ECE 3640 - Discrete-Time Signals and Systems

z-transform
Jake Gunther

Spring 2015

Department of Electrical & Computer Engineering

 z-transform

∞
X
X(z) = x[n]z −n
n=−∞

ROC = {z ∈ C : |X(z)| < ∞}

• need to specify both algebraic formula for X(z) and the ROC

• the ROC never contains poles

• zeros may lie in the ROC

• ROC is always circular

 z-transform is a “surface”

This figure was taken from Manolakis & Ingle “Applied Digital Signal Processing,”, Cambridge University Press.

• actually |X(z)|, ∠X(z), <{X(z)}, ={X(z)} are surfaces

• for poles |X(z)| → ∞

• for zeros |X(z)| = 0

 finite duration signals

This figure was taken from Proakis & Manolakis “Digital Signal Processing: Principles, Algorithms and Applications,”, 3rd edition, Prentice Hall.
 infinite duration signals

This figure was taken from Proakis & Manolakis “Digital Signal Processing: Principles, Algorithms and Applications,”, 3rd edition, Prentice Hall.
 example

x[n] = anu[n] x[n] = −anu[−n − 1]

1 z 1 z
X(z) = = X(z) = =
1 − az −1 z − a 1 − az −1 z − a
pole: z = a pole: z = a
zero: z=0 zero: z=0
ROC: |z| > |a| ROC: |z| < |a|
(convergence outside a circle) (convergence inside a circle)
(causal (right-sided) signal) (anticausal (left-sided) signal)

• two different signals can give the same formula for X(z)

• they will have different ROCs

 first order poles

This figure was taken from Proakis & Manolakis “Digital Signal Processing: Principles, Algorithms and Applications,”, 3rd edition, Prentice Hall.
 signal and pole examples

This figure was taken from Manolakis & Ingle “Applied Digital Signal Processing,”, Cambridge University Press.
 example

x[n] = nanu[n] x[n] = −nanu[−n − 1]

az −1 az az −1 az
X(z) = = X(z) = =
(1 − az −1)2 (z − a)2 (1 − az −1)2 (z − a)2
pole: z = a (2nd order pole) pole: z = a (2nd order pole)
zero: z=0 zero: z=0
ROC: |z| > |a| ROC: |z| < |a|
(convergence outside a circle) (convergence inside a circle)
(causal (right-sided) signal) (anticausal (left-sided) signal)

• two different signals can give the same formula for X(z)

• they will have different ROCs

 second order poles

This figure was taken from Proakis & Manolakis “Digital Signal Processing: Principles, Algorithms and Applications,”, 3rd edition, Prentice Hall.
 example

x[n] = rn cos(ω0n)u[n], r > 0, 0 ≤ ω0 < 2π

1 n jω0n n −jω0 n


= r e +r e u[n]
2
1 jω0 n −jω0 n


= (re ) + (re ) u[n]
2
1
= (pn + (p∗)n) u[n], p = rejω0
2

   
1 1 1 1
X(z) = +
2 1 − pz −1 2 1 − p∗z −1
1 − r cos(ω0)z −1 z(z − r cos(ω0))
= =
(1 − pz −1)(1 − p∗z −1) (z − p)(z − p∗)

ROC : {z ∈ C : |z| > |p|} ∩ {z ∈ C : |z| > |p|} = {z ∈ C : |z| > r}

 complex conjugate poles

This figure was taken from Proakis & Manolakis “Digital Signal Processing: Principles, Algorithms and Applications,”, 3rd edition, Prentice Hall.
 two-sided exponential

n n 1 1
x[n] = a u[n] − b u[−n − 1] ↔ X(z) = +
1 − az −1 1 − bz −1
ROC : |a| < |z| < |b|

This figure was taken from Manolakis & Ingle “Applied Digital Signal Processing,”, Cambridge University Press.
• the ROC of a sum of transforms is the intersection of the ROCs

• must have |a| < |b|, otherwise X(z) does not exist

• sequence is stable if and only if |a| < 1 < |b|, then unit circle in ROC
 z-transform pairs

This table was taken from Manolakis & Ingle “Applied Digital Signal Processing,”, Cambridge University Press.
 z-transform is linear

x[n] ↔ X(z)
y[n] ↔ y(z)

z[n] = ax[n] + by[n] ↔ Z(z) = aX(z) + bY (z)

ROCZ at least ROCX ∩ ROCY

 example

2z 2 + 0.8z − 2.2
X(z) = 2 , |z| > 0.8
z + 0.3z − 0.4
−0.4z + 1
Y (z) = 2 , |z| > 0.8
z + 0.5z − 0.24
2z 2 − 1.8z − 0.2
Z(z) = X(z) + Y (z) = 2 , |z| > 0.5
z − 0.8z + 0.15

• Matlab: roots([1 0.3 -0.40]) = [-0.8, 0.5]

• Matlab: roots([1 0.5 -0.24]) = [-0.8, 0.3]

• Matlab: roots([1 -0.8 0.15]) = [ 0.3, 0.5]

• the ROC of Z(z) is larger than the intersection of the ROCs of X(z) and Y (z)
because of pole-zero cancellation when two rational functions are summed
 z-transform of time shift

x[n] ↔ X(z), ROCZ

y[n] = x[n − k] ↔ Y (z) = z −k X(z)

• ROCY = ROCX with consideration for z = 0 and z = ∞

• this is easy to derive using a single change of variables (m = n − k)

X X X
−n −n
Y (z) = y[n]z = x[n − k]z = z[m]z −mz −k = z −k X(z)
n n m
 example

x[n] = u[n] ↔ X(z) = 1 + z −1 + z −2 + z −3 + · · · , 0 < |z|

x[n] = u[n − 1] ↔ z −1X(z) = z −1 + z −2 + z −3 + z −4 + · · · , 0 < |z|
x[n] = u[n + 1] ↔ z 1X(z) = z 1 + 1 + z −1 + z −2 + · · · , 0 < |z| < ∞

x[n] = u[−n] ↔ X(z) = · · · + z 3 + z 2 + z + 1, |z| < ∞

x[n] = u[−(n + 1)] ↔ z 1X(z) = · · · + z 4 + z 3 + z 2 + z, |z| < ∞
x[n] = u[−(n − 1)] ↔ z −1X(z) = · · · + z 2 + z 1 + 1 + z −1, 0 < |z| < ∞

• sometimes the ROC does not change

• sometimes the ROC changes by removing the point at z = 0 or z = ∞

 z-transform of difference equation
• take z-transforms of both sides of the difference quation using linearity and the
time-shifting property

N
X M
X
a[k]y[n − k] = b[k]x[n − k]
k=0 k=0
N
X M
X
a[k]z −k Y (z) = b[k]z −k X(z)
k=0 k=0
N
! M
!
X X
a[k]z −k Y (z) = b[k]z −k X(z)
k=0 k=0

• now rearrange and define the transfer (system) function

PM −k ∞
Y (z) b[k]z B(z) X
H(z) = k=0
= PN = = h[n]z −n,
X(z) k=0 a[k]z
−k A(z) n=−∞

where h[n] is the impulse response

 PN −k
• finite poles are roots pk , k = 1, 2, · · · , N of A(z) = k=0 a[k]z

PM −k
• finite zeros are roots zk , k = 1, 2, · · · , M of B(z) = k=0 b[k]z

• may also have poles or zeros at z = ∞

• counting poles and zeros at ∞, a rational function has equal numbers of poles
and zeros

• the transfer function may also be written in factored form exposing the poles and
zeros

M
X M
Y
b[k]z −k (1 − zk z −1)
k=0 b[0] k=1
H(z) = N
= · N
X a[0] Y
a[k]z −k (1 − pk z −1)
k=0 k=1

• Matlab roots, poly and conv functions

 z-transform of convolution

x[n] ↔ X(z)
y[n] ↔ y(z)

z[n] = x[n] ∗ y[n] ↔ Z(z) = X(z) · Y (z)

ROCZ at least ROCX ∩ ROCY

• this property is easy to derive using one change of variables

 example
determine the output of LTI system described by y[n] = 12 y[n − 1] + x[n] driven by
x[n] = 10 cos(πn/4)u[n] and y[−1] = 0

1 10(1 − √12 z −1)

H(z) = 1 −1 , X(z) = √
1− 2z 1 − 2z −1 + z −2

10(1 − √12 z −1)

Y (z) = H(z)X(z) = 1 −1
√
(1 − 2 z )(1 − 2z −1 + z −2)
◦ ◦
−1.9 6.78e−j28.7 6.78ej28.7
= 1 −1 + −1
+
1 − 2z 1−e jπ/4 z 1 − e−jπ/4z −1

  n 
1 ◦ ◦
y[n] = −1.9 + 6.78ej28.7 ejπn/4 + 6.78ej28.7 ejπn/4 u[n]
2
  n 
1
= −1.9 + 13.56 cos(πn/4 − 28.7◦) u[n]
2

• partial fraction expansion was used (Matlab: residuez)

 example (continued)
[r,p,k]=residuez(10*[1,−1/sqrt(2)],conv([1,−1/2],[1,−sqrt(2),1]));
n = [0:20];
y = −1.9074*(0.5).ˆn + 13.56*cos(pi*n/4−28.7*pi/180);
plot(n,y,'b'); hold on;
y = filter(1,[1,−0.5],10*cos(pi*n/4));
stem(n,y,'r'); hold off;
 exponential modulation

x[n] ↔ X(z), ROCX

y[n] = anx[n] ↔ Y (z) = X(a−1z), ROCY = |a|ROCX

• let p be a pole of X(z): X(p) → ∞

• then Y (ap) = X(a−1(ap)) = X(p) → ∞, so ap is a pole of Y (z)

• the ROC stretches (|a| > 1), shrinks (|a| < 1), or stays the same (|a| = 1)

• let a = ejω0 , then pole at p gets rotated to ejω0 p, but the ROC stays the same
 differentiation property

x[n] ↔ X(z)
dX(z)
nx[n] ↔ −z
dz

proof:

∞ ∞
dX(z) X X
= x[n](−n)z −n−1 = (−z −1) (nx[n])z −n
dz n=−∞ n=−∞
∞
dX(z) X
−z = (nx[n])z −n
dz n=−∞

• the ROC does not change

 example

n 1
a u[n] ↔ −1
, |z| > |a|
1 − az
d 1 az −1
nanu[n] ↔ −z −1
= −1 2
, |z| > |a|
dz 1 − az (1 − az )
 more properties

• complex conjugate

x[n] ↔ X(z), ROCX

y[n] = x∗[n] ↔ Y (z) = X ∗(z ∗), ROCY = ROCX

• time-reversal

x[n] ↔ X(z), ROCX

y[n] = x[−n] ↔ Y (z) = X(z −1), ROCY = 1/ROCX

1 1
• if ROCX : r1 < |z| < r2, then ROCY : r2 < |z| < r1
 more properties

• initial value theorem for causal x[n],

x[0] = lim X(z)

z→∞

• final value theorem for causal x[n],

lim x[n] = lim (1 − z −1)X(z)

n→∞ z→1

assuming the limits exist

 you should be able to ...

• derive the time shifting property

• derive the time-reversal property

• derive the conjugation property

• derive the convolution property

• compute z-transforms of common signals

• ... and levitate

 inverse z-transform

Z
1
x[n] = X(z)z n−1dz, C ∈ ROC
2πj C:CCW
 observe

N
X
x[n] = ck pnku[n], distinct pk
k=1
N
X ck
X(z) =
1 − pk z −1
k=1
N
Y N −1
PN
(1 − pmz −1)
X
k=1 ck b[k]z −k
m=1,m6=k k=0
= N
= N
, a0 = 1
Y X
(1 − pk z −1) a[k]z −k
k=1 k=0

• this is a proper rational function

• given a rational function with distinct poles, we can reverse this procedure to find
x[n]

• use partial fraction expansion (Matlab: residuez)

 example

1 + z −1 z(z + 1)
X(z) = 1 −1 =
(1 − z )(1 − 2 z ) (z − 1)(z − 12 )
−1

1
• poles at p1 = 1 and p2 = 2

• zeros at z1 = 0 and z2 = −1

• two distinct poles gives three different possibilities for the ROC

• PFE:
[R,P,K]=residuez([1,1],conv([1,-1],[1,-0.5]));
[R,P,K]=residuez([1,1],poly([1,0.5]));
(residue,pole) pairs = (R=4,P=1) and (R=-3, P=0.5)

4 −3
X(z) = +
1 − z −1 1 − 12 z −1
 example continued

1 + z −1 4 −3
X(z) = = +
(1 − z −1)(1 − 21 z −1) 1 − z −1 1 − 12 z −1

with two distinct poles p1 = 1, p2 = 21 , there are three possible regions of convergence

1 1 n


1. left-sided sequence, ROC : |z| < 2, x[n] = 3 2 u[−n − 1] − 4u[−n − 1]

1 1 n


2. two-sided sequence, ROC : 2 < |z| < 1, x[n] = −3 2 u[n] − 4u[−n − 1]

1 n


3. right-sided sequence, ROC : 1 < |z|, x[n] = −3 2 u[n] + 4u[n]
 example continued

none of these sequences are stable because none of the ROCs contain the unit circle
 example
x[n] is causal and

1 + z −1 A A∗
X(z) = −1 −2
= −1
+ ∗ −1
, |z| > |p|
1 − z + 0.5z 1 − pz 1−p z

x[n] = (Apn + A∗p∗n) u[n] = 2<{Apn}u[n]

[R,P,K]=residuez([1,1],[1,-1,0.5]);
(residue,pole) pairs = (R, P ) = ( 21 − j 32 , 12 + j 12 ) and (R, P ) = ( 12 + j 23 , 12 − j 21 )
√
10 −j71.56◦ 1 π
A= e , p = √ ej 4
2 2
√  n
10 1 π 
x[n] = √ cos n − 71.56◦ u[n]
2 2 4

is this a stable sequence? does the ROC include the unit circle?
 example continued

√  n
10 1 π
◦

x[n] = √ cos n − 71.56 u[n]
2 2 4
 inverse z-transform of rational functions

PM −k
k=0 b[k]z
X(z) = PN
−k
k=0 a[k]z

• assume that A(z) has distinct roots (distinct poles)

• PFE of X(z) has the form

M −N N
X X Ak
X(z) = Ck z −k +
1 − pk z −1
k=0 k=1

• first sum (direct term) exists when M ≥ N

• then a causal signal x[n] has the simple form

M
X −N N
X
x[n] = Ck δ[n − k] + Ak pnku[n]
k=0 k=1
• but there are N + 1 possible inverses x[n], each corresponding to a different ROC

• at most one of the inverses x[n] is stable

• if a pole on the unit circle, then none of the x[n] are stable
 LTI systems

• transfer function and impulse response of LTI system

PM −k ∞
b[k]z X
H(z) = PN k=0
= h[n]z −n, ROCH
k=0 a[k]z −k n=−∞

• system is stable if ROC contains the unit circle

• system is causal if ROC lies outside the outtermost pole

• system is causal and stable if all poles lie inside the unit circle

• zeros of H(z) can be anywhere

 LTI systems example
is a causal realization of the system below stable?

1 − z −2
H(z) =
1 + 0.9z −1 + 0.6z −2 + 0.05z −3

[R,P,K]=residuez([1,0,-1],[1,0.9,0.6,0.05]);
zplane([1,0,-1],[1,0.9,0.6,0.05]);
roots([1,0.9,0.6,0.05])
the last one produces

-0.4022 + 0.6011i
-0.4022 - 0.6011i
-0.0956

abs(roots([1,0.9,0.6,0.05]))
this produces

0.7233
0.7233
0.0956
 LTI systems example continued

zplane([1,0,-1],[1,0.9,0.6,0.05]);
freqz([1,0,-1],[1,0.9,0.6,0.05]);
 complex conjugate roots
M
X
b[k]z −k M −N K1 K
2
X X Ak X bk,0 + bk,1z −1
H(z) = k=0
= Ck z −k + −1
+
XN 1 − pk z 1 + ak,1z −1 + ak,2z −2
a[k]z −k k=0 k=1 k=1

k=0

M
X −N K1
X K2
X
Ck δ[n − k] Ak pnku[n] 2|Bk |rkn cos(ωk n + θk )u[n]
k=0 k=0 k=0
(direct terms) (real distinct poles) (complex conjugate poles)

Bk Bk∗ bk,0 + bk,1z −1

+ = , Bk = |Bk |ejθk , pk = rk ejωk
1 − pk z −1 1 − p∗k z −1 1 + ak,1z −1 + ak,2z −2
bk,0 = 2|Bk | cos θk ak,1 = −2rk cos ωk
bk,1 = −2rk |Bk | cos(ωk − θk ) ak,2 = rk2
 symmetries

• if x[n] = x∗[n] (i.e. x[n] is real), then X(z) = X ∗(z ∗) (prove this)

• if X(z0) = 0, then X(z0∗) = 0

• zeros off the real axis occur in complex conjugate pairs

• ex: x[n] = δ[n] + δ[n − 2] ↔ X(z) = 1 + z −2 = (1 − jz −1)(1 + jz −1)

• x=randn(1,5); roots(x) returns

0.5565 + 0.6921i
0.5565 - 0.6921i
-1.3343
-0.6746

as expected, zeros are real or occur in complex conjugate pairs

 symmetries

• if x[n] = x[−n] (i.e. x[n] is even), then X(z) = X(z −1) (prove this)

• if X(z0) = 0, then X(z0−1) = 0

• zeros occur in reciprocal pairs

• example:
 
−35 31 −35
x[n] = 1, , , ,1
6 3 6
35 1 31 35 −1
2
X(z) = z − z + − z + z −2
6 3 6
   
1 1
= z 2 1 − 2z −1 1 − z −1 1 − 3z −1 1 − z −1



2 3
 symmetries

• if h[n] = h[−n] = h∗[n] = h∗[−n] (i.e. real and even),

then H(z0) = H(z0∗) = H(z0−1) = H(z0−∗) = 0
• in general zeros occur in 4-tuples (z0, z0∗, z0−1, z0−∗)
• unit circle zeros appear in conjugate pairs (z0, z0∗)
• real zeros occur in reciprocal pairs (z0, z0−1)
• zeros at ±1 can appear alone

h = -0.8320
2.2262
-3.6809
5.1558
-5.7986
6.7314
-5.7986
5.1558
-3.6809
2.2262
-0.8320
 one-sided z-transform

• z-transform of x[n]u[n]

• x[n] for n < 0 is ignored

∞
X
X +(z) = x[n]z −n ROC=exterior of circle
n=0

• most z-transform properties carry over to the one-sided z-transform except for
the time-shifting property

x[n] ↔ X +(z)
x[n − 1] ↔ x[−1] + z −1X +(z)
x[n − 2] ↔ x[−2] + x[−1]z −1 + z −2X +(z)
k
X
x[n − k] ↔ x[−m]z −k+m + z −k X +(z)
m=1
 zero-input and zero-state response of LTI system

• difference equation with non-zero initial condition

y[n] = ay[n − 1] + bx[n], n ≥ 0, y[−1] 6= 0

• take one-sided z-transform of both sides

Y +(z) = ay[−1] + z −1Y +(z) + bX +(z)

• solve for Y +(z) yields

ay[−1] b
Y +(z) = −1
+ −1
X +
(z)
|1 −{z
az } |1 − az {z }
zero-input zero-state
 zero-input and zero-state response of LTI system
consider a step input

+ 1
x[n] = u[n] ↔ X (z) =
1 − z −1

then the output is

ay[−1] b
Y +(z) = +
1 − az −1 (1 − az −1)(1 − z −1)
ay[−1] b/(1 − a) ab/(1 − a)
= −1
+ −1
−
1 − az 1−z 1 − az −1
n+1 b n+1


y[n] = y[−1]a + 1−a , n≥0
| {z } 1 − a
zero-input | {z }
zero-state