Software for Partial Fraction Expansion

Figure 6.3 illustrates the use of residuezJ.5) for performing a partial fraction expansion on the transfer function

$\displaystyle H(z) \eqsp \frac{1 + 0.5^3 z^{-3}}{1 + 0.9^5z^{-5}}
$

The complex-conjugate terms can be combined to obtain two real second-order sections, giving a total of one real first-order section in parallel with two real second-order sections, as discussed and depicted in §3.12.

Figure 6.3: Use of residuez to perform a partial fraction expansion of an IIR filter transfer function $ H(z)=B(z)/A(z)$.

 
B = [1 0 0 0.125];
A = [1 0 0 0 0 0.9^5];
[r,p,f] = residuez(B,A)
% r =
%   0.16571
%   0.22774 - 0.02016i
%   0.22774 + 0.02016i
%   0.18940 + 0.03262i
%   0.18940 - 0.03262i
%
% p =
%   -0.90000
%   -0.27812 - 0.85595i
%   -0.27812 + 0.85595i
%    0.72812 - 0.52901i
%    0.72812 + 0.52901i
%
% f = [](0x0)

Example 2

For the filter

$\displaystyle H(z)$ $\displaystyle \isdef$ $\displaystyle \frac{2+6z^{-1}+6z^{-2}+2z^{-3}}{1-2z^{-1}+z^{-2}}$ (7.20)
  $\displaystyle =$ $\displaystyle (2+10z^{-1}) + z^{-2}\left[\frac{8}{1-z^{-1}} + \frac{16}{(1-z^{-1})^2}\right]
\protect$ (7.21)

we obtain the output of residuedJ.6) shown in Fig.6.4. In contrast to residuez, residued delays the IIR part until after the FIR part. In contrast to this result, residuez returns r=[-24;16] and f=[10;2], corresponding to the PFE

$\displaystyle H(z) = 10+2z^{-1}-\frac{24}{1-z^{-1}} + \frac{16}{(1-z^{-1})^2},$ (7.22)

in which the FIR and IIR parts have overlapping impulse responses.

See Sections J.5 and J.6 starting on page [*] for listings of residuez, residued and related discussion.

Figure 6.4: Use of residued to perform a partial fraction expansion of an IIR filter transfer function $ H(z)=B(z)/A(z)$.

 
B=[2 6 6 2]; A=[1 -2 1];
[r,p,f,m] = residued(B,A)
% r =
%    8
%   16
%
% p =
%   1
%   1
%
% f =
%    2  10
%
% m =
%   1
%   2


Polynomial Multiplication in Matlab

The matlab function conv (convolution) can be used to perform polynomial multiplication. For example:

B1 = [1 1];   % 1st row of Pascal's triangle
B2 = [1 2 1]; % 2nd row of Pascal's triangle
B3 = conv(B1,B2) % 3rd row
% B3 = 1  3  3  1
B4 = conv(B1,B3) % 4th row
% B4 = 1  4  6  4  1
% ...
The matlab conv(B1,B2) is identical to filter(B1,1,B2), except that conv returns the complete convolution of its two input vectors, while filter truncates the result to the length of the ``input signal'' B2.7.10 Thus, if B2 is zero-padded with length(B1)-1 zeros, it will return the complete convolution:
B1 = [1 2 3];
B2 = [4 5 6 7];
conv(B1,B2)
% ans = 4  13  28  34  32  21
filter(B1,1,B2)
% ans = 4  13  28  34
filter(B1,1,[B2,zeros(1,length(B1)-1)])
% ans = 4  13  28  34  32  21


Polynomial Division in Matlab

The matlab function deconv (deconvolution) can be used to perform polynomial long division in order to split an improper transfer function into its FIR and strictly proper parts:

B = [ 2 6 6 2]; % 2*(1+1/z)^3
A = [ 1 -2 1];  % (1-1/z)^2
[firpart,remainder] = deconv(B,A)
% firpart =
%   2  10
% remainder =
%    0    0   24   -8
Thus, this example finds that $ H(z)$ is as written in Eq.$ \,$(6.21). This result can be checked by obtaining a common denominator in order to recalculate the direct-form numerator:
Bh = remainder + conv(firpart,A)
%  = 2 6 6 2

The operation deconv(B,A) can be implemented using filter in a manner analogous to the polynomial multiplication case (see §6.8.8 above):

firpart = filter(B,A,[1,zeros(1,length(B)-length(A))])
%       = 2 10
remainder = B - conv(firpart,A)
%         =  0 0 24 -8
That this must work can be seen by looking at Eq.$ \,$(6.21) and noting that the impulse-response of the remainder (the strictly proper part) does not begin until time $ n=2$, so that the first two samples of the impulse-response come only from the FIR part.

In summary, we may conveniently use convolution and deconvolution to perform polynomial multiplication and division, respectively, such as when converting transfer functions to various alternate forms.

When carrying out a partial fraction expansion on a transfer function having a numerator order which equals or exceeds the denominator order, a necessary preliminary step is to perform long division to obtain an FIR filter in parallel with a strictly proper transfer function. This section describes how an FIR part of any length can be extracted from an IIR filter, and this can be used for PFEs as well as for more advanced applications [].


Separating the Transfer Function Numerator and Denominator

From Eq.$ \,$(6.5) we have that the transfer function of a recursive filter is a ratio of polynomials in $ z$:

$\displaystyle H(z) = \frac{B(z)}{A(z)} \protect$ (8.4)

where

\begin{eqnarray*}
B(z) &=& b_0 + b_1 z^{-1}+ \cdots + b_M z^{-M}\\
A(z) &=& 1 + a_1 z^{-1}+ \cdots + a_N z^{-N}.
\end{eqnarray*}

By elementary properties of complex numbers, we have

\begin{eqnarray*}
G(\omega) &=& \frac{\left\vert B(e^{j\omega T})\right\vert}{\...
...a(\omega) &=& \angle B(e^{j\omega T}) - \angle A(e^{j\omega T}).
\end{eqnarray*}

These relations can be used to simplify calculations by hand, allowing the numerator and denominator of the transfer function to be handled separately.


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Frequency Response in Matlab
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Summary of the Partial Fraction Expansion