{\mathrm{2\pi /T}}\left(\omega \right)$
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.
$y\left(n\right)=\{\begin{array}{cc}x(n/M)\hfill & <n>M=0(\text{or}\phantom{\rule{5pt}{0ex}}n\phantom{\rule{5pt}{0ex}}=\mathrm{kM})\hfill \\ 0\hfill & \text{otherwise}\hfill \end{array}$
For the finite length sequence case we calculate the DFT of the stretched
or up–sampled sequence by
$${C}_{s}\left(k\right)=\sum _{n=0}^{MN-1}y\left(n\right)\phantom{\rule{0.166667em}{0ex}}{W}_{MN}^{nk}$$
$${C}_{s}\left(k\right)=\sum _{n=0}^{MN-1}x(n/M)\phantom{\rule{0.166667em}{0ex}}{\u2a3f}_{M}\left(n\right)\phantom{\rule{0.166667em}{0ex}}{W}_{MN}^{nk}$$
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
$${C}_{s}\left(\omega \right)=\sum _{n=-\infty}^{\infty}x(n/M)\phantom{\rule{0.166667em}{0ex}}{\u2a3f}_{M}\left(n\right)\phantom{\rule{0.166667em}{0ex}}{e}^{-j\omega n}=\sum _{m}x\left(m\right)\phantom{\rule{0.166667em}{0ex}}{e}^{-j\omega Mm}$$
$$=C\left(M\omega \right)$$
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:
$$Y\left(z\right)=\sum _{n=-\infty}^{\infty}y\left(n\right)\phantom{\rule{0.166667em}{0ex}}{z}^{-n}=\sum _{n}x(n/M)\phantom{\rule{0.166667em}{0ex}}{\u2a3f}_{M}\left(n\right)\phantom{\rule{0.166667em}{0ex}}{z}^{-n}=\sum _{m}x\left(m\right)\phantom{\rule{0.166667em}{0ex}}{z}^{-Mm}$$
$$=X\left({z}^{M}\right)$$
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
${\u2a3f}_{M}\left(n\right)$ ), 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:
$$\sum _{n=0}^{M-1}{e}^{-j2\pi nl/M}=\{\begin{array}{cc}M\hfill & \text{if n is an integer multiple of}M\hfill \\ 0\hfill & \text{otherwise.}\hfill \end{array}$$
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
$x\left(n\right)$ .
${C}_{d}\left(k\right)=\sum _{n=0}^{N-1}x\left(n\right){x}_{M}\left(n\right){W}_{L}^{nk}$
$=\sum _{n=0}^{N-1}x\left(n\right)\left[\frac{1}{M}\sum _{l=0}^{M-1}{e}^{-j2\pi nl/M}\right]{e}^{-j2\pi nk/N}$
$=\frac{1}{M}\sum _{l=0}^{M-1}\sum _{n=0}^{N-1}x\left(n\right){e}^{j2\pi (k+Ll)n/N}$
$=\frac{1}{M}\sum _{l=0}^{M-1}C(k+Ll)$
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
$$y\left(n\right)=\frac{1}{N}\sum _{k=0}^{L-1}\sum _{l=0}^{M-1}C(k+Ll)\phantom{\rule{0.166667em}{0ex}}{W}_{N}^{-Mn(k+Ll)}.$$
$$=\frac{1}{N}\sum _{k=0}^{L-1}\left[\sum _{l=0}^{M-1},C,(k+Ll)\right]{W}_{N}^{-Mnk}.$$
From the term in the brackets, we have
$${C}_{s}\left(k\right)=\sum _{l=0}^{M-1}C(k+Ll)$$
as was obtained in
[link] .
Now consider still another derivation using shah functions. Let
$${x}_{s}\left(n\right)={\u2a3f}_{M}\left(n\right)\phantom{\rule{0.166667em}{0ex}}x\left(n\right)$$
From the convolution property of the DFT we have
$${C}_{s}\left(k\right)=L\phantom{\rule{0.166667em}{0ex}}{\u2a3f}_{L}\left(k\right)*C\left(k\right)$$
therefore
$${C}_{s}\left(k\right)=\sum _{l=0}^{M-1}C(k+Ll)$$
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.
$${C}_{s}\left(\omega \right)=\sum _{m=-\infty}^{\infty}x\left(Mm\right)\phantom{\rule{0.166667em}{0ex}}{e}^{-j\omega Mm}$$
$$=\sum _{n}x\left(n\right)\phantom{\rule{0.166667em}{0ex}}{\u2a3f}_{M}\left(n\right)\phantom{\rule{0.166667em}{0ex}}{e}^{-j\omega n}$$
$$=\sum _{n}x\left(n\right)\left[\frac{1}{M},\sum _{l=0}^{M-1},{e}^{-j2\pi nl/M}\right]{e}^{-j\omega n}$$
$$=\frac{1}{M}\sum _{l=0}^{M-1}\sum _{n}x\left(n\right)\phantom{\rule{0.166667em}{0ex}}{e}^{-j(\omega -2\pi l/M)n}$$
$$=\frac{1}{M}\sum _{l=0}^{M-1}C(\omega -2\pi l/M)$$
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] .
$${C}_{s}\left(\omega \right)=\frac{1}{M}\sum _{l=0}^{M-1}C(\omega /M-2\pi l/M).$$
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
$\omega =({\omega}_{1}+2\pi l)/M$ giving
$$x\left(Mn\right)=\frac{1}{2\pi}\sum _{l=0}^{M-1}{\int}_{-\pi}^{\pi}C({\omega}_{1}/M+2\pi l/M)\phantom{\rule{0.166667em}{0ex}}{e}^{j({\omega}_{1}/M+2\pi l/M)Mn}\phantom{\rule{0.166667em}{0ex}}d{\omega}_{1}$$
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
$${x}_{s}\left(n\right)={\u2a3f}_{M}\left(n\right)\phantom{\rule{0.166667em}{0ex}}x\left(n\right)$$
which has as a DTFT
$${C}_{s}\left(\omega \right)=\left(\frac{2\pi}{M}\right)\phantom{\rule{0.166667em}{0ex}}{\u2a3f}_{2\pi /M}\left(\omega \right)*C\left(\omega \right)$$
$$=\frac{2\pi}{M}\sum _{l=0}^{M-1}C(\omega +2\pi l/M)$$
which after compressing becomes
$${C}_{s}=\frac{2\pi}{M}\sum _{l=0}^{M-1}C(\omega /M+2\pi l/M)$$
which is same as
Equation 9 from Chebyshev of Equal Ripple Error Approximation Filters .
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
$${X}_{s}\left(z\right)=\sum _{n}x\left(Mn\right)\phantom{\rule{0.166667em}{0ex}}{z}^{-Mn}=\sum _{n}x\left(n\right)\phantom{\rule{0.166667em}{0ex}}{\u2a3f}_{M}\left(n\right)\phantom{\rule{0.166667em}{0ex}}{z}^{-n}$$
$$=\sum _{n}x\left(n\right)\phantom{\rule{0.166667em}{0ex}}\sum _{l=0}^{M-1}{e}^{j2\pi nl/M}\phantom{\rule{0.166667em}{0ex}}{z}^{-n}$$
$$=\sum _{l=0}^{M-1}\sum _{n}x\left(n\right){\left\{{e}^{j2\pi l/M},\phantom{\rule{0.166667em}{0ex}},z\right\}}^{-n}$$
$$=\sum _{l=0}^{M-1}X\left({e}^{-j2\pi l/M}\phantom{\rule{0.166667em}{0ex}}z\right)$$
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.