Part A: Signal Processing Chapter 3: Digital Signal Processing

Transcription

1 Prt A: Signl Processing Chpter 3: Digitl Signl Processing

2 Chpter 3: 3. Introduction to Digitl Signl Processing 3. Anlogue to Digitl Conversion Process 3.3 Quntistion nd Encoding 3.4 Smpling of Anlogue Signls 3.5 Alising 3.6 Digitl to Anlogue Conversion 3.7 Introduction to digitl filters 3.7. Non-Recursive Digitl filters 3.7. Recursive Digitl filters 3.8 Digitl filter Relistion 3.8. Prllel Relistion 3.8. Cscde Relistion 3.9 Mgnitude nd Phse Responses 3.0 Minimum/Mximum/Mixed Phse Systems 3. All-pss Filters 3. Second Order Resonnt Filter 3.3 Stbility of Second Order Filter 3.4 Digitl Oscilltors 3.5 Notch Filters Problem Sheet A3 Digitl Signl Processing

3 Chpter 3: Digitl Signl Processing DSP 3.0 Introductions Digitl Signl Processing is rpidly developing technology for scientists nd engineers. In the 990s the digitl signl processing revolution strted, both in terms of the consumer boom in digitl udio, digitl telecommunictions nd the wide used of technology in industry. Due to the vilbility of low cost digitl signl processors, mnufcturers re producing plug-in DSP bords for PCs, together with high-level tools to control these bords.

4 There re mny res where DSP technology is now being used nd the current prolifertion of such technology will open up further pplictions. Audiologists nd speech therpists re exposed to DSP systems for both testing person s level of hering nd subsequently DSP hering id filtering. The professionl music industry uses spectrum nlysers, digitl filtering, smpling conversion filters etc nd is one of the biggest users nd exploiters of DSP technology.

5 In summry, DSP is pplied in the re of control nd power systems, biomedicl engineering, instrumenttion test nd mesurement, utomotive engineering, telecommunictions, mobile communiction, speech nlysis nd synthesis, udio nd video processing, seismic, rdr nd sonr processing nd neurl computing.

6 There re mny dvntges to using DSP techniques for vriety of pplictions, these include: high relibility nd reproducibility flexibility nd progrmmbility the bsence of component drift problem compressed storge fcility

7 DSP hrdwre llows for progrmmble opertions. Through softwre, one cn esily modify the signl processing functions to be performed by the hrdwre. For ll these resons, there hs been vst growth in DSP theory & pplictions over the pst decde.

8 3. Introductions to DSP An nlogue signl processing system is shown in Fig 3., in which both the input signl nd output signl re in nlogue form Anlogue Input Signl xt st nt signl noise Anlogue signl processor e.g. low-pss filter Anlogue Output Signl st Figure 3.. A generl description of nlogue systems whose input nd output re in nlogue form.

9 A digitl signl processing system Figure 3. provides n lterntive method for processing the nlogue signl. nlogue prefilter or ntilising filter nlogue to digitl converter db 3 db xn xt A/D x t Converter Lowpss filtered signl Smpling frequency Digitl Signl Processor discrete-time signl sn D/A converter Digitl to nlogue converter Figure 3.. A generl process of converting nlogue signls into digitl signls nd bck to nlogue form. db st reconstruction filter nlogue filter sme s the pre-filter

10 Note: The A/D converter converts the nlogue input signl into digitl form. The D/A converter converts the processed signl bck into nlogue form. The reconstruction filter smooths out the outputs of the D/A nd removes unwnted high frequency components. The nlogue input filter is used to bnd-limit the nlogue input signl prior to digitistion to reduce lising see lter The hert of the system in Figure 3. is the digitl signl processor which my be bsed on DSP chip such s Texs instruments TMS 30C60

11 The digitl signl processor my implement one of the severl DSP lgorithms, for exmple digitl filtering low-pss filter mpping the input x[n] into the output s[n]. Digitl signl processor implies tht the input signl must be in digitl form before it cn be processed.

12 3. Anlogue to digitl conversion Process Before ny DSP lgorithm cn be performed, the signl must be in digitl form. The A/D conversion process involves the following steps: The signl Bnd-limited is first smpled, converting the nlogue signl into discrete-time signl The mplitude of ech smple is quntised into one of B levels where B is the number of bits used to represent smple in the A/D converter The discrete mplitude levels re represented or encoded into distinct binry words ech of length B bits.

13 A prcticl representtion of the A/D conversion process is shown in Figure 3.3. x t Anlogue Signl bndlimited Smple & Hold close & open the switch t f s H B A/D converter Quntiser Figure 3.3. Anlogue to digitl conversion process. T encoder Logic circuit B bits digitl output x[n]

14 Smple nd hold S/H tkes snpshot of the nlogue signl every T sec nd then holds tht vlue constnt for T secs until the next snpshot is obtined. x t S/H output T f s T Input signl Figure 3.4. An exmple of smple nd hold process to convert nlogue signls into digitl signls. t t

15 Exmple : V x[n] 6V 0V -6V -V T smpling period Smples Smpled Signl Anlogue Signl Figure 3.5. An exmple of smpling nlogue signls to discrete-time signls. The smpling period is T. n

16 Exmple: 4-bit B 4 A/D converter bipolr digitl nlogue Input-output chrcteristic of 4-bit quntiser liner two s complement nottion

17 3.3 Quntistion nd encoding [] Before conversion to digitl, the nlogue smple is ssigned one of B vlues see Fig 3.6. This process, termed quntition, introduces n error, which cnnot be removed.

18 Quntistion Level Quntistion Level LSB encoder output smpling instnts Figure 3.6. Quntistion of discrete-time signls. 0 3-bit A/D Converter Unipolr 3 bits code output

19 A bit A/D converter bipolr with n input voltge rnge of ±0V will hve lest significnt bit LSB of 0V mv 0 bits bits -0V mv resolution 0V levels 4096 Resolution step-sie 0V ΔV 4. 9mV

20 Note: level n level n v level n- V/ V/ smpling instnt V ΔV Quntistion error one hlf of n V LSB 4.9 mv /.45 mv

21 For n A/D converter with Binry digits the number of quntistion level is B, nd the intervl between levels, tht is the quntistion step sie ΔV resolution is given by ΔV V B V-full scle rnge of the A/D converter with bipolr signl inputs. The mximum quntistion error, for the cse where the vlues re rounded up or down. V B

22 For sine wve input of mplitude A, the quntistion step sie becomes ΔV The quntistion error e for ech smple, is normlly ssumed to be rndom nd uniformly ΔV distributed in the intervl ± with ero men. In this cse, the quntistion noise power or vrince is given by A B -A A A e ctul mplitude - quntised mplitude

23 σ ΔV e P e de ΔV e Hence, ΔV constnt σ ΔV e ΔV ΔV ΔV e de for uniform quntistion Note : Uniform quntistion - ll steps re of equl sie

24 For the sine wve input, the verge A A signl power is, ie. rms vlue The signl-to-quntistion noise power rtio SQNR in decibels is SQNR 0log A ΔV 3 0log 0log B A B A/ SQNR 6.0B.76 db

25 The SQNR increses with the number of bits, B. In mny DSP pplictions, n A/D converter resolution between nd 6 bits is dequte. Number of Bits Levels db Thus, the signl-to-quntistion noise rtio increses pproximtely 6dB for ech bit. 8 SQNR 8.7 db 5.3 db 3.6 db 37.7 db

26 Summry: Anlogue to Digitl Converter Exmple: xt 0 to 5 volts -bit A/D unipolr Step sie or Resolution Conversion time sy 35 μs -bit 5 volts Smpling frequency x[n]

27 In prctice, the A/D is proceeded by smple nd hold S/H which freees the signl during conversion. Two prmeters relted to S/H: perture time 5 ns for exmple cquisition time μs for exmple Then the mximum frequency tht cn be converted becomes f s 7kH mx mximum smpling frequency for the bove A/D converter.

28 3.4 Smpling of Anlogue Signl Suppose tht n nlogue signl xt is smpled every T seconds, then t the output of the smpler we obtin discrete-time signl xn xt t nt. xt xω A/D Smpling freq. f s T x[n] X ωt f f s

29 x t X ω e ω t d ω tnt x [] n ω ω nt X e d ω Inverse Fourier Trnsform 3.

30 Now e nωt is periodic function of period. Eqution 3. becomes [] ω ω ω d e X n x nt 3. [] T d e T k X T T d e T k X T T T d e X n x nt nt T k k k k nt k ω ω ω ω ω ω ω ω ω

31 Let ωt x [] n We hve T By compring 3. nd 3.3, We obtin Digitl spectrum k X ω k X e T n [] n x n X e d 3.3 d Inverse Fourier Trnsform for discrete signl X X ω k < < T k T Anlogue spectrum 3. Repets itself 3.4

32 It is seen tht X is periodic with period. The digitl spectrum is repetition of the nlogue spectrum.

33 3.5 Alising Figure 3.7 illustrtes the reltionship between the digitl spectrum X nd the nlogue spectrum Xω for the cse Xω 0, ω > f T or f > s.

34 Cse : Xω 0, ω > T smpling theorem holds Xω A -/T /T ω X A/T Anlogue Spectrum ω f s Figure 3.7. Above : Frequency response of n nlogue signl. Below : Frequency response of the smpled nlogue signl. f s f s Digitl Spectrum

35 Note: ω corresponds to or f T The digitl spectrum is the sme s the originl nlogue spectrum nd repets t multiples of the smpling frequency f s. f s

36 ω > Cse : Xω 0,, but Xω 0, 3 T T T Xω A X ω > lising Figure 3.8. Above: Frequency response of n nlogue signl whose highest frequency component is lrger thn the smpling frequency. Below: Frequency response of the smpled nlogue signl. The overlpped region represents lising. T A/T 3 T 3 ω T

37 If the smpling frequency, f s is not sufficiently high, the spectrum centred on f s will fold over or lis into the bse bnd frequencies Fig 3.8. Eqution 3.4 tells us tht lising cn only be voided if the nlogue signl is bnd limited such tht Xω 0,. ω > This results in the fmilir smpling theorem. The minimum smpling frequency for which eqution 3.4 holds is clled the Nyquist frequency. f s f T f T

38 Note: It should be noted tht even if Xω is not strictly bnd limited, tht it hs some negligible energy outside, smll enough T cn be chosen T so tht the overlp of the components of the summtion in eqution 3.4 is below prescribed level. This is importnt when smpling frequency is to be selected for prticulr signl.

39 Digitl-to-Anlogue Conversion D/A Signl recovery The D/A conversion process is employed to convert the digitl signl into n nlogue form fter it hs been digitlly processed. The reson for such conversion my be for exmple, to generte n udio signl to drive loudspeker or to sound n lrm. The D/A process is shown in Figure 3.9. A register is used to buffer the D/A s input to ensure tht its output remins the sme until the D/A is fed the next digitl input.

40 Note: The inputs to the D/A re series of impulses, while the output of the DAC hs stircse shpe s ech impulse is held for time T sec. y[n] Digitl Signl Processor y[n] 8 or bits T-Smpling period n D/A yˆ t yˆ t T Smpling period Low pss filter t yt reconstruction filter or smoothing filter Figure 3.9. Conversion process from digitl signls to nlogue signls.

41 The D/A shown in Figure 3.9 is referred to s ero-order hold. By compring its output nd its input y[n], it is evident tht for ech digitl code fed into the D/A, its output is held for time T. The result is the chrcteristic stircse shpe the D/A output. The D/A output pproximtes the nlogue signl by series of rectngulr pulses whose height is equl to the corresponding vlue of the signl pulse. yˆ t

42 Just consider one pulse. h t 0 t T 0 otherwise ht T t

43 The corresponding frequency response is sin ] [ 0 0 T T e T T e e e T e e e e w e dt e dt e t h H T T T T T T wt wt T t T t t ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω

44 The mgnitude of Hω is plotted in Figure sin x x 7 7 Hω T ω Figure 3.0. Mgnitude response of rectngulr pulse.

45 In the frequency domin, the stircse ction of the DAC introduces type of distortion known s the ωt or perture distortion, where. Y x input to the D/A yˆ ω sin x D/A output x ω sin x x

46 The mplitude of the output signl spectrum is multiplied by the function, which cts like lowpss filter, with the high frequencies hevily ttenuted. sin x x The effect is due to the holding ction of the DAC nd, in signl recovery, introduces n mplitude distortion. sin x x

47 For ero-order hold, the function flls to bout 4dB t hlf the smpling frequency giving n verge error of bout 36.4%. Aperture error cn be eliminted by equlition. In prctice this cn be chieved by first pplying the signl, before converting it to nlogue, through digitl filter whose mplitude-frequency response hs shpe. sin x x x sin x f s

48 Reconstruction Filter [4] The output of the D/A converter contins unwnted high frequency t multiples of the smpling frequency s well s the desired frequency components. The role of the output filter is to smooth out the steps in the D/A output thereby removing the unwnted high frequency components. In generl, the requirements of the nti-imging filter re similr to those of the nti-lising filter.

49 Idel D/A Converter y[n] ^yt Idel D/A Impulse not squre pulses s in the cse of n non-idel D/A T T T H ω T Idel Lowpss filter yt

50 Consider Just n impulse δ t T H ω ω T T ht t t h t sin ht

51 If we pply the signl to the input of the filter, we obtin y t h t* yˆ t sin t / T * t / T yˆ t n y n δ t y[n] cn be written in this form. y t n nt sin t / T y[] n δ t nt * t / T

52 Using the property we cn obtin x t * δ t t0 x t t0 y t n t nt sin [] T y n 3.5 t nt T

53 The originl signl cn be obtined by dding together n infinite number of pulses. The n th sin x x pulse here is shifted through distnce nt with respect to the origin nd multiplied weighted by fctor y[n]. sin x x This recovery process is clled interpoltion. Figure 3. shows the implementtion of eqution 3.5.

54 y- y- -T y0 0 T y T yx xt originl signl y Figure 3.. Ech discrete-time smple is multiplied by shifted sinc function. Summing these sinc functions will produce the originl nlogue signl. t

55 The signl xt is reconstructed from the x[ n] y[ n] smples of by summtion of weighted nd shifted sin x x pulse.

56 Exmple: Consider the nlogue signl [] xt 3 cos50t 0 sin 300t cos 00t Wht is the Nyquist rte for this signl? The frequencies present in the signl bove re f 5H; f 50 H; f 3 50 H Hence, f mx 50 H f smpling > f mx 300 H The Nyquist rte is f N f mx 300 H.

57 Note: Consider xt 0 sin 300t fs f 300 H x [] n 0sin 300nT x[n] 300 0sin n f s 0sin n

58 x We re smpling the nlogue sinusoid t its ero-crossing points nd hence we miss the signl completely. The sitution will not occur if the sinusoid is offset by some phse here. In such cse we hve t sin t φ x nd [] n 0sin n φ 0[ sin n cos φ cos n sin φ ] 0cos n sin φ T, where f s 300H. f s for n 0,,,..

59 Since cosn - n, x[ n] n 0 sin φ If 0 or, the smples of the sinusoid tken t the Nyquist rte re not ll ero. Note: xt A cosf 0 t is continuous-time sinusoidl signl x n f s Acos fs f0 f f 0 s n 3.6

60 On the other hnd, if the sinusoids, [where f k f 0 kf s, k ±, ±, ±3,.] 3.7 re smpled t rte f s, it is cler tht the frequency f k is outside the fundmentl frequency rnge ; consequently the smpled signl is t f A t x k cos 0 s f s f f [] kn n f f A n f kf f A n f f A n x s s s s k cos cos cos 0 0 [] 3.8 cos 0 n f f A n x s

61 x [] 0 n Acos n Which is identicl to the discrete-time signl in eqution 3.6. If we re given sequence x[n] there is n mbiguity s to which continuous-time signl xt these vlues represent. We cn sy the frequencies f k f 0 kf s re indistinguishble from the frequency f 0 fter smpling nd hence they re lises of f 0. f f s

62 Note: f s smpling frequency f s corresponds to f s is the highest frequency tht cn be represented uniquely with smpling rte f s f s is clled hlf the smpling frequency or folding frequency. ωt f f s

63 Exmple : Consider the nlogue signl x t 3 cos 000t 5sin 6000t 0cos 000t Wht is the Nyquist rte for this signl? The frequencies existing in the nlogue signl re: f kh; f 3 kh; f 3 6 kh Thus f mx 6 kh nd ccording to the smpling theorem, f s > f mx kh The Nyguist rte is kh.

64 b Assume now tht we smple this signl xt using smpling rte f s 5 KH smples/sec. Wht is the discrete-time signl obtined fter smpling? f s 5000H 500 f s xt 3 cos 000t 5 sin 3000t 0 cos 6000t [] n n n n n n n n n n n n n x 5 0 cos 5 5sin 5 3cos 5 0 cos 5 5sin 5 3cos cos 5 3 5sin 5 3cos cos sin cos

65 [] x n 3cos n 5sin n 5 5 The sme result cn be obtined using eqution 3.7. Second Method: We hve f k f 0 kf s ; f 0 f k kf s f s 5 kh cn be obtined by subtrcting from f k n integer multiple of f s such tht s f s. f s. 5 f f 0 kh

66 The frequency f 000 H is < 500 H nd thus it is not ffected by lising. However, the other two frequencies f & f 3 re bove the folding frequency nd they will be chnged by the lising effect. f ' f f s kh f 3 ' f 3 f s kh x[] n 3cos n 5sin n 0 cos n This is greement with the result obtined before. f s

67 c Wht is the nlogue signl yt we cn reconstruct from the smples if we use idel interpoltion. Since only frequency components t kh nd kh re present in the smpled signl, the nlogue signl we cn recover is, yt3cos000t-5sin4000t which is obviously different from the originl signl xt. The distortion of the originl nlogue signl ws cused by the lising effect, due to the low smpling rte used.

68 Exmple: An nlogue signl xt sin480t3sin70t is smpled 600 times per second. [] Determine the Nyguist smpling rte for xt b Determine the folding frequency or hlf the smpling frequency c Wht re the frequencies, in rdins, in the resulting discrete time signl x[n]? d If x[n] is pssed through n idel D/A converter wht is the reconstructed signl yt

69 xt sin 40t 3sin 360t f 40 H fmx 360 H f 360 H F Nyquist f mx 70 H f s b f s 600 H f fold or 300 H

70 c [] n n n n n n n n n t x n x nt t 5 4 sin 5 4 3sin 5 4 sin 5 4 3sin 5 4 sin 5 6 3sin 5 4 sin sin sin

71 Second Method We hve f k f 0 kf s nd therefore f 0 f k k f s f 40 H nd is < 300 H f 360 H nd is 300 H f s f s 0 f s -not ffected by lising -ffected by lising < f s Alised frequency f 0 f k k f s H f

72 [] n n n n x 5 4 sin sin sin t t y n f n nt t n n y s 480 sin sin d

73 Note : bit rte f s no of bits 8000 smples/sec 96,000 smples/sec xt -bits A/D f s 8,000 kh x[n]

74 b In the cse of PCM, speech signls re filtered to remove effectively ll frequency components bove 3.4 kh nd the smpling rte is 8000 smples per sec speech signl xt 8-bitscompressed PCM 8-bit persmple kh A/D converter f s 8000 H 8000 smples/sec Bit rte bits per second smpling frequency bits/smple 8000 smples/second 8-bits/smple 64,000 bits/sec

75 c 6 bit f s 44.kH CD 6 bit f s 44. kh CD Reder bit rte bits/sec Mbits/sec 6 bit D/A lowpss filter AMP

76 3.7 Introduction to Digitl Filters There re two types of digitl filters: Recursive there is t lest one feedbck pth in the filter Non-recursive no feedbck pths A liner time invrint discrete LTD system described by the following eqution is commonly clled digitl filter: y M [ n] x[ n k] b y[ n k] 3.9 k 0 k Feed forwrd L k k Feedbck

77 y M [ n] x[ n k] b y[ n k] 3.9 k 0 k where x[n] is the input signl, y[n] is the output signl. 0,,,..., M ; b, b, b 3,..., b L re constnts filter coefficients. These coefficients determine the chrcteristics of the system. when b k 0 the filter is sid to be non-recursive type k when b k 0 recursive type. L k

78 3.7. Non Recursive Digitl Filters FIR If b k 0, then the clcultion of y[n] does not require the use of previously clculted smples of the output see eqution 3.9. This is recognised s convolution sum. [] [ ] [] [ ] [ ] k n x n x n x k n x n y M M k k L 0 0 [] [][ ] M k k n x k h n y 0

79 y Therefore the impulse response is identicl to the coefficients, tht is, M h [] n 0 n 0 Any filter tht hs n impulse response of finite durtion is clled Finite Impulse Response FIR filter. n M otherwise [] n x[ n k] x[] n x[ n ] L x[ n k] k 0 k 0 h0 h M hm

80 Exmple : [] [ ] [ ] [ ] 0 0 X Y H n x n x n x n y This is non-recursive second order FIR filter Property: A property of the FIR filter is tht it will lwys be stble.

81 Stbility requires tht there should be no poles outside the unit circle. This condition is utomticlly stisfied since there re no poles t ll outside the origin. In fct, ll poles re locted t the origin. b Another property of non-recursive filter is tht we mke filters with exctly liner phse chrcteristics [4] Note: The bility to hve n exctly liner phse response is one of the most importnt properties of LTD system filter. When signl psses through filter, it is modified in mplitude nd/or phse. The nture nd extent of the modifiction of the signl is dependent on the mplitude nd phse chrcteristics of the filter.

82 The phse dely or group dely of the filter provides useful mesure of how the filter modified the phse chrcteristic of the signl. If we consider signl tht consists of severl frequency components eg. speech wveform the phse dely of the filter is the mount of time dely ech frequency component of the signl suffers in going through the filter. phse _ dely T p φ 3.0 [the negtive of the phse ngle divided by frequency]

83 The group dely on the other hnd is the verge time dely the composite signl suffers t ech frequency s it psses from the input to the output of the filter. group _ dely Tg dφ 3. [the negtive of the derivtive of the phse with respect to frequency]

84 φ φ - - Figure 3.. Phse response of liner phse filter A constnt group dely mens tht signl components t different frequencies receive the sme dely in the filter. A liner phse filter gives sme time dely to ll frequency components of the input signl. A filter with nonliner phse chrcteristic will cuse phse distortion in the signl tht psses through it.

85 This is becuse the frequency components in the signl will ech be delyed by n mount not proportionl to frequency, thereby ltering their hrmonic reltionship. Such distortion is undesirble in mny pplictions, for exmple music, video etc. A filter is sid to hve liner phse response if its phse response stisfies one of the following reltionships: φ φ b 3. where nd b re constnts.

86 Exmple Two filter structures re shown below. Show tht both filters hve liner phse. x[n] x[n] y[n] y[n]

87 - - - x[n] y[n] [] [] [ ] [ ] cos e e e e e e H H n x n x n x n y - - x[n] y[n] [] [ ] [ ] sin sin e e e e e e e H H n x n x n y phse: φ / -, liner phse phse: φ -, liner phse

88 3.7. Recursive Digitl filter IIR Every recursive digitl filter must contin t lest one closed loop. Ech closed loop contins t lest one dely element. y M [] n [ ] [ ] k x n k bk y n k k 0 k For Recursive digitl filters b k 0. Let 0 0, k 0 for k > 0, b b & b k 0 for k > y H [ n] x[ n] b y[ n ] 0 b 0 IIR filter L

89 A recursive filter is n infinite impulse response filter IIR. Exmples : H H H 0 b b b b nd order FIR filter Zeros only nd order IIR filter ll pole filter IIR filter Poles nd Zeros

90 Note: Exmple : The difference eqution is: y[n] x[n] y[n-]. The DC gin of H cn be obtined by substituting 0. If dc gin is undesirble, introduce constnt gin fctor of -, so tht H becomes Poles nd eros cn be rel or imginry H 0 < < H 0 0 e [ n] x[ n] y[ n ] H y dc gin

91 Exmple: Consider lowpss filter y[n- ] bx[n], 0 < < y[n] i Determine b so tht H0 ii Determine the 3dB bndwidth here for the normlised filter in prt i

92 b i Y Y - b X H H H b e b H 0 H e cos we hve H0 b - cos sin sin cos

93 Second Method: * cos e e H H H 0 0 H c H H hlf-power point H 0 db 3 db c cos H

94 c cos cos c 4 H c

95 Exmple: Consider filter described by H c where & c re constnts. c Show tht the mgnitude response H is unity for ll.

96 ] [ ] [ * e e c c e e c c ce e c ce e c H H H H - This is n ll-pss filter.

97 3.8 Digitl filter Relistion [] [ ] [ ] [ ] structure poles L k k k eros M k k k L k k k M k k k L k k k M k k k L k k M k k b b X Y H Y b X Y k n y b k x n y structure eros M k k k poles L k k k b

98 X M -b -b -b L Figure 3.3. Structure or Direct Form Y

99 X -b -b -b L M Figure 3.4. Structure or Direct Form II Y

100 In the cse when L M, we hve Cnonic form relistion. X -b -b -b L A discrete-time filter is sid to be cnonic if it contins the minimum numbers of dely elements necessry to relise the ssocited frequency response. 0 M Figure 3.5. Cnonic form Y

101 3.8. Prllel Relistion [] L L structure prllel k k i i L L M M H H H H H b b b H _ use prtil frction to obtin H i Y H H H k X Figure 3.6. Prllel structure

102 3.8. Cscde Relistion [] L L structure cscde k k i i L L M M H H H H H H b b b H _ 3 0 ˆ... ˆ ˆ ˆ ˆ Product of lower order trnsfer function ie. st or nd order sections The cscde structure is the most populr form Y X Fig 3.7. Cscde structure ˆ H ˆ H ˆ H k

103 Exmple: A prllel relistion of third order system H is given by D C B A H H 0 H H

104 H H

105 x[n] y[n]

106 Exmple: A cscde relistion of third-order system is given by H

107 x[n] y[n] Cscde

108 Exmple: Implement the following system in the cscde, direct form II nd prllel structures. All coefficients re rel.. H b x[n] - - cscde structure b y[n] -

109 H b b x[n] y[n] --b - b - Direct form II

110 H A B b b b b x[n] b - b - - b b b y[n] Prllel structure

111 b H 3 x[n] - cscde structures No prllel structure exists becuse prtil frction expnsion cnnot be performed. - y[n] -

112 H 3 3 x[n] y[n] Direct Form II

113 cscde b b b b prllel b H c. b b H y[n] x[n] - - prllel structure b - b -

114 x[n] H b -b b b - -b - y[n] b b b

115 3.9 Mgnitude nd Phse Response [] We cn show tht the mgnitude response is n even function of frequency The phse response is n odd function of frequency

116 Exmple: Clculte the mgnitude nd phse response of the 3-smple verger given by [] otherwise n n h 0 3 n n n n n h n h H [ ] 3 H

117 [ ] [ ] 3 3 e e e e e H H [ ] cos 3 H Precutions must be tken when determining the phse response of filter hving rel-vlued trnsfer function, becuse negtive rel vlues produce n dditionl phse of rdins.

118 For exmple, let us consider the following liner-phse form of the trnsfer function H e -k B rel-vlued function of tht cn tke positive nd negtive vlues. H H B cos k B sin k B cos k B sin k Let phse ngle is φ : tnφ B sin k B cos k tn k

119 tnφ tn-k φ -k phse ngle The phse function φ includes liner phse term nd lso ccommodtes for the sign chnges in B. Since - cn be expressed s, phse umps of ± will occur t frequencies where B chnges sign. If B > 0, then φ -k. If B < 0, then φ -k ±.. or φ -k

120 Let us get bck to our exmple ω ω ω φ φ < < < < < ± > nd ] cos [ 3 ] cos [ 3 H H H H

121 The pproprite sign of must be chosen to mke φ n odd function of frequency. H - - -/3 /3 φ - - -/3 /3 - Even function Odd function

122 Exmple: Find the mgnitude nd phse response of the following: 0,,, 0 n h h h h φ ] cos [ B e H e e e e e H H The mplitude function is never negtive therefore there is no phse umps of ±

123 Even function Odd function H - φ - -

124 Exmple : δ δ n n 0 H n 0 otherwise cse φ - -

125 h[n] δ[n-k] H -k H e -k δn-k B k n Note: When phse exceeds ± rnge ump of ± is needed to bring the phse bck into ± rnge. φ -k H - φ - -

126 Phse Jumps: From the previous exmples, we note tht there re two occsions for which the phse function experiences discontinuities or umps. A ump of ± occurs to mintin the phse function within the principl vlue rnge of [- nd ] A ump of ± occurs when B undergoes chnge of sign The sign of the phse ump is chosen such tht the resulting phse function is odd nd, fter the ump, lies in the rnge [- nd ].

127 Exmple: Mgnitude nd phse response of cusl 3-smple verge. [] [ ] φ φ < < < < < ± < < > ] cos [ 3 ; ] cos [ otherwise 0 0 for 3 B B B H e H e e e e e H H n n h B φ - H - -/ / /3 -/3 Phse is undefined t points H 0.

128 Exmple: Determine nd sketch the mgnitude nd phse response of the following filters: i ii iii y y [] n x[] n x[ n ] [] n x[] n x[ n 8] [] n x[ n 4] y

129 sin sin ] [ ] [ ] [ e e e e e e e e e e e H H X X Y H - / -/ φ - i

130 φ 4 sin 4 sin B e e e e e e e H X Y X X Y φ H /4 / 3/ -/ / /4 / 3/ ii

131 iii H Y H H y - [ n] x[ n 4] 4 X H e 4 4 φ -4 /4 -

132 Exmple: Determine nd sketch the mgnitude nd phse response of st order recursive filter IIR filter [] [ ] [ ] n y n x n y phse e e e H e H H X Y φ φ cos sin tn cos sin cos cos cos sin tn cos sin cos cos

133 Mgnitude: H H H * [H * is the complex conugte] e e H Even Symmetry Odd Symmetry Assuming 0 < < φ - Non-liner phse H cos

134 Exmples : k0. H Low - Pss Filter The gin k 0 cn be selected s, so tht the filter hs unity gin t 0. b. H k0 In this cse, for unity gin t 0.

135 The ddition of ero t - further ttenutes the response of the filter t high frequencies H H H - Low-Pss Filter

136 c We cn obtin simple high-pss filters by reflecting folding the pole-ero loctions of the low-pss filters bout the imginry xis in the -plne. H - 3 H 3 High pss filter

137 d e H y H 4 [] n x[] n x[ n ] 5 y [] n x[] n x[ n ] - Low-Pss Filter High-Pss Filter H 5 H 5 -cos

138 f g H H H H 6 4-cos 6 H 7 H 7 8-cos 3

139 3.0 Minimum-phse, Mximum-phse nd Mixed phse systems [] Let us consider two FIR filters: H H ρ ρ -0.5

140 H is the reverse of the system H. This is due to the reciprocl reltionship between the eros of H & H. The mgnitude chrcteristics for the two filters re identicl becuse the roots of H & H re reciprocl. cos 4 5 & H H e H e H

141 Phse: φ tn φ tn φ - - sin sin cos cos φ - -

142 Note: If we reflect ero ρ tht is inside the unit circle into ero outside the unit circle the mgnitude chrcteristic of the system is unltered, but the phse response chnges. We observe tht the phse chrcters φ begins t ero phse t frequency 0 nd termintes t ero phse t the frequency ω. Hence the net phse chnge. φ φ 0 0 ρ Minimum phse filter

143 On the other hnd, the phse chrcteristic for the filter with the ero outside the unit circle undergoes net phse chnge φ φ 0 rdins As consequence of these different phse chrcteristics, we cll the first filter minimumphse system nd the second system is clled mximum-phse system. If filter with M eros hs some of its eros inside the unit circle nd the remining outside the unit circle, it is clled mixed-phse system.

144 A minimum-phse property of FIR filter crries over to IIR filter. Let us consider H B A is clled minimum phse if ll its poles nd eros re inside the unit circle. Minimum phse Re

145 If ll the eros lie outside the unit circle, the system is clled mximum phse. Re Mximum phse

146 If eros lie both inside nd outside the unit circle, the system is clled mixed-phse. Re Mixed phse

147 Note: For given mgnitude response, the minimum-phse system is the cusl system tht hs the smllest mgnitude phse t every frequency. Tht is, in the set of cusl nd stble filters hving the sme mgnitude response, the minimumphse response exhibits the smllest devition from ero phse.

148 Exmple: Consider fourth-order ll-ero filter contining double complex conugte set of eros locted t 0.7e ± 4. The minimum-phse, mixed phse nd mximum phse system pole-ero ptterns hving identicl mgnitude response re shown below. 4 Minimum-phse ρ0.7 4 ρ /ρ mixed-phse mximum-phse 4

149 The mgnitude response nd the phse response of the three systems re shown below: The minimumphse system seems to hve the phse with the smllest devition from ero t ech frequency H φ minimum phse -4 mximum phse mixed-phse In the cse liner phse

150 Exmple: A third order FIR filter hs trnsfer function G given by From G, determine the trnsfer function of n FIR filter whose mgnitude response is identicl to tht of G nd hs minimum phse response. G 6 5 G G 3 >

151 lm 3 The Minimum phse filter P k Re

152 %Exercise. Minimum nd mximum phse filter %We exmine the properties of minimum nd mximum phse filter %We strt with mximum phse filter from the exmple in pge 64. eros_mxphse [3/ -4/3-5/]' % eros position %Note the trnspose opertion since the inputs re eros nd poles. %If the inputs re row vectors, they will be treted s polynomil coefficents. poles_mxphse [0 0 0] %poles re locted t origin %After specifying the poles nd eros position, we now plot their position. figure %crete new figure No. plneeros_mxphse, poles_mxphse %We lso find the numertor nd denomintor coefficients of the trnsfer function Imginry Prt Rel Prt 3

153 % in order to plot the mgnitude nd phse responses. numertor_mxphse polyeros_mxphse denomintor_mxphse ; % since ll poles re t origin figure % crete new figure No. freqnumertor_mxphse, denomintor_mxphse Phse degrees Mgnitude db Normlied Angulr Frequency rds/smple Normlied Angulr Frequency rds/smple

154 % We now consider the minimum phse filter, which is derived from the mximum % phse filter bove. % Recll tht the minimum phse filter hs ll eros inside the unit circle, % tht is, their vlues re less or equl to. eros_minphse./ eros_mxphse % tke the reciprocl poles_minphse poles_mxphse % no chnge to the poles % After specifying the poles nd eros position, we now plot their position. figure3 % crete new figure No.3 plneeros_minphse, poles_minphse % % We lso find the numertor nd denomintor coefficients of the trnsfer function Imginry Prt Rel Prt 3

155 % in order to plot the mgnitude nd phse responses. numertor_minphse polyeros_minphse denomintor_minphse ; % since ll poles re t origin figure4 % crete new figure No.4 freqnumertor_minph se, denomintor_minphse % Mgnitude db Phse degrees Normlied Angulr Frequency rds/smple Normlied Angulr Frequency rds/smple

156 3. All-Pss Filters [] An ll-pss filter is one whose mgnitude response is constnt for ll frequencies, but whose phse response is not identiclly ero. [The simplest exmple of n ll-pss filter is pure dely system with system function H -k] A more interesting ll-pss filter is one tht is described by where 0 nd ll coefficients re rel. L L L L L L H L L 0

157 If we define the polynomil A s 0 0 e L L k k k H H H A A H A i.e. ll pss filter,

158 Furthermore, if 0 is pole of H, then is ero of H {ie. the poles nd eros re reciprocls of one nother}. The figure shown below illustrtes typicl pole-ero ptterns for single-pole, singleero filter nd two-pole, two-ero filter. 0 All-pss filter H 0 r 0 /r, - 0 All pss filter < for stbility 0 /r, 0 r, - 0

159 We cn esily show tht the mgnitude response is constnt. Phse response: * cos cos e e e e H H H H H e φ cos sin tn cos sin cos e e e e H

160 - φ When 0 < <, the ero lies on the positive rel xis. The phse over 0 is positive, t 0 it is equl to nd decreses until ω, where it is ero. When -< < 0, the ero lies on the negtive rel xis. The phse over 0 is negtive, strting t 0 for 0 nd decreses to - t ω.

161 3. A second Order Resonnt Filter x[n] H -b -b - - b y[n] p p re re 0 b b b r cos 0 r p 0 r cos 0 r sin A p 0 0 r sin All pole system hs poles only without counting the eros t the origin 0

162 B cos r r r e e r H re re p p b b H 0, cos r b r b 0 b b Cos f s f resonnt frequency Compring A nd B, we obtin

163 3.3 Stbility of second-order filter Consider two-pole resonnt filter given by b & b re coefficients This system hs two eros t the origin nd poles t b b b b H 4, b b b p p ±

164 The filter is stble if the poles lies inside the unit circle i.e. p < & p < For stbility b <. If b then the system is n oscilltor Mrginlly stble Assume tht the poles re complex i.e. b 4b < 0 b < 4b nd If b 4b 0 then we get rel roots. b < ± b, b > 0

165 The stbility conditions define region in the coefficient plne b, b which is in the form of tringle see below The system is only stble if nd only if the point b, b lie inside the stbility tringle. prbol b b b b Complex Conugte Poles Rel Poles b 4 - b b b b

166 Stbility Tringle If the two poles re rel then they must hve vlue between - nd for the system to be stble. 0 0 nd 4 nd 4 4 nd > < < > < < < ± < < ± < b b b b b b b b b b b b b b b b b b b b b b b

167 prbol b b b The region below the prbol b > 4b corresponds to rel nd distinct poles. b Complex Conugte Poles Rel Poles b 4 The points on the prbol b 4b result in rel nd equl double poles. The points bove the prbol correspond to complex-conugte poles. - b b b b

168 3.4 Digitl Oscilltors A digitl oscilltor cn be mde using second order discrete-time system, by using pproprite coefficients. A difference eqution for n oscillting system is given by p [ n] Acos n From the tble of -trnsforms we know tht the -trnsform of p[n] bove is P cos cos

169 Let P Y X cos cos Tking inverse -trnsform on both sides, we obtin y [] n cosy[ n ] y[ n ] x[ n] cosx[ n ] No Input term for n oscilltor x[n] 0, x[n-] 0 So the eqution of the digitl oscilltor becomes y [] n cosy[ n ] y[ n ]

170 % Exercise. Digitl oscilltor % We now implement digitl oscilltor which uses some initil conditions. thet pi/8 % ngulr frequency of the oscilltor mplitude % mplitude of the oscilltor number_of_smples 00 % The oscilltion depends on the initil conditions. % Tht is, we cn mke sine or cosine wve by specifying the right initil conditions. % These re initil conditions for creting sine wve. % Remove the `%' symbol to use them y -mplitude * sinthet, y -mplitude * sin*thet % These re initil conditions for creting cosine wve. % Remove the `%' symbol to use them y mplitude * costhet, y mplitude * cos*thet y eros,number_of_smples; % initiliing oscilltion smples weight * costhet % Notice the oscilltion smples re evluted without clling the commnds `cos' or `sin'. % It only requires multipliction nd dditionsubtrction. for n :number_of_smples yn weight * y - y; y y; % updting previous smples y yn; end

171 figure4 stemy hold on plotmplitude*costhet*0: number_of_smples-,'r-.' hold off % % Notice tht if the first set of initil conditions re selected, the oscilltion strts % t ero level. If the second set is selected, it strts t the mplitude level. % Other initil conditions cn be selected which will mke the oscilltion to strt t % certin phse. % % Experiment with different vlues of `thet', `mplitude'

172 So the eqution of the digitl oscilltor becomes y [] n cosy[ n ] y[ n ] nd its structure is shown below. b cos b - y[n] A cosn - y[n-] - y[n-]

173 To obtin y[n] Acosn, use the following initil conditions: y[0] A cos0. A y[-] A cos-. A cos The frequency cn be tuned by chnging the coefficient b b is constnt. The resonnt frequency of the oscilltor is, b b cos For n oscilltor b b

174 Exmple: A digitl sinusoidl oscilltor is shown below. x[n] -b Assuming 0 is the resonnt frequency of the digitl oscilltor, find the vlues of b nd b for sustining the oscilltion. - - y[n] A sinn

175 0 cos r r K r e e r K re re K b b K H b - r cos 0 ; b r For oscilltion b r b - cos 0

176 b. Write the difference eqution for the bove figure. Assuming x[n] Asin 0 δ[n], nd y- y- 0. Show, by nlysing the difference eqution, tht the ppliction of n impulse t n 0 serves the purpose of beginning the sinusoidl oscilltion, nd prove tht the oscilltion is self-sustining therefter.

177 y y [] n b [ ] [ ] [ ] y n y n x n [] n cos y[ n ] y[ n ] Asin δ [] n 0 0 n 0 y[0] cos0 y[-] y[-] A sin 0 δ[0] n y[0] A sin y[] cos 0 y[0] y[-] A sin 0 δ[] y[] cos 0, Asin 0 A sin 0

178 n y[] cos 0 y[] y[0] A sin 0 δ[] cos 0 Asin 0 A sin 0 A cos 0 [ sin 0 cos 0 ] - sin 0 A sin 0 [4 cos 0 ] A[3sin 0 4 sin 3 0 ] where sin3 0 3sin 0 4 sin 3 0 y[] A sin3 0 nd so forth.

179 By setting the input to ero nd under certin initil conditions, sinusoidl oscilltion cn be obtined using the structure shown bove. Find these initil conditions. [ n] cosy[ n ] y[ n ] x[ n] y x[n] 0 for n oscilltor n 0 y[0] cos 0 y[-] y[-] for oscilltion, y[-] 0 no cosine terms y[0] -y[-] y[-] -Asin0 sine term is equired y[0] 0--A sin 0 A sin 0 Initil conditions: y[-] 0; y[-] -Asin 0

180 Sine nd cosine oscilltors [] Sinusoidl oscilltors cn be used to deliver the crrier in modultors. In modultion schemes, both sines nd cosines oscilltors nd needed. A structure tht delivers sines nd cosines simultneously is shown below: sin -sin - cos cos - y[n] cosn x[n] sinn

181 Proof: Trigonmetric eqution for cosn is: cosn cosncos -sinn sin Let y[n] cosn nd x[n] sinn y[n] cos y[n] sin x[n] Replce n by n- y[n] cos y[n-] sin x[n-] Similrly sinn sin cosn sinn cos x[n] sin y[n] x[n] cos Replce n n- x[n] sin y[n-] x[n-] cos A B Using equtions A & B bove, the structure shown bove cn be obtined.

182 Exercise: An oscilltor is given by the following coupled difference equtions expressed in mtrix form. y y c s [] n [] n cos 0 sin 0 Drw the structure for the relistion of this oscilltor, where 0 is the oscilltion frequency. If the initil conditions y c [-] Acos 0 nd y s [-] -Asin 0, obtin the outputs y c [n] & y s [n] using the bove difference equtions. sin 0 y cos 0 y c s [ n ] [ ] n

183 y y c s [] n cos [ ] [ ] 0 yc n sin 0 ys n [ n] sin y [ n ] cos y [ n ] -sin 0 0 sin 0 c - cos 0 cos 0-0 s y c [n] y s [n]

184 n0 y s [0] sin 0 A cos 0 cos 0 -Asin 0 0 n0 y c [0] cos 0 Acos 0 - sin 0 -Asin 0 A n y c [] cos 0.A - sin 0.0 Acos 0 n y s [] A sin 0 0 A sin 0 n y c [] cos 0 y c [] - sin 0 y s [] cos 0 A cos 0 -sin 0 A sin 0 A cos 0 n n y c [n] A cos n 0 similrly y s [n] A sinn 0

185 Exercise: For the structure shown below, write down the pproprite difference equtions nd hence stte the function of this structure. - - y [n] - cos - sin y [n]

186 3.5 Notch filters [4] When ero is plced t given point on the -plne, the frequency response will be ero t the corresponding point. A pole on the other hnd produces pek t the corresponding frequency point. Poles tht re close to the unit circle give rise lrge peks, where s eros close to or on the unit circle produces troughs or minim. Thus, by strtegiclly plcing poles nd eros on the -plne, we cn obtin smple low pss or other frequency selective filters notch filters.

187 Exmple: Obtin, by the pole-ero plcement method, the trnsfer function of smple digitl notch filter see figure below tht meets the following specifictions: [4] Notch Frequency: 50H 3db width of the Notch: ±5H Smpling frequency: 500 H The rdius, r of the poles is determined by : 0 Hf f H Δf r f s

188 To reect the component t 50H, plce pir of complex eros t points on the unit circle corresponds to H. i.e. t ngles of 360 ± 36 ± To chieve shrp notch filter nd improved mplitude response on either side of the notch frequency, pir of complex conugte eros re plced t rdius r <. r Δf 0 f s

189 cos cos e e e e e e e e H

190 Problem Sheet A3 [5,] Q. The frequency response of n idel differentitor is given by H This response is periodic with period. The quntity τ is the dely of the system in smples. e τ

191 . Sketch the mgnitude nd phse responses of the system for -. H > 0 rg < 0 H rg H 0 0 τ τ mgnitude Phse

192 b Find the impulse response h[n] of the system s sum of sine nd cosine function. Ans: Note: [ n τ ] sin h n n τ e k d k e k k cos n τ n τ

193 Q. For the system in Figure 8, sketch the output y[n] when the input x[n] is δ[n] nd H is n idel lowpss filter s follows: x[n] H H 0 ω[n] 0 < Figure 8 - n y[n]

194 n n n sin ω [ ] sin n n n y n y[n] n Ans:

195 Q3 Show tht both digitl filters given below hve the sme mgnitude response : i y ii y m [] n c [ ] ix n i im m [] n c [ ] ix n m i im y[n] output; x[n] input; c i coefficients

196 Compute the 3dB bndwidth of the following filters Which filter hs smller 3dB bndwidth?,0 ; < < H H Ans:. cos 4 cos : filter Ans nd c c c c <

197 c Find the mgnitude response for the system function H nd comment on your result H 3 3 Ans: H llpss filter Drw the cnonic relition of the system H.

198 Q4. x[n] Determine the mgnitude nd phse response of the multi-pth chnnel y[n] x[n] x[n-m]. At wht frequencies H 0? b Determine nd sketch the mgnitude nd phse response of the system shown below y[n]

199 c Assuming tht the digitl filter G is to be relied using the cscde structure, drw suitble block digrm nd develop the difference equtions. G

200 d Determine the frequency response H of the ldder filter shown below: X Ans : - - H - b be be be Y

201 e Determine the frequency response H of the lttice filter x[n] - Ans : H T b e b -b e y[n] T

202 f Show tht the filter structure shown below hs liner phse chrcteristic eqution given by: φ -. x[n] y[n]

203 Ans g Determine the trnsfer function of the system shown below. Check the stbility of 4 the system when r 0.9 nd. x[n] r sin 0 - r cos 0 r sin -r sin 0 0 : H r cos0 r 0 r cos 0, y[n] - b 0.64 stble _ b.3 filter

204 Q5. Consider the following cusl IIR trnsfer function: H Is H stble function? If it is not stble, find stble trnsfer function G such tht G H. Is there ny trnsfer function hving the sme mgnitude response s H? 3 4 Ans : Unstble, G 3 G H A, A { Allpss filter} H A

205 c Anlyse the digitl structure given below nd determine its trnsfer function. H Y X i Is this cnonic structure? ii Wht should be the vlue of the multiplier coefficient K so tht H hs unity gin t 0? iii Wht should be the vlue of the multiplier coefficient K so tht H hs unity gin t? iv Is there difference between these two vlues of K? If not, why not? X K β α Y

206 X Ans: - - H α - - β α β - K α β α α β β i 4 delys noncnonic ii& iii K iv no ll pss filter - Y

207 Q6: A first-order digitl filter is described by H where, 0 <, b < ssume i Determine k, so tht the mximum vlue of H is equl to. ii Compute the 3-dB bndwidth of the filter H. b k Ans c : cos k b b

208 iii Drw cnonic relition of the system function H. b A two-pole low-pss filter hs the system function H b 0 Determine the vlues of k 0 nd b such tht the frequency response H stisfy the conditions H 0 Ans: k0.46, b 0.3 k ; H 4

209 c A third order FIR filter hs trnsfer function G given by G from G, determine the trnsfer function of n FIR filter whose mgnitude response is identicl to tht of G nd hs mximum phse response.

210 Q7. For the system shown in the next slide, Express y [n] in terms of y [n-], y [n-], nd x[n]; do the sme for y [n]. b Assume A cos 0 ; B sin 0 ; y - cos- 0 ; y - sin- 0. If x[n] 0, show tht : y [n] cosn 0 nd y [n] sinn 0 c Clculte for rbitrry A nd B the system function H Y X nd H Y X

211 A B d Tke nd drw the poles-eros plot for H nd H. A B e Tke nd x[n]δ[n]. Clculte the impulse response h [n] for - n 0. A x[n] - - B -B A y [n] y [n]

212 Q8. Obtin prllel relition for the following H Implement the prllel relition of H which you hve obtined. b Obtin prllel relition for the following trnsfer function H 4 H

213 c A digitl filter is represented by H Does this trnsfer function represent n FIR or n IIR filter? Write difference eqution for H using the direct form. Implement prllel relition of H.

214 Q9. The trnsfer function of discrete-time system hs poles t 0.5, 0. ±0. nd eros t - nd. i Sketch the pole-ero digrm for the system ii Derive the system trnsfer function H from the pole-ero digrm. iii Develop the difference eqution. iv Drw the block digrm of the discrete system. Ans : H

215 b A notch filter is given by Determine the frequency response t dc, f s nd. f s smpling frequency. 0.8 Sketch the frequency response in the intervl 0 f f s Ans:.8 H H f s 4.8

216 Q0. A digitl filter is shown below. x[n] - - b i Determine the system function H for the bove structure. H 0 ii With 0 ; ; b.5 nd b -0.75, determine the pole-ero pttern of H nd indicte if the system is stble or not. b b b 0 y[n]