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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 polynomials 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 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:
Q6.1 By running Program P6_1 with num = [2 10 23 34 31 16 4] and den = [1] we arrive at
the following second-order factors:
h [0] = 2
In other words, with regards to Eq. (6.3) on p. 92 of the Lab Manual, we have
H 1 ( ) z = 2 1 ( + 3 z −^1 + 2 z −^2 )( 1 + z −^1 + 2 z −^2 )( 1 + z −^1 +0.5 z −^2 )
The block-diagram of the cascade realization obtained from these factors is given below:
H 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:
h [0] = 6
11
13
The block-diagram of the cascade realization obtained from these factors is given below:
H 2 (z) is a Type I linear-phase transfer function with odd length and even symmetry.
The block-diagram of the cascade realization obtained from 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 =
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 =
β = α 13 =^123
The block-diagram of the cascade realization obtained from these factors is given below:
A copy of Program P6_2 is given below:
% 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:
Q6.5 By running Program 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 H 1 (z) in z –1^ given 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
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 )
a jb a jb^ a^ ac^ bd^ z
c jd z c jd z cz c d z
− − − (^) − −
+ − −^ +
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
a jb a jb az^ ac^ bd^ z az^ ac^ bd^ z
z c jd z c jd z cz c d z cz c d z
−^ −^ − − (^) − −
+ − −^ +^ −^ +
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
z
H z
z z
z
z z
2 2
z
z z
z z z z z
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
z z z z z
H z
z z z z z z z z
z z z z z
z z z z z
− − − − − − − − − − − − − − − −
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:
The block-diagram of the parallel-form I realization of H 1 (z) is thus as indicated below:
Plugging into the complex pole pair formulas derived in Q6.5, we have
( ) ( )
( ) ( )
1 (^2 1 2 )
1
1 2 2 2
z
H z
z z
z
z z
−
− −
−
− −
( ) ( )
1
1 2 2 2
1 1 1 1 2 1 2 1
z
z z
z z z
z z z z z z
−
− −
− − − − − − − −
With relation to (6.10) on p. 96 of the Lab Manual, the Parallel Form I parameters are:
and the partial-fraction expansion of H 2 (z) in z given by:
The residues and poles returned for Parallel From II are:
Parallel Form II
Residues are 0.5159 + 0.2062i 0.5159 - 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
z
H z
z z
z
z z
( ) ( )
2 2 2
2 2 2 1
z
z z
z z z
z z z z z z
z −
2 1 2 1 2 1 2 1 2 1 2
z z z z z
z z z z z z
− − − − − − − − − − −
With relation to (6.11) on p. 96 of the Lab Manual, the Parallel Form II parameters are:
The block-diagram of the parallel-form II realization of H 2 (z) is thus as indicated below:
Project 6.4 Realization of an Allpass Transfer function
Answers:
Q6.7 Using Program P4_4 we arrive at the following values of {k i} for A 5 (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 make the numbers look nicer). So you may be bothered by the fact that the
help for poly2rc says that if d0 isn’t one then everything will get scaled. That’s
true, but it’s just fine here because the numerator and denominator both get
scaled by 16; in other words, the reflection coefficients are not affected by the
scaling. The result of calling ploy2rc is:
k(5) = 0.0625 k(4) = 0.2196 k(3) = 0. k(2) = 0.6837 k(1) = 0.
The block-diagram of the cascaded lattice realization of A 5 (z) is thus as shown below:
From the values of {k i} we conclude that the transfer function A 5 (z) is - STABLE, since
ki < for all 1 ≤ i ≤ 5.
Q6.8 Using Program P4_4 we arrive at the following values of {k i} for A 6 (z):
k(6) = 0.0278 k(5) = 0.1344 k(4) = 0. k(3) = 0.5922 k(2) = 0.7711 k(1) = 0.
The block-diagram of the cascaded lattice realization of A 6 (z) is thus as shown below.
From the values of {k i} we conclude that the transfer function A 6 (z) is – STABLE. All of the
reflection coefficients have squared-magnitudes strictly less than unity.
Q6.9 Using zp2sos we obtain the following factors of A 5 (z):
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
factors as:
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
z z z z z
A z
z z z z z
z z z z z
z z z z z
z z z z z
z z z
− (^) − − − −
− − − − −
− − − − −
− − − − −
− − − − − − − −
z −^ + z −
The block-diagram of the canonic cascade realization of A 5 (z) using Type 1 and 2 allpass
sections is thus as indicated below:
NOTE: Since a canonic realization is called for, it is required to use the structures
Type 1B given in Fig. 6.9(b) on page 97 of the Lab manual and Type 2A given in Fig.
6.10(a) on page 98. The other structures given in the other parts of these figures are
direct , but not canonical , since they use a number of delay elements that exceeds the
order of the section.
NOTE 2: In this problem we are required to use Type 1 and Type 2 allpass sections.
The use of Type 3 sections such as the one shown in Fig. 6.11(a) is therefore not
allowed. In this regard, you should contrast (6.14) and (6.15) on p. 97 of the Lab
The total number of multipliers in the final structure is 6z.
Project 6.5 Cascaded Lattice Realization of an IIR Transfer function
A copy of Program P6_3 is given below:
% 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)
Answers :
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
given below:
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
structure as given below:
From the lattice parameters obtained using Program P6_3 we conclude that the transfer
function H 1 (z) is – STABLE, since all the lattice parameters have squared magnitudes
strictly less than unity.
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
given below:
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
as given below:
From the lattice parameters obtained using Program P6_3 we conclude that the transfer
function H 2 (z) is – STABLE, since all the lattice parameters have squares strictly less
than unity in magnitude.
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.
The transfer function obtained is EQUIVALENT to H 1 (z) of Q6.3; as demonstrated above
the numerator and denominator coefficient vectors returned by latc2tf are equal to 1/
times the values shown in (6.27).
Q6.14 Using this program we arrive at the lattice parameters and the feed-forward multiplier
coefficients (vectors k and alpha) of the Gray-Markel realization of the transfer function
H 2 (z) of Q6.4 as given below:
k =
alpha =
-0.
-0. -0.
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 0.05555555555556 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.
The transfer function obtained is EQUIVALENT to H 2 (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).
