Partial fraction expansion#

We expand $F(z)$ into a summation of terms whose inverse is known. These terms have the form:

where $k$ is a constant, and $r_i$ and $p_i$ represent the residues and poles respectively, and can be real or complex¹.

Notes

1. If complex, the poles and residues will be in complex conjugate pairs

Step 1: Make Fractions Proper#

• Before we expand $F(z)$ into partial fraction expansions, we must first express it as a proper rational function.

• This is done by expanding $F(z)/z$ instead of $F(z)$

• That is we expand

Step 2: Find residues#

• Find residues from

Step 3: Map back to transform tables form#

• Rewrite $F(z)/z$:

We will work through an example in class.

[Skip next slide in Pre-Lecture]

Example 1#

Karris Example 9.4: use the partial fraction expansion to compute the inverse z-transform of

MATLAB solution for example 1#

See example1.mlx. (Also available as example1.m.)

Uses MATLAB functions:

• collect – expands a polynomial

• sym2poly – converts a polynomial into a numeric polymial (vector of coefficients in descending order of exponents)

• residue – calculates poles and zeros of a polynomial

• ztrans – symbolic z-transform

• iztrans – symbolic inverse z-transform

• stem – plots sequence as a “lollipop” diagram

clear all
cd matlab
format compact

open example1

syms z n
assume(n,'integer')

The denoninator of $F(z)$

Dz = (z - 0.5)*(z - 0.75)*(z - 1);

Multiply the three factors of Dz to obtain a polynomial

Dz_poly = collect(Dz)

Make into a rational polynomial#

$z^2$

num = [0, 1, 0, 0];

$z^3-9/4z^2-13/8z-3/8$

den = sym2poly(Dz_poly)

Compute residues and poles#

[r,p,k] = residue(num,den);

Print results#

• fprintf works like the c-language function where "%4.2f" means print a floating point number with four significant digits and 2 places of decimals.

fprintf('\n')
fprintf('r1 = %4.2f\t', r(1)); fprintf('p1 = %4.2f\n', p(1));...
fprintf('r2 = %4.2f\t', r(2)); fprintf('p2 = %4.2f\n', p(2));...
fprintf('r3 = %4.2f\t', r(3)); fprintf('p3 = %4.2f\n', p(3));

Symbolic proof#

% z-transform
fn = 2*(1/2)^n-9*(3/4)^n + 8;
Fz = ztrans(fn)

% inverse z-transform
iztrans(Fz)

Sequence#

n = 0:15;
sequence = subs(fn,n);
stem(n,sequence)
title('Discrete Time Sequence f[n] = 2*(1/2)^n-9*(3/4)^n + 8');
ylabel('f[n]')
xlabel('Sequence number n')

Example 2#

Karris example 9.5: use the partial fraction expansion method to to compute the inverse z-transform of

MATLAB solution for example 2#

See example2.mlx. (Also available as example2.m.)

Uses additional MATLAB functions:

• dimpulse – computes and plots a sequence $f[n]$ for any range of values of $n$

open example2

Results for example 2#

Example 3#

Karris example 9.6: use the partial fraction expansion method to to compute the inverse z-transform of

MATLAB solution for example 3#

See example3.mlx. (Also available as example3.m.)

open example3

Results for example 3#

Inverse Z-Transform by the Inversion Integral#

The inversion integral states that:

where $C$ is a closed curve that encloses all poles of the integrant.

This can (apparently) be solved by Cauchy’s residue theorem!!

Fortunately (:-), this is beyond the scope of this module!

See Karris Section 9.6.2 (pp 9-29—9-33) if you want to find out more.

Inverse Z-Transform by the Long Division#

To apply this method, $F(z)$ must be a rational polynomial function, and the numerator and denominator must be polynomials arranged in descending powers of $z$.

We will work through an example in class.

[Skip next slide in Pre-Lecture]

Example 4#

Karris example 9.9: use the long division method to determine $f[n]$ for $n=0,1,\mathrm{and}2$, given that

MATLAB solution for example 4#

See example4.mlx. (also available as example4.m.)

open example4

Results for example 4#

sym_den =
 
z^3 - (3*z^2)/2 + (11*z)/16 - 3/32
 

fn =

    1.0000
    2.5000
    5.0625
    ....

Partial Fraction Expansion#

Advantages

• Most familiar.

• Can use MATLAB residue function.

Disadvantages

• Requires that $F(z)$ is a proper rational function.

Inversion Integral#

Advantage

• Can be used whether $F(z)$ is rational or not

Disadvantages

• Requires familiarity with the Residues theorem of complex variable analaysis.

Long Division#

Advantages

• Practical when only a small sequence of numbers is desired.

• Useful when z-transform has no closed-form solution.

Disadvantages

• Can use MATLAB dimpulse function to compute a large sequence of numbers.

• Requires that $F(z)$ is a proper rational function.

• Division may be endless.

Answer to Example 1#

Answer to Example 2#

Answer to Example 3#

Answer to Example 4#

$f[0]=1$, $f[1]=5/2$, $f[2]=81/16$, ….