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 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 terms to zero in each length- M block (multiply x ( n ) by ⨿ M ( n ) ), 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 ( 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 We now calculate the DFT of the down--sampled signal. C d ( k ) = ∑ m = 0 L − 1 x ( M m ) W L m k where N = L M and k = 0 , 1 , ⋯ , L − 1 . This done by masking x ( n ) . 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 = 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 ) The compression or removal of the masked terms is achieved in the frequency domain by using k = 0 , 1 , ⋯ , L − 1 . This is a length- L DFT of the samples of x ( n ) which is the sum of M shifted versions of C ( k ) , the DFT of x ( n ) . Unless C ( k ) is sufficiently bandlimited, this causes aliasing and 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 . The sampled signal gives y ( n ) = x ( M n ) = 1 N ∑ k = 0 N − 1 C ( k ) W N − M n k . for 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 ) . = 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 ) as was obtained in (). Now consider still another derivation using shah functions. Let x s ( n ) = ⨿ M ( n ) x ( n ) >From the convolution property of the DFT we have C s ( k ) = L ⨿ L ( k ) * C ( k ) therefore C s ( k ) = ∑ l = 0 M − 1 C ( k + L l ) which again is the same as in (). 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 = ∑ 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 = 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 ) which shows the aliasing caused by the masking (sampling without compression). We now give the effects of compressing x s ( n ) which is a simple scaling of ω . This is the inverse of the stretching results in (). C s ( ω ) = 1 M ∑ l = 0 M − 1 C ( ω / M − 2 π 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 ( n ) = 1 2 π ∫ − π π C ( ω ) e j ω n ⅆ ω which for the down--sampled signal becomes x ( M n ) = 1 2 π ∫ − π π C ( ω ) e j ω M n ⅆ ω The integral broken into the sum of M sections using a change of variables of 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 ⅆ ω 1 which shows the transform to be the same as given in (). Still another approach which uses the shah function can be given by x s ( n ) = ⨿ M ( n ) x ( n ) which has as a DTFT 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 ) which after compressing becomes C s = 2 π M ∑ l = 0 M − 1 C ( ω / M + 2 π l / M ) which is same as (). Now we consider the effects of down--sampling on the z-transform of a signal. X ( z ) = ∑ n = − ∞ ∞ x ( n ) z − n Applying this to the sampled signal gives X s ( z ) = ∑ n x ( M n ) z − M n = ∑ n x ( n ) ⨿ M ( n ) 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 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 ) . 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.