1、1 Name: SOLUTION Section: Laboratory Exercise 6 DIGITAL FILTER STRUCTURES 6.1 REALIZATION OF FIR TRANSFER FUNCTIONS Project 6.1 Cascade Realization Note: program P6_1.m cannot be called directly as suggested in Q6.1 below. This is because tf2zp requires the length of the numerator and denominator po
2、lynomials to be the same. Thus, it is necessary to MODIFY P6_1.m as shown below. A copy of the MODIFIED Program P6_1 is given below: % Program P6_1A % Conversion of a rational transfer function % to its factored form. % MODIFIED to make the numerator and denominator coefficient vectors % the same le
3、ngth for calling tf2zp. num = input(Numerator coefficient vector = ); den = input(Denominator coefficient vector = ); b,a = eqtflength(num,den); % make lengths equal z,p,k = tf2zp(b,a); sos = zp2sos(z,p,k) Answers: Q6.1 By running Program P6_1 with num = 2 10 23 34 31 16 4 and den = 1 we arrive at t
4、he following second-order factors: 0 2h = 113 = 212 = 121 = 222 = 131 = 230.5 =In other words, with regards to Eq. (6.3) on p. 92 of the Lab Manual, we have ( )( )( )12 12 1 21() 21 3 2 1 2 1 0.5H zzzzzzz =+ 2 The block-diagram of the cascade realization obtained from these factors is given below: H
5、1(z) is NOT a linear-phase transfer function, because the coefficients do not have the required symmetry. Q6.2 By running Program P6_1 with num = 6 31 74 102 74 31 6 and den = 1 we arrive at the following second-order factors: 0 6h =11156 =211 = 122 = 223 = 1323 = 2313 = The block-diagram of the cas
6、cade realization obtained from these factors is given below: H2(z) is a Type I linear-phase transfer function with odd length and even symmetry. 3 The block-diagram of the cascade realization of H2(z) with only 4 multipliers is shown below: 6.2 REALIZATION OF IIR TRANSFER FUNCTIONS Project 6.2 Casca
7、de Realization Answers: Q6.3 By running Program P6_1 with num = 3 8 12 7 2 2 and den = 16 24 24 14 5 1 we arrive at the following second-order factors: The result of running the modified program P6_1 is the following: sos = 0.1875 -0.0625 0 1.0000 0.5000 0 1.0000 2.0000 2.0000 1.0000 0.5000 0.2500 1
8、.0000 1.0000 1.0000 1.0000 0.5000 0.5000 In terms of the parameters 0,p ,jk and jk given in Eq. (6.8) of the Lab Manual, this corresponds to the following: 0316p =1113 =210 = 1112 = 210 = 122 = 222 = 1212 = 2214 = 131 = 231 = 1312 = 2312 = 4 The block-diagram of the cascade realization obtained from
9、 these factors is given below: Q6.4 By running Program P6_1 with num = 2 10 23 34 31 16 4 and den = 36 78 87 59 26 7 1 we arrive at the following second-order factors: The result of running the modified program P6_1 is the following: sos = 0.0556 0.1667 0.1111 1.0000 0.5000 0.2500 1.0000 1.0000 2.00
10、00 1.0000 0.6667 0.3333 1.0000 1.0000 0.5000 1.0000 1.0000 0.3333 In terms of the parameters 0,p ,jk and jk given in Eq. (6.8) of the Lab Manual, this corresponds to the following: 0118p =113 = 212 = 1112 = 2114 = 121 = 222 = 1223 = 2213 = 131 = 2312 = 131 = 2313 = The block-diagram of the cascade r
11、ealization obtained from these factors is given below: 5 A copy of Program P6_2 is given below: % Program P6_2 % Parallel Form Realizations of an IIR Transfer num = input(Numerator coefficient vector = ); den = input(Denominator coefficient vector = ); r1,p1,k1 = residuez(num,den); r2,p2,k2 = residu
12、e(num,den); disp(Parallel Form I) disp(Residues are);disp(r1); disp(Poles are at);disp(p1); disp(Constant value);disp(k1); disp(Parallel Form II) disp(Residues are);disp(r2); disp(Poles are at);disp(p2); disp(Constant value);disp(k2); Project 6.3 Parallel Realization Answers: Q6.5 By running Program
13、 P6_2 with num = 3 8 12 7 2 2 and den = 16 24 24 14 5 1 we arrive at the partial-fraction expansion of H1(z) in z1given by: For Parallel Form I, the program returns: Parallel Form I Residues are -0.4219 + 0.6201i -0.4219 - 0.6201i 0.3438 - 2.5079i 0.3438 + 2.5079i 2.3438 Poles are at -0.2500 + 0.661
14、4i -0.2500 - 0.6614i -0.2500 + 0.4330i -0.2500 - 0.4330i -0.5000 Constant value -2 Note that the complex poles occur in conjugate pairs with resides that are also conjugates. Thus, for a pair of conjugate poles at cjd+ and cjd with residues ajb+ and ,ajb we get a pair of terms in the Partial Fractio
15、n Expansion given by (read the help for residuez if this isnt clear to you) () ()( )()111122222.1112aacbdzajb ajbc jdz c jdzcz c d z+=+ +6 For example, for the first pole pair returned for Parallel Form I above, we have 0.4219,a = 0.6201,b = 0.2500,c = and 0.6614.d = Thus, the partial fraction expan
16、sion in 1zis given by (to within roundoff) ()( )()( )112212122122( 0.4219) 2 ( 0.4219)( 0.25) (0.6201)(0.6614)() 21 2( 0.25) 0.25 0.66142(0.3438) 2 (0.3438)( 0.25) ( 2.5079)(0.4330)1 2( 0.25) 0.25 0.4330zHzzzzzz+= + + + + +111112 122.343810.52.3438 0.8438 1.0312 0.6876 2.34371 0.5 1 0.5 0.5 1 0.5 0.
17、25zzzzzzzz+ += + + +Comparing this partial fraction expansion to Eq. (6.10) on p. 96 of the Lab Manual, we have the following values for the Parallel Form I parameters: 02 = 012.3438 = 110 = 110.5 =210 = 020.8438 =121.0312 = 120.5 =220.5 = 030.6876 = 132.3437 = 130.5 =230.25 =and the partial-fractio
18、n expansion of H1(z) in z given by: For Parallel Form II, the program returns: Parallel Form II Residues are -0.3047 - 0.4341i -0.3047 + 0.4341i 1.0000 + 0.7758i 1.0000 - 0.7758i -1.1719 Poles are at -0.2500 + 0.6614i -0.2500 - 0.6614i -0.2500 + 0.4330i -0.2500 - 0.4330i -0.5000 Constant value 0.187
19、5 7 The complex poles again occur in conjugate pairs with residues that are also conjugates. Read the help for residue if you are unclear on how to put this together into the partial fraction expansion. Thus, for a pair of conjugate poles at cjd+ and cjd with residues ajb+ and ,ajb we get a pair of
20、terms in the partial fraction expansion of the form ()()( )()()()1222222 122222 2 2.21az ac bd az ac bd zajb ajb zzcjd zcjd zzczcd czcdz+ += =+ + +For example, for the first pole pair returned for Parallel Form II above, we have 0.3047,a = 0.4341,b = 0.2500,c = and 0.6614.d = Thus, the partial fract
21、ion expansion in z is given by (to within roundoff)()( )()( )12222222( 0.3047) 2 ( 0.3047)( 0.25) ( 0.4341)(0.6614)( ) 0.18752( 0.25) 0.25 0.66142(1) 2 (1)( 0.25) (0.7758)(0.4330)2( 0.25) 0.25 0.43301.1719zHzzzzzz +=+ + + + +220.51.1719 0.6094 0.4219 2 0.17180.1875 .0.5 0.5 0.5 0.5 0.25zzzz zz zz+ =
22、+ + + +Multiplying each fraction in this expression times appropriate powers of 1zon top and bottom then gives: 1221 12 22 211 12 21.1719 0.6094 0.4219 2 0.1718( ) 0.18750.5 0.5 0.5 0.5 0.251.1719 0.6094 0.4219 2 0.17180.1875 .10.5 10.5 0.5 10.5 0.25zzzzzHzz zzz zzz zzzzzz + =+ + + + += + + +Compari
23、ng this partial fraction expansion to Eq. (6.11) on p. 96 of the Lab Manual, we have the following values for the Parallel Form II parameters: 00.1875 = 111.1719 = 210 = 110.5 =210 = 120.6094 =220.4219 = 120.5 =220.5 = 132 = 230.1718 = 130.5 =230.25 =8 The block-diagram of the parallel-form I realiz
24、ation of H1(z) is thus as indicated below: 9 The block-diagram of the parallel-form II realization of H1(z) is thus as indicated below: Q6.6 By running Program P6_2 with num = 2 10 23 34 31 16 4 and den = 36 78 87 59 26 7 1 we arrive at the partial-fraction expansion of H2(z) in z1given by: Followin
25、g the same procedure as in Q6.5, the residues and poles returned for Parallel From I are: Parallel Form I Residues are -0.5952 - 0.7561i -0.5952 + 0.7561i -0.5556 - 2.2785i -0.5556 + 2.2785i -0.8214 + 4.3920i -0.8214 - 4.3920i Poles are at -0.5000 + 0.2887i -0.5000 - 0.2887i -0.3333 + 0.4714i -0.333
26、3 - 0.4714i -0.2500 + 0.4330i -0.2500 - 0.4330i Constant value 4 10 Plugging into the complex pole pair formulas derived in Q6.5, we have ()( )()()1222121222( 0.5952) 2 ( 0.5952)( 0.5) ( 0.7561)(0.2887)() 41 2( 0.5) 0.5 0.28872( 0.5556) 2 ( 0.5556)( 0.3333) ( 2.2785)(0.4714)1 2( 0.3333) 0.3333 0.471
27、4zHzzzz+=+ + + + +()( )1221211 112 12 12( 0.8214) 2 ( 0.8214)( 0.25) (4.3920)(0.4330)1 2( 0.25) 0.25 0.43301.1905 0.1587 1.1111 1.7778 1.6429 4.214341 0.3333 1 0.6667 0.3333 1 0.5 0.25zzzzzz zz z + + + + =+ + + + + + +2.With relation to (6.10) on p. 96 of the Lab Manual, the Parallel Form I paramete
28、rs are: 04 = 011.1905 = 110.1587 = 111 = 210.3333 = 021.1111 = 121.7778 = 120.6667 =220.3333 = 031.6429 = 134.2143 = 130.5 = 230.25 = and the partial-fraction expansion of H2(z) in z given by: The residues and poles returned for Parallel From II are: Parallel Form II Residues are 0.5159 + 0.2062i 0.
29、5159 - 0.2062i 1.2593 + 0.4976i 1.2593 - 0.4976i -1.6964 - 1.4537i -1.6964 + 1.4537i Poles are at -0.5000 + 0.2887i -0.5000 - 0.2887i -0.3333 + 0.4714i -0.3333 - 0.4714i -0.2500 + 0.4330i -0.2500 - 0.4330i Constant value 0.0556 11 Plugging into the complex pole pair formulas derived in Q6.5, we have
30、 ()( )()()22222222(0.5159) 2 (0.5159)( 0.5) (0.2062)(0.2887)( ) 0.05562( 0.5) 0.5 0.28872(1.2593) 2 (1.2593)( 0.3333) (0.4976)(0.4714)2( 0.3333) 0.3333 0.4714zHzzzzzz+=+ + + + +()( )22222 212( 1.6964) 2 ( 1.6964)( 0.25) ( 1.4537)(0.4330)2( 0.25) 0.25 0.43301.0317 0.3968 2.5185 0.3704 3.3929 0.41070.
31、05560.3333 0.6667 0.3333 0.5 0.251.0317 00.0556zzzzzz z z z zz+ + +=+ + + + + + +=+21 1212 12.3968 2.5185 0.3704 3.3929 0.4107.1 0.3333 1 0.6667 0.3333 1 0.5 0.25zz zzzz zz + + + + +With relation to (6.11) on p. 96 of the Lab Manual, the Parallel Form II parameters are: 00.0556 = 111.0317 = 210.3968
32、 = 111 = 210.3333 = 122.5185 = 220.3704 = 120.6667 =220.3333 = 133.3929 = 230.4107 = 130.5 = 230.25 = 12 The block-diagram of the parallel-form I realization of H2(z) is thus as indicated below: 13 The block-diagram of the parallel-form II realization of H2(z) is thus as indicated below: Project 6.4
33、 Realization of an Allpass Transfer function Answers: Q6.7 Using Program P4_4 we arrive at the following values of ki for A5(z): All that is required for this problem is to call poly2rc with the coefficients of the denominator polynomial. The first coefficient should be a “1” and it is not here (to
34、make the numbers look nicer). So you may be bothered by the fact that the help for poly2rc says that if d0 isnt one then everything will get scaled. Thats true, but its just fine here because the numerator and denominator both get scaled by 16; in other words, the reflection coefficients are not aff
35、ected by the scaling. The result of calling ploy2rc is: k(5) = 0.0625 k(4) = 0.2196 k(3) = 0.4811 k(2) = 0.6837 k(1) = 0.6246 The block-diagram of the cascaded lattice realization of A5(z) is thus as shown below: 14 From the values of ki we conclude that the transfer function A5(z) is - STABLE, sinc
36、e 21ik 16*num2 ans = Columns 1 through 4 3.00000000000000 8.00000000000000 12.00000000000000 7.00000000000000 Columns 5 through 6 2.00000000000000 -2.00000000000000 16*den2 ans = Columns 1 through 4 16.00000000000000 24.00000000000000 24.00000000000000 14.00000000000000 Columns 5 through 6 5.0000000
37、0000000 1.00000000000000 The transfer function obtained is EQUIVALENT to H1(z) of Q6.3; as demonstrated above the numerator and denominator coefficient vectors returned by latc2tf are equal to 1/16 times the values shown in (6.27). Q6.14 Using this program we arrive at the lattice parameters and the
38、 feed-forward multiplier coefficients (vectors k and alpha) of the Gray-Markel realization of the transfer function H2(z) of Q6.4 as given below: k = 0.81093584641352 0.77112772506402 0.59215187769984 0.37169052478550 0.13436293436293 0.02777777777778 alpha = -0.01112037033486 0.02345313662512 -0.01
39、456452038379 -0.04739265773254 0.15199485199485 0.20370370370370 0.11111111111111 The parameters obtained using this program are THE SAME as those obtained in Q6.12. Using the function latc2tf we obtain the following transfer function from the vectors k and alpha: num2 = Columns 1 through 4 0.055555
40、55555556 0.27777777777778 0.63888888888889 0.94444444444444 Columns 5 through 7 0.86111111111111 0.44444444444444 0.11111111111111 den2 = Columns 1 through 4 1.00000000000000 2.16666666666667 2.41666666666667 1.63888888888889 Columns 5 through 7 20 0.72222222222222 0.19444444444444 0.02777777777778
41、36*num2 ans = Columns 1 through 4 2.00000000000000 10.00000000000000 22.99999999999999 33.99999999999999 Columns 5 through 7 31.00000000000000 16.00000000000000 4.00000000000000 36*den2 ans = Columns 1 through 4 36.00000000000000 77.99999999999997 87.00000000000000 58.99999999999999 Columns 5 throug
42、h 7 25.99999999999999 7.00000000000000 1.00000000000000 The transfer function obtained is EQUIVALENT to H2(z) of Q6.4; the numerator and denominator coefficients returned by latc2tf in this question are equal to 1/36 times the original ones appearing in (6.28). 21 Project 6.6 Parallel Allpass Realiz
43、ation of an IIR Transfer function Answers: Q6.15 Using zplane we obtain the pole-zero plot of G(z) as shown below: -1 -0.5 0 0.5 1-1-0.8-0.6-0.4-0.200.20.40.60.813Real PartImaginaryPartNext using roots we obtain the pole locations of G(z) as given below: p2 = roots(den) p2 = 0.25221914966490 + 0.745
44、15855654836i 0.25221914966490 - 0.74515855654836i 0.47166170067020 Theta = angle(p2)*180/pi Theta = 71.30021430520364 -71.30021430520364 0 22 Making use of the pole-alteration property we thus arrive at the two allpass sections A0(z) and A1(z) as given below: Ordering the poles by increasing angle i
45、mplies that the pair of conjugate poles should be associated with A0(z), whereas the real pole should be associated with A1(z). Thus, we have ( ) ( )11012,( ) 1 0.2522 0.7452 1 0.2522 0.74521 0.5044 0.6189denominatorAz jz jzzz = + = +11.( ) 1 0.4717denominatorAz z= The numerator polynomials for A0(z
46、) and A1(z) then follow from the allpass property; i.e., they must be the mirror image polynomials of the respective denominator polynomials. We have 120 120.6189 0.5044() ,1 0.5044 0.6189zzAzzz+=+11 1.0.4714()1 0.4714zAzz+=The power-complementary transfer function H(z) is therefore given by 0112 11
47、23122.1() () ()21 0.6189 0.5044 0.471421 0.5044 0.6189 1 0.47140.5453 1.01713 1.01713 0.54531 0.9761 0.8568 0.2919Hz A z Azzz zzzzzzz =+=+The order of A0(z) is N=2. The order of A1(z) is N=1. 23 The block-diagram of a 3-multiplier realization of G(z) and H(z) using Type 1 and Type 2 allpass structures is as indicated below: From the numerator of A0(z), it follows that d1 = -0.5044. Comparing the above expression for A0(z) with (6.14) on p. 97 of the Lab manual, we then solve for d2 = -1.2270. The value
