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# m29 - Downsampling, Subsampling and Decimation

Module by: C. Sidney Burrus. E-mail the author

Summary: Down sampling, subsampling, and decimation of discrete time signals can be studied using Fourier transforms.

## 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 by removing 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 t h M t h sample of the original sequence. This process is best seen as done in two steps. The first is to mask off the terms to be removed by setting M 1 M 1 terms to zero in each length- M M block (multiply x ( n ) x ( n ) by ⨿ M ( n ) ⨿ M ( n ) ), then that sequence is compressed or shortened by removing the M 1 M 1 zeroed terms.

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

n = 0 M 1 e j 2 π n l / M = { M if  n  is an integer multiple of  M 0 otherwise n = 0 M 1 e j 2 π n l / M = { M if  n  is an integer multiple of  M 0 otherwise
(1)

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

C d ( k ) = m = 0 L 1 x ( M m ) W L m k C d ( k ) = m = 0 L 1 x ( M m ) W L m k
(2)
where N = L M N = L M and k = 0 , 1 , , L 1 k = 0 , 1 , , L 1 . This done by masking x ( n ) x ( n ) . C d ( k ) = n = 0 N 1 x ( n ) ⨿ M ( n ) W N n k C d ( k ) = n = 0 N 1 x ( n ) ⨿ M ( n ) W N n k = n = 0 N 1 x ( n ) [ 1 M l = 0 M 1 e j 2 π n l / M ] e j 2 π n k / N = n = 0 N 1 x ( n ) [ 1 M l = 0 M 1 e j 2 π n l / M ] e j 2 π n k / N = 1 M l = 0 M 1 n = 0 N 1 x ( n ) e j 2 π ( k + L l ) n / N = 1 M l = 0 M 1 n = 0 N 1 x ( n ) e j 2 π ( k + L l ) n / N
= 1 M l = 0 M 1 C ( k + L l ) = 1 M l = 0 M 1 C ( k + L l )
(3)
The compression or removal of the masked terms is achieved in the frequency domain by using k = 0 , 1 , , L 1 k = 0 , 1 , , L 1 . This is a length- L L DFT of the samples of x ( n ) x ( n ) which is the sum of M M shifted versions of C ( k ) C ( k ) , the DFT of x ( n ) x ( n ) . Unless C ( k ) C ( k ) is sufficiently bandlimited, this causes aliasing and x ( n ) x ( n ) 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 ( n ) = 1 N k = 0 N 1 C ( k ) W N n k . x ( n ) = 1 N k = 0 N 1 C ( k ) W N n k .
(4)
The sampled signal gives y ( n ) = x ( M n ) = 1 N k = 0 N 1 C ( k ) W N M n k . y ( n ) = x ( M n ) = 1 N k = 0 N 1 C ( k ) W N M n k . for n = 0 , 1 , , L 1 n = 0 , 1 , , L 1 . This sum can be broken down by y ( n ) = 1 N k = 0 L 1 l = 0 M 1 C ( k + L l ) W N M n ( k + L l ) . y ( n ) = 1 N k = 0 L 1 l = 0 M 1 C ( k + L l ) W N M n ( k + L l ) . = 1 N k = 0 L 1 [ l = 0 M 1 C ( k + L l ) ] W N M n k . = 1 N k = 0 L 1 [ l = 0 M 1 C ( k + L l ) ] W N M n k . >From the term in the brackets, we have C s ( k ) = l = 0 M 1 C ( k + L l ) C s ( k ) = l = 0 M 1 C ( k + L l ) as was obtained in (Equation 3).

Now consider still another derivation using shah functions. Let

x s ( n ) = ⨿ M ( n ) x ( n ) x s ( n ) = ⨿ M ( n ) x ( n )
(5)
>From the convolution property of the DFT we have C s ( k ) = L ⨿ L ( k ) * C ( k ) C s ( k ) = L ⨿ L ( k ) * C ( k ) therefore
C s ( k ) = l = 0 M 1 C ( k + L l ) C s ( k ) = l = 0 M 1 C ( k + L l )
(6)
which again is the same as in (Equation 3).

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

C s ( ω ) = m = x ( M m ) e j ω M m C s ( ω ) = m = x ( M m ) e j ω M m
(7)
= n x ( n ) ⨿ M ( n ) e j ω n = n x ( n ) ⨿ M ( n ) e j ω n = n x ( n ) [ 1 M l = 0 M 1 e j 2 π n l / M ] e j ω n = n x ( n ) [ 1 M l = 0 M 1 e j 2 π n l / M ] e j ω n = 1 M l = 0 M 1 n x ( n ) e j ( ω 2 π l / M ) n = 1 M l = 0 M 1 n x ( n ) e j ( ω 2 π l / M ) n
= 1 M l = 0 M 1 C ( ω 2 π l / M ) = 1 M l = 0 M 1 C ( ω 2 π l / M )
(8)
which shows the aliasing caused by the masking (sampling without compression). We now give the effects of compressing x s ( n ) x s ( n ) which is a simple scaling of ω ω . This is the inverse of the stretching results in ((Reference)).
C s ( ω ) = 1 M l = 0 M 1 C ( ω / M 2 π l / M ) . C s ( ω ) = 1 M l = 0 M 1 C ( ω / M 2 π l / M ) .
(9)

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

x ( n ) = 1 2 π π π C ( ω ) e j ω n ω x ( n ) = 1 2 π π π C ( ω ) e j ω n ω
(10)
which for the down--sampled signal becomes x ( M n ) = 1 2 π π π C ( ω ) e j ω M n ω x ( M n ) = 1 2 π π π C ( ω ) e j ω M n ω The integral broken into the sum of M M sections using a change of variables of ω = ( ω 1 + 2 π l ) / M ω = ( ω 1 + 2 π l ) / M giving x ( M n ) = 1 2 π l = 0 M 1 π π C ( ω 1 / M + 2 π l / M ) e j ( ω 1 / M + 2 π l / M ) M n ω 1 x ( M n ) = 1 2 π l = 0 M 1 π π C ( ω 1 / M + 2 π l / M ) e j ( ω 1 / M + 2 π l / M ) M n ω 1 which shows the transform to be the same as given in ((Reference)).

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

x s ( n ) = ⨿ M ( n ) x ( n ) x s ( n ) = ⨿ M ( n ) x ( n )
(11)
which has as a DTFT C s ( ω ) = ( 2 π M ) ⨿ 2 π / M ( ω ) * C ( ω ) C s ( ω ) = ( 2 π M ) ⨿ 2 π / M ( ω ) * C ( ω )
= 2 π M l = 0 M 1 C ( ω + 2 π l / M ) = 2 π M l = 0 M 1 C ( ω + 2 π l / M )
(12)
which after compressing becomes C s = 2 π M l = 0 M 1 C ( ω / M + 2 π l / M ) C s = 2 π M l = 0 M 1 C ( ω / M + 2 π l / M ) which is same as ((Reference)).

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

X ( z ) = n = x ( n ) z n X ( z ) = n = x ( n ) z n
(13)
Applying this to the sampled signal gives
X s ( z ) = n x ( M n ) z M n = n x ( n ) ⨿ M ( n ) z n X s ( z ) = n x ( M n ) z M n = n x ( n ) ⨿ M ( n ) z n
(14)
= n x ( n ) l = 0 M 1 e j 2 π n l / M z n = n x ( n ) l = 0 M 1 e j 2 π n l / M z n = l = 0 M 1 n x ( n ) { e j 2 π l / M z } n = l = 0 M 1 n x ( n ) { e j 2 π l / M z } n = l = 0 M 1 X ( e j 2 π l / M z ) = l = 0 M 1 X ( e j 2 π l / M z ) which becomes after compressing
= l = 0 M 1 X ( e j 2 π l / M z 1 / M ) . = l = 0 M 1 X ( e j 2 π l / M z 1 / M ) .
(15)
This concludes our investigations of the effects of down--sampling a discrete--time signal and we discover much the same aliasing properties as in sampling a continuous--time signal. We also saw some of the mathematical steps used in the development.

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