Digital Transmission of Analog Signals: PCM, DPCM and DM

Transcription

1 A T CHAPTER 6 Digital Tranmiion of Analog Signal: PCM, DPCM and DM 6.1 Introduction Quite a few of the information bearing ignal, uch a peech, muic, video, etc., are analog in nature; that i, they are function of the continuou variable t and for any t = t 1, their value can lie anywhere in the interval, ay A to A. Alo, thee ignal are of the baeband variety. If there i a channel that can upport baeband tranmiion, we can eaily et up a baeband communication ytem. In uch a ytem, the tranmitter could be a imple a jut a power amplifier o that the ignal that i tranmitted could be received at the detination with ome minimum power level, even after being ubject to attenuation during propagation on the channel. In uch a ituation, even the receiver could have a very imple tructure; an appropriate filter (to eliminate the out of band pectral component) followed by an amplifier. If a baeband channel i not available but have acce to a paband channel, (uch a ionopheric channel, atellite channel etc.) an appropriate CW modulation cheme dicued earlier could be ued to hift the baeband pectrum to the paband of the given channel. Intereting enough, it i poible to tranmit the analog information in a digital format. Though there are many way of doing it, in thi chapter, we hall explore three uch technique, which have found widepread acceptance. Thee are: Pule Code Modulation (PCM), Differential Pule Code Modulation (DPCM) mywbut.com 1

2 and Delta Modulation (DM). Before we get into the detail of thee technique, let u ummarize the benefit of digital tranmiion. For implicity, we hall aume that information i being tranmitted by a equence of binary pule. i) During the coure of propagation on the channel, a tranmitted pule become gradually ditorted due to the non-ideal tranmiion characteritic of the channel. Alo, variou unwanted ignal (uually termed interference and noie) will caue further deterioration of the information bearing pule. However, a there are only two type of ignal that are being tranmitted, it i poible for u to identify (with a very high probability) a given tranmitted pule at ome appropriate intermediate point on the channel and regenerate a clean pule. In thi way, we will be completely eliminating the effect of ditortion and noie till the point of regeneration. (In long-haul PCM telephony, regeneration i done every few kilometer, with the help of regenerative repeater.) Clearly, uch an operation i not poible if the tranmitted ignal wa analog becaue there i nothing like a reference waveform that can be regenerated. ii) Storing the meage in digital form and forwarding or redirecting them at a later point in time i quite imple. iii) Coding the meage equence to take care of the channel noie, encrypting for ecure communication can eaily be accomplihed in the digital domain. iv) Mixing the ignal i eay. All ignal look alike after converion to digital form independent of the ource (or language!). Hence they can eaily be multiplexed (and demultiplexed) 6. The PCM ytem Two baic operation in the converion of analog ignal into the digital i time dicretization and amplitude dicretization. In the context of PCM, the former i accomplihed with the ampling operation and the latter by mean of quantization. In addition, PCM involve another tep, namely, converion of mywbut.com

3 quantized amplitude into a equence of impler pule pattern (uually binary), generally called a code word. (The word code in pule code modulation refer to the fact that every quantized ample i converted to an R -bit code word.) Fig. 6.1: A PCM ytem Fig. 6.1 illutrate a PCM ytem. Here, mt ( ) i the information bearing meage ignal that i to be tranmitted digitally. mt ( ) i firt ampled and then quantized. The output of the ampler i m( nt ) m( t) ampling period and n i the appropriate integer. f =. T i the t = nt 1 = i called the ampling T rate or ampling frequency. The quantizer convert each ample to one of the value that i cloet to it from among a pre-elected et of dicrete amplitude. The encoder repreent each one of thee quantized ample by an R -bit code word. Thi bit tream travel on the channel and reache the receiving end. With f a the ampling rate and R -bit per code word, the bit rate of the PCM ytem i Rf R = bit/ec. The decoder convert the R -bit code word into T the correponding (dicrete) amplitude. Finally, the recontruction filter, acting on thee dicrete amplitude, produce the analog ignal, denoted by mt ( ). If there are no channel error, then mt ( ) mt ( ). mywbut.com 3

4 6.3 Sampling We hall now develop the ampling theorem for lowpa ignal. Theoretical bai of ampling i the Nyquit ampling theorem which i tated below. Let a ignal x () t be band limited to W Hz; that i, X( f) = 0 for f > W. Let ( ) () x nt = x t, < n < repreent the ample of x () t at t = nt uniform interval of T econd. If T 1 W { } x () t exactly from the et of ample, ( ) x nt., then it i poible to recontruct { } In other word, the equence of ample ( ) complete time behavior of ( ) x t. Let f 1 =. Then f T x nt can provide the = W i the minimum ampling rate for x () t. Thi minimum ampling rate i called the Nyquit rate. Note: If x () t i a inuoidal ignal with frequency f 0, then f > f 0. f = f 0 i not adequate becaue if the two ample per cycle are at the zero croing of the tone, then all the ample will be zero! We hall conider three cae of ampling, namely, i) ideal impule ampling, ii) ampling with rectangular pule and iii) flat-topped ampling Ideal impule ampling Conider an arbitrary lowpa ignal x ( t ) hown in Fig. 6.(a). Let x t x t t nt n = (6.1a) () = () δ( ) mywbut.com 4

5 n x () t ( t nt ) (6.1b) = δ n = x ( nt) ( t nt) (6.1c) = δ = where δ () t i the unit impule function of ection x ( ) t, hown in red in Fig. 6.(b) conit of a equence of impule; the weight of the impule at t nt = i equal to x ( nt ). x ( ) t i zero between two adjacent impule. Fig. 6.: (a) A lowpa ignal x ( t ) (b) x ( t ), ampled verion of x ( t ) It i very eay to how in the frequency domain that x ( ) complete information of x ( t ). A defined in Eq. 6.1(a), x ( ) x () t and δ( t nt ) convolution. That i, n t preerve the t i the product of. Hence, the correponding Fourier relation i mywbut.com 5

6 1 n X ( f) = X( f) δ f T n = T (6.a) 1 n = X f T T (6.b) n n = 1 = T X( f nf ) (6.c) = From Eq. 6.(c), we ee that X ( f ) i a uperpoition of X( f ) and it hifted verion (hifted by multiple of f, the ampling frequency) caled by 1. Thi i hown in Fig Let X( f ) be a triangular pectrum a hown in Fig. 6.3(a). T Fig. 6.3: Spectra of x ( t ) and x ( t ) (a) X( f ) (b) X ( ) (c) X ( ) f, f f, f = W (d) X ( f ), f > < W W mywbut.com 6

7 From Fig. 6.3(b) and 6.3(c), it i obviou that we can recover ( ) x t from x () t by paing x () t through an ideal lowpa filter with gain T and bandwidth W, a hown in Fig Fig. 6.4: Recontruction of x ( t ) from x ( t ) Of coure, with repect to Fig. 6.3(b), which repreent the over-ampled cae, recontruction filter can have ome tranition band which can fit into the gap between f = W and f = ( f W). However, when f < W, (under-ampled cae) we ee that pectral lobe overlap reulting in ignal ditortion, called aliaing ditortion. In thi cae, exact ignal recovery i not poible and one mut be willing to tolerate the ditortion in the recontructed ignal. (To avoid aliaing, the ignal x () t i firt filtered by an anti-aliaing filter band-limited to f W and the filtered ignal i ampled at the rate of f ample per econd. In thi way, even if a part of the ignal pectrum i lot, the remaining pectral component can be recovered without error. Thi would be a better option than permitting aliaing. See Example 6.1.) It i eay to derive an interpolation formula for x ( t ) in term of it ample x ( nt ) when the recontruction filter i an ideal filter and f W. Let H( f ) repreent an ideal lowpa filter with gain T and bandwidth W ' = f where mywbut.com 7

8 f W f W. Then, ( ) i, ht () T W ' inc( Wt ' ) ht the impule repone of the ideal lowpa filter = and WT ' =. A x ( t) x ( t) h( t) () = ( ) in ( ) = 1, we have ' (6.3a) n = x t x nt c W t nt If the ampling i done at the Nyquit rate, then W ' = W and Eq. 6.3(a) reduce to n x () t = x c( W t n) W in (6.3b) n = That i, the impule repone of the ideal lowpa filter, which i a in c ( ) function, act a the interpolating function and given the input, { x ( nt ) ( t nt )} δ, it interpolate the ample and produce x () t for all t. impule at Note that x () t repreent a equence of impule. The weight of the t = nt i equal to x ( nt ). In order that the ampler output be equal to x ( nt ), we require conceptually, the impule modulator to be followed by a unit that convert impule into a equence of ample value which are baically a equence of number. In [1], uch a cheme ha been termed a an ideal C- to-d converter. For implicity, we aume that the output of the ampler { } repreent the ample equence ( ) x nt. To recontruct () { } x t from x( nt ), we have to perform the invere operation, namely, convert the ample equence to an impule train. Thi ha been termed a an ideal D-to-C converter in [1]. We will aume that the recontruction filter in Fig. 6.1 will take care of thi apect, if neceary. mywbut.com 8

9 6.3. Sampling with a rectangular pule train A it i not poible in practice to generate impule, let look at a more practical method of ampling, namely, ampling with a rectangular pule train. (Note that an impule i a limiting cae of a rectangle pule a explained in Sec ) 6.5(b). Let Let y () t repreent the periodic rectangular pule train a hown in Fig. p () () () x t = x t y t (6.4a) p Fig. 6.5: Sampling with a rectangular pule train (a) x ( t ), (b) the pule train, (c) the ampled ignal mywbut.com 9

10 Then, X ( f) X( f) Y ( f) = (6.4b) But, from exercie 1.1, we have Hence, A p τ p n = T ( ) = in ( τ) δ( ) Y f c nf f nf τ T n = X ( f) = inc( nf τ) X( f nf ) (6.5) n inc τ T τ τ τ = + inc X( f + f) + X( f) + inc X( f f) + T T T i only a cale factor that depend on n, we find that X( f ) and it hifted replica are weighted by different factor, unlike the previou cae where all the weight were equal. A ketch of X ( ) and, f > W and X( f ) of Fig. 6.3(a). f i hown in Fig. 6.6 for τ = T 1 10 Fig. 6.6: Plot of Eq. 6.5 In the Fig. 6.6, in 0.1 α 1 = = , 10 c ( ) in 0. α = = etc. From 10 c ( ) Eq. 6.5 and Fig. 6.6, it i obviou that each lobe of X ( ) f i multiplied by a different number. However a the cale factor i the ame for a given pectral mywbut.com 10

11 lobe of X ( f ), there i no ditortion. Hence x ( t ) can be recovered from x ( ) Eq. 6.4(a). t of From Fig. 6.5(c), we ee that during the time interval τ, when yp () t = 1, x () t follow the hape of ( ) exact canning. x t. Hence, thi method of ampling i alo called Flat topped ampling Conider the cheme hown in Fig Fig. 6.7: Flat topped ampling ZOH i the zero order hold (Example 1.15) with the impule repone () t ht () τ ga = τ x t i the ame a the one given by Eq A y() t x () t h() t =, (6.6a) y() t will be a hown (in red color) in Fig mywbut.com 11

12 Fig. 6.8: Sampled waveform with flat top pule Holding the ample value contant provide time for ubequent coding operation. However, a large value of τ would introduce noticeable ditortion and exact recovery of x () t i not poible unle y ( t ) i paed through an equalizer, a will be een hortly. Taking the Fourier tranform of Eq. 6.6(a), we have ( ) ( ) ( ) Y f = X f H f (6.6b) where ( ) j πf τ H f =τe inc( f τ ) A X( f) = X( f nf) Let Y ( f) 1 T n, we have τ Y f X f nf e c f T (6.7) n j πf τ ( ) = ( ) in ( τ) 0 denote the term for n = 0 in Eq That i, τ jπf τ Y0 ( f) = e X( f) inc( f τ ) (6.8) T By holding the ample contant for a fixed time interval, we have introduced a τ delay of and more importantly X( f ) i multiplied with inc( f ) τ, which i a function of the variable f ; that each pectral component of X( f ) i multiplied by a different number, which implie pectral ditortion. Auming that the delay of mywbut.com 1

13 τ i not eriou, we have to equalize the amplitude ditortion caued and thi can be achieved by multiplying Y ( f) 0 by 1 H( f). Of coure, if τ < 0.1, then the amplitude ditortion may not be ignificant. The ditortion caued by holding the pule contant i referred to a the aperture effect. T Underampling and the problem of aliaing When f < W, we have een that there i pectral overlap (Fig. 6.3d) and the reulting ditortion i called the aliaing ditortion. Let u try to undertand thi in ome detail. Fig. 6.9: A part of the aliaed pectrum Let X ( ) f repreent the aliaed pectrum and let u conider the three 1 T lobe from it, namely, X( f), X( f f ) and X( f f ) 1 T 1 T +, indicated by 1,, and 3 repectively in Fig The left ide of thee triangular pectra are hown in red. A can be een from the figure, the complex exponential with the (negative) frequency f 1 ' i interfering with the frequency f 1. That i, e j πf ' 1 t i giving rie to the ame et of ample a that j π e ft 1, for the T choen. We mywbut.com 13

14 hall make thi clear by conidering the ampling of a coine ignal. Let the ignal to be ampled be the coine ignal, () = co ( π 4) x t t It pectrum conit of two impule, one at 4 Hz and the other at - 4 Hz a hown in Fig. 6.10(a), in which the negative frequency component ha been hown with a broken line. Let x ( t ) be ampled at the rate of 6 ample/ec. Fig. 6.10: Sampling of a coine ignal with aliaing (Note that the Nyquit rate i greater than 8 ample per econd.) The reulting pectrum i hown in Fig. 6.10(b). Notice that in X ( ) f, there are pectral component at ± Hz. The impule with a broken line at Hz i originally - 4 Hz component. In other word, ( ) j t e π e when ampled with T ( ) j π 4 t e ( ) e j n e j n = 3 = 3 e i giving rie to ame et of ample a 1 = ec. Thi i eay to verify. 6 4π 4π jπ 4 t j π n n t = 6 mywbut.com 14

15 π j n 3 = e = e ( ) t jπ t = n 6 Similarly, the et of value obtained from j t e 8 π for t = nt i the ame a thoe from e j 4 πt (Notice that in Fig. 6.10(b), there i a olid line at f = ). In other word, a 4 Hz coine ignal become an alia of Hz coine ignal. If there were to be a Hz component in X( f ), the 4 Hz will interfere with the Hz component and thereby leading to ditortion due to under ampling. In fact, a frequency component at f = f i the alia of another lower frequency component at f = f 1 if f = kf ± f, where k i an integer. Hence, with f = 6, we find that the 4 Hz 1 component i an alia the of Hz component, becaue 4 = 6 -. Similarly, co π( 8)t a 8 = 6 +. A few of the aliae of the Hz component with f = 6 have been hown on the aliaing diagram of Fig Fig. 6.11: Aliaing diagram with f = 6 A een from the figure, upto 3 Hz there i no aliaing problem. Beyond that, the pectrum fold back, making a 4 Hz component align with a Hz component. With repeated back and forth folding, we find that 8 Hz, 10 Hz, 14 Hz, etc. are aliae of Hz and all of them will contribute to the output at Hz, when X ( f ) i filtered by an ILPF with cutoff at f 3 = Hz. mywbut.com 15

16 Note: Recontruction of the analog ignal i done by placing an ideal LPF with a cutoff at f. If ( t ) co 8π were to be ampled at 6 ample/ec, the output of the lowpa filter (with a cutoff at 3 Hz) i coine ignal with the frequency of Hz! Fig. 6.1: Aliaing picture in the time domain The concept of aliaing can alo be illutrated in the time domain. Thi i hown in Fig The red dot repreent the ample from the two coine ignal (one at 4 Hz and the other at Hz) at t =, 0,,,. By extrapolation, we find that co( 8π t ) will give rie to ame et of ample a ( t ) co 4π at t n =, n = 0, ± 1, ± etc. 6 We will now demontrate the effect of aliaing on voice and muic ignal. There are two voice recording: one i a poken entence and the other i part of a Sankrit loka rendered by a peron with a very rich voice. The muic ignal pertain to a re ga ma pa da ne... ound generated by playing a flute. (The recording of the flute ignal wa done on a cell phone.) 1) a. The ignal pertaining to the poken entence God i ubtle but not maliciou, i ampled at 44.1 kp and recontructed from thee mywbut.com 16

17 ample. b. The recontructed ignal after down ampling by a factor of 8. ) a. Lat line of a Sankrit loka. Sampling rate 44.1 kp. b. Recontructed ignal after down ampling by a factor of 8. c. Recontructed ignal after down ampling by a factor of 1. 3) a. Flute muic ampled at 8 kp. b. Recontructed ignal after down ampling by a factor of. 600 Magnitude pectrum of original verion magnitude frequency (Hz) Fig. 6.13(a): Spectrum of the ignal at (a) mywbut.com 17

18 Magnitude pectrum of aliaed verion (downampling factor = 8) magnitude frequency (Hz) Fig. 6.13(b): Spectrum of the ignal at (b) 140 Magnitude pectrum of aliaed verion (downampling factor = 1) magnitude frequency (Hz) Fig. 6.13(c): Spectrum of the ignal at (c) mywbut.com 18

19 With repect to 1b, we find that there i noticeable ditortion when the word maliciou i being uttered. A thi word give rie to pectral component at the higher end of the voice pectrum, fold back of the pectrum reult in evere aliaing. Some ditortion i to be found elewhere in the entence. Aliaing effect i quite evident in the voice output (b) and (c). (It i much more viible in (c).) We have alo generated the pectral plot correponding to the ignal at (a), (b) and (c). Thee are hown in Fig. 6.13(a), (b) and (c). From Fig. 6.13(a), we ee that the voice pectrum extend all the way upto 6 khz with trong pectral component in the range 00 Hz to 1500 Hz and 700 Hz to 3300 Hz. (Sampling frequency ued i 44.1 khz.) Fig. 6.13(b) correpond to the down ampled (by a factor of 8) verion of (a); that i, ampling frequency i about 5500 ample/ec. Spectral component above 750 Hz will get folded over. Becaue of thi, we find trong pectral component in the range Hz. Fig. 6.13(c) correpond to a ampling rate of 3600 ample/ec. Now pectrum fold over take place with repect to 1800 Hz and thi can be eaily een in Fig. 6.13(c). A it i difficult to get rid of aliaing once it et in, it i better to pa the ignal through an anti-aliaing filter with cutoff at f and then ample the output of the filter at the rate of f ample per econd. Thi way, we are ure that all the pectral component upto f f will be preerved in the ampling proce. Example 6.1 illutrate that the recontruction mean quare error i le than or equal to the error reulting from ampling without the anti-aliaing filter. mywbut.com 19

20 Example 6.1 An analog ignal x () t i paed through an anti-aliaing filter with cutoff at f c and then ampled with f f c X f i not band limited to f c.) Let y1 () t be the ignal recontructed from thee ample and =. (Note that ( ) f 1 1 ( () ()) e = x t y t dt Now the anti-aliaing filter i withdrawn and x ( t ) i ampled directly with = f a before. Let y () t be the ignal recontructed from thee ample c ( ) e x t y t dt. Show that e e1. and let = () () Let X( f ) be a hown in the figure 6.14(a), with f N being the highet frequency in it. Y ( f) 1 i a hown in Fig. 6.14(b). A can be hown from the figure, Y ( f) that part of ( ) 1 doe not have any aliaing but X f for f c f f N i miing. Thi introduce ome ditortion and the energy of the error ignal, ( () ()) e = y t x t dt 1 1 ( ) ( ) = Y f X f df 1 But Y ( f) 1 ( ), c X f f f = 0, f > f c. f c Hence, e1 = X( f) df + X( f) df (6.9a) f c mywbut.com 0

21 Fig. 6.14: (a) Original pectrum, X( f ) (b) Spectrum of the recontructed ignal after x ( t ) i paed through an anti-aliaing filter before ampling (c) Spectrum of the recontructed ignal with aliaing It i evident from Fig. 6.14(c), that y ( t ) uffer from aliaing and fc fc = + + f f ( ) ( ) ( ) ( ) (6.9b) e X f df X f df Y f X f df c c mywbut.com 1

22 for f fc. Comparing Eq. 6.9(a) and (b), we find that Eq. But Y ( f) X( f) 6.9(b) ha an extra term which i greater or equal to zero. Hence e e. 1 Example 6. Find the Nyquit ampling rate for the ignal () in ( 00 ) in ( 1000 ) x t = c t c t. ( t) inc 00 ha a rectangular pectrum in the interval f 100 Hz and ( ) ( ) ( ) inc 1000t = inc 1000t inc 1000t ha a triangular pectrum in the frequency range f 1000 Hz. Hence X( f ) ha pectrum confined to the range f 1100 Hz. Thi implie the Nyquit rate i 00 ample/ec. Example 6.3 a) Conider the bandpa ignal x ( t ) with X( f ) a hown in Fig Let x () t be ampled at the rate of 0 ample per econd. Sketch X ( ) within a cale factor), for the frequency range 0 f 30 Hz. f (to b) Let x () t be now ampled a the rate of 4 ample per econd. Sketch ( ) X f (to within a cale factor) for 0 f 30 Hz. Fig. 6.15: X( f ) for example 6.3 mywbut.com

23 a) Let P denote X( f ) for f 0 and let N denote X( f ) for f < 0. (N i hown with a broken line.) Conider the right hift of X( f ) in multiple of 0 Hz, which i the ampling frequency. Then P will be hifted away from the frequency interval of interet. We have to only check whether hifted N can contribute to pectrum in the interval that thi will not happen. Thi implie, in X ( ) hape a X( f ) for That i, for X( f ). 0 f 30 0 f 30. It i eay to ee f, pectrum ha the ame 0 f 30. Thi i alo true for the left hift of X( f )., X ( ) f, to within a cale factor, i the ame a b) Let f = 4. Conider again right hift. When N i hifted to the right by f 48 =, it will occupy the interval ( ) = 18 to ( ) That i, we have the ituation hown in Fig = 8. Fig. 6.16: The two lobe contributing to the pectrum for 0 f 30 mywbut.com 3

24 Let the um of (a) and (b) be denoted by Y( f ). Then, Y( f ) i a hown in Fig Fig. 6.17: Spectrum for 0 f 30 when f = 4 For 30 f 0, Y( f ) i the mirror image of the pectrum hown in Fig Notice that for the bandpa ignal, increaing f ha reulted in the ditortion of the original pectrum. Example 6.4 Let x () t = co( 800π t) + co( 1400π t). ( ) x t i ampled with the rectangular pule train x () t hown in Fig Find the pectral component in p the ampled ignal in the range.5 khz to 3.5 khz. Fig. 6.18: xp ( t ) of the Example 6.4 mywbut.com 4

25 X( f ) ha pectral component at ± 400 Hz and ± 700 Hz. The impule in Xp ( f ) occur at ± 1 khz, ± khz, ± 4 khz, ± 5 khz, etc. (Note the abence of pectral line in Xp ( f ) at f = ± 3 khz, ± 6 khz, etc.) Convolution of X( f ) with the impule in Xp ( f ) at khz will give rie to pectral component at.4 khz and.7 khz. Similarly, we will have pectral component at 4 ( 0.4) = 3.6 khz and 4 ( 0.7) = 3.3 khz. Hence, in the range.5 khz to 3.5 khz, ampled pectrum will have two component, namely, at.7 khz and 3.3 khz. Exercie A inuoidal ignal () = co ( π 10 ) x t A t i ampled with a uniform impule train at the rate of 1500 ample per econd. The ampled ignal i paed through an ideal lowpa filter with a cutoff at 750 Hz. By ketching the appropriate pectra, how that the output i a inuoid at 500 Hz. What would be the output if the cutoff frequency of the LPF i 950 Hz? Exercie 6. Let x () t be the um of two coine ignal. X( f ) i hown in Fig Let x () t be ampled at the rate of 300 ample per econd. Let x () t be paed through an ideal LPF with a cutoff at 150 Hz. a) Sketch the pectrum at the output of the LPF. b) Write the expreion for the output of the LPF. Fig. 6.19: X( f ) for the Exercie 6. mywbut.com 5

26 6.4 Quantization Uniform quantization An analog ignal, even it i limited in it peak-to-peak excurion, can in general aume any value within thi permitted range. If uch a ignal i ampled, ay at uniform time interval, the number of different value the ample can aume i unlimited. Any human enor (uch a ear or eye) a the ultimate receiver can only detect finite intenity difference. If the receiver i not able to ditinguih between two ample amplitude, ay v 1 and v uch that v1 v <, then we can have a et of dicrete amplitude level eparated by and the original ignal with continuou amplitude, may be approximated by a ignal contructed of dicrete amplitude elected on a minimum error bai from an available et. Thi enure that the magnitude of the error between the actual ample and it approximation i within and thi difference i irrelevant from the receiver point of view. The realitic aumption that a ignal mt ( ) i (eentially) limited in it peak-to-peak variation and any two adjacent (dicrete) amplitude level are eparated by will reult in a finite number (ay L ) of dicrete amplitude for the ignal. The proce of converion of analog ample of a ignal into a et of dicrete (digital) value i called quantization. Note however, that quantization i inherently information loy. For a given peak-to-peak range of the analog ignal, maller the value of, larger i the number of dicrete amplitude and hence, finer i the quantization. Sometime one may reort to omewhat coare quantization, which can reult in ome noticeable ditortion; thi may, however be acceptable to the end receiver. The quantization proce can be illutrated graphically. Thi i hown in Fig. 6.0(a). The variable x denote the input of the quantizer, and the variable mywbut.com 6

27 y repreent the output. A can be een from the figure, the quantization proce implie that a traight line relation between the input and output (broken line through the origin) of a linear continuou ytem i replaced by a taircae characteritic. The difference between two adjacent dicrete value,, i called the tep ize of the quantizer. The error ignal, that i, difference between the input and the quantizer output ha been hown in Fig. 6.0(b). We ee from the figure that the magnitude of the error i alway le than or equal to. (We are auming that the input to the quantizer i confined to the range 7 7 to. Fig. 6.0: (a) Quantizer characteritic of a uniform quantizer (b) Error ignal mywbut.com 7

28 Let the ignal mt ( ), hown in Fig. 6.1(a), be the input to a quantizer with the quantization level at 0, and 4 ± ±. Then, m ( ) q t, the quantizer output i the waveform hown in red. Fig. 6.1(b) how the error ignal () () () et = mt m t a a function of time. Becaue of the noie like appearance q of et (), it i a common practice to refer to it a the quantization noie. If the input to the quantizer i the et of (equipaced) ample of mt ( ), hown in Fig. 6.1(c), then the quantizer output i the ample equence hown at (d). Fig. 6.1: (a) An analog ignal and it quantized verion (b) The error ignal mywbut.com 8

29 Fig. 6.1: (c) Equipaced ample of mt ( ) (d) Quantized ample equence For the quantizer characteritic hown in Fig. 6.0(a), the maximum value of the quantizer output ha been hown a 3 and the minimum a ( 3 ). Hence, if the input x i retricted to the range ± 3.5 (that i, mywbut.com 9

30 x max 7 = xmin, the quantization error magnitude i le than or equal to. Hence, (for the quantizer hown), ± 3.5 i treated a the overload level. where The quantizer output of Fig. 6.0(a) can aume value of the form Hi ± Hi = 0,1,,. A quantizer having thi input-output relation i aid to be of the mid-tread type, becaue the origin lie in the middle of a tread of a taircae like graph. A quantizer whoe output level are given by Hi, where ± Hi = 1, 3, 5, i referred to a the mid-rier type. In thi cae, the origin lie in the middle of the riing part of the taircae characteritic a hown in Fig. 6.. Note that overload level ha not been indicated in the figure. The difference in performance of thee two quantizer type will be brought out in the next ubection in the context of idle-channel noie. The two quantizer type decribed above, fall under the category of uniform quantizer becaue the tep ize of the quantizer i contant. A non-uniform quantizer i characterized by a variable tep ize. Thi topic i taken up in the ub-ection mywbut.com 30

31 Fig. 6.: Mid-rier type quantizer 6.4. Quantization Noie Conider a uniform quantizer with a total of L quantization level, ymmetrically located with repect to zero. Let x repreent the quantizer input, and y QZ( x) = denote it output. Conider a part of the quantizer operating range hown in Fig Let I k be the interval defined by { + } I = x < x x 1, k = 1,,, L. k k k Then y y x = k, if k I ; we can alo expre y k a yk = x + q, if x Ik where q denote the quantization error, with q. mywbut.com 31

32 Fig. 6.3: Input interval and the correponding output level of a quantizer Let u aume that the input x i the ample value of a random variable X with zero mean and variance σ X. When the quantization i fine enough, the ditortion produced by the quantization affect the performance of a PCM ytem a though it were an additive, independent ource of noie with zero mean and mean quare value determined by the quantizer tep ize. Let the random variable Q denote the quantization error, and let q denote it ample value. Auming Q i uniformly ditributed in the range,, we have f Q ( q) where f ( ) Q 1, q = 0, otherwie q i the PDF of the quantization error. Hence, the variance of the quantization error, Q σ i mywbut.com 3

33 σ Q = (6.10) 1 The recontructed ignal at the receiver output, can be treated a the um of the original ignal and quantization noie. We may therefore define an output ignalto-quantization noie ratio ( SNR) 0, q a ( SNR X X ) = = 0, q σ σ 1σ Q A can be expected, Eq. 6.11(a) tate the reult for a given σ X (6.11a), maller the value of the tep ize, larger i the ( SNR) 0,. Note that σ q Q i independent of the input PDF provided overload doe not occur. ( SNR) 0, q i uually pecified in db. ( SNR) ( ) = ( SNR) in db 10 log 0, q 10 0, q (6.11b) Idle Channel Noie Idle channel noie i the coding noie meaured at the receiver output with zero input to the tranmitter. (In telephony, zero input condition arie, for example, during a ilence in peech or when the microphone of the handet i covered either for ome conultation or deliberation). The average power of thi form of noie depend on the type of quantizer ued. In a quantizer of the mid-rier type, zero input amplitude i coded into one of two innermot repreentation level, ±. Auming that thee two level are equiprobable, the idle channel noie for mid-rier quantizer ha a zero mean with the mean quared value = 4 On the other hand, in a quantizer of the mid-tread type, the output i zero for zero input and the ideal channel noie i correpondingly zero. Another difference between the mid-tread and mid-rier quantizer i that, the number of quantization level i odd in the cae of the former where a it i even for the latter. But for mywbut.com 33

34 thee minor difference, the performance of the two type of quantizer i eentially the ame. Example 6.5 Let a baeband ignal x ( t ) be modeled a the ample function of a zero mean Gauian random proce X( t ). x ( t ) i the input to a quantizer with the overload level et at ± 4σ X, where X σ i the variance of the input proce (In other word, we aume that the ( ) σx P x t 4 0). If the quantizer output i coded with R -bit equence, find an expreion for the ( SNR) 0,. q With R -bit per code word, the number of quantization level, For calculating the tep ize, we hall take x x( t) x Step ize max 8σX 8σ = = = X L L R ( SNR X X ) = = 0, q = σ σ σ Q 1 R ( 1σX ) 64σ X max = a 4 σ X. max R L =. 3 R = 16 ( SNR) ( ) 0, q in db = 6.0R 7. (6.1) Thi formula tate that each bit in the code word of a PCM ytem contribute 6 db to the ignal-to-noie ratio. Remember that thi formula i valid, provided, i) The quantization error i uniformly ditributed. ii) The quantization i fine enough (ay n 6 ) to prevent ignal correlated pattern in the error waveform. mywbut.com 34

35 iii) The quantizer i aligned with the input amplitude range (from 4σ X to 4σ X ) and the probability of overload i negligible. Example 6.6 Let x ( t ) be modeled a the ample function of a zero mean tationary proce () X t with a uniform PDF, in the range ( a, a). Let u find the ( ) q SNR 0, auming an R -bit code word per ample. Now the ignal x () t ha only a finite upport; that i, xmax = a. It variance, Step ize X a σ =. 3 = = a R a ( R ) 1 SNR a σm 3 a 1 1 Hence, ( ) = = 0, q ( R 1) = R ( SNR) ( ) 0, q in db = 6.0R (6.13a) Even in thi cae, we find the 6 db per bit behavior of the ( SNR) 0,. q Example 6.7 Let mt (), a inuoidal ignal with the peak value of A be the input to a uniform quantizer. Let u calculate the ( SNR) 0, q auming R -bit code word per ample. mywbut.com 35

36 Step ize = Signal power ( SNR) = = A R 0, q A = 3 ( R ) R ( SNR) ( ) 0, q A 1 4 A in db = 6.0R (6.13b) We now make a general tatement that ( SNR) 0, q of a PCM ytem uing a uniform quantizer can be taken a 6 R + α, where R i the number of bit/code x word and α i a contant that depend on the ratio rm σx =, a hown xmax xmax below. = x max R xmax Q R σ = = 1 3 X X R ( SNR) = = 3 0, q σ σ σ x Q max 10 log X 10 ( SNR) R 0 log σ = + + 0, q 10 xmax = 6.0 R + α (6.14a) where log σx 10 xmax α = + (6.14b) mywbut.com 36

37 Example 6.8 Let mt ( ) be a real bandpa ignal with a non-zero pectrum (for f > 0) only in the range f L to f H, with f H f L. Then, it can be hown that, the minimum ampling frequency 1 to avoid aliaing i, fl 1 + B fh ( f ) = B = (6.15) min fl k 1 + B Where B = fh fl, k = f H and x B denote the integer part of x. Let mt () be a bandpa ignal with f L = 3 MHz and f H = 5 MHz. Thi ignal i to be ent uing PCM on a channel whoe capacity i limited to 7 Mbp. Aume that the ample of mt ( ) are uniformly ditributed in the range - V to V. a) Show that the minimum ampling rate a given by Eq i adequate to avoid aliaing. b) How many (uniform) quantization level are poible and what are thee level o that the error i uniform with in c) Determine the ( SNR) 0, q ±. (in db) that can be achieved. a) Eq give ( ) f min a ( ) f min = = 5 10 ample/ec Let M( f ), the pectrum of mt ( ) be a hown in Fig Note that a ampling frequency f ( f ) f H > could reult in aliaing (ee Example 6.3) unle min f, in which cae the ampling of a bandpa ignal reduce to that of ampling a lowpa ignal. mywbut.com 37

38 Fig. 6.4: Bandpa pectrum of Example 6.8 A f 6 = 5 10 ample/ec, M( f ) i to be repeated at interval of Hz. Shifting N to the right by 5 MHz will create a pectral lobe between 0 to MHz wherea hifting by 10 MHz will reult in a pectral lobe occupying the range 5 to 7 MHz. That i, right hift of N will not interfere with P. Similarly, left hift of P (in multiple of 5 MHz) will not fall in the range - 5 MHz to - 3 MHz. Hence, there i no aliaing in the ampling proce. b) A the ampling rate i ample/ec and the capacity of the channel i limited to 7Mbp; it i poible to end only one bit per ample; that i, we have only a two-level quantization. The quantizer level are at ± 1 o that tep ize i V and the error i uniform in the range ± 1. c) A the ignal i uniformly ditributed in the range (, ), we have ( ) 4 4 σ M = =. 1 3 Variance of the quantization noie, ( ) 1 σ Q = = 1 3 Hence ( ) q 4 SNR 3 0, = = ( ) ( ) = = SNR db 10 log db 0, q mywbut.com 38

39 between When the aumption that quantization noie i uniformly ditributed ±, i not valid, then we have to take a more baic approach in calculating the noie variance. Thi i illutrated with the help of Example 6.9. Example 6.9 Conider the quantizer characteritic hown in Fig. 6.5(a). Let X be the input to the quantizer with f ( ) a) the value of A b) the total quantization noie variance, Q σ X x a hown at Fig. 6.5(b). Find c) I it i the ame a? 1 Fig. 6.5(a): Quantizer characteritic (b) the input PDF (Example 6.9) mywbut.com 39

40 4 X x d x = 1, A 4 a) A f ( ) = 1 4 b) f ( x) X 1 1 x, x 4 = , otherwie Let u calculate the variance of the quantization noie for x 0. Total variance i twice thi value. For x > 0, let 4 σ ' = x f x dx + x f x dx ( 1) ( ) ( 3) ( ) Q X X 0 Carrying out the calculation, we have ' Q 1 σ = and hence, Q 1 σ = 6 3 c) A 4 1 = =, in thi cae we have σ Q the ame a Exercie 6.3 The input to the quantizer of Example 6.9 (Fig. 6.5(a)) i a random variable with the PDF, f X ( x) x Ae, x 4 = 0, otherwie Find the anwer to (a), (b) and (c) of Example 6.9. Anwer: σ Q = Exercie 6.4 A random variable X, which i uniformly ditributed in the range 0 to 1 i quantized a follow: QZ ( x) QZ ( x) = 0, 0 x 0.3 = 0.7, 0.3 < x 1 Show that the root-mean quare value of the quantization i mywbut.com 40

41 Exercie 6.5 A ignal mt () with amplitude variation in the range mp to m p i to be ent uing PCM with a uniform quantizer. The quantization error hould be le than or equal to 0.1 percent of the peak value m p. If the ignal i band limited to 5 khz, find the minimum bit rate required by thi cheme. An: 5 10 bp Non-uniform quantization and companding In certain application, uch a PCM telephony, it i preferable to ue a non-uniform quantizer, where in the tep ize i not a contant. The reaon i a follow. The range of voltage covered by voice ignal (ay between the peak of a loud talk and between the peak of a fairly weak talk), i of the order of 1000 to 1. For a uniform quantizer we have een that ( SNR X ) = where i a contant. Hence ( SNR) 0, q 0, q 1 σ i decided by ignal power. For a, peron with a loud tone, if the ytem can provide an ( SNR) 0, q of the order of 40 db, it may not be able to provide even 10 db ( SNR) 0, q for a peron with a oft voice. A a given quantizer would have to cater to variou peaker, (in a commercial etup, there i no dedicated CODEC 1 for a given peaker), uniform quantization i not the proper method. A non-uniform quantizer with the feature that tep ize increae a the eparation from the origin of the input -output characteritic i increaed, would provide a much better alternative. Thi i becaue larger ignal amplitude are ubjected to coarer quantization and the weaker paage are ubjected to fine quantization. By chooing the quantization characteritic properly, it would be poible to obtain an acceptable ( SNR) 0, q over a wide range of input ignal variation. In other word, uch a 1 CODEC tand for COder-DECoder combination mywbut.com 41

42 quantizer favor weak paage (which need more protection) at the expene of loud one. The ue of a non-uniform quantizer i equivalent to paing the baeband ignal through a compreor and then applying the compreed ignal to a uniform quantizer. In order to retore the ignal ample to their correct relative level, we mut of coure ue a device in the receiver with a characteritic complimentary to the compreor. Such a device i called an expander. A nonuniform quantization cheme baed on thi principle i hown in Fig In the cheme hown, C( x ) denote the output of the compreor for the input x. The characteritic C( x ) i a monotonically increaing function that ha odd ymmetry, C( x) = C( x). Ideally, the compreion and expanion law are exactly invere o that except for the effect of quantization, the expander output i equal to the compreor input. In the cheme of Fig. 6.6, C 1 (). denote the expander characteritic. The combination of a COMpreor and an expander i called a compander. We hall now derive the compreor characteritic C (.), which i capable of giving rie to a fairly contant ( SNR) 0, q over a wide range of the input ignal variation. Fig. 6.6: Non-uniform quantization through companding Let the input ignal x be bounded in the range ( x to x ) characteritic C( x ) analytically define non-uniform interval max max. The k via uniform interval of x max L (Fig. 6.7). L i the number of quantization level and i aumed to be large. Note that the uniform quantizer in Fig. 6.7 i the mid-rier mywbut.com 4

43 variety. When x I 6, C( x) (, ) I ' 6 and the input to the UQ will be in the range. Then the quantizer output i 3. When x Ik, the compreor characteritic C( x ), may be approximated by a traight line egment with a lope equal to xmax L, where k i the width of the interval I k. That i, k ( ) dc x x C ' ( x) = max, x I dx L k k A C ( x) ' x = 0 and k i the larget at x i maximum at the origin, the equivalent tep ize i mallet at =. If L i large, the input PDF f ( ) x max X x can be treated to be approximately contant in any interval I k, k = 1,, L. Note that we are auming that the input i bounded in practice to a value x max, even if the PDF hould have a long tail. We alo aume that fx ( ) i f ( x) = f ( x). Let f ( x) contant = f ( y ) x I X X, k, where X X k xk xk yk = = length of I = x x and k k k + 1 k x i ymmetric; that k + 1 P = P x I = f x dx, and, Then, [ ] ( ) k k X x x k k L = 1 P k = 1 Let q k denote the quantization error when x That i, ( ) q = QZ x x, x I k k I. k = y x, x I k k mywbut.com 43

44 Fig. 6.7: A typical compreion characteritic and non-uniform quantization level Let the correponding random variable be uniformly ditributed in the range k, k. Q k. We can aume Q k to be Q k = k 1 σ Q = variance of the total quantization noie Note that if But k L L k PQ k k Pk k = 1 k = 1 1 = = k x LC ' = for all k, then σ Q =, a expected. 1 max ( x), x I k Hence σ L x Q P max k k = 1 ' 4 1 L 1 C ( x) mywbut.com 44

45 x L max P k C ' ( x ) L k = 1 3 x x L k + 1 max 3L k = 1 x k fx ( x) C ' ( x) dx xmax max X ( ) ' ( ) L x x f x C x dx (6.16a) 3 max Thi approximate formula for Q σ i referred to a Bennet formula. ( SNR) x fx ( x) d x σx 3L = 0, q σ x Q x max max ' X ( ) ( ) xmax f x C x dx xmax x fx ( x) d x L xmax xmax max ' X ( ) ( ) x 3 (6.16b) x f x C x d x max From the above approximation, we ee that it i poible for u to have a contant ( SNR) 0, i.e. independent of σ q X, provided C ' ( x) 1 Kx = where K i a contant, or ( ) C ' x = K x Then, ( ) q SNR A C ( x) ' 0, = K 3L xmax 1 =, we have Kx ln x ( ) C x = log x.) e ln x = + α, with x > 0 and α being a contant. (Note that K mywbut.com 45

46 The contant α i arrived at by uing the condition, when x C( x) = x max. Therefore, Hence, C( x) ln x ln x = + xmax K K ln x α = x max max. K max 1 x = ln + xmax, x > 0 K xmax A C( x) = C( x) we have 1 x C x ln xmax gn x K x max where, gn( x) ( ) = + ( ) 1, x > 0 = 1, x < 0 = x max,, (6.17) A the non-uniform quantizer with the above C( x ) reult in a contant ( SNR) 0, q x, it i alo called robut quantization. The above C( x ) i not realizable becaue C( x) ( ) x lim 0. We want lim C x 0. That i, we have to modify the theoretical companding law. We 0 dicu two cheme, both being linear at the origin. (i) A-law companding: ( ) C x Ax x 1 gn ( x), lna xmax A Ax = 1+ ln xmax 1 x x max gn ( x), 1 1+ lna A xmax (6.18) The contant A decide the compreion characteritic, with A = 1 providing a linear I-O relationhip. The behavior of C( x ) for A = and are hown in Fig mywbut.com 46

47 Fig. 6.8: A-law compreion characteritic A = i the European commercial PCM tandard which i being followed in India. (ii) µ -law companding: x ln 1 +µ xmax C( x) = x max gn x ln( 1 +µ ) C( x ) i again quai-logarithmic. For mall value of x ( x 0) µ x <<, C( x) x max µ ln 1 ( +µ ) ( ) (6.19) >, uch that x, which indicate a linear relationhip between the input x and output C( x ). Note that C( x) C( x) =. If x i uch that µ x >>, then C( x ) i logarithmic, given by x max ( ) C x xmax µ x ln gn ln( 1 +µ ) xmax ( x) mywbut.com 47

48 µ i a contant, whoe value decide the curvature of C( x ) ; µ = 0 correpond to no compreion and a the value of µ increae, ignal compreion increae, a hown in Fig Fig. 6.9: µ -law compreion characteritic µ= 55 i the north American tandard for PCM voice telephony. Fig compare the performance of the 8-bit µ -law companded PCM with that of an 8 bit uniform quantizer; input i aumed to have the bounded Laplacian PDF which i typical of peech ignal. The ( SNR) 0, q for the uniform quantizer i baed on Eq A can be een from thi figure, companding reult in a nearly contant ignal-to-quantization noie ratio (within 3 db of the maximum value of 38 db) even when the mean quare value of the input change by about 30 db (a factor of 3 10 ). G c i the companding gain which i indicative of the improvement in SNR for mall ignal a compared to the uniform quantizer. In Fig. 6.30, tep ize, G c ha been hown to be about 33 db. Thi implie that the mallet min i about 3 time maller than the tep ize of a correponding mywbut.com 48

49 uniform quantizer with identical x max and L, the number of quantization level. min would be maller by a factor more than 3, a a G c of 30 db would give rie to 1 3 min =. ( uniform ) The A - law characteritic, with A = ha a imilar performance a that of µ = 55 law, with a companding gain of about 4 db; that i, for low amplitude ignal, a uniform quantizer would require four additional bit to have the ame ( SNR 0, performance of the A = companded quantizer. (See Exercie 6.9) ) q ) q Fig. 6.30: ( SNR 0, performance of µ -law companded and uniform quantization A number of companie manufacture CODEC chip with variou pecification and in variou configuration. Some of thee companie are: Motorola (USA), National emiconductor (USA), OKI (Japan). Thee IC include the anti-aliaing bandpa filter (300 Hz to 3.4 khz), receive lowpa filter, pin electable A -law / µ -law option. Lited below are ome of the chip number manufactured by thee companie. Motorola: MC to MC mywbut.com 49

50 National: TP3070, TP3071 and TP3070-X OKI: MSM7578H / 7578V / 7579 Detail about thee IC can be downloaded from their repective webite. Example 6.10 A random variable X with the denity function f ( ) X x hown in Fig i given a input to a non-uniform quantizer with the level ± q 1, ± q and ± q 3. Fig. 6.31: Input PDF of Example 6.10 q 1, q and q 3 are uch that q q 1 fx ( x) dx = fx ( x) dx = = fx ( x) d x = (6.0) q q q 3 a) Find q 1, q and q 3. b) Sugget a compreor characteritic that hould precede a uniform quantizer uch that Eq. 6.0 i atified. I thi unique? q q a) X ( ) X ( ) q f x dx = f x dx = q q1 Hence, ( 1 ) 0 x d x = 1 1 mywbut.com 50

51 Solving, we obtain q 1 = Similarly we have equation q 1 fx ( x) dx = (6.1a) q3 1 fx ( x) dx = (6.1b) 6 q From Eq. 6.1(a), we get q = 0.9 and olving Eq. 6.1(b) for q 3 yield q 3 = b) A uniform quantizer will have level at 1 ±, 6 1 ± and 5 ±. 6 Let C( x ) be the compreor characteritic which i a non-decreaing function of it argument and i anti-ymmetrical about x = 0. Any compreion characteritic that doe the following mapping will take care of the requirement of the problem. x C( x) A uch, C( x ) i not unique. Example 6.11 given by Conider a companded PCM cheme uing the µ -law characteritic a mywbut.com 51

52 ( +µ x ) ln 1 C( x) = gn ( x), µ > 0, x 1 ln 1 ( +µ ) Find the ( SNR) 0, q expected of thi ytem when the input X i uniformly ditributed in the range ( 1, 1). Uing Bennet formula, Eq. 6.16(a), let u compute σ Q. From the C( x ) given, we have ( ) dc x dx = C ' = ( x) µ 1 ln 1 + µ 1 + µ x ( ) ( +µ ) C ' ln 1 ( x) = + µ x µ ( 1 ) Q σ = But f ( x) X 1 f X ( x ) C ' ( x ) 3L dx 0 1, x 1 = 0, otherwie Carrying out the integration, ( ) ln 1 +µ µ σ Q = µ 3L µ 3 1 X σ = 3 ( SNR) 0, q µ L µ = µ ln( 1 +µ ) 3 1 If µ 100, then µ >> µ and 1 3 3L µ >>, and ( SNR ) 0, q ln( 1 +µ ) mywbut.com 5

53 Example 6.1 Let ( SNR) 0, q of a µ -law companded PCM be approximated a ( SNR) 0, q ln 1 3L ( +µ ) We will how that ( SNR) 0, (in db) follow the q ( R ) 6 + α rule. Let ( SNR) R L =. Then 0, q 3 ln 1 R C 4 Then ( ) ( ) 0, q ( +µ ), where C R ( ) = 3 ln 1 SNR in db = 10 log C + 6R 10 ( +µ ) =α+ 6R Example 6.13 A muic ignal band-limited to 15 khz i ampled at the rate of ample/ec and i ent uing 8 bit µ -law (µ = 55) companded PCM. ( ) q SNR 0, of thi ytem wa found be inadequate by atleat 10 db. If the ampling rate i now reduced to improvement in ( SNR) 0, q ample/ec, let u find the expected, when the bit rate i not to exceed the previou cae. With f 3 the tranmitted bit rate a = ample/ec, and auming 8 bit/ample, we have = bit/ec. With ( ) f 3 = 35 10, number of bit/ample that can be ued mywbut.com 53

54 R = = A R ha to be an integer, it can be taken a 10. A R ha increaed by two bit, we can expect 1 db improvement in SNR. Exercie 6.6 Conider the three level quantizer (mid-tread type) hown in Fig The input to the quantizer i a random variable X with the PDF f ( ) X x give by f X ( x) 1, x 1 4 = 1, 1 < x 3 8 Fig, 6.3: Quantizer characteritic for the Exercie 6.5 a) Find the value of A uch that all the quantizer output level are equiprobable. (Note that A ha to be le than 1, becaue with A = 1, we have P QZ( x) 1 = 0 = ) b) Show that the variance of the quantization noie for the range x A, i mywbut.com 54

55 Exercie 6.7 Let the input to µ -law compander be the ample function x ( ) proce X() t co( f t ) m j t of a random = π + Θ where Θ i uniformly ditributed in the range ( 0, π ). Find the ( SNR) 0, q Note that ample of x () t will have the PDF, 1, x 1 fx ( x ) = π 1 x 0, otherwie j expected of the cheme. Anwer: ( SNR) 0, q 3 µ L µ 4µ = ln( 1 ) +µ π 1 Note that if µ >> µ >> 1, we have ( ) 3L SNR 0, q ln ( 1 +µ ) Exercie 6.8 Show that for value of x uch that Ax A-law PCM i given by ( SNR) q 0, = 6R + α where α = log [ 1 + ln A] 10 >> x max, ( SNR) 0, q of the Exercie 6.9 Let C( x ) denote the compreion characteritic. Then ( ) dc x dx x 0 i called the companding gain, G c. Show that a) G c (A-law) with A = i (and hence 0 log10 G c 4 db) b) G c (µ -law) with µ= 55 i 46.0 (and hence 0 log10 G c 33 db.) mywbut.com 55

56 Exercie 6.10 Show that for the µ -law, maximum tep ize minimum tep ize lim C ' ( x) max x 0 = = = µ + min lim C ' x xmax ( x) ( 1). 6.5 Encoding The encoding operation convert the quantized ample into a form that i more convenient for the purpoe of tranmiion. It i a one-to-one repreentation of the quantized ample by uing code element or ymbol of the required length per ample. In the binary code, each ymbol may be either of two ditinct value or kind, which are cutomarily denoted a 0 and 1. In a ternary code, each ymbol may be one of three ditinct value; imilarly for the other code. By far, the mot popular from the point of view of implementation are the binary code. With R - binary digit (bit) per ample, we can have R ditinct code word and we require R (number of quantization level), o that it i poible for u to maintain a one-to-one relationhip between the code word and the quantization level. Let u identify the R -bit equence a b R b R b b b. In the natural binary code, thi equence repreent a number (or level) N, where R R R ( ) R ( ) N = b + b + + b + b (6.) Natural binary code reult when the codeword ymbol or digit are aigned to N, with N lited in an increaing or decreaing (decimal) order; that i, though the quantized ample could be either poitive or negative, we imply mywbut.com 56