Z Transform 1

©Dr. James S. Kang

Professor
ECE Department
Cal Poly Pomona

1
 Definition
• The z transform of a discrete-time signal has the same role as the
Laplace has on a continuous time signal.
• The z transform is defined as
∞

𝑋𝑋 𝑧𝑧 = � 𝑥𝑥 𝑘𝑘 𝑧𝑧 −𝑘𝑘
𝑘𝑘=−∞
• z is a complex variable. z is represented in polar coordinates as
𝑧𝑧 = 𝑟𝑟𝑒𝑒 𝑗𝑗𝜃𝜃

2
 Relation to Discrete-Time Fourier Transform
• If 𝑧𝑧 = 𝑟𝑟𝑒𝑒 𝑗𝑗𝜃𝜃 is substituted to the definition of z transform, we get
∞ ∞

𝑋𝑋 𝑧𝑧 = � 𝑥𝑥 𝑘𝑘 (𝑟𝑟𝑒𝑒 𝑗𝑗𝜃𝜃 )−𝑘𝑘 = � 𝑥𝑥 𝑘𝑘 𝑟𝑟 −𝑘𝑘 𝑒𝑒 −𝑗𝑗𝑗𝑗𝜃𝜃

𝑘𝑘=−∞ 𝑘𝑘=−∞
• The z transform is the discrete-time Fourier transform (DTFT) of
𝑥𝑥 𝑘𝑘 𝑟𝑟 −𝑘𝑘 .
• Multiplying x(k) by 𝑟𝑟 −𝑘𝑘 may make the summation to converge.

3
 Z transform of Kronecker Delta Function
• The z transform of Kronecker delta function δ(k) is given by
∞

𝑍𝑍 𝛿𝛿(𝑘𝑘) = � 𝛿𝛿 𝑘𝑘 𝑧𝑧 −𝑘𝑘 = 𝛿𝛿 0 𝑧𝑧 −0 = 1
𝑘𝑘=−∞
• Notice that 𝛿𝛿 𝑘𝑘 is 1 for k = 0 and zero for k ≠ 0.
• Compare with the Laplace transform of Dirac delta function:
L[𝛿𝛿 𝑡𝑡 ] = 1.

4
 Z transform of Unit Step Function
• The z transform of unit step
∞
function u(k) is given by
𝑍𝑍 𝑢𝑢(𝑘𝑘) = � 𝑢𝑢 𝑘𝑘 𝑧𝑧 −𝑘𝑘 = 1 + 𝑧𝑧 −1 + 𝑧𝑧 −2 + ⋯
𝑘𝑘=−∞
1 𝑧𝑧 2
Region of Convergence

= −1
=
1 − 𝑧𝑧 𝑧𝑧 − 1 1
if |z-1|< 1 or |z| > 1 or r > 1. r > 1 represents

Imaginary
outside the unit circle in the complex z plane. 0

In this region, the z transform is finite. This -1

region is called region of convergence (ROC). -2

-2 -1 0 1 2
• A pole at z = 1 and a zero at z = 0. Real

5
 EXAMPLE
• Find the z transform of u(-k)
∞
and specify ROC.
𝑍𝑍 𝑢𝑢(−𝑘𝑘) = � 𝑢𝑢 −𝑘𝑘 𝑧𝑧 −𝑘𝑘 = 1 + 𝑧𝑧1 + 𝑧𝑧 2 + ⋯
𝑘𝑘=−∞
1
=
1 − 𝑧𝑧
if |z|< 1 or r < 1. r < 1 represents inside the unit circle in the complex z
plane.
• If the sequence is right-handed, ROC is outside of a circle of certain
radius. If the sequence is left-handed, ROC is inside of a circle of
certain radius.

6
 EXAMPLE
• Find the z transform of aku(k) and specify Region of Convergence
ROC. 2
∞

𝑍𝑍 𝑎𝑎𝑘𝑘 𝑢𝑢(𝑘𝑘) = � 𝑎𝑎𝑘𝑘 𝑢𝑢 𝑘𝑘 𝑧𝑧 −𝑘𝑘 1

𝑘𝑘=−∞
=1+ 𝑎𝑎𝑎𝑎 −1 + 𝑎𝑎2 𝑧𝑧 −2
+⋯

Imaginary
1 𝑧𝑧 0
= =
1 − 𝑎𝑎𝑧𝑧 −1 𝑧𝑧 − 𝑎𝑎
if |az-1|< 1 or |z| > |a| or r > |a|. r > |a| -1
represents outside the circle of radius |a| in the
complex z plane.
-2
• A pole at z = a and a zero at z = 0. -2 -1 0 1 2
Real

7
 EXAMPLE
• Find the z transform of x(k)=a∞
-ku(-k) and specify ROC.

𝑋𝑋 𝑧𝑧 = 𝑍𝑍 𝑎𝑎−𝑘𝑘 𝑢𝑢(−𝑘𝑘) = � 𝑎𝑎−𝑘𝑘 𝑢𝑢 −𝑘𝑘 𝑧𝑧 −𝑘𝑘 = 1 + 𝑎𝑎𝑎𝑎1 + 𝑎𝑎2 𝑧𝑧 2 + ⋯

𝑘𝑘=−∞
1
=
1 − 𝑎𝑎𝑎𝑎
if |az|< 1 or |z| < 1/|a| or r < 1/|a|. r < 1/|a| represents inside the circle of
radius 1/|a| in the complex z plane.
• A pole at z = 1/a and a zero at z = ∞.
1 1
• If z is replaced by z-1 in 𝑍𝑍 𝑎𝑎𝑘𝑘 𝑢𝑢(𝑘𝑘) = , we get . This is called time
1−𝑎𝑎𝑧𝑧 −1 1−𝑎𝑎𝑎𝑎
reversal property.

8
 Region of Convergence (ROC)
• Finite length sequence: X(z) converges everywhere except at z = 0
and/or z = ∞.
• EXAMPLE (a) x(k) = δ(k) + 2δ(k-1) + δ(k-2), X(z) = 1 + 2z-1 + z-2
ROC: entire z plane except z = 0.
(b) x(k) = 2δ(k+1) + δ(k) + 2δ(k-1), X(z) = 2z + 1 + 2z-1
ROC: entire z plane except z = 0 and z = ∞.
(c) x(k) = δ(k+2) + 4δ(k+1) + 3δ(k), X(z) = z2 + 4z + 3
ROC: entire z plane except z = ∞.

9
 Region of Convergence (ROC)
• Right-handed sequence: The right-handed sequences converge
outside of a circle centered at the origin with certain radius with
possible exception at z = ∞.
EXAMPLE Z[aku(k)] = 1/(1-az-1), ROC is|z| > |a|.
• Left-handed sequence: The left-handed sequences converge inside of
a circle centered at the origin with certain radius with possible
exception at z = 0.
EXAMPLE Z[a-ku(-k)] = 1/(1-az), ROC is|z| < 1/|a|.

10
 Region of Convergence (ROC)
• Two-sided sequences: The two-sided sequences converge between two
concentric circles.
• EXAMPLE Find the z transform, ROC, and poles and zeros of
x(k) = (0.5)ku(k) + (0.8)-ku(-k)
1 1 𝑧𝑧 1
𝑋𝑋 𝑧𝑧 = + = +
1 − 0.5𝑧𝑧 −1 1 − 0.8𝑧𝑧 𝑧𝑧 − 0.5 −0.8 𝑧𝑧 − 1
0.8
𝑧𝑧 1.25 𝑧𝑧 2 − 2.5𝑧𝑧 + 0.625 (𝑧𝑧 − 0.2818)(𝑧𝑧 − 2.2182) 1 − 2.5𝑧𝑧 −1 + 0.625𝑧𝑧 −2
= − = = =
𝑧𝑧 − 0.5 𝑧𝑧 − 1.25 𝑧𝑧 2 − 1.75𝑧𝑧 + 0.625 (𝑧𝑧 − 0.5)(𝑧𝑧 − 1.25) 1 − 1.75𝑧𝑧 −1 + 0.625𝑧𝑧 −2
• ROC: 0.5 < |z| < 1.25
• Poles: z = 0.5, z = 1.25, zeros: z = 0.2818, z = 2.2182

11
 Linearity Property
• If Z[x1(k)] = X1(z) and Z[x2(k)] = X2(z), then
Z[a1x1(k) + a2x2(k)] = a1X1(z) + a2X2(z).
• The ROC is the intersection of the two ROCs.

12
 Time Shifting Property
• If Z[x(k)] = X(z), then
𝑍𝑍 𝑥𝑥(𝑘𝑘 − 𝑘𝑘𝑑𝑑 ) = 𝑧𝑧 −𝑘𝑘𝑑𝑑 𝑋𝑋(𝑧𝑧)
• Proof Let m = k – kd. Then, k = m + kd, When k→∞, m→∞, When k→-
∞, m→-∞ ∞ ∞

𝑍𝑍 𝑥𝑥(𝑘𝑘 − 𝑘𝑘𝑑𝑑 ) = � 𝑥𝑥 𝑘𝑘 − 𝑘𝑘𝑑𝑑 𝑧𝑧 −𝑘𝑘 = � 𝑥𝑥 𝑚𝑚 𝑧𝑧 −𝑚𝑚−𝑘𝑘𝑑𝑑

𝑘𝑘=−∞ 𝑚𝑚=−∞

= 𝑧𝑧 −𝑘𝑘𝑑𝑑 � 𝑥𝑥 𝑚𝑚 𝑧𝑧 −𝑚𝑚 = 𝑧𝑧 −𝑘𝑘𝑑𝑑 𝑋𝑋(𝑧𝑧)

𝑚𝑚=−∞

13
 EXAMPLE
• Find the z-transform of the following signals.
(a) x(k) = u(-k+3) (b) x(k) = (0.6)k-3u(k-4) (c) x(k) = (0.8)k+2u(k+3)
1 𝑧𝑧 −3
• (a) x(k) = u(-(k-3)), Z[u(-k)] = , X(z) =
1−𝑧𝑧 1−𝑧𝑧
𝑧𝑧 −4
• (b) x(k) = (0.6)k-4+1u(k-4) = (0.6)(0.6)k-4u(k-4) = 0.6
1−0.6𝑧𝑧 −1
𝑧𝑧 3
• (c) x(k) = (0.8) k+3-1u(k+3) = (0.8) (0.8) u(k+3) = 1.25
-1 k+3
1−0.8𝑧𝑧 −1

14
 EXAMPLE
syms k
x=(0.6)^(k-3)*heaviside(k-4)
X=ztrans(x)
X=simplify(X)
X=vpa(X,8)

X = (3*(1/((5*z)/3 - 1) + 1))/(5*z^4)
X = 3/(z^3*(5*z - 3))
X = 3.0/(z^3*(5.0*z - 3.0))
X=3.0/(5*z^4*(1-0.6*z^(-1)))=0.6*z^(-4)/(1-0.6*z^(-1))

15
 Exponential Multiplication Property
𝑧𝑧
• If Z[x(k)] = X(z), then Z[akx(k)] = 𝑋𝑋 = 𝑋𝑋 𝑎𝑎−1 𝑧𝑧 .
𝑎𝑎
• Replace 𝑧𝑧 by 𝑎𝑎−1 𝑧𝑧
• Replace 𝑧𝑧 −1 by 𝑎𝑎𝑧𝑧 −1
• Proof ∞ ∞

𝑋𝑋 𝑧𝑧 = � 𝑥𝑥 𝑘𝑘 𝑎𝑎𝑘𝑘 𝑧𝑧 −𝑘𝑘 = � 𝑥𝑥 𝑘𝑘 𝑎𝑎−1 𝑧𝑧 −𝑘𝑘 = 𝑋𝑋 𝑎𝑎−1 𝑧𝑧

𝑘𝑘=−∞ 𝑘𝑘=−∞

1
• 𝑍𝑍 𝑢𝑢 𝑘𝑘 =
1−𝑧𝑧 −1

16
 EXAMPLE
𝑒𝑒 𝑗𝑗𝜃𝜃𝑜𝑜𝑘𝑘 +𝑒𝑒 −𝑗𝑗𝜃𝜃𝑜𝑜 𝑘𝑘 1 1
• Z[cos(θok)u(k)] = 𝑍𝑍 𝑢𝑢(𝑘𝑘) = 𝑍𝑍 𝑒𝑒 𝑗𝑗𝜃𝜃𝑜𝑜𝑘𝑘 𝑢𝑢 𝑘𝑘 + 𝑍𝑍 𝑒𝑒 −𝑗𝑗𝜃𝜃𝑜𝑜𝑘𝑘 𝑢𝑢 𝑘𝑘
2 2 2
1 1
•= 2 � + 2 �
1−𝑧𝑧 −1 𝑧𝑧 −1 =𝑒𝑒 𝑗𝑗𝜃𝜃𝑜𝑜 𝑧𝑧 −1 1−𝑧𝑧 −1 𝑧𝑧 −1 =𝑒𝑒 −𝑗𝑗𝜃𝜃𝑜𝑜 𝑧𝑧 −1

1 1 1 1 0.5 1−𝑒𝑒 −𝑗𝑗𝜃𝜃𝑜𝑜 𝑧𝑧 −1 +0.5 1−𝑒𝑒 𝑗𝑗𝜃𝜃𝑜𝑜 𝑧𝑧 −1

= + =
2 1−𝑒𝑒 𝑗𝑗𝜃𝜃𝑜𝑜 𝑧𝑧 −1 2 1−𝑒𝑒 −𝑗𝑗𝜃𝜃𝑜𝑜 𝑧𝑧 −1 1−𝑒𝑒 𝑗𝑗𝜃𝜃𝑜𝑜 𝑧𝑧 −1 1−𝑒𝑒 −𝑗𝑗𝜃𝜃𝑜𝑜 𝑧𝑧 −1

𝑒𝑒𝑗𝑗𝜃𝜃𝑜𝑜 +𝑒𝑒−𝑗𝑗𝜃𝜃𝑜𝑜 −1
1− 2
𝑧𝑧
• =
1−𝑒𝑒 𝑗𝑗𝜃𝜃𝑜𝑜 𝑧𝑧 −1 1−𝑒𝑒 −𝑗𝑗𝜃𝜃𝑜𝑜 𝑧𝑧 −1

1−𝑐𝑐𝑐𝑐𝑐𝑐 𝜃𝜃𝑜𝑜 𝑧𝑧 −1 1−𝑐𝑐𝑐𝑐𝑐𝑐 𝜃𝜃𝑜𝑜 𝑧𝑧 −1

• = =
1− 𝑒𝑒 𝑗𝑗𝜃𝜃𝑜𝑜 +𝑒𝑒 −𝑗𝑗𝜃𝜃𝑜𝑜 𝑧𝑧 −1 +𝑧𝑧 −2 1−2𝑐𝑐𝑐𝑐𝑐𝑐 𝜃𝜃𝑜𝑜 𝑧𝑧 −1 +𝑧𝑧 −2

17
 Z Transform of sin(θok)u(k)
𝑒𝑒 𝑗𝑗𝜃𝜃𝑜𝑜 𝑘𝑘 −𝑒𝑒 −𝑗𝑗𝜃𝜃𝑜𝑜 𝑘𝑘 1 1
• Z[sin(θok)u(k)] = 𝑍𝑍 2𝑗𝑗
𝑢𝑢(𝑘𝑘) =
2𝑗𝑗
𝑍𝑍 𝑒𝑒 𝑗𝑗𝜃𝜃𝑜𝑜 𝑘𝑘 𝑢𝑢(𝑘𝑘) − 𝑍𝑍
2𝑗𝑗
𝑒𝑒 −𝑗𝑗𝜃𝜃𝑜𝑜 𝑘𝑘 𝑢𝑢(𝑘𝑘)
1 1 1 1 −0.5𝑗𝑗 1−𝑒𝑒 −𝑗𝑗𝜃𝜃𝑜𝑜 𝑧𝑧 −1 +0.5𝑗𝑗 1−𝑒𝑒 𝑗𝑗𝜃𝜃𝑜𝑜 𝑧𝑧 −1
•= − =
2𝑗𝑗 1−𝑒𝑒 𝑗𝑗𝜃𝜃𝑜𝑜 𝑧𝑧 −1 2𝑗𝑗 1−𝑒𝑒 −𝑗𝑗𝜃𝜃𝑜𝑜 𝑧𝑧 −1 1−𝑒𝑒 𝑗𝑗𝜃𝜃𝑜𝑜 𝑧𝑧 −1 1−𝑒𝑒 −𝑗𝑗𝜃𝜃𝑜𝑜 𝑧𝑧 −1

−0.5𝑗𝑗 + 0.5𝑗𝑗𝑗𝑗 −𝑗𝑗𝜃𝜃𝑜𝑜 𝑧𝑧 −1 + 0.5𝑗𝑗 − 0.5𝑗𝑗𝑗𝑗 𝑗𝑗𝜃𝜃𝑜𝑜 𝑧𝑧 −1 −0.5𝑗𝑗 𝑒𝑒 𝑗𝑗𝜃𝜃𝑜𝑜 − 𝑒𝑒 −𝑗𝑗𝜃𝜃𝑜𝑜 𝑧𝑧 −1

= 𝑗𝑗𝜃𝜃 −1 −𝑗𝑗𝜃𝜃 −1
=
1 − 𝑒𝑒 𝑧𝑧 𝑜𝑜 1 − 𝑒𝑒 𝑜𝑜 𝑧𝑧 1 − 𝑒𝑒 𝑗𝑗𝜃𝜃𝑜𝑜 𝑧𝑧 −1 1 − 𝑒𝑒 −𝑗𝑗𝜃𝜃𝑜𝑜 𝑧𝑧 −1

𝑒𝑒 𝑗𝑗𝜃𝜃𝑜𝑜 − 𝑒𝑒 −𝑗𝑗𝜃𝜃𝑜𝑜 −1
𝑧𝑧 𝑠𝑠𝑠𝑠𝑠𝑠 𝜃𝜃𝑜𝑜 𝑧𝑧 −1
2𝑗𝑗
= =
𝑗𝑗𝜃𝜃
1 − 𝑒𝑒 𝑧𝑧 𝑜𝑜 −1 1 − 𝑒𝑒 −𝑗𝑗𝜃𝜃𝑜𝑜 𝑧𝑧 −1 1 − 2𝑐𝑐𝑐𝑐𝑐𝑐 𝜃𝜃𝑜𝑜 𝑧𝑧 −1 + 𝑧𝑧 −2

18
 EXAMPLE
𝑘𝑘 1−𝑐𝑐𝑐𝑐𝑐𝑐 𝜃𝜃𝑜𝑜 𝑎𝑎𝑧𝑧 −1
• 𝑍𝑍 𝑎𝑎 𝑐𝑐𝑐𝑐𝑐𝑐 𝜃𝜃𝑜𝑜 𝑘𝑘 𝑢𝑢(𝑘𝑘) =
1−2𝑐𝑐𝑐𝑐𝑐𝑐 𝜃𝜃𝑜𝑜 𝑎𝑎𝑧𝑧 −1 +𝑎𝑎2 𝑧𝑧 −2
𝑘𝑘 𝑠𝑠𝑠𝑠𝑠𝑠 𝜃𝜃𝑜𝑜 𝑎𝑎𝑎𝑎 −1
• 𝑍𝑍 𝑎𝑎 𝑠𝑠𝑠𝑠𝑠𝑠 𝜃𝜃𝑜𝑜 𝑘𝑘 𝑢𝑢(𝑘𝑘) =
1−2𝑐𝑐𝑐𝑐𝑐𝑐 𝜃𝜃𝑜𝑜 𝑎𝑎𝑎𝑎 −1 +𝑎𝑎2 𝑧𝑧 −2
𝑐𝑐𝑐𝑐𝑐𝑐 𝜙𝜙 −𝑐𝑐𝑐𝑐𝑐𝑐 𝜃𝜃𝑜𝑜 −𝜙𝜙 𝑧𝑧 −1
• 𝑍𝑍 𝑐𝑐𝑐𝑐𝑐𝑐 𝜃𝜃𝑜𝑜 𝑘𝑘 + 𝜙𝜙 𝑢𝑢(𝑘𝑘) =
1−2𝑐𝑐𝑐𝑐𝑐𝑐 𝜃𝜃𝑜𝑜 𝑧𝑧 −1 +𝑧𝑧 −2
𝑠𝑠𝑠𝑠𝑠𝑠 𝜙𝜙 +𝑠𝑠𝑠𝑠𝑠𝑠 𝜃𝜃𝑜𝑜 −𝜙𝜙 𝑧𝑧 −1
• 𝑍𝑍 𝑠𝑠𝑠𝑠𝑠𝑠 𝜃𝜃𝑜𝑜 𝑘𝑘 + 𝜙𝜙 𝑢𝑢(𝑘𝑘) =
1−2𝑐𝑐𝑐𝑐𝑐𝑐 𝜃𝜃𝑜𝑜 𝑧𝑧 −1 +𝑧𝑧 −2
𝑐𝑐𝑐𝑐𝑐𝑐 𝜙𝜙 −𝑐𝑐𝑐𝑐𝑐𝑐 𝜃𝜃𝑜𝑜 −𝜙𝜙 𝑎𝑎𝑎𝑎 −1
• 𝑍𝑍 𝑎𝑎𝑘𝑘 𝑐𝑐𝑐𝑐𝑐𝑐 𝜃𝜃𝑜𝑜 𝑘𝑘 + 𝜙𝜙 𝑢𝑢(𝑘𝑘) =
1−2𝑐𝑐𝑐𝑐𝑐𝑐 𝜃𝜃𝑜𝑜 𝑎𝑎𝑎𝑎 −1 +𝑎𝑎2 𝑧𝑧 −2
𝑠𝑠𝑠𝑠𝑠𝑠 𝜙𝜙 +𝑠𝑠𝑠𝑠𝑠𝑠 𝜃𝜃𝑜𝑜 −𝜙𝜙 𝑎𝑎𝑎𝑎 −1
• 𝑍𝑍 𝑎𝑎𝑘𝑘 𝑠𝑠𝑠𝑠𝑠𝑠 𝜃𝜃𝑜𝑜 𝑘𝑘 + 𝜙𝜙 𝑢𝑢(𝑘𝑘) =
1−2𝑐𝑐𝑐𝑐𝑐𝑐 𝜃𝜃𝑜𝑜 𝑎𝑎𝑎𝑎 −1 +𝑎𝑎2 𝑧𝑧 −2

19
 Time Reversal Property
• If Z[x(k)] = X(z), then Z[x(-k)] = X(z-1). Replace z by z-1. Replace z-1 by z.
• EXAMPLE
1−𝑐𝑐𝑐𝑐𝑐𝑐 𝜃𝜃𝑜𝑜 𝑧𝑧
• Z[cos(θok)u(-k)] =
1−2𝑐𝑐𝑐𝑐𝑐𝑐 𝜃𝜃𝑜𝑜 𝑧𝑧+𝑧𝑧 2
𝑠𝑠𝑠𝑠𝑠𝑠 𝜃𝜃𝑜𝑜 𝑧𝑧
• Z[sin(−θok)u(−k)] =
1−2𝑐𝑐𝑐𝑐𝑐𝑐 𝜃𝜃𝑜𝑜 𝑧𝑧+𝑧𝑧 2
• Conjugate property
• If Z[x(k)] = X(z), then Z[x*(k)] = X*(z*).

20
 Differentiation in the Z Domain
𝑑𝑑𝑑𝑑(𝑧𝑧)
• If Z[x(k)] = X(z), then 𝑍𝑍 𝑘𝑘𝑘𝑘(𝑘𝑘) = −𝑧𝑧 . In general,
𝑑𝑑𝑑𝑑
𝑑𝑑 𝑚𝑚 𝑋𝑋(𝑧𝑧)
• 𝑍𝑍 𝑘𝑘 𝑚𝑚 𝑥𝑥(𝑘𝑘) = (−𝑧𝑧)𝑚𝑚
𝑑𝑑𝑧𝑧 𝑚𝑚
• Proof Taking the derivative on both sides of 𝑋𝑋 𝑧𝑧 = ∑∞
𝑘𝑘=−∞ 𝑥𝑥 𝑘𝑘 𝑧𝑧 −𝑘𝑘 ,

𝑑𝑑𝑋𝑋 𝑧𝑧 1 ∞
• = ∑∞
𝑘𝑘=−∞ −𝑘𝑘𝑘𝑘 𝑘𝑘 𝑧𝑧 −𝑘𝑘−1
= − ∑𝑘𝑘=−∞ 𝑘𝑘𝑥𝑥 𝑘𝑘 𝑧𝑧 −𝑘𝑘
𝑑𝑑𝑑𝑑 𝑧𝑧
𝑑𝑑𝑑𝑑(𝑧𝑧)
• Thus, 𝑍𝑍 𝑘𝑘𝑘𝑘(𝑘𝑘) = −𝑧𝑧
𝑑𝑑𝑑𝑑

21
 EXAMPLE
• Find z transform of (a) kaku(k) (b) k2aku(k) (d) k3aku(k)
𝑑𝑑 1 𝑑𝑑
• 𝑍𝑍 𝑘𝑘𝑎𝑎𝑘𝑘 𝑢𝑢(𝑘𝑘) = −𝑧𝑧 = −𝑧𝑧 1 − 𝑎𝑎𝑧𝑧 −1 −1
𝑑𝑑𝑑𝑑 1−𝑎𝑎𝑎𝑎 −1 𝑑𝑑𝑑𝑑
𝑎𝑎𝑧𝑧 −1
• = −𝑧𝑧 −1 1 − 𝑎𝑎𝑧𝑧 −1 −2 −1 −1 𝑎𝑎𝑧𝑧 −2 =
1−𝑎𝑎𝑧𝑧 −1 2
𝑑𝑑 𝑎𝑎𝑧𝑧 −1 𝑎𝑎𝑧𝑧 −1 +𝑎𝑎2 𝑧𝑧 −2
• 𝑍𝑍 𝑘𝑘 2 𝑎𝑎𝑘𝑘 𝑢𝑢(𝑘𝑘) = −𝑧𝑧 =
𝑑𝑑𝑑𝑑 1−𝑎𝑎𝑎𝑎 −1 2 1−𝑎𝑎𝑎𝑎 −1 3
𝑑𝑑 𝑎𝑎𝑧𝑧 −1 +𝑎𝑎2 𝑧𝑧 −2 𝑎𝑎𝑧𝑧 −1 +4𝑎𝑎2 𝑧𝑧 −2 +𝑎𝑎3 𝑧𝑧 −3
• 𝑍𝑍 𝑘𝑘 3 𝑎𝑎𝑘𝑘 𝑢𝑢(𝑘𝑘) = −𝑧𝑧 =
𝑑𝑑𝑑𝑑 1−𝑎𝑎𝑎𝑎 −1 3 1−𝑎𝑎𝑎𝑎 −1 4

22
 EXAMPLE
• Find z transform of (a) ku(k) (b) k2u(k) (d) k3u(k)
𝑑𝑑 1 𝑑𝑑
• 𝑍𝑍 𝑘𝑘𝑢𝑢(𝑘𝑘) = −𝑧𝑧 = −𝑧𝑧 1 − 𝑧𝑧 −1 −1
𝑑𝑑𝑑𝑑 1−𝑧𝑧 −1 𝑑𝑑𝑑𝑑
𝑧𝑧 −1
• = −𝑧𝑧 −1 1 − 𝑧𝑧 −1 −2 −1 −1 𝑧𝑧 −2 =
1−𝑧𝑧 −1 2
𝑑𝑑 𝑧𝑧 −1 𝑧𝑧 −1 +𝑧𝑧 −2
• 𝑍𝑍 𝑘𝑘 2 𝑢𝑢(𝑘𝑘) = −𝑧𝑧 =
𝑑𝑑𝑑𝑑 1−𝑧𝑧 −1 2 1−𝑧𝑧 −1 3
𝑑𝑑 𝑧𝑧 −1 +𝑧𝑧 −2 𝑧𝑧 −1 +4𝑧𝑧 −2 +𝑧𝑧 −3
• 𝑍𝑍 𝑘𝑘 3 𝑢𝑢(𝑘𝑘) = −𝑧𝑧 =
𝑑𝑑𝑑𝑑 1−𝑧𝑧 −1 3 1−𝑧𝑧 −1 4

23
 EXAMPLE
• Find the z transform of x(k) = kak-2u(k-3).
• X(z) = Z[kak-2u(k-3)] = Z[(k-3+3)ak-3+1u(k-3)]
• X(z) = a Z[(k-3)ak-3u(k-3)] + 3a Z[ak-3u(k-3)]
𝑎𝑎𝑧𝑧 −1 𝑧𝑧 −3 𝑧𝑧 −3 𝑎𝑎2 𝑧𝑧 −4 3𝑎𝑎𝑎𝑎 −3
• 𝑋𝑋 𝑧𝑧 = 𝑎𝑎 + 3𝑎𝑎 = +
1−𝑎𝑎𝑎𝑎 −1 2 1−𝑎𝑎𝑎𝑎 −1 1−𝑎𝑎𝑎𝑎 −1 2 1−𝑎𝑎𝑎𝑎 −1
syms k z a
x=k*a^(k-2)*heaviside(k-3)
X=ztrans(x)

X = -(a*(2*a - 3*z))/(z^2*(a - z)^2)

24
 Convolution Property
• If Z[x1(k)] = X1(z) and Z[x2(k)] = X2(z), then Z[x1(k)*x2(k)] = X1(z)X2(z).
• Convolution in the time domain results in multiplication in the z
domain.
• Proof 𝑦𝑦 𝑘𝑘 = ∑∞ 𝑚𝑚=−∞ ℎ 𝑘𝑘 − 𝑚𝑚 𝑥𝑥(𝑚𝑚)
• 𝑌𝑌 𝑧𝑧 = 𝑍𝑍 𝑦𝑦(𝑘𝑘) = ∑∞ ∑ ∞
𝑘𝑘=−∞ 𝑚𝑚=−∞ ℎ 𝑘𝑘 − 𝑚𝑚 𝑥𝑥(𝑚𝑚) 𝑧𝑧 −𝑘𝑘
∞
• = ∑∞ 𝑚𝑚=−∞ 𝑥𝑥(𝑚𝑚) ∑ 𝑘𝑘=−∞ ℎ 𝑘𝑘 − 𝑚𝑚 𝑧𝑧 −𝑘𝑘

• From time shifting property, ∑∞ 𝑘𝑘=−∞ ℎ 𝑘𝑘 − 𝑚𝑚 𝑧𝑧 −𝑘𝑘 = 𝐻𝐻(𝑧𝑧)𝑧𝑧 −𝑚𝑚

• 𝑌𝑌 𝑧𝑧 = ∑∞ 𝑚𝑚=−∞ 𝑥𝑥(𝑚𝑚)𝐻𝐻(𝑧𝑧)𝑧𝑧 −𝑚𝑚 = 𝐻𝐻 𝑧𝑧 ∑∞

𝑚𝑚=−∞ 𝑥𝑥 𝑚𝑚 𝑧𝑧 −𝑚𝑚

• 𝑌𝑌 𝑧𝑧 = 𝐻𝐻 𝑧𝑧 𝑋𝑋 𝑧𝑧
25
 Convolution Property
• Since the output y(k) of a linear, time-invariant (LTI) system is given by
the convolution of the impulse response h(k) of the system and the
input signal x(k), y(k) = h(k)*x(k), the z transform Y(z) of the output of
a linear, time-invariant system is given by the product of the transfer
function H(z), which is the z transform of the impulse response h(k),
and the z-transform X(z) of the input Y(z) = H(z)X(z). H(z) = Y(z)/X(z).
LTI System

x(k) h(k) y(k) = h(k)∗x(k)

X(z) H(z) Y(z) = H(z)X(z)

26
 Multiplication Property
• If Z[x1(k)] = X1(z) and Z[x2(k)] = X2(z), then
1 𝑧𝑧 1
𝑍𝑍 𝑥𝑥1 𝑘𝑘 𝑥𝑥2 (𝑘𝑘) = � 𝑋𝑋1 (𝑣𝑣)𝑋𝑋2 𝑑𝑑𝑑𝑑
2𝜋𝜋𝜋𝜋 𝐶𝐶 𝑣𝑣 𝑣𝑣
• Correlation Property: If Z[x1(k)] = X1(z) and Z[x2(k)] = X2(z), then
∞

𝑍𝑍 � 𝑥𝑥1 (𝑚𝑚)𝑥𝑥2 (𝑚𝑚 − 𝑘𝑘) = 𝑋𝑋1 (𝑧𝑧)𝑋𝑋2 𝑧𝑧 −1

𝑚𝑚=−∞
• Partial Sum: If Z[x(k)] = X(z), then
𝑘𝑘
𝑋𝑋(𝑧𝑧)
𝑍𝑍 � 𝑥𝑥(𝑚𝑚) =
1 − 𝑧𝑧 −1
𝑚𝑚=−∞

27
 Initial Value Theorem
• 𝑥𝑥 𝑘𝑘𝑜𝑜 = lim 𝑧𝑧 𝑘𝑘𝑜𝑜 𝑋𝑋 𝑧𝑧
𝑧𝑧⟶∞
• If ko = 0, 𝑥𝑥 0 = lim 𝑋𝑋 𝑧𝑧
𝑧𝑧⟶∞
• Examples
1 1 1
• Z[u(k)] = , 𝑥𝑥 0 = lim −1 = lim 1 = 1, u(0) = 1
1−𝑧𝑧 −1 𝑧𝑧⟶∞ 1−𝑧𝑧 𝑧𝑧⟶∞ 1−𝑧𝑧
1 1 1
• Z[aku(k)] = , 𝑥𝑥 0 = lim = lim 𝑎𝑎 = 1, a0u(0) = 1
1−𝑎𝑎𝑧𝑧 −1 𝑧𝑧⟶∞ 1−𝑎𝑎𝑧𝑧 −1 𝑧𝑧⟶∞ 1− 𝑧𝑧

28
 Final Value Theorem
• 𝑥𝑥 ∞ = lim 𝑧𝑧 − 1 𝑋𝑋(𝑧𝑧)
𝑧𝑧→1
• Examples
𝑧𝑧 𝑧𝑧
• Z[u(k)] = , 𝑥𝑥 ∞ = lim (𝑧𝑧 − 1) = lim 𝑧𝑧 = 1, u(∞) =1
𝑧𝑧−1 𝑧𝑧⟶1 𝑧𝑧−1 𝑧𝑧⟶1
𝑧𝑧 𝑧𝑧
• Z[a u(k)] =
k , 𝑥𝑥 ∞ = lim (𝑧𝑧 − 1) = 0, a∞u(∞) = 0, a < 1.
𝑧𝑧−𝑎𝑎 𝑧𝑧⟶1 𝑧𝑧−𝑎𝑎

29