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2.161 Signal Processing: Continuous and Discrete

Fall 2008

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 Massachusetts Institute of Technology
Department of Mechanical Engineering

2.161 Signal Processing - Continuous and Discrete

Fall Term 2008

Lecture 141
Reading:

• Proakis & Manolakis, Chapter 3 (The z-transform)

• Oppenheim, Schafer & Buck, Chapter 3 (The z-transform)

1 The Discrete-Time Transfer Function

Consider the discrete-time LTI system, characterized by its pulse response {hn }:

c o n v o lu tio n

{fn } {y } = { f Ä h }
n n n
L T I s y s t e m

Z Z
h n
F (z ) Y (z ) = F (z )H (z )

m u ltip lic a tio n

We saw in Lec. 13 that the output to an input sequence {fn } is given by the convolution
sum:
∞ ∞
yn = fn ⊗ hn = fk hn−k = hk fn−k ,
k=−∞ k=−∞

where {hn } is the pulse response. Using the convolution property of the z-transform we have
at the output
Y (z) = F (z)H(z)
where F (z) = Z {fn }, and H(z) = Z {hn }. Then

Y (z)
H(z) =
F (z)

is the discrete-time transfer function, and serves the same role in the design and analysis
of discrete-time systems as the Laplace based transfer function H(s) does in continuous
systems.
1 c D.Rowell 2008
copyright 

14–1
 In general, for LTI systems the transfer function will be a rational function of z, and may
be written in terms of z or z −1 , for example
N (s) b0 + b1 z −1 + b2 z −2 + . . . + bM z −M
H(z) = =
D(s) a0 + a1 z −1 + a2 z −2 + . . . + aN z −N
where the bi , i = 0, . . . , m, ai , i = 0, . . . , n are constant coefficients.

2 The Transfer Function and the Difference Equation

As defined above, let
Y (z) b0 + b1 z −1 + b2 z −2 + . . . + bM z −M
H(z) = =
F (z) a0 + a1 z −1 + a2 z −2 + . . . + aN z −N
and rewrite as
   
a0 + a1 z −1 + a2 z −2 + . . . + aN z −N Y (z) = b0 + b1 z −1 + b2 z −2 + . . . + bM z −M F (z)

If we apply the z-transform time-shift property

Z {fn−k } = z −k F (z)

term-by-term on both sides of the equation, (effectively taking the inverse z-transform)

a0 yn + a1 yn−1 + a2 yn−2 + . . . + aN yn−N = b0 fn + b1 fn−1 + b2 fn−2 + . . . + bM fn−M

and solve for yn

1 1
yn = − (a1 yn−1 + a2 yn−2 + . . . + aN yn−N ) + (b0 fn + b1 fn−1 + b2 fn−2 + . . . + bM fn−M )
a0 a0
 −ai
N    bi

M  
= yn−i + fn−i

i=1
a0 i=0
a0

which is in the form of a recursive linear difference equation as discussed in Lecture 13.

The transfer function H(z) directly defines the computational dif­

ference equation used to implement a LTI system.

Example 1
Find the difference equation to implement a causal LTI system with a transfer
function
(1 − 2z −1 )(1 − 4z −1 )
H(z) =
z(1 − 12 z −1 )
Solution:
z −1 − 6z −2 + 8z −3
H(z) =
1 − 12 z −1

14–2
 from which
1
yn − yn−1 = fn−1 − 6fn−2 + 8fn−3 ,
2
or
1
yn = yn−1 + (fn−1 − 6fn−2 + 8fn−3 ).
2

The reverse holds as well: if we are given the difference equation, we can define the system
transfer function.

Example 2
Find the transfer function (expressed in powers of z) for the difference equation

yn = 0.25yn−2 + 3fn − 3fn−1

and plot the system poles and zeros on the z-plane.

Solution: Taking the z-transform of both sides

Y (z) = 0.25z −2 Y (z) + 3F (z) − 3z −1 F (z)

and reorganizing

Y (z) 3(1 − z −1 ) 3z(z − 1)

H(z) = = = 2
F (z) 1 − 0.25z −2 z − 0.25

which has zeros at z = 0, 1 and poles at z = −0.5, 0.5:

Á {z }

z - p la n e

x o x o  {z }
-0 .5 0 .5 1

14–3

3 Introduction to z-plane Stability Criteria

The stability of continuous time systems is governed by pole locations - for a system to be
BIBO stable all poles must lie in the l.h. s-plane. Here we do a preliminary investigation of
stability of discrete-time systems, based on z-plane pole locations of H(z).
Consider the pulse response hn of the causal system with
z 1
H(z) = =
z−a 1 − az −1
with a single real pole at z = a and with a difference equation
yn = ayn−1 + fn .
Á {z }

z - p la n e

a < -1 -1 < a < 0 0 < a < 1 a > 1

o x  {z }
1

p o le lo c a tio n

Clearly the pulse response is 

1 n=0
hn =
an n≥1
The nature of the pulse response will depend on the pole location:
0 < a < 1: In this case hn = an will be a decreasing function of n and limn→∞ hn = 0 and
the system is stable.
a = 1: The difference equation is yn = yn−1 + fn (the system is a summer and the impulse
response is hn = 1, (non-decaying). The system is marginally stable.
a > 1: In this case hn = an will be a increasing function of n and limn→∞ hn = ∞ and the
system is unstable.
−1 < a < 0: In this case hn = an will be a oscillating but decreasing function of n and
limn→∞ hn = 0 and the system is stable.
a = −1: The difference equation is yn = −yn−1 + fn and the impulse response is hn = (−1)n ,
that is a pure oscillator. The system is marginally stable.
a < −1: In this case hn = an will be a oscillating but increasing function of n and limn→∞ |hn | =
∞ and the system is unstable.

14–4

This simple demonstration shows that this system is stable only for the pole position −1 <
a < 1. In general for a system
M
k=1 (z − zk )
H(z) = K N
k=1 (z − pk )
having complex conjugate poles (pk ) and zeros (zk ) :
A discrete-time system will be stable only if all of the poles of its
transfer function H(z) lie within the unit circle on the z-plane.

4 The Frequency Response of Discrete-Time Systems

Consider the response of the system H(z) to an infinite complex exponential sequence
fn = A ej ωn = A cos(ωn) + jA sin(ωn),
where ω is the normalized frequency (rad/sample). The response will be given by the con­
volution

∞ 
∞
 
yn = hk fn−k = hk A ej ω(n−k)
k=−∞ k=−∞


∞
−j ωk
= A hk e ej ωn
k=−∞
j ω j ωn
= AH(e )e
where the frequency response function H(ej ω ) is
H(ej ω ) = H(z)|z=ej ω
that is

The frequency response function of a LTI discrete-time system is

H(z) evaluated on the unit circle - provided the ROC includes the
unit circle. For a stable causal system this means there are no poles
lying on the unit circle.
Á {z }

0 < w < p z - p la n e
j1

w i n c r e a s i n g e jw

w w = 0
w = p  {z }
w = -p -1 1

- j1
- p < w < 0

14–5

Alternatively, the frequency response may be based on a physical frequency Ω associated

with an implied sampling interval ΔT , and

H(ej ΩΔT ) = H(z)|z=ej ΩΔT

which is again evaluated on the unit circle, but at angle ΩΔT .

Á {z }
0 < W < p /D T z - p la n e
j1
W in c r e a s in g e jW D T
N y q u is t fr e q u e n c y
W = 0
W = p /D T W D T
 {z }
W = -p /D T -1 1

- j1
- p /D T < W < 0

From the definition of the DTFT based on a sampling interval ΔT


∞
∗
H (jΩ) = hn e−mjnΩΔT = H(z)|z= e−mjnΩΔT
n=0

we can define the mapping between the imaginary axis in the s-plane and the unit-circle in
the z-plane
s = j Ωo ←→ z = ej Ωo ΔT
jW Á {z }
N y q u is t fr e q u e n c y s - p la n e m a p p in g
z - p la n e
jp / D T j1
jW o D T
e
N y q u is t fr e q u e n c y
jW o
W = 0
" p r im a r y " s tr ip W = p /D T W D T
 {z }
s
W = -p /D T -1 1

- j1
- jp / D T

The periodicity in H( ej ΩΔT ) can be clearly seen, with the “primary” strip in the s-plane
(defined by −π/ΔT < Ω < π/ΔT ) mapping to the complete unit-circle. Within the primary
strip, the l.h. s-plane maps to the interior of the unit circle in the z-plane, while the r.h.
s-plane maps to the exterior of the unit-circle.

14–6

 Aside: We use the argument to differentiate between the various classes of transfer
functions:
H(s) H(jΩ) H(z) H( ej ω )
   
Continuous Continuous Discrete Discrete
Transfer Frequency Transfer Frequency
Function Response Function Response

5 The Inverse z-Transform

The formal definition of the inverse z-transform is as a contour integral in the z-plane,
 ∞
1
F (z)z n−1 dz
2πj −∞
where the path is a ccw contour enclosing all of the poles of F (z).

Cauchy’s residue theorem states

 ∞ 
1
F (z) dz = Res [F (z), pk ]
2πj −∞ k

where F (z) has N distinct poles pk , k = 1, . . . , N and ccw path lies in the
ROC.
For a simple pole at z = zo

Res [F (z), zo ] = lim (z − zo )F (z),

z→zo

and for a pole of multiplicity m at z = zo

1 dm−1
Res [F (z), zo ] = lim (z − zo )m F (z)
z→zo (m − 1)! dz m−1

The inverse z-transform of F (z) is therefore

 
fn = Z −1 {F (z)} = Res F (z)z n−1 , pk .
k

Example 3
A first-order low-pass filter is implemented with the difference equation

yn = 0.8yn−1 + 0.2fn .

Find the response of this filter to the unit-step sequence {un }.

14–7
Solution: The filter has a transfer function
Y (z) 0.2 0.2z
H(z) = = =
F (z) 1 − 0.8z −1 z − 0.8

The input {fn } = {un } has a z-transform

z
F (z) =
z−1
so that the z-transform of the output is

0.2z 2
Y (z) = H(z)U (z) =
(z − 1)(z − 0.8)

and from the Cauchy residue theorem

 
yn = Res Y (z)z n−1 , 1 + Res Y (z)z n−1 , 0.8
= lim(z − 1)Y (z)z n−1 + lim (z − 0.8)Y (z)z n−1
z→1 z→0.8
n+1 n+1
0.2z 0.2z
= lim + lim
z − 0.8
z→1 z→0.8 z−1
= 1 − 0.8n+1

which is shown below

y n

0 .8

0 .6

0 .4

0 .2

0
0 2 4 6 8 1 0 1 2 1 4 1 6 1 8
n ( s a m p le s )

Example 4
Find the impulse response of the system with transfer function

1 z2 z2
H(z) = = =
1 + z −2 z2 + 1 (z + j 1)(z − j 1)

14–8

Solution: The system has a pair of imaginary poles at z = ±j 1. From the

Cauchy residue theorem
 
hn = Z −1 {H(z)} = Res H(z)z n−1 , j 1 + Res H(z)z n−1 , −j 1
z n+1 z n+1
= lim + lim
z→j1 z + j 1 z→−j 1 z − j 1

1 1

= (j 1)n+1 − (−j 1)n+1

j2 j2
n  
j
= 1 + (−1)n+1
2
0 n odd
hn = n/2
(−1) n even
= cos(nπ/2)
where we note that the system is a pure oscillator (poles on the unit circle) with
a frequency of half the Nyquist frequency.
y n
1

0 .5

2 6 1 0 1 4 1 8
0
4 8 1 2 1 6 n

-0 .5

-1

Example 5
Find the impulse response of the system with transfer function
1 z2 z2
H(z) = = =
1 + 2z + z −2 z 2 + 2z + 1 (z + 1)2
Solution: The system has a pair of coincident poles at z = −1. The residue at
z = −1 must be computed using
1 dm−1
Res [F (z), zo ] = lim (z − zo )m F (z).
z→zo (m − 1)! dz m−1

With m = 2, at z = −1,
 1 d
Res H(z)z n−1 , −1 = lim (z − 1)2 H(z)z n−1
z→−1 (1)! dz

d n+1
= lim z
z→−1 dz

= (n + 1)(−1)n

14–9
 The impulse response is

hn = Z −1 {H(z)} = Res H(z)z n−1 , −1 = (n + 1)(−1)n .

h n
2 0

1 0

0
2 4 6 8 1 0 1 2 1 4 1 6 1 8 n

-1 0

-2 0

Other methods of determining the inverse z-transform include:

Partial Fraction Expansion: This is a table look-up method, similar to the method
used for the inverse Laplace transform. Let F (z) be written as a rational function of
z −1 :
M −k
k=0 bi z
F (z) = N
−k
k=0 ai z
M −1
k=1 (1 − ci z )
= N
k=1 (1 − di z )
−1

If there are no repeated poles, F (z) may be expressed as a set of partial fractions.


N
Ak
F (z) =
k=1
1 − dk z −1

where the Ak are given by the residues at the poles

Ak = lim (1 − dk z −1 )F (z).
z→dk

Since
Z 1
(dk )n un ←→
1 − dk z −1


N
fn = Ak (dk )n un .
k=1

14–10

 Example 6
Find the response of the low-pass filter in Ex. 3 to an input

fn = (−0.5)n

Solution: From Ex. 3, and from the z-transform of {fn },

1 0.2
F (z) = , H(z) =
1 − 0.5z −1 1 − 0.8z −1
so that
0.2
Y (z) =
(1 + − 0.8z −1 )
0.5z −1 )(1
A1 A2
= +
1 + 0.5z −1 1 − 0.8z −1
Using residues
0.2 0.1
A1 = lim =
1 − 0.8z
z→−0.5 −1 1.3
0.2 0.16
A2 = lim =
z→0.8 1 + 0.5z −1 1.3

and
   
0.1 −1 1 0.16 −1 1
yn = Z + Z
1.3 1 + 0.5z −1 1.3 1 − 0.8z −1
0.1 0.16
= (−0.5)n + (0.8)n
1.3 1.3

Note: (1) If F (z) contains repeated poles, the partial fraction method must be ex­
tended as in the inverse Laplace transform.
(2) For complex conjugate poles – combine into second-order terms.

Power Series Expansion: Since


∞
F (z) = fn z −n
n=−∞

if F (z) can be expressed as a power series in z −1 , the coefficients must be fn .

14–11

 Example 7
Find Z −1 {log(1 + az −1 )}.

Solution: F (z) is recognized as having a power series expansion:


∞
(−1)n+1 an
−1
F (z) = log(1 + az ) = z −n for |a| < |z|
n=1
n
Because the ROC defines a causal sequence, the samples fn are
⎧
⎨0 for n ≤ 0
fn = (−1)n+1 an
⎩ for n ≥ 1.
n

Polynomial Long Division: For a causal system, with a transfer function written as
a rational function, the first few terms in the sequence may sometimes be computed
directly using polynomial division. If F (z) is written as
N (z −1 )
F (z) = = f0 + f1 z −1 + f2 z −2 + f2 z −2 + · · ·
D(z −1 )
the quotient is a power series in z −1 and the coefficients are the sample values.

Example 8
Determine the first few terms of fn for
1 + 2z −1
F (z) =
1 − 2z −1 + z −2
using polynomial long division.
Solution:
1 + 4z −1 + 7z −2 + · · ·

1 − 2z −1 + z −2 1 + 2z −1
1 − 2z −1 + z −2
4z −1 − z −2
4z −1 − 8z −2 + 4z −3
7z −2 − 4z −3
so that
1 + 2z −1
F (z) = = 1 + 4z −1 + 7z −2 + · · ·
1 − 2z −1 + z −2
and in this case the general term is
fn = 3n + 1 for n ≥ 0.

14–12
In general, the computation can become tedious, and it may be difficult to recognize
the general term from the first few terms in the sequence.

14–13