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ENSC380
Lecture 29
Objectives:
• Inverse z-Transform
• Synthetic division
• Partial fraction expansion
• Solving difference equations
• Relationship between z and Laplace transforms
• Bilateral z-transform
Atousa Hajshirmohammadi, SFU
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Inverse z-Transform
• The inversion integral for z-transform is:
1
x[n] =
j2π
I
X(z)z n−1 dz
C
H
Where C indicated the contour integral in the complex plane, and is beyond
the scope of this course.
• Here we will look at two other methods:
• Synthetic division: This method results in directly finding x[0], x[1], x[2], . . ..
• Partial fraction expansion: This method results in a summation of simple
z-transforms whose inverses can be looked up from the table.
Atousa Hajshirmohammadi, SFU
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Synthetic Division
• Let’s look at an example: Find the inverse z-transform of
X(z) =
z2
2
15 2
z + 17
z
2
36
z3 −
z3 −
−
1
18
• Synthetically divide the numerator by the denominator:
• The result of the division is an infinite series. Compare it with
P
X(z) =
∞
0
x[n]z −n . This results in:
• This method can be used if you need to find the first few samples of x[n].
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Partial Fraction Expansion
• This method is similar to the same method for inverse Laplace transform; X(z)
of the previous example can be written as:
z 2 (z − 1/2)
X(z) =
(z − 2/3)(z − 1/3)(z − 1/4)
• The fraction is not “proper” because the degree of the numerator is not less
than the degree of the denominator. But we can write: X(z) = zX1 (z), where
X1 (z) is proper and can be expanded into partial fractions:
X1 (z) =
• Multiply by z to get x(z) =
4/5
2
9/5
+
−
z − 2/3
z − 1/3
z − 1/4
4z
5
z−2/3
+
2z
z−1/3
−
9z
5
Z
. Use αn u[n] ←→
z−1/4
z
z−α
to
find x[n]:
Atousa Hajshirmohammadi, SFU
Solving Difference Equations
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• Example: Solve this difference equation using z-transform.
y[n + 2] −
3
1
1
y[n + 1] + y[n] = ( )n ,
2
2
4
y[0] = 10 y[1] = 4
• Recall the time-shifting (advance) property of z-transform:
• Using this property find the z-transform of both sides of the equation:
• Find Y (z) and the inverse of it y[n] using the partial fraction expansion method.
• The result is y[n] = 16 ( 1 )n + 4( 1 )n +
3 4
2
2
3
u[n], which solves the difference
equation and agrees with the given initial conditions.
Atousa Hajshirmohammadi, SFU
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Laplace and z-Transform
• Consider the CT signal x(t), its sampled DT signal x[n] = x(nTs ), and its
P
impulse samples CT signal xδ (t) =
x[n]δt − nTs .
• Now we can find the z-transform of x[n] and the Laplace transform of xδ (t):
• The relationship is then:
Xδ (s) = X(z)|z→esTs
• We also need to see how the relationship z = esTs maps a point or a region in
the s-plane to the z-plane,
Atousa Hajshirmohammadi, SFU
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s-Plane to z-Plane
Let’s see how the line s = jω = j2πf maps to the z-plane through the relationship
z = esTs .
Atousa Hajshirmohammadi, SFU
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s-Plane to z-Plane (Cont.)
Now we see how does a region in the s-plane map to the z-plane through the
relationship z = esTs .
Atousa Hajshirmohammadi, SFU
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Bilateral z-Transform
• We know that the bi-lateral z-transform of x[n] is:
X(z) =
∞
X
x[n]z
−n
= X(z) =
−∞
∞
X
x[n]z
−n
+ X(z) =
0
−1
X
x[n]z −n
−∞
P0
−n and X (z) = X(z) =
−n . Then we can
• Let Xc (z) = P ∞
x[n]z
ac
0
−∞ x[n]z
write:
X(z) = Xc (z) + Xac [n] − x[0]
• It can be easily seen that
Xac (z) =
∞
X
x[−n](1/z)−n
0
• This gives us a method to find the bilateral transform, using unilateral
transform.
• Some of the properties of bilateral z-transform are different from the unilateral
transform. See Text,pp.786-787.
Atousa Hajshirmohammadi, SFU