ELEN E4810: Digital Signal Processing
Topic 4: The Z Transform
1. The Z Transform
2. Inverse Z Transform
Dan Ellis
2013-10-02
1
1. The Z Transform


Powerful tool for analyzing & designing
DT systems
Generalization of the DTFT:
G(z) = Z{g[n]} =

n
g[n]z

Z Transform
n=

z is complex...


Dan Ellis
z = ej! → DTFT
z = r·ej! →
n g[n]r
n  jn
2013-10-02
e
DTFT of
r-n·g[n]
2
Region of Convergence (ROC)

Critical question:

n
Does summation G(z) = n= x[n]z
converge (to a finite value)?

In general, depends on the value of z
Im{z}
→ Region of Convergence:
Portion of complex z-plane
for which a particular G(z)
λ
will converge

ROC
|z| > λ
Dan Ellis
2013-10-02
Re{z}
z-plane
3
ROC Example

e.g. x[n] =
∏nµ[n]

 X(z) = 

n=0
n n
 z
-2 -1
1 2 3 4
n
1 “closed form”
when
=
1
1  z |∏z-1| < 1
Im{z}
ß converges only for |∏z-1| < 1
Re{z}
λ


(previous slide)
i.e. ROC is |z| > |∏|
|∏| < 1 (e.g. 0.8) - finite energy sequence
|∏| > 1 (e.g. 1.2) - divergent sequence,
infinite energy, DTFT does not exist
but still has ZT when |z| > 1.2 (in ROC)
Dan Ellis
2013-10-02
4
M
About ROCs


ROCs always defined in terms of |z|
→ circular regions on z-plane
(inside circles/outside circles/rings)
If ROC includes
Im{z}
unit circle (|z| = 1),
→ g[n] has a DTFT
Re{z}
1
(finite energy
Unit circle
sequence)
z-plane
lies in ROC
→ DTFT OK
Dan Ellis
2013-10-02
5
Another ROC example

Anticausal (left-sided) sequence:
n
x [n] =  µ [n 1] -5 -4 -3 -2 -1
X ( z ) = n (  µ [ n  1]) z
n
n
=  n =  z = m =1  z
1
1
1
=  z
=
1
1
1   z 1  z
1

n
n

m
m
1 2 3 4
n
ROC:
|λ| > |z|
Same ZT as ∏nµ[n], different sequence?
Dan Ellis
2013-10-02
6
ROC is necessary!
A closed-form expression for ZT
must specify the ROC:
x[n] = ∏nµ[n]  X(z) = 1
Im
1
1  z
n
ROC |z| > |∏|
-4 -3 -2 -1
1 2 3 4
Re
|λ|
n
1
x[n] = -∏ µ[-n-1]  X(z) =
1
1  z
z-plane
-4 -3 -2 -1
1 2 3 4 n
ROC |z| < |∏|
DTFTs?
 A single G(z) expression can match
several sequences with different ROCs

Dan Ellis
2013-10-02
7
Rational Z-transforms
G(z) expression can be any function;
rational polynomials are important
class:
1
(M 1)
M
P (z ) p0 + p1z +…+ pM 1z
+ pM z
G (z) =
=
D(z ) d0 + d1z 1 +…+ d N1z (N1) + d N z N



Dan Ellis
By convention, expressed in terms of z-1
– matches ZT definition
(Reminiscent of LCCDE expression...)
2013-10-02
8
Factored rational ZTs
Numerator, denominator can be
factored:
1
M
M
p0  M
1


z
z p0  =1 (z    )
=1

G (z) =
=
N
N
N
1
z d0  =1 (z    )
d0  =1 1    z

(
(


Dan Ellis
)
)
{≥} are roots of numerator
→ G(z) = 0 → {≥} are the zeros of G(z)
{∏} are roots of denominator
→ G(z) = ∞ → {∏} are the poles of G(z)
2013-10-02
9
Pole-zero diagram

Can plot poles and zeros on
complex z-plane:
Im{z}
o
×
×
o
o
poles ∏
(cpx conj for real g[n])
o
× 1
Re{z}
zeros ≥
z-plane

(Value of) expression determined by roots
Dan Ellis
2013-10-02
10
Z-plane surface

G(z): cpx function of a cpx variable

Can calculate value over entire z-plane
ROC
not
shown!!
Dan Ellis
2013-10-02
11
M
ROCs and sidedness
1
Two sequences have: G(z) =
1
1  z

ROC |z| > |∏| → g[n] = ∏nµ[n]
RIGHT-SIDED
10
8
6
2
n
1
|λ|
4
( |∏| < 1 )
0.5
0
0
−1
−0.5
ROC |z| < |∏| → g[n] = -∏nµ[-n-1]
LEFT-SIDED
n
−0.5
0
−1
0.5
z-plane

1
Each ZT pole → region in ROC outside
or inside |∏| for R/L sided term in g[n]

Dan Ellis
Overall ROC is intersection of each term’s
2013-10-02
12
ZT is Linear
G(z) = Z{g[n]} =  g[n]z

n
Z Transform
n
y[n] = Æg[n] + Øh[n]
Y(z) = ß(Æg[n]+Øh[n])z-n
Linear 
= ßÆg[n]z-n + ßØh[n]z-n = ÆG(z)+ØH(z)

n
n
y[n]
=


µ
[n]
+


Thus, if
1 1
2 2 µ[n]
1
2
then
Y (z) =
+
1
1   1z
1   2 z 1
Dan Ellis
2013-10-02
ROC:
|z|>|λ1|,|λ2|
13
ROC intersections


1
1
+
Consider G(z) =
1
1
1   1z
1  2z
with |∏1| < 1 , |∏2| > 1 ...
no ROC specified
Two possible sequences for ∏1 term...
-∏1 µ[-n-1]
n

n
or
∏1nµ[n]
or
∏2nµ[n]
n
Similarly for ∏2 ...
-∏2nµ[-n-1]
n
→ 4 possible g[n] seq’s and ROCs ...
Dan Ellis
2013-10-02
14
n
ROC intersections: Case 1
G(z) =
1
1
+
1   1 z 1 1   2 z 1
g[n] = ∏1nµ[n] + ∏2nµ[n]
both right-sided:
n
ROC: |z| > |∏1| and |z| > |∏2|
Im
Im
Re
|λ1| |λ2|
Dan Ellis
2013-10-02
15
ROC intersections: Case 2
G(z) =
1
1
+
1   1 z 1 1   2 z 1
g[n] = -∏1nµ[-n-1] - ∏2nµ[-n-1]
both left-sided:
n
ROC: |z| < |∏1| and |z| < |∏2|
Im
Im
Re
|λ1| |λ2|
Dan Ellis
2013-10-02
16
ROC intersections: Case 3
G(z) =
1
1
+
1   1 z 1 1   2 z 1
g[n] = ∏1nµ[n] - ∏2nµ[-n-1]
two-sided:
n
ROC: |z| > |∏1| and |z| < |∏2|
Im
Im
Re
|λ1| |λ2|
Dan Ellis
2013-10-02
17
ROC intersections: Case 4
G(z) =
1
1
+
1   1 z 1 1   2 z 1
g[n] = -∏1nµ[-n-1] + ∏2nµ[n]
two-sided:
n
ROC: |z| < |∏1| and |z| > |∏2| ?
Im
no ROC
...
Re
|λ1| |λ2|
Dan Ellis
2013-10-02
18
ROC intersections

Note: Two-sided exponential
g[n] = 
n
 < n < 
n
=  µ [n] +  µ [n 1]
n
ROC
|z| > |Æ|

n
ROC
|z| < |Æ|
No overlap in ROCs
→ ZT does not exist
Im
Re
|α|
(does not converge for any z)
Dan Ellis
2013-10-02
19
ZT of LCCDEs

LCCDEs have solutions of form:
yc [n] =  i  i µ [n] + ...
(same
∏s)
i
Hence ZT Yc ( z ) =
+
1
1  i z
Each term ∏in in g[n] corresponds to a
pole ∏i of G(z) ... and vice versa
n



LCCDE sol’ns are right-sided
ROCs are |z| > |∏i| outside
circles
Dan Ellis
2013-10-02
20
Z-plane and DTFT

Slice between surface and unit cylinder
(|z| = 1 z = ej!) is G(ej!), the DTFT
|G(ej!)|
z = ej!
0
Dan Ellis
2013-10-02
º
! / rad/samp
21
Some common Z transforms
g[n]
±[n]
µ[n]
Ænµ[n]
ROC
∀z
|z| > 1
|z| > |Æ|
G(z)
1
1
1z 1
1
1z 1
0n)µ[n]
1r cos( 0 ) z 1
1 2 2
12r cos( 0 ) z +r z
|z| > r
rnsin(!0n)µ[n]
r sin ( 0 ) z 1
12r cos
 z1+r
2 z2
( 0)
|z| > r
rncos(!
poles at z = re±j!
0
Dan Ellis
2013-10-02
×
×
sum of
rnej! n + rne-j! n
0
0
“conjugate pole
pair”
22
Z Transform properties
w/ROC Rg
g[n]
G(z)
g*[n]
G*(z*)
Rg
Time reversal g[-n]
G(1/z)
1/Rg
Time shift
g[n-n0]
z-n0G(z)
Rg (0/∞?)
Exp. scaling
Æng[n]
G(z/Æ)
|Æ|Rg
ng[n]
dG(z)
z
dz
Rg (0/∞?)
Conjugation
Diff. wrt z
Dan Ellis
2013-10-02
23
Z Transform properties
g[n] Convolution g[n] ∗ h[n]
Modulation g[n]h[n]
ROC
G(z)
1
2 j
at least
G(z)H(z)
( )
Rg∩Rh
1
z
C G
(v)
H v v dv
at least
RgRh

Parseval:
 g[n]h [n] =
*
1
2 j
n=
Dan Ellis
2013-10-02
( )
1
1
C G (v)H v v dv
*
24
ZT Example

x [n] = r cos( 0 n)µ [n] ; can express as
n
j 0
1µ n 

re
2 [ ]
(
) + (re
n
 j 0 n 
)
 = v[n] + v* [n]

v[n] = 1/2µ[n]Æn ; Æ = rej!
→ V(z) = 1/(2(1- rej! z-1))
ROC: |z| > r
0
0

( )
Hence, X (z ) = V (z ) + V z
*
= 21
=
Dan Ellis
(
*
1
+
1re j 0 z 1 1re  j 0 z 1
1
)
1r cos( 0 ) z 1
12r cos( 0 ) z 1 +r 2 z 2
2013-10-02
25
Another ZT example
y[n] = (n +1) µ [n]
n
= x[n] + nx[n]
where x[n] = Æn µ[n]
dX
z
(
)
1

z
X (z) =
dz
1
1  z
( |z| > |Æ| )
d 1 
z 1
= z 
=
1 
dz 1  z  (1  z 1 )2
1
z 1
1
repeated
 Y (z) =
+
=
1
1 2
1 2 root - IZT
(1  z ) (1  z )
ROC |z| > |Æ| 1  z
Dan Ellis
2013-10-02
26
2. Inverse Z Transform (IZT)

Forward z transform was defined as:
G(z) = Z{g[n]} =

 g[n]z
n
n=

3 approaches to inverting G(z) to g[n]:



Dan Ellis
Generalization of inverse DTFT
Power series in z (long division)
Manipulate into recognizable
pieces (partial fractions)
2013-10-02
the useful
one
27
IZT #1: Generalize IDTFT

If z = re
j
then
G(z) = G(re ) =
j

z=
rej!
so
g[n]r
d! = dz/jz
⇤
n
n
g[n]r
=
=
⌅
1
2
1
2 j
⇤
e
j n
= DTFT g[n]r
j⇥
G re
n
G
(z)
z
C
⇥
ej⇥n d
1
r
n
IDTFT
dz

Re
Any closed contour around origin will do
Cauchy: g[n] = ß[residues of G(z)zn-1]
Dan Ellis
2013-10-02
⇥
Im
Counterclockwise
closed contour at |z| = r
within ROC

n
28
IZT #2: Long division

n
 Since G(z) = 
g[n]z
n=
if we could express G(z) as a simple
power series G(z) = a + bz-1 + cz-2 ...
then can just read off g[n] = {a, b, c, ...}

Typically G(z) is right-sided (causal)
P(z)
and a rational polynomial G (z ) =
D(z)

Can expand as power series through
long division of polynomials
Dan Ellis
2013-10-02
29
IZT #2: Long division

Procedure:




Express numerator, denominator in
descending powers of z (for a causal fn)
Find constant to cancel highest term
→ first term in result
Subtract & repeat → lower terms in result
Just like long division for base-10
numbers
Dan Ellis
2013-10-02
30
IZT #2: Long division
1

1+ 2z
e.g. H (z ) =
1+ 0.4z 1  0.12z 2
1
2
Result
3
1+1.6z  0.52z + 0.4z ...
1+ 0.4z 1  0.12z 2 ) 1+ 2z 1
1
2
1+ 0.4z  0.12z
1
2
1.6z + 0.12z
1
2
3
1.6z + 0.64z  0.192z
0.52z 2 + 0.192z 3
...
Dan Ellis
2013-10-02
31
IZT#3: Partial Fractions

Basic idea: Rearrange G(z) as sum of
terms recognized as simple ZTs

especially
1
1  z 1
  n µ [n]
or sin/cos forms

P(z)
i.e. given products
1  z 1 1  z 1 
A
B
+
+
rearrange to sums
1
1
1  z
1  z
Dan Ellis
(
2013-10-02
)(
)
32
Partial Fractions

Note that:
A
B
C
+
+
=
1
1
1
1  z
1  z
1  z
(
)(
order 2 polynomial
u + vz-1 + wz-2
) (
)(
) (
1
1
1
1


z
1


z
1


z
(
)(
)(
)
)(
A 1  z 1 1  z 1 + B 1  z 1 1  z 1 + C 1  z 1 1  z 1
order 3 polynomial


Can do the reverse i.e.
P(z)
go from
to
 N=1 (1    z 1 )

=11   z 1

N
if order of P(z) is less than D(z)
Dan Ellis
2013-10-02
)
else cancel
w/ long div.
33
Partial Fractions

order N-1
Procedure:
N
P(z)

F(z) = N
=
1
1
=1
 =1 (1    z )
1  z
 f [n] = 
no repeated
poles!

N
 (  ) µ [n]
=1 
n
1

=
1


z
where  (
)F (z) z=

i.e. evaluate F(z) at the pole (cancels term in
denominator)
but multiplied by the pole term
→ dominates = residue of pole
Dan Ellis
2013-10-02
34
Partial Fractions Example
1
1+ 2z
Given H (z ) =
(again)
1+ 0.4z 1  0.12z 2
factor:
1
1+ 2z
1
2
=
=
+
1
1
1
1
1+
0.6z
1

0.2z
1+ 0.6z 1  0.2z

(

)(
where:
1 = (1+ 0.6z
1
1
)
)H (z) z=0.6
 2 = 1+ 2z 1
= 2.75
1+ 0.6z z=0.2
Dan Ellis
1+ 2z 1
=
= 1.75
1
1  0.2z z=0.6
2013-10-02
35
Partial Fractions Example

1.75
2.75
Hence H (z ) =
+
1
1+ 0.6z
1  0.2z 1

If we know ROC |z| > |Æ| i.e. h[n] causal:
 h[n] = (1.75)(0.6) µ [n] + (2.75)(0.2) µ [n]
n
Dan Ellis
n
= –1.75{ 1 -0.6 0.36 -0.216 ...}
+2.75{ 1 0.2 0.04 0.008 ...}
same as
= {1 1.6 -0.52 0.4 ...}
long division!
2013-10-02
36