. Fix-rate Signal Processing Fix rate filters - same number of input and output samples Filter x(n) 8 samples y(n) 8 samples y(n) = h(n) * x(n) Figure 1

. Multirate Signal Processing Multirate filters - different numbers of input and output samples Filter x(n) 8 samples y(n) 4 samples Figure 2

. Multirate Signal Processing Filter Decimation - M>NM samplesN samples Filter Interpolation - M<NM samplesN samples Figure 3

Decimator x(n)x(n) M y(n)y(n) Basic application - reduce bit-rate by discarding samples Consequence - Distortion and Aliasing error Figure 4

Interpolator x(n)x(n) M y(n)y(n) Basic application - Insert samples between missing gaps Consequence - Restore the number of samples before decimation Figure 4

Decimation x(n)x(n) M y(n)y(n) e.g. M = 2x(n)x(n)x’(n) y(n)y(n) Figure 5

Decimation x(n)x(n)x’(n) x’(n) = x(n) n = 0, +M, +2M, +3M, +4M,... 0 otherwise (1)

Decimation (1) i(n) is a periodic impulse train that can be expressed as (2) i(n)i(n) 0M2M-2M-M Figure 6

Decimation (1) (2) (3)

Decimation x(n)x(n) M y(n)y(n) e.g. M = 2x(n)x(n)x’(n) y(n)y(n)

Decimation x(n)x(n)x’(n)y(n)y(n) y(n) = x’(Mn)(4) According to equation (3) hence (5)

Decimation Distortion due to Decimation can be seen in the frequency domain Consider the z transform of x(n) and x’(n) (6) According to equation (3)

Decimation According to equation (3) (7)

Decimation According to equation (4),y(n) = x’(Mn) (8)where p = Mn (9)

Key equations of Decimation x(n)x(n)x’(n)y(n)y(n) y(n) = x’(Mn)

Key equations of Decimation x(n)x(n)x’(n)y(n)y(n) Convert to DFT with z = e j  z transform

Spectral Changes in Decimation x(n)x(n)x’(n)y(n)y(n) 0  /4  /4  /2  /4  /4  /2  /4  /4  /2  /4  Figure 7a

Spectral Changes in Decimation x(n)x(n)x’(n)y(n)y(n) 0  /4  /4  /2  /4  /4  /2  /4  /4  /2  /4  0  /4  /4  /2  /4  /4  /2  /4  /4  /2  /4  images M=4 Figure 7a Figure 7b

Spectral Changes in Decimation x(n)x(n)x’(n)y(n)y(n) 0  /4  /4  /2  /4  /4  /2  /4  /4  /2  /4  0  /4  /4  /2  /4  /4  /2  /4  /4  /2  /4  0  /4  /4  /2  /4  /4  /2  /4  /4  /2  /4  M=4 Figure 7a Figure 7b Figure 7c

Spectral Changes in Decimation x(n)x(n)x’(n)y(n)y(n) 0  /4  /4  /2  /4  /4  /2  /4  /4  /2  /4  0  /4  /4  /2  /4  /4  /2  /4  /4  /2  /4  0  /4  /4  /2  /4  /4  /2  /4  /4  /2  /4  M=4 Figure 7a Figure 7b Figure 7c

Spectral Changes in Decimation x(n)x(n)x’(n) 1. Retaining one out of M samples in x(n) generates M replicated images of the original spectrum. 2. The spacing of images is 2  /M 3. Decimation by a factor of M stretches the width of the spectrum by M times x’(n)y(n)y(n)

Interpolation x(n)x(n) M y(n)y(n) e.g. M = 2x’(n) y(n)y(n) e.g. M = 3x’(n) y(n)y(n) Figure 8

Interpolation x(n)x(n) M y(n)y(n) (10) y(n) = x(n/M) n = 0, +M, +2M, +3M,... 0 otherwise (11) (12)

Spectral Changes in Interpolation x(n)x(n) M y(n)y(n) (13)

Spectral Changes in Interpolation 0  /4  /4  /2  /4  /4  /2  /4  /4  /2  /4  Figure 9a x(n)x(n) M y(n)y(n) 0  /4  /4  /2  /4  /4  /2  /4  /4  /2  /4  M=4 Figure 9b

Spectral Changes in Interpolation 0  /4  /4  /2  /4  /4  /2  /4  /4  /2  /4  Figure 9a x(n)x(n) M y(n)y(n) 0  /4  /4  /2  /4  /4  /2  /4  /4  /2  /4  M=4 Figure 9b

Spectral Changes in Interpolation x(n)x(n)y(n)y(n) 2. M images separating from each other by a spacing of 2  /M are generated 1. Interpolation by a factor of M compresses the width of the spectrum by M times

Decimation & Interpolation (M=4) M M Input Sequence Output Sequence Figure

Decimation & Interpolation (M=4) 0  /4  /4  /2  /4  /4  /2  /4  /4  /2  /4  M=4 M M Bandwidth -  /8 Figure 11

Decimation & Interpolation It seems that the bitrate can be reduced simply by decimation

Decimation & Interpolation But something is wrong,what’s the problem?