Solution of ECE 315 Test 7 F08
{ (
)
X1 k = 3 sinc k / 2 k }
, T0 = TF = 1
()
() () ( )
x ( t ) is periodic and one period of x ( t ) = t + 1 , 2 < t < 2 and T
x 2 t is periodic and one period of x 2 t = t1 t rect t / 5 , 2.5 < t < 2.5 and T0 = TF = 5
3
3
0
= TF = 4
|X4[k]|
6
4
2
0
-10
0
10
4
Phase of X [k]
k
2
{X
4
}
k = 0 , k > 10 , TF = T0 = 10
0
-2
-10
0
10
k
()
FS
X n k .
For each signal x n t Circle all correct answers. If none of the answers is correct, circle none.
1.
2.
3.
4.
5.
6.
7.
Which continuous-time signals are even functions? 1
Which continuous-time signals are not even but can be made even by adding or subtracting a constant?
none
Which continuous-time signals are odd functions? 2, 4
Which continuous-time signals are not odd but can be made odd by adding or subtracting a constant? 3
Which continuous-time signals have an average value of zero? 1, 2, 4
Which of the continuous-time signals are square waves? 1
What is the average signal power of x1 t ?
()
Using
TF = T0
and
(1 / w) rect (t / w) (t ) f sinc ( wkf )
FS
( ) ()
T0
0
0
{ (
)
TF is arbitrary
FS
1
k FS
3 2 rect 2t 1 t 1
X1 k = 3 sinc k / 2 k 8.
}
, T0 = TF = 1
This is a square wave alternating between +3 and –3. Its square is a constant +9. Therefore its average
signal power is 9.
What is the average signal power of x 4 t ?
()
By Parseval’s theorem, the average signal power is the sum of the squares of the magnitudes of the
impulse strengths in the harmonic function. For this signal
(
)
P4 = 2 52 + 2 12 + 2 42 + 2 32 = 2 25 + 1 + 16 + 9 = 102
Solution of ECE 315 Test 7 F08
{ (
)
X1 k = 4 sinc k / 2 k }
, T0 = TF = 1
()
()
() ( )
x ( t ) is periodic and one period of x ( t ) = t + 1 , 2 < t < 2 and T
x 2 t is periodic and one period of x 2 t = t 21 t rect t / 5 , 2.5 < t < 2.5 and T0 = TF = 5
3
3
0
= TF = 4
|X4[k]|
6
4
2
0
-10
0
10
4
Phase of X [k]
k
2
{X
4
}
k = 0 , k > 10 , TF = T0 = 10
0
-2
-10
0
10
k
()
FS
X n k .
For each signal x n t Circle all correct answers. If none of the answers is correct, circle none.
1.
2.
3.
4.
5.
6.
7.
Which continuous-time signals are even functions? 1, 2
Which continuous-time signals are not even but can be made even by adding or subtracting a constant?
none
Which continuous-time signals are odd functions? 4
Which continuous-time signals are not odd but can be made odd by adding or subtracting a constant? 3
Which continuous-time signals have an average value of zero? 1, 4
Which of the continuous-time signals are square waves? 1
What is the average signal power of x1 t ?
()
Using
TF = T0
and
(1 / w) rect (t / w) (t ) f sinc ( wkf )
FS
( ) ()
T0
0
0
{ (
)
TF is arbitrary
FS
1
k FS
4 2 rect 2t 1 t 1
X1 k = 4 sinc k / 2 k 8.
}
, T0 = TF = 1
This is a square wave alternating between +4 and –4. Its square is a constant +16. Therefore its average
signal power is 16.
What is the average signal power of x 4 t ?
()
By Parseval’s theorem, the average signal power is the sum of the squares of the magnitudes of the
impulse strengths in the harmonic function. For this signal
(
)
P4 = 2 62 + 2 12 + 2 42 + 2 22 = 2 36 + 1 + 16 + 4 = 114
Solution of ECE 315 Test 7 F08
{ (
)
X1 k = 10 sinc k / 2 k }
, T0 = TF = 1
()
() () ( )
x ( t ) is periodic and one period of x ( t ) = t + 1 , 2 < t < 2 and T
x 2 t is periodic and one period of x 2 t = t1 t rect t / 5 , 2.5 < t < 2.5 and T0 = TF = 5
3
3
0
= TF = 4
|X4[k]|
4
2
0
-10
0
10
4
Phase of X [k]
k
2
{X
4
}
k = 0 , k > 10 , TF = T0 = 10
0
-2
-10
0
10
k
()
FS
X n k .
For each signal x n t Circle all correct answers. If none of the answers is correct, circle none.
1.
2.
3.
4.
5.
6.
7.
Which continuous-time signals are even functions? 1
Which continuous-time signals are not even but can be made even by adding or subtracting a constant?
none
Which continuous-time signals are odd functions? 2, 4
Which continuous-time signals are not odd but can be made odd by adding or subtracting a constant? 3
Which continuous-time signals have an average value of zero?
1, 2, 4
Which of the continuous-time signals are square waves? 1
What is the average signal power of x1 t ?
()
Using
TF = T0
(1 / w) rect (t / w) (t ) f sinc ( wkf )
and
FS
( ) ()
T0
0
0
{ (
)
TF is arbitrary
FS
1
k FS
10 2 rect 2t 1 t 1
X1 k = 10 sinc k / 2 k 8.
}
, T0 = TF = 1
This is a square wave alternating between +10 and –10. Its square is a constant +100. Therefore its
average signal power is 100.
What is the average signal power of x 4 t ?
()
By Parseval’s theorem, the average signal power is the sum of the squares of the magnitudes of the
impulse strengths in the harmonic function. For this signal
(
)
P4 = 2 32 + 2 42 + 2 12 + 2 32 = 2 9 + 16 + 1 + 9 = 70