Intro
Recap
Problems
EE123 Discussion Section 3
Frank Ong
February 4, 2015
Frank Ong
EE123 Discussion Section 3
Intro
Recap
Problems
Plan
I
Recap of Z-transform
I
Practice problems
I
Distribute SDRs
Frank Ong
EE123 Discussion Section 3
Intro
Recap
Problems
Z-transform
X (z) =
1
X
x[n]z
n
n= 1
I
Simpler notations ( z instead of e j! )
I
Generalizes the Fourier transform to more signals (e.g. the
unit step, exploding exponentials)
I
Region of convergence, Zeros and Poles
Frank Ong
EE123 Discussion Section 3
EE 123 Spring 2015
Discussion Section 03
Frank Ong
Page 1 of 2
EE 123 Spring 2015
Discussion Section 03
Frank Ong
Page 2 of 2
Intro
Recap
Problems
Z-transform Inversion in this class
Three methods to invert z-transform:
I
Inspection / Properties of z-transform
an u[n] ,
1
az
I
1
Partial fraction expansion
I
Long division / Power series
Frank Ong
1
{z : |z| > a}
EE123 Discussion Section 3
Intro
Recap
Problems
Question 1
I
Find the inverse z-transform of the following, assuming
right-sided:
X (z) =
Frank Ong
1
1
3
1
4z
+ 18 z
2
EE123 Discussion Section 3
Intro
Recap
Problems
Solution 1
I
Example: Partial Fraction Expansion
1
X (z) =
3
1 4 z 1 + 18 z 2
1
=
1
1
1 4z
1 12 z
A1
A2
=
+
1
1
1 4z
1 12 z
Frank Ong
1
1
EE123 Discussion Section 3
Intro
Recap
Problems
Solution 1
I
Example: Partial Fraction Expansion
1
X (z) =
3
1 4 z 1 + 18 z 2
1
=
1
1
1 4z
1 12 z
A1
A2
=
+
1
1
1 4z
1 12 z
1
1
Find A1 and A2
A1 =
(1
A2 =
(1
Frank Ong
1
z
4
1
1
z
2
1
)X (z)
=
z= 14
)X (z)
=2
z= 12
EE123 Discussion Section 3
1
Intro
Recap
Problems
Solution 1
Partial fraction expansion:
X (z) =
1
1
1
1
4z
+
2
1
1
1
2z
From the tables:
Frank Ong
EE123 Discussion Section 3
Intro
Recap
Problems
Solution 1
Partial fraction expansion:
X (z) =
1
1
1
1
4z
From the tables:

1
1
x[n] =
( )n + 2( )n u[n]
4
2
Frank Ong
+
2
1
1
1
2z
because right sided
EE123 Discussion Section 3
Intro
Recap
Problems
Question 2
I
Find the inverse z-transform of the following system:
X (z) =
I
1 + 2z
1 32 z
1
2
+z
1 + 1z
2
2
ROC = {z : |z| > 1}
1
Ak
dk z
Hint: If M < N,
X (z) =
N
X
k=1
If M
1
N,
X (z) =
M
XN
Br z
r =0
Frank Ong
r
+
N
X
k=1
1
Ak
dk z
EE123 Discussion Section 3
1
Intro
Recap
Problems
Solution 2
I
Find the impulse response of the following system:
X (z) =
1 + 2z
1 32 z
Frank Ong
1
+z
1 + 1z
2
2
2
EE123 Discussion Section 3
Intro
Recap
Problems
Solution 2
I
Find the impulse response of the following system:
X (z) =
1 + 2z
1 32 z
Frank Ong
1
+z
1 + 1z
2
2
2
M=N=2
EE123 Discussion Section 3
Intro
Recap
Problems
Solution 2
I
Find the impulse response of the following system:
+z 2
M=N=2
1 + 1z 2
2
A1
A2
= B0 +
+
1
1
1
z 1
1 2z
X (z) =
1 + 2z
1 32 z
Matching coefficients: A1 =
Frank Ong
1
9, A2 = 8, B0 = 2
EE123 Discussion Section 3
Intro
Recap
Problems
Solution 2
I
Find the impulse response of the following system:
+z 2
M=N=2
1 + 1z 2
2
A1
A2
= B0 +
+
1
1
1
z 1
1 2z
X (z) =
1 + 2z
1 32 z
Matching coefficients: A1 =
1
9, A2 = 8, B0 = 2
ROC = {z : |z| > 1}
1
) x[n] = 2 [n] 9( )n u[n] + 8u[n]
2
Frank Ong
EE123 Discussion Section 3
Intro
Recap
Problems
Question 3
I
What is the inverse z-transform of
X (z) = log(1 + az
I
Hint: use the property nx[n] ,
Frank Ong
1
),
|z| > |a|
z dXdz(z)
EE123 Discussion Section 3
Intro
Recap
Problems
Solution 3
I
X (z) = log(1 + az
1 ),
|z| > |a|
Frank Ong
EE123 Discussion Section 3
Intro
Recap
Problems
Solution 3
I
X (z) = log(1 + az
1 ),
|z| > |a|
dX (z)
az 2
=
dz
1 + az 1
Frank Ong
EE123 Discussion Section 3
Intro
Recap
Problems
Solution 3
I
X (z) = log(1 + az
1 ),
|z| > |a|
dX (z)
az 2
=
dz
1 + az 1
I
Use the di↵erentiation property
nx[n] ,
z
Frank Ong
dX (z)
az 1
=
dz
1 + az 1
EE123 Discussion Section 3
Intro
Recap
Problems
Solution 3
I
X (z) = log(1 + az
1 ),
|z| > |a|
dX (z)
az 2
=
dz
1 + az 1
I
Use the di↵erentiation property
nx[n] ,
I
z
dX (z)
az 1
=
dz
1 + az 1
Use the shift property
nx[n] = a( a)n
Frank Ong
1
u[n
1]
EE123 Discussion Section 3
Intro
Recap
Problems
Solution 3
I
X (z) = log(1 + az
1 ),
|z| > |a|
dX (z)
az 2
=
dz
1 + az 1
I
Use the di↵erentiation property
nx[n] ,
I
z
dX (z)
az 1
=
dz
1 + az 1
Use the shift property
nx[n] = a( a)n
I
n
x[n] = ( 1)n+1 an u[n
X(z))
1
u[n
1]
1] (This is actually the power series of
Frank Ong
EE123 Discussion Section 3
Intro
Recap
Problems
Question 4
I
What is the ROC and inverse z-transform of
X (z) = z 2 (1
1
z
2
Frank Ong
1
)(1 + z
1
)(1
z
EE123 Discussion Section 3
1
)
Intro
Recap
Problems
Solution 4
X (z) = z 2 (1
1
z
2
Frank Ong
1
)(1 + z
1
)(1
z
EE123 Discussion Section 3
1
)
Intro
Recap
Problems
Solution 4
X (z) = z 2 (1
I
1
z
2
1
)(1 + z
1
)(1
z
Expanding the terms
X (z) = z 2
Frank Ong
1
z
2
1
1+ z
2
1
EE123 Discussion Section 3
1
)
Intro
Recap
Problems
Solution 4
X (z) = z 2 (1
I
1
z
2
)(1 + z
1
)(1
z
1
)
Expanding the terms
X (z) = z 2
I
1
1
z
2
1
1+ z
2
1
By inspection
X (z) = [n + 2]
Frank Ong
1
[n + 1]
2
[n] +
1
[n
2
EE123 Discussion Section 3
1]
Intro
Recap
Problems
Question 5
I
Let x[n] be causal, i.e., x[n] = 0 for n < 0. And assume X (z)
is rational.
I
Can limz!1 X (z) be infinite? That is, can there be a pole at
z = 1?
Frank Ong
EE123 Discussion Section 3
Intro
Recap
Problems
Solution 5
I
The answer is no as long as x[0] 6= 1 because of the initial
value theorem:
I
Proof:
X (z) =
1
X
x[n]z
n
n= 1
Frank Ong
EE123 Discussion Section 3
Intro
Recap
Problems
Solution 5
I
The answer is no as long as x[0] 6= 1 because of the initial
value theorem:
I
Proof:
X (z) =
1
X
n
x[n]z
n= 1
I
Since x[n] = 0:
X (z) =
1
X
x[n]z
n
n=0
Frank Ong
EE123 Discussion Section 3
Intro
Recap
Problems
Solution 5
I
The answer is no as long as x[0] 6= 1 because of the initial
value theorem:
I
Proof:
X (z) =
1
X
n
x[n]z
n= 1
I
Since x[n] = 0:
X (z) =
1
X
x[n]z
n
n=0
lim X (z) = x[0]
z!1
Frank Ong
EE123 Discussion Section 3
Intro
Recap
Problems
Question 6
Which of the followings could be causal and stable? Find the ROC
whenever it can be causal and stable
I
(1 z 1 )2
(1 14 z 1 )
I
(z
(z
I
1 3z
(1 2z
2
1)
I
1
(1 2z
1)
1 6
)
4
1 5
)
2
4z 2
(1 2z 1 )
Frank Ong
EE123 Discussion Section 3
Intro
Recap
Problems
Solution 6
I
(1 z 1 )2
(1 14 z 1 )
I
(z
(z
I
1 3z
(1 2z
2
1)
Could not be causal and stable, because the
outermost pole is outside the unit circle.
I
1
(1 2z
1)
1 6
)
4
1 5
)
2
Could be. ROC = {z : |z| > 14 }
Could not be causal, because limz!1 X (z) = 1
4z 2
(1 2z 1 )
= 1 + 2z
Frank Ong
1
Always causal and stable
EE123 Discussion Section 3