Student Identification Number (No name please)
Solution of ECE 315 Final Examination Su09
1.
2.
Find the numerical values of the constants in the z -transform pairs below.
(a)
( ) n u
[ ]
Z
A z z
+ a
, z
> b
(b)
− 2
⎡⎣ ( − 0.4
) n
⎡⎣ n −
Cc n ⎤⎦ u
[ ]
Z
− (
7 / 6
) ( − 0.3
) n ⎤⎦ u
[ ]
Z
3 z 2
(
7 / 3
) ⎡⎣ ( −
0.3
) n or
− (
6 / 7
) ( −
0.4
) n ⎤⎦ u
[ ]
Z
( ) n u
[ ]
Z z z
+
2
, z
>
2
( +
1
)
+
2.1
z
+
0.36
, z
>
0.4
1
3
(
( +
1
) z
+
0.4
) ( z
+
0.3
)
=
1
3
(
( +
1
) z + 0.4
) ( z + 0.3
)
= z
3
⎡
⎣⎢
−
6 z
+
0.4
+ z
3
⎡
⎣⎢
−
6 z + 0.4
+
7 z
+
0.3
⎤
⎦⎥
, z
>
0.4
7 z + 0.3
⎤
⎦⎥
, z > 0.4
(c)
( ) n u
[
− n + 8
]
Z A z b z + a
, z < c
0.3
n u
[ − n
−
1
] =
0.3
n u ⎡⎣ − ( n
+
1
)
⎤⎦
( )
9
0.3
n
−
9 u ⎡⎣ − ( n − 8
) ⎤⎦ = ( )
9 z z
−
0.3
, z
<
0.3
0.3
n
−
9 u
[
− n + 8
]
Z
( )
9 z
−
9 z z
−
0.3
= −
1.97
×
10
−
5 z
−
0.3
z
−
8
(d) Aa n
− c ⎡⎣ (
π ( − c
) )
+
B
(
π b n
− c
) ) ⎤⎦ u
[ n
− c
]
Z z
−
1
1.5625
z − 1 z 2 +
0.64
, z
>
0.8
, z < 0.3
Find the numerical region of convergence of the z transform of x
[ ] =
12 0.85
) n
( π n / 10
) u
[ − n
−
1
] +
3 0.4
n
+
2 u
[ n
+
2
] or, if it does not exist, state that fact and explain why.
ROC is 0.4
< z
<
0.85

3. A periodic discrete-time signal with fundamental period N = 3 has the values
[ ] =
7 , x 2 of
[ ]
.
= −
3 , x 3
=
1 . If x
[ ] ← DFT →
X
[ ]
, find the numerical magnitude and angle (in radians)
4.
X
[ ]
=
N
−
1 ∑ n
=
0 x
[ ] e
− j 2
π kn / N =
2 ∑ n
=
0 x
[ ] e
− j 2
π kn / 3
[ ] =
2 ∑ x
X 1 n
=
0
[ ] =
1
+
7 e
− j 2
[ ]
π
/ 3 e
− j 2
π n / 3 =
−
3 e
− j 4
π
/ 3
[ ] + x 1 e
− j 2
π
/ 3 + x 2 e
− j 4
π
/ 3
= −
1
− j 8.66
=
8.7178
∠ −
1.6858
Given that x n
[ ]
is periodic and real-valued with fundamental period 8, and x
[ ] ← DFT →
X
16
[ ]
, x
[ ] ← DFT →
X
8
[ ]
, X
8
[ ] = −
2
+ j 5
(a) What is the numerical value of X
16
[ ]
?
The change of period property for the DFT is
If x n ← ⎯⎯ → X ⎡⎣ ⎤⎦ then x n
X
16
[ ] =
←
2 X
8 mN
→
⎪
⎩⎪ m X ⎡⎣ k / m ⎤⎦ , k / m an integer
0 , otherwise
[ ] = −
4
+ j 10
(b) What is the numerical value of X
8
[ ]
?
For real-valued x
[ ]
X k
=
X * − k . Also, X
8
X
8
[ ]
= X
8
[ ]
= X *
8
[ ]
= − 2 − j 5 .
[ ]
is periodic with period 8 so that
5. Let x
= ⎢
⎢
⎢
⎢
⎢
⎡
⎢
1 e
− j
π
/ 3 e
− j 2
π
/ 3 e
−
1
− j 4
π
/ 3 e
− j 5
π
/ 3
⎥
⎥
⎥
⎥
⎤
⎥
⎥
and y
= ⎢
⎢
⎢
⎢
⎡
⎢
⎢ e
− j 4
π
/ 3 e
− j 8
π
/ 3 e
−
1 j 16
π
/ 3 e
− j 20
π
/ 3
1
⎥
⎥
⎥
⎥
⎤
⎥
⎥
. Are they orthogonal? Explain how you know.
Explanation: x H y
=
⎡⎣
1 e j
π
/ 3 e j 2
π
/ 3 −
1 e j 4
π
/ 3 e j 5
π
/ 3
⎤⎦
⎢
⎢
⎢
⎢
⎢
⎡
⎢ e
− j 4
π
/ 3 e
− j 8
π
/ 3 e
−
1 j 16
π
/ 3 e
− j 20
π
/ 3
1
⎥
⎥
⎥
⎥
⎤
⎥
⎥
= e
− j 4
π
/ 3 x H y
=
0 . The inner product is zero. Therefore they are orthogonal.
+ e
− j 7
π
/ 3 + e j 2
π
/ 3 − e
− j 16
π
/ 3 + e
− j 16
π
/ 3 + e j 5
π
/ 3

Solution of ECE 315 Final Examination Su09
1. Find the numerical values of the constants in the z -transform pairs below.
(a)
(b)
( −
1.5
) n
⎡⎣ n u
[ ]
Z
A z z
+ a
+
Cc n ⎤⎦ u
[ ]
Z
, z
> b
(
4 z 2
−
1.5
) n u
[ ]
Z z z + 1.5
, z
>
1.5
( + 1
)
+
2.8
z
+
0.4
, z
>
0.5
(
2 / 3
) ⎡⎣ ( −
0.2
) n − (
5 / 8
) ( −
0.5
) n ⎤⎦ u
[ ]
Z or
1
4
(
( + 1
) z + 0.5
) ( z + 0.2
)
= z
4
⎡
⎣⎢
−
5 / 3 z + 0.5
+
− (
5 / 12
) ⎡⎣ ( −
0.5
) n − (
8 / 5
) ( −
0.2
) n ⎤⎦ u
[ ]
Z z
4
⎡
⎣⎢
−
5 / 3 z
+
0.5
+
8 / 3 z
+
0.2
⎤
⎦⎥
, z
>
0.5
(c)
0.6
n
( ) n u
[
− n + 4
]
Z A z b z
+ a u
[ − n
−
1
] =
0.6
n u
⎡⎣ − ( n
+
1
) ⎤⎦
, z < c z z
−
0.6
, z
<
0.6
8 / 3 z + 0.2
⎤
⎦⎥
, z
>
0.5
( )
5
0.6
n
−
5 u ⎡⎣ − ( n
−
4
)
⎤⎦ = ( )
5
0.6
n
−
5 u
[ − n
+
4
]
Z
( )
5 z
−
5 z z
−
0.6
= −
0.778
z
−
4 z
−
0.6
, z
<
0.6
(d) Aa n
− c ⎡⎣ (
π ( − c
) )
+
B
(
π b n
− c
) ) ⎤⎦ u
[ n
− c
]
Z z
−
1
1.2346
z
−
1 z 2 +
0.81
, z
>
0.9
2. Find the numerical region of convergence of the z transform of x
[ ] =
9 0.7
n
( π n / 10
) u
[ − n
−
1
] +
3 0.6
n
+
2 u
[ n
+
2
] or, if it does not exist, state that fact and explain why.
ROC is 0.6
< z < 0.7

3. A periodic discrete-time signal with fundamental period N = 3 has the values
[ ] =
5 , x 2 of
[ ]
.
=
2 , x 3
= −
1 . If x
[ ] ← DFT →
X
[ ]
, find the numerical magnitude and angle (in radians)
4.
X
[ ] =
N
−
1 ∑ n
=
0 x
[ ] e
− j 2
π kn / N =
2 ∑ n
=
0 x
[ ] e
− j 2
π kn / 3
[ ]
=
2 ∑ x
[ ] e
− j 2
π n / 3 =
[ ]
+ x 1 e
− j 2
π
/ 3 + x 2 e
− j 4
π
/ 3
X 1 n
=
0
[ ] = −
1
+
5 e
− j 2
π
/ 3 +
2 e
− j 4
π
/ 3 = −
4.5
− j 2.5981
=
5.1962
∠ −
2.618
Given that x n
[ ]
is periodic and real-valued with fundamental period 8, and x
[ ] ← DFT →
X
16
[ ]
, x
[ ]
← DFT → X
8
[ ]
, X
8
[ ]
= − 1 + j 2
(a) What is the numerical value of X
16
[ ]
?
The change of period property for the DFT is
If x n ← ⎯⎯ → X ⎡⎣ ⎤⎦ then x n
X
16
[ ] =
←
2 X
8 mN
→
⎪
⎩⎪ m X ⎡⎣ k / m ⎤⎦ , k / m an integer
0 , otherwise
[ ] = −
2
+ j 4
(b) What is the numerical value of X
8
[ ]
?
For real-valued x
[ ]
X k
=
X * − k . Also, X
8
X
8
[ ]
= X
8
[ ]
= X *
8
[ ]
= − 1 − j 2 .
[ ]
is periodic with period 8 so that
5. Let x =
⎡
⎢
⎢
⎢
⎢
⎢
1 e j
π
/ 3 e j 2
π
/ 3 e
− 1 j 4
π
/ 3 e j 5
π
/ 3
⎤
⎥
⎥
⎥
⎥
⎥
and y =
⎡
⎢
⎢
⎢
⎢
⎢ e
− j 4
π
/ 3 e
− j 8
π
/ 3
1 e
− j 16
π
/ 3 e
− j 20
π
/ 3
1
⎤
⎥
⎥
⎥
⎥
⎥
. Are they orthogonal? Explain how you know.
Orthogonal Not Orthogonal
Explanation: x H y
= ⎡⎣
1 e
− j
π
/ 3 e
− j 2
π
/ 3 −
1 e
− j 4
π
/ 3 e
− j 5
π
/ 3
⎤⎦
⎢
⎢
⎢
⎢ e
− j 4
π
/ 3 e
− j 8
π
/ 3
1 e
− j 16
π
/ 3 e
− j 20
π
/ 3
1
⎥
⎥
⎥
⎥
= e x H y = 0 The inner product is zero. Therefore they are orthogonal.
− j 4
π
/ 3 + e
− j 9
π
/ 3 + e
− j 2
π
/ 3 − e
− j 16
π
/ 3 + e
− j 24
π
/ 3 + e
− j 5
π
/ 3

Solution of ECE 315 Final Examination Su09
1.
2.
Find the numerical values of the constants in the z -transform pairs below.
(a)
(b)
( −
2.5
) n n u
[ ]
+ Cc n
Z A
⎤⎦ u
[ ]
Z z z + a
, z
> b
( − 2.5
5 z 2
) n u
[ ]
Z z z + 2.5
( +
1
)
+ 3.5
z + 0.3
, z > 0.6
, z > 2.5
(
9 / 25
) ⎡⎣ ( − 0.1
) n or
− (
4 / 9
) ( − 0.6
) n ⎤⎦ u
[ ]
Z
− (
4 / 25
) ⎡⎣ ( −
0.6
) n − (
9 / 4
) ( −
0.1
) n ⎤⎦ u
[ ]
Z
(c)
( ) n u
[ − n
+
6
]
Z A z b z
+ a n u
[
− n − 1
]
= 0.1
n u ⎡⎣ − ( n + 1
) ⎤⎦
1
5
(
( +
1
) z
+
0.6
) ( z
+
0.1
)
= z
5
⎡
⎣⎢
−
4 / 5 z
+
0.6
+
, z
< c z z
−
0.1
, z < 0.1
( )
7 n
−
7
1
5
(
( +
1
) z
+
0.6
) ( z
+
0.1
)
= z
5
⎡
⎣⎢
−
4 / 5 z
+
0.6
+
9 / 5 z
+
0.1
⎤
⎦⎥
, z > 0.6
u ⎡⎣ − ( n
−
6
)
⎤⎦ = ( )
7
( ) n
−
7 u
[ − n
+
6
]
Z
( )
7 z
−
7
9 / 5 z
+
0.1
⎤
⎦⎥
, z
>
0.6
z z
−
0.1
= −
10
−
7 z
− z
−
6
0.1
, z
<
0.1
(d) Aa n
− c ⎡⎣ (
π ( − c
) )
+ B
(
π b n − c
) ) ⎤⎦ u
[ n − c
]
Z z
−
1
2.0408
z
−
1 z 2 +
0.49
, z > 0.7
z
−
1
2.0408
z 2 z −
+
0.49
1
=
2.0408
z
−
2 z 2 z 2
In the damped sinusoid forms
− 0.49
z
+
0.49
α =
=
2.0408
0.7 and
Ω
0 z
−
2
⎛
⎝⎜ z
= π
/ 2.
2 z 2
+
0.49
− z 2
0.49
z
⎞
+
0.49
⎠⎟
, z
>
0.7
( ) n
−
2 ⎡⎣ cos
(
π ( n
−
2
)
/ 2
)
−
0.7 sin
(
π ( n
−
2
)
/ 2
) ⎤⎦ u
[ n
−
2
]
Z
2.0408
z
−
2
⎛
⎝⎜ z 2 z 2
+
0.49
− z 2
0.49
z
+
0.49
⎞
⎠⎟
, z
>
0.7
Find the numerical region of convergence of the z transform of x
[ ] =
12 1.2
n
( π n / 10
) u
[ − n
−
1
] +
3 0.9
n
+
2 u
[ n
+
2
] or, if it does not exist, state that fact and explain why.
ROC is 0.9
< z
<
1.2

3. A periodic discrete-time signal with fundamental period N = 3 has the values
[ ] =
7 , x 2 of
[ ]
.
=
9 , x 3
=
1 . If x
[ ] ← DFT →
X
[ ]
, find the numerical magnitude and angle (in radians)
4.
X
[ ] =
N
−
1 ∑ n
=
0 x
[ ] e
− j 2
π kn / N =
2 ∑ n
=
0 x
[ ] e
− j 2
π kn / 3
[ ] =
2 ∑ x
[ ] e
− j 2
π n / 3 = [ ] + x 1 e
− j 2
π
/ 3 + [ ] e
− j 4
π
/ 3
X 1 n
=
0
[ ] =
1
+
7 e
− j 2
π
/ 3 +
9 e
− j 4
π
/ 3 = −
7
+ j 1.732
=
7.2111
∠
2.899
Given that x n
[ ]
is periodic and real-valued with fundamental period 8, and x
[ ]
← DFT → X
16
[ ]
, x
[ ] ← DFT →
X
8
[ ]
, X
8
[ ] =
5
+ j 2
(a) What is the numerical value of X
16
[ ]
?
The change of period property for the DFT is
If x n ← ⎯⎯ → X ⎡⎣ ⎤⎦ then x n
X
16
[ ]
=
←
2 X
8 mN
→
⎪
⎩⎪ m X ⎡⎣ k / m ⎤⎦ , k / m an integer
0 , otherwise
[ ]
= 10 + j 4
(b) What is the numerical value of X
8
[ ]
?
For real-valued x
[ ]
X k
=
X * − k . Also, X
8
X
8
[ ] =
X
8
[ ] =
X *
8
[ ] =
5
− j 2 .
[ ]
is periodic with period 8 so that
5. Let x
= ⎢
⎢
⎢
⎢
⎢
⎡
⎢
1 e
− j
π
/ 3 e
− j 2
π
/ 3 e
−
1
− j 4
π
/ 3 e
− j 5
π
/ 3
⎥
⎥
⎥
⎥
⎤
⎥
⎥
and y
= ⎢
⎢
⎢
⎢
⎡
⎢
⎢ e
− j 4
π
/ 3 e
− j 8
π
/ 3 e
−
1 j 16
π
/ 3 e
− j 20
π
/ 3
1
⎥
⎥
⎥
⎥
⎤
⎥
⎥
. Are they orthogonal? Explain how you know.
Orthogonal
Explanation:
Not Orthogonal x H y
= ⎡⎣
1 e j
π
/ 3 e j 2
π
/ 3 −
1 e j 4
π
/ 3 e j 5
π
/ 3
⎤⎦
⎢
⎢
⎢
⎢ e
− j 4
π
/ 3 e
− j 8
π
/ 3
1 e
− j 16
π
/ 3 e
− j 20
π
/ 3
1
⎥
⎥
⎥
⎥
= e
− j 4
π
/ 3 x H y = 0 . The inner product is zero. Therefore they are orthogonal.
+ e
− j 7
π
/ 3 + e j 2
π
/ 3 − e
− j 16
π
/ 3 + e
− j 16
π
/ 3 + e j 5
π
/ 3