Module 5 - Multirate Signal Processing

1 Module 5 - Multirate Signal ProcessingPatrick A. NaylorDigital Signal Processing of Multirate Signal processingFundamentalsdecimationinterpol ationResampling by rational fractionsMultirate identitiesPolyphase representationsMaximally decimated filter banksaliasingamplitude and phase distortionperfect reconstruction conditionsDigital Signal Processing single-rate DSP systems, all data is sampled at the same rateno change of rate within the Multirate DSP systems, sample rates are changed (or are different)within the systemMultirate can offer several advantagesreduced computational complexityreduced transmission data Signal Processing : Audio sample rate conversionrecording studios use 192 kHzCD uses kHzwideband speech coding using 16 kHzmaster from studio must be rate-converted by a Signal Processing .

2 Oversampling ADCC onsider a Nyquist rate ADC in which the Signal is sampled at the desiredprecision and at a rate such that Nyquist s sampling criterion is for audio is 20 Hz< f <20 kHzAntialiasing filter required has very demanding specification|H(j )|= 0 dB, f <20 kHz|H(j )|<96 dB, f high order analogue filter such as elliptic filters that have verynonlinear phase characteristicshard to design, expensive and bad for audio Signal Processing Rate Conversion Anti-aliasing Signal Processing oversampling the Signal at, say, 64 times the Nyquist rate butwith lower precision. Then use Multirate techniques to convert sample rateback to kHz with full (over-sampled) sampling rate is 64 simple antialiasing filter|H(j )|= 0 dB, f <20 kHz|H(j )|<96 dB, f ( 64) be implemented by simple filter (eg.)

3 RC network)Recover desired sampling rate by downsampling Signal Processing Conversion Antialiasing FilterDigital Signal Processing SystemThis is a simplified versionIn these lectures we will study blocks likeG(z)and 64 Digital Signal Processing : Subband CodingConsider quantizing the samples of a speech Signal . How manybits arerequired?In general, 16 bits precision per sample is normally used gives an adequate dynamic practice, certain frequency bands are less important perceptuallybecause they contain less significant informationbands with less information or lower perceptual importancemay bequantized with lower precision - fewer the spectrum of the Signal into several subbands thenallocatebits to each band Signal Processing bits per sample, 10 kHz sampling frequency gives160 kbits/sDivide into 2 bands: high frequency and low frequency frequencies of speech are less important to use only 8 bits per sampleThe sampling frequency can be reduced by a factor of 2 sincebandwidth is halved, still satisfying Nyquist 16 + 5 8 = 120kbits/s4.

4 3 compressionReconstructed Signal has no noticeable reduction is Signal Processing Multirate OperationsDownsampling by a factorMfilter and M-fold decimatorUpsampling by a factorLL-fold expander and filterDigital Signal Processing DecimatorFor an input sequencex(n), select only the samples which occur at integermultiples ofM. The other samples are thrown (n) =x(M n)Aliasing will occur inyD(n)unlessx(n)is sufficiently bandlimitedloss of Signal Processing 2 Digital Signal Processing ExpanderFor an input sequencex(n), insertL 1zeros between each (n) =x(M n)x(n)can always be recovered fromyE(n)no loss of information, no Signal Processing 2 Digital Signal Processing Domain View of the ExpanderFrom the definition of the z-transformYE(z) = n= yE(n)z n= k= yE(kL)z kL= k= x(k)z kL=X(zL)For frequency response writez=ej givingYE(ej ) =X(ej L)YEis a compressed version ofXMultiple images ofX(ej )are created inYE(ej )

5 Between = 0and = 2 Digital Signal Processing use the expander for interpolation, a lowpass filter is applied afterthe expander to remove the images (shaded).Digital Signal Processing Domain View of the DecimatorFrom the definition of the z-transformYD(z) = n= yD(n)z n= n= x(M n)z nLetx1(n) ={x(n)ifnis an integer multiple ofM0otherwiseThenYD(z) = n= x1(M n)z n= k= x1(k)z k/Msincex1(k) = 0unlesskis a multiple Signal Processing (z) =X1(z1/M) =1MM 1 k=0X(z1/MWkM)as will be shown on the next slide and usingWkM=e j2 k/MFor frequency response writez=ej to giveYD(ej ) =1MM 1 k=0X(ej( 2 k)/M)Digital Signal Processing arrive at the previous expression forYD(z)by considering a newsequencecM(n) ={1ifnis an integer multiple ofM0otherwiseand then writingx1(n) =cM(n)x(n)Further consideration ofcM(n)tells us thatcM(n)is the inverse Fouriertransform of unity and can be writtencM(n)}}

6 =1MM 1 k=0W knMDigital Signal Processing (z) =1MM 1 k=0 n= x(n)W knMz n=1MM 1 k=0 n= x(n)(WkMz) n=1MM 1 k=0X(zWkM)from the definition of the z-transform. So finallyYD(z) =1MM 1 k=0X(z1/MWkM)Digital Signal Processing doesYD(z) =1M M 1k=0X(z1/MWkM)represent?stretching ofX(ej )toX(ej /M)creatingM 1copies of the stretched versionsshifting each copy by successive multiples of2 and superimposing(adding) all the shifted copiesdividing the result byMDigital Signal Processing use a decimation process we must first bandlimit the signalto| |< Signal Processing Signal Processing