EE480.3 Digital Control Systems
Part 2. z-transform
Kunio Takaya
Electrical and Computer Engineering
University of Saskatchewan
January 14, 2008
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Contents
1 z-Transform 3
2 z-transform of simple functions 3
3 Properties of the z-transform 8
4 Continue to discrete time systems 23
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1 z-Transform
The definition of the z-Transform is
X(z) =+∞∑
k=−∞
x(k)z−k.
where, x(k) is a discrete time sequence (sampled data). When x(k)
is defined for k ≥ 0, i.e. causal, one sided z-transform is given by
X(z) =+∞∑
k=0
x(k)z−k.
The variable z is complex, so is X(z).
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2 z-transform of simple functions
δ function
δ(k) =
1 if k = 0
0 otherwise
Z[δ(k)] =+∞∑
k=0
δ(k)z−k = 1
unit step function
u(k) =
1 if k ≥ 0
0 if k < 0
Z[u(k)] =+∞∑
k=0
u(k)z−k = z0 + z−1 + z−2 + +z−3 · · ·
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Since the sum of n terms of a geometric progression is given by the
formula,n−1∑
k=0
ark = a1− rn
1− r
Z[u(k)] = limn→∞
1− z−n
1− z−1=
1
1− z−1.
This z-trnasform is valid only for jzj < 1, in the region of
convergence in the z-plane.
Decaying exponential function
f(t) = e−at t ≥ 0, sampled with a sampling period T
⇒ f(k) = e−aTk, k ≥ 0
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Z[f(k)] =∞∑
k=0
e−aTkz−k =∞∑
k=0
(e−aT z−1)k
=1
1− e−aT z−1
The region of convergence is e−aT < jzj. If a > 0, the system is
stable and the pole at z = e−aT < 1 is inside the unit circle. If
a < 0, the system is unstable and the pole at z = e−aT > 1 is
outside the unit circle.
Damped cosine wave
f(k) = e−aTk cosωTk, k ≥ 0
Z[f(k)] =∞∑
k=0
e−aTk cosωTk z−k
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=1
2
∞∑
k=0
(e−aT+jωT z−1)k +1
2
∞∑
k=0
(e−aT−jωT z−1)k
=1
2
1
1− e−aT+jωT z−1+
1
2
1
1− e−aT−jωT z−1
=1− e−aT cosωTz−1
1− 2e−aT cosωTz−1 + e−2aT z−2
=z2 − e−aT cosωTz
z2 − 2e−aT cosωTz + e−2aT
The region of convergence is e−aT < jzj.
• Exercise: Find the z-transform of a discrete sequence
f(k) = 2−k for k ≥ 0
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3 Properties of the z-transform
Shift Operations - One step delay with u(k)
Z[x(k − 1)u(k)] =∞∑
k=0
x(k − 1)z−k
=∞∑
k=−1
x(k)z−(k+1)
=∞∑
k=0
x(k)z−k−1 + x(−1)z0
= z−1∞∑
k=0
x(k)z−k + x(−1)
= z−1Z[x(k)u(k)] + x(−1)
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Shift Operations - One step delay with u(k − 1)
Z[x(k − 1)u(k − 1)] =∞∑
k=1
x(k − 1)z−k
=∞∑
k=0
x(k)z−(k+1)
= z−1∞∑
k=0
x(k)z−k
= z−1Z[x(k)u(k)]
Shift Operations - m step delay with u(k)
Z[x(k −m)u(k)] =∞∑
k=0
x(k −m)z−k
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=∞∑
k=−m
x(k)z−(k+m)
=∞∑
k=0
x(k)z−k−m +−1∑
k=−m
x(k)z−k−m
= z−m∞∑
k=0
x(k)z−k +m∑
k=1
x(−k)zk−m
= z−mZ[x(k)u(k)] +m∑
k=1
x(−k)zk−m
Shift Operations - m step delay with u(k −m)
Z[x(k −m)u(k −m)] =∞∑
k=m
x(k −m)z−k
=∞∑
k=0
x(k)z−(k+m)
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= z−mZ[x(k)u(k)]
Shift Operations - m step time advance with u(k +m)
Z[x(k +m)u(k +m)] =∞∑
k=−m
x(k +m)z−k
=∞∑
k=0
x(k)z−(k−m)
= z+mZ[x(k)u(k)]
Initial Value theorem
limz→∞
X(z) = limz→∞
+∞∑
k=0
x(k)z−k = x(0)
Final Value theorem
Z[x(k + 1)u(k)]−Z[x(k)u(k)]
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= zX(z)− zx(0)−X(z)
=∞∑
k=0
x(k + 1)z−k −∞∑
k=0
x(k)z−k
zX(z)−X(z) = zx(0) +∞∑
k=0
x(k + 1)z−k −∞∑
k=0
x(k)z−k
limz→1
(z − 1)X(z) = x(0) +∞∑
k=0
x(k + 1)−∞∑
k=0
x(k)
= x(∞)
Derivative
dX(z)
dz=
d
dz
∞∑
k=0
x(k)z−k
=∞∑
k=0
x(k)d
dzz−k
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=∞∑
k=0
−x(k)kz−k−1
= −z−1Z[kx(k)]
Z[kx(k)] = −z dX(z)
dz
Example:
Z[ku(k)] = −z ddz
1
1− z−1
= −z −z−2
(1− z−1)2
=z−1
(1− z−1)2=
z
(z − 1)2
Convolution
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Consider multiplying G(z) and X(z).
G(z)X(z)
=∞∑
k=0
g(k)z−k ×∞∑
k=0
x(k)z−k
= [g(0) + g(1)z−1 + g(2)z−2 + g(3)z−3 + · · ·]
× [x(0) + x(1)z−1 + x(2)z−2 + x(3)z−3 + · · ·]
= g(0)x(0) + [g(0)x(1) + g(1)x(0)]z−1
+ [g(0)x(2) + g(1)x(1) + g(2)x(0)]z−2
+ [g(0)x(3) + g(1)x(2) + g(2)x(1) + g(32)x(0)]z−3
+n∑
k=0
g(k)x(n− k) z−n + · · ·
=∞∑
n=0
n∑
k=0
g(k)x(n− k) z−n
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= Z[n∑
k=0
g(k)x(n− k)] = Z[n∑
k=0
x(k)g(n− k)]
G(z)
Impulse ResponseInput Sequence
X(z)
Output Sequence
Y(z)=G(z)X(z)
Figure: Convolution and Transfer Function
When an input sequence x(k) is applied to a system having an
impulse response g(k), the response y(k) of the system is given the
convolution sum.
y(k) =k∑
n=0
x(n)g(k − n)
The equivalent expression in the z-transform is Y (z) = G(z)X(z).
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Inverse z-Transform
Partial Fraction Method
The z-transform of an exponential sequence x(k) = ak is given by
X(z) =1
1− az−1=
z
z − a.
Any sequence that starts with a non-zero value at k = 0 usually
has the same order in the numerator and denominator of the
z-transform. To find the inverse z-transform, one must take partial
fraction expansion onX(z)
zinstead of X(z) itself.
X ′(z) =X(z)
z=
B(z)
(z − a1)k(z − a2)(z − a3) · · ·
=c11
(z − a1)k+
c12
(z − a1)k−1+ · · ·+ c1k
(z − a1)
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+c2
(z − a2)+
c3(z − a3)
+ · · ·
Where, unknown coefficients are given by
ci = (z − ai) X ′(z)jz=ai
c1n =1
(n− 1)!
dn−1
dzn−1(z − a1)kX ′(z)
∣
∣
∣
∣
z=a1
After partial fraction expansion, find the expanded form of X(z) by
multiplying X ′(z) by z, then find the inverse z-transform
term-by-term by table look-up.
Example
E(z) =1
(z − 1)(z − 2)
E(z)
z=
1
2
1
z− 1
z − 1+
1
2
1
z − 2
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E(z) =1
2− z
z − 1+
1
2
z
z − 2
e(k) = (1
2δ(k)− 1k +
1
22k)u(k)
• Exercise: Find the inverse z-transform of
Y (z) =1
(1− z−1)(1− 2z−1)
Use of MATLAB for partial fraction expansion
>> [r,p,k]=residue([1],[1, -3, 2, 0])
r =
0.5000
-1.0000
0.5000
p =
2
1
18
0
k =
[]
Difference Equation
y(k) + a1y(k − 1) + a2y(k − 2) + · · ·+ aky(0)
= b0u(k) + b1u(k − 1) + · · ·+ bku(0)
Where, y(k) is output sequence, and u(k) is input sequence. In
control systems, b0 is often 0, as input u(k) does not immediately
affect output y(k). Take the z-transform of this difference equation
considering u(k) = 0 and y(k) = 0 for k < 0.
Z f[y(k) + a1y(k − 1) + a2y(k − 2) + · · ·+ aky(0)]u(k)g
= Z f[b0u(k) + b1u(k − 1) + · · ·+ bku(0))]u(k)g
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Since Z[y(k − i)] = z−iY (z) and Z[u(k − i)] = z−iU(z),
Y (z) + a1z−1Y (z) + a2z
−2Y (z) + · · ·+ akz−kY (z)
= b0U(z) + b1z−1U(z) + b2z
−2U(z) + · · ·+ bkz−kU(z)
This leads to a transfer function of the difference equation.
G(z) =Y (z)
U(z)=b0 + b1z
−1 + b2z−2 + · · ·+ bkz
−k
1 + a1z−1 + a2z−2 + · · ·+ akz−k
The output Y (z) for a given input U(z) is given by
Y (z) = G(z)U(z) =b0 + b1z
−1 + b2z−2 + · · ·+ bkz
−k
1 + a1z−1 + a2z−2 + · · ·+ akz−kU(z)
Example
A difference equation,
m(k) = e(k)− e(k − 1)−m(k − 1)
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is driven by an input sequence
e(k) =
1, k ≥ 0 even
0, k ≥ 0 odd
Using the z-Transform,
M(z) =z − 1
z + 1E(z)
E(z) =1
1− z−2=
z2
z2 − 1
Thus, the solution in the z-Transform is
M(z) =z2
z2 + 2z + 1=
z2
(z + 1)2
= 1− 2z−1 + 3z−2 − 4z−3 + · · ·
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Assignment No. 2
Solve following problems related to the z-transform fundamentals
in the textbook pp. 78-79.
1. 2-1
2. 2-2 (a)(b)
3. 2-3 (c)
4. 2-4
5. 2-8 (a)(e) Do only the partial fraction expansion method.
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4 Continue to discrete time systems
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