Stability – Z Transforms
Absolutely summable
A discrete-time signal x is said to be absolutely summable if

 x(n)
exists and is finite. The “absolutely summable” refers to the use of absolute values in the
n  
summation.
BIBO stability
A system is said to be bounded input-bounded output stable (BIBO stable or just stable) if the output
signal is bounded for all input signals that are bounded.
Consider a discrete-time system with input x and output y. The input is said to be bounded if there is a real
number M < ∞ such that
x(k )  M for all k.
An output is bounded if there is a real number N < ∞ such that
y(k )  N for n.
The system is stable if for any input bounded by M, there is some bound N on the output.
Theorem:
A discrete –time LTI system is stable if and only if its impulse response is absolutely summable.
Proof of if:
Consider a discrete-time LTI system with impulse response h. The output y corresponding to the input x is
given by the convolution sum,

n  Integers , y ( n ) 
 h(m) x(n  m)
(1)
m  
Suppose that the input is bounded with bound M. Then, applying the triangle inequality, we see that
y (n) 

 h(m) x(n  m)  M
m  

 h(m)
m  
Thus, if the impulse response is absolutely summable, then the output is bounded with bound

N=M
 h(m)
m  
Proof of only if:
To show that the system is not stable, we need to find one bounded input for which the output either
does not exist or is not bounded. Such an input is given by
 n  Integers,
h( n )
x(n) 
, h(n )  0
h( n )
 0, h( n )  0
The input is clearly bounded, with bound M=1. Plugging this input to the convolution sum (1) and evaluating
at n=0, we get
y ( 0) 



( h( m)) 2
  h(m)
m   h ( m )
m  
 h(m) x ( m)  
m  
But since by assumption that the impulse response in not absolutely summable, y(0) does not exist or is not
finite, so the system is not stable.
Z-transform
Consider a discrete-time signal x that is not absolutely summable. The scaled signal xr is given by
 n  Integers , xr (n)  x(n)r  n ,
for some real number r≥0. Often, this signal is absolutely summable when r is chosen appropriately


The Z-transform of x is defined as
X ( z) 
 x(m) z
m
,
m  
with region of convergence RoC(x)
 Complex is defined by
RoC(x) = {z=rejω
 Complex | x(n)r-n is absolutely summable }
Example:
h(n)=0, n  0 [Bank Account]
= an-1, n > 0
a > 1 is not absolutely summable.
Z-transform of Impulse Response is

H ( z)   z a
n
n 1


 z a
n
0
n 1
z
1

(z
1
a)
( n 1)
1
z
1

(z
0
Thus, hr(n)=h(n)r-n is absolutely summable if r > a i.e. ROC(h) = {z=rejω
1
a)
m
z 1

1  ( z 1a )
 Complex | r > a }
If reiθ1 is in RoC(h), so is reiθ2 because of | reiθ1|=| reiθ2|=r.
There are three RoC structures.
[a] Anticausal RoC
[b] Two-sided RoC
[c] Causal-or right sided RoC
RoC for u(n) = 0 when n<0
= 1 when n ≥1 ,
And its Z-transform is
1
1  z 1
.
A discrete time sequence x(n) is causal if x(n) = 0 for all n <0. The RoC is the set of complex numbers z=rejω
where the following series converges

 x(m)r
m
. But if x is causal, then
m  


m  
m 0
 x ( m ) r m   x ( m ) r m
If this series converges for some given r, then it must also converge for any r’>r [because m ≥ 0, r’-m<r-m].
Thus, if z  RoC, the RoC must include all points in the complex palne on the circle passing through z and
every point outside that circle.
A discrete time sequence is anti-causal if x(n) = 0 for all n > 0. By a similar argument, if z  RoC, the RoC
must include all points in the complex plane on the circle passing through z and every point inside that circle.
Fig [c] shows the RoC of a signal which is neither causal nor anti-causal.
A discrete-time LTI system with impulse response h is stable if and only if the transfer
function H, which is the Z-transform of h, has a region of convergence that includes the unit

circle. Note that putting z = 1 in H(z) gives that
 h(m)  H (1) .
m 
For any system we’ll consider its Z-transform will be a rational polynomial in z.
H ( z) 
A( z )
B( z )
Causal LTI is stable if and only if the poles lie within the unit circle [unless cancelled by zeros –
Pole zero cancellation might be difficult when we take into account finite word length effects]
Linearity of Z-transform:
Suppose x and y have Z-transforms X(z) and Y(z), a and b are two complex constants, and
w=ax+by
Then the Z transform of w is
 z  RoC(w), W(z) = aX(z) + bY(z)
The region of convergence of w
RoC(w)
 RoC(x)  RoC(y)
Linearity is extremely useful because it makes it easy to find the Z transform of complicated signals that can
be expressed as a linear combination of signals with known Z transforms.
e.g. if x(n)=δ(n)+0.9 δ(n-4)+0.8δ(n-5)
X(z) =1+0.9z-4+0.8z-5
RoC z > 0
Delay
For any integer N (positive or negative) and signal x, let y=DN(x) be the signal given by
 n  Integers, y(n)=x(n-N)
Suppose x has a Z-transform X(z) with domain RoC(x). Then RoC(y) = RoC(x) and
 z  RoC(y),

Y(z) =
 y (m) z
m

 x(m  N ) z
=
m  
m
 z N X ( z)
m  
Thus if a signal is delayed by N samples, it’s Z transform is multiplied by z-N.
Convolution
Suppose x and h have Z transforms X(z) and H(z). Let
y=x h
Then
 z  RoC(y), Y(z)=X(z)H(z)
This follows from using the definition of convolution,
 n  Integers, y(n) =

 x(m)h(n  m) in the definition of the z-transform
m 
Y ( z) 

 y ( n ) z n 
n  



x ( m ) z  m h ( n  m ) z ( n  m ) 
n  m  


  x(m) z
m
h ( l ) z l  X ( z ) H ( z )
l   m  
The Z-transform of y converges absolutely at least at the values of z where both X and H converge absolutely.
Thus, RoC(y)  RoC(x)  RoC(h)
We are interested in systems characterized by difference equations
N 1
M 1
y (n)   ai y (n  i )   bi x(n  i )
i 0
j 0
N 1
M 1
i 0
i 1
Y ( z )   ai z iY ( z )   bi z i X ( z )
M 1
Y ( z) 
b z
i
i
i 1
N 1
1   a i z i
i 1