Need for the unilateral z-transform
We have discussed the mathematics of the bilateral z-transform :
In our problems n represents time. We want to solve initial condition problems in which the output
is specified prior to a particular time 
say and calculated from onward. For convenience
we take 
. There is nothing in the bilateral z-transform that makes n=0 special. We make
n=0 special by defining a unilateral z-transform :
(subscript u to emphasize unilateral)
To use 
rather than F(z) do we need to recalculate our various theorems and tables
of z-transforms ? In principle yes we do . However the outputs of the causal systems we study
generally have a unit step function factor ie f(n)u(n) and for such functions the values of n<0
make no contribution to the z-transform :
An important case where differs from F(z) is the shifting theorem.
Shifting theorem for the unilateral z-transform
Substitute r=n-m so n=r+m
We recall that in the corresponding bilateral transform shifting theorem the first term on the
right is absent. We use the unilateral transform version of the shifting theorem in the
initial condition solutions of DE's. If the output is y(n) , refers only to the solution
y(n) for n>0. The first term on RHS of the theorem is where we specify the initial conditions
y(-1) , y(-2) , 
, y(-N).
Solution of the DE
Take the z-transform, applying the unilateral shifting theorem :
Collect terms in Y(z) on LHS and arrange in powers of z on RHS :
Write :
The term 
contains the initial conditions for x and y .
We see that Y(z) is in the form of sums of polynomial fractions. We then do the inverse
transform to find y(n) for n>0 .
Time domain and transform methods
Example solution of a DE by z-transform
with initial conditions y(-1)=y(-2)=1 and an input sequence x(n)=u(n) .
Taking the unilateral z-transform gives :
x(-1)=0 because x(n)=u(n) and y(-1)=y(-2)=1 the given initial conditions.
Multiply by 
to generate positive powers of z :
Making a common denominator and multiplying out :
Make a proper fraction by dividing through by z and factorize the denominator :
The PFE expansion then is of the form :
To find 
multiply by z+0.80 :
Evaluate this at z=-0.80 to get 
Similarly 
and 
Take the causal function inverse transform :
Check the PFE with MATLAB function residue on 
:
b=[0.6 0.8 -0.4] a=[1 -0.7 -0.7 0.40] [r,p,k]=residue(b,a)
