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Material Type: Assignment; Professor: Havlicek; Class: Digital Signal Processing; Subject: ELECTRICAL AND COMPUTER ENGINEERING; University: University of Oklahoma; Term: Unknown 1989;
Typology: Assignments
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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 polynomials to be the same. Thus, it is necessary to MODIFY P6_1.m as shown 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 length 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:
h [0] = 2
In other words, with regards to Eq. (6.3) on p. 92 of the Lab Manual, we have
h [0] = 6
11
13
The result of running the modified program P6_1 is the following:
sos =
In terms of the parameters p 0 (^) , α (^) jk , and β (^) jk given in Eq. (6.8) of the Lab Manual, this corresponds to the following:
0
p =
% Program P6_ % 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] = residue(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:
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
Poles are at
-0.2500 + 0.6614i -0.2500 - 0.6614i -0.2500 + 0.4330i -0.2500 - 0.4330i -0.
Constant value
Note that the complex poles occur in conjugate pairs with resides that are also conjugates. Thus, for a pair of conjugate poles at c + jd and c − jd with residues a + jb and a − jb ,we get a pair of terms in the
Partial Fraction Expansion given by (read the help for residuez if this isn’t clear to you)
1 (^1 1 1 2 2 )
− − − (^) − −
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 c + jd and c − jd with residues a + jb and a − jb ,we get a pair of terms in the
partial fraction expansion of the form
2 1 2 2 2 2 2 1 2 2 2
−^ −^ − − (^) − −
For example, for the first pole pair returned for Parallel Form II above, we have a = −0.3047, b = −0.4341, c = − 0.2500, and d = 0.6614. Thus, the partial fraction expansion in z is given by (to within roundoff)
(^1 22 )
2 2 2
2 2
Multiplying each fraction in this expression times appropriate powers of z −^1 on top and bottom then gives:
1 2 2 (^1 1 2 2 2 ) 1 1 2 1 2 1 1 2 1 2
− − − − − − − − − − − − − − − −
Comparing 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:
Plugging into the complex pole pair formulas derived in Q6.5, we have
( ) ( )
( ) ( )
1 (^2 1 2 )
1
1 2 2 2
−
− −
−
− −
( ) ( )
1
1 2 2 2
1 1 1 1 2 1 2 1
−
− −
− − − − − − − −
With relation to (6.10) on p. 96 of the Lab Manual, the Parallel Form I parameters are:
Parallel Form II
Residues are 0.5159 + 0.2062i 0.5159 - 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
Plugging into the complex pole pair formulas derived in Q6.5, we have
( ) ( )
( ) ( )
(^2 22 )
2 2 2
( ) ( )
2 2 2
2 2 2 1
2 1 2 1 2 1 2 1 2 1 2
− − − − − − − − − − −
With relation to (6.11) on p. 96 of the Lab Manual, the Parallel Form II parameters are:
Project 6.4 Realization of an Allpass Transfer function
Answers:
k(5) = 0.0625 k(4) = 0.2196 k(3) = 0. k(2) = 0.6837 k(1) = 0.
k(6) = 0.0278 k(5) = 0.1344 k(4) = 0. k(3) = 0.5922 k(2) = 0.7711 k(1) = 0.
sos =
0.0625 0.1250 0 1.0000 0.5000 0 1.0000 2.0000 4.0000 1.0000 0.5000 0. 1.0000 1.0000 2.0000 1.0000 0.5000 0.
From the above factors we arrive at the decomposition of A 5 (z) into its low-order allpass
1 1 1 1 2 1 1 16 8 (^5 1 1 1 1 1 2 1 1 ) 2 2 4 2 2 1 1 1 1 1 2 1 1 1 2 2 4 2 2 2 1 1 1 1 1 2 1 1 1 2 2 2 4 2 2 1 1 1 1 1 2 1 1 1 2 2 4 2 2 2 1 1 1 1 1 2 1 2 2 4
− (^) − − − −
− − − − −
− − − − −
− − − − −
− − − − − − − −
The block-diagram of the canonic cascade realization of A 5 (z) using Type 1 and 2 allpass
Project 6.5 Cascaded Lattice Realization of an IIR Transfer function
% Program P6_ % Gray-Markel Cascaded Lattice Structure % k is the lattice parameter vector % alpha is the vector of feedforward multipliers format long % Read in the transfer function coefficients num = input('Numerator coefficient vector = '); den = input('Denominator coefficient vector = '); N = length(den)-1; % Order of denominator polynomial k = ones(1,N); a1 = den/den(1); alpha = num(N+1:-1:1)/den(1); for ii = N:-1:1, alpha(N+2-ii:N+1) = alpha(N+2-ii:N+1)-alpha(N-ii+1)a1(2:ii+1); k(ii) = a1(ii+1); a1(1:ii+1) = (a1(1:ii+1)-k(ii)a1(ii+1:-1:1))/(1-k(ii)*k(ii)); end disp('Lattice parameters are');disp(k) disp('Feedforward multipliers are');disp(alpha)
Q6.11 Using Program P6_3 we arrive at the lattice parameters and the feed-forward multiplier coefficients of the Gray-Markel realization of the causal IIR transfer function H 1 (z) of Q6.3 as
Lattice parameters are
Columns 1 through 4 0.62459686089013 0.68373782742919 0.48111942348398 0.
Column 5
Feedforward multipliers are
Columns 1 through 4 -0.12500000000000 0.31250000000000 0.16053921568627 0.
Columns 5 through 6 -0.09085169508677 -0.
From these parameters we obtain the block-diagram of the corresponding Gray-Markel
From the lattice parameters obtained using Program P6_3 we conclude that the transfer
Q6.12 Using Program P6_3 we arrive at the lattice parameters and the feed-forward multiplier coefficients of the Gray-Markel realization of the causal IIR transfer function H 2 (z) of Q6.4 as
Lattice parameters are
Columns 1 through 4 0.81093584641352 0.77112772506402 0.59215187769984 0.
Columns 5 through 6 0.13436293436293 0.
Feedforward multipliers are
Columns 1 through 4 0.11111111111111 0.20370370370370 0.15199485199485 -0.
Columns 5 through 7 -0.01456452038379 0.02345313662512 -0.
From these parameters we obtain the block-diagram of the corresponding Gray-Markel structure
From the lattice parameters obtained using Program P6_3 we conclude that the transfer
16*num ans = Columns 1 through 4 3.00000000000000 8.00000000000000 12.00000000000000 7. Columns 5 through 6 2.00000000000000 -2.
16*den ans = Columns 1 through 4 16.00000000000000 24.00000000000000 24.00000000000000 14. Columns 5 through 6 5.00000000000000 1.
Q6.14 Using this program we arrive at the lattice parameters and the feed-forward multiplier
k =
alpha =
-0.
-0. -0.
num2 = Columns 1 through 4 0.05555555555556 0.27777777777778 0.63888888888889 0. Columns 5 through 7 0.86111111111111 0.44444444444444 0.
den2 = Columns 1 through 4 1.00000000000000 2.16666666666667 2.41666666666667 1. Columns 5 through 7
36*num ans = Columns 1 through 4 2.00000000000000 10.00000000000000 22.
Columns 5 through 7 31.00000000000000 16.00000000000000 4.
36*den
ans = Columns 1 through 4 36.00000000000000 77.99999999999997 87.
Columns 5 through 7 25.99999999999999 7.00000000000000 1.