2.
1.
Solution of EECS 315 Test 1 F10
Find the numerical magnitude and phase in radians of this function. (If the phase is undefined just write "undefined"). e
(
15
+ j 4
) f
at f
= −
0.2
0.04979
∠ − 0.8
Find the numerical fundamental periods of these signals.
(a) x
[ ] =
13
δ
16
[ ] −
11
δ
20
[ ]
Fundamental period is LCM of 16 and 20 which is 80
(b) x
[ ]
= 4 sin 18
π n / 24
) − 3cos 10
π n / 16
) x
[ ] =
(
π n
(
9 / 24
) )
−
(
π n
(
5 / 16
) )
=
(
π n
(
3 / 8
) )
−
(
π n
(
5 / 16
) )
Fundamental period is LCM of 8 and 16 which is 16.
(a) Is x
( )
even, odd or neither?
Neither
(b) Is y
( )
even, odd or neither?
Even
Even Odd Neither
Even Odd Neither

4.
6.
5.
If x
( ) =
5
δ
(
4
( t
−
2
) )
, find the numerical value of
−
3
∫
8 x t dt .
∫
8
−
3 x
( ) dt
=
8
−
3
∫
5
δ
(
4
( t
−
2
) ) dt
8
= (
5 / 4
) δ ( t
−
2
) dt
−
3
∫ =
5 / 4
If x
[ ]
=
5
δ
12
⎡⎣
4
( n
−
2
) ⎤⎦ , find the numerical value of
8 ∑ n
= −
3 x
[ ]
. n
8 ∑
= −
3 x
[ ] = n
8 ∑
= −
3
5
δ
12
⎡⎣ 4
( n
−
2
)
⎤⎦ =
5
δ
12 n
8 ∑
= −
3
⎡⎣ 4
( n
−
2
)
⎤⎦ = ( +
0
+
1
+
0
+
0
+
1
+
0
+
0
+
1
+
0
+
0
+
1
) =
20
The amplitude scaling factor is A and the time scaling and shifting between the original signal and the scaled and shifted signal is in the form t
→ t − t w
0 . Find numerical values of A , t
0
and w .
Original Scaled and Shifted
2
1
-1
0
5
4
3
-2
-3
-4
-5
2
1
0
-1
5
4
3
-2
-3
-4
-5
A = 0.4 , t
0
= -1 , w = -2

9.
7.
8.
If x
( ) =
3 t 2 +
9 t
+
4 and x e
( )
is the even part of x
( )
, find the numerical value of x e
( )
. x e
( ) = 3 t 2 + 4 ⇒ x e
( ) = 79
Find the numerical signal energy of x
[ ] =
0.8
2 n
( u
[ n
+
2
] − u
[ n
−
1
] )
.
E x
=
∞
∑ n
= −∞
0.8
2 n
( u
[ n
+
2
] − u
[ n
−
1
] )
2 =
0 ∑ n
= −
2
0.8
4 n =
0.8
−
8 +
0.8
−
4 +
0.8
0 =
9.4019
If one period of x
( )
is described by x t
= rect 2 t
−
3 ,
−
1
< t
<
1 , find the numerical average signal power of x
( )
.
P x
=
1
T
∫
T x
( ) 2 dt
=
1
2
∫
1
−
1

( ) even
−
3
2 dt
=
1
∫
0
⎡⎣ rect 2
( ) +
9
− ( ) ⎤⎦ dt
P x
=
1/ 4
0
∫ dt
+
9
1
0
∫ dt
−
6
1/ 4
0
∫ dt
=
1 / 4
+
9
−
6 / 4
=
31 / 4
=
7.75

Solution of EECS 315 Test 1 F10
2.
1. e
(
12
+ j 3
) f
Find the numerical magnitude and phase in radians of this function. (If the phase is undefined just write "undefined").
at f
= −
0.2
0.09072
∠ −
0.6
Find the numerical fundamental periods of these signals.
(a) x
[ ]
=
13
δ
32
[ ]
−
11
δ
40
[ ]
Fundamental period is LCM of 32 and 40 which is 160
(b) x
[ ] =
4 sin 18
π n / 12
) −
3cos 10
π n / 8
) x
[ ]
=
(
π n
(
9 / 12
) )
−
(
π n
(
5 / 8
) )
=
(
π n
(
3 / 4
) )
−
(
π n
(
5 / 8
) )
Fundamental period is LCM of 4 and 8 which is 8.
3. A continuous-time function is defined by x
( ) =
4 u
( ) − t
and y
( )
is the generalized derivative of x
( )
.
(a) (1 pt) Is x
( )
even, odd or neither?
Neither
(b) Is y
( )
even, odd or neither?
Even

4.
6.
5.
If x
( ) =
9
δ
(
4
( t
−
2
) )
, find the numerical value of
−
3
∫
8 x t dt .
∫
8
−
3 x
( ) dt
=
8
−
3
∫
9
δ
(
4
( t
−
2
) ) dt
8
= (
9 / 4
) δ ( t
−
2
) dt
−
3
∫ =
9 / 4
If x
[ ]
=
3
δ
9
⎡⎣
3
( n
−
2
) ⎤⎦ , find the numerical value of
8 ∑ n
= −
3 x
[ ]
. n
8 ∑
= −
3 x
[ ] = n
8 ∑
= −
3
3
δ
9
⎡⎣ 3
( n
−
2
)
⎤⎦
8
=
3
δ n
∑
= −
3
9
⎡⎣ 3
( n
−
2
)
⎤⎦ = ( +
0
+
1
+
0
+
0
+
1
+
0
+
0
+
1
+
0
+
0
+
1
) =
12
The amplitude scaling factor is A and the time scaling and shifting between the original signal and the scaled and shifted signal is in the form t
→ t − t
0 w
. Find numerical values of A , t
0
and w .
Original Scaled and Shifted
3
2
1
0
5
4
-1
-2
-3
-4
-5
3
2
1
0
5
4
-1
-2
-3
-4
-5
A = 0.5 , t
0
= -2 , w = -2

9.
7.
8.
If x
( ) =
5 t 2 −
4 t
+
2 and x e
( )
is the even part of x
( )
, find the numerical value of x e
( )
. x e
( ) = 5 t 2 + 2 ⇒ x e
( ) = 127
Find the numerical signal energy of x
[ ] =
0.9
2 n
( u
[ n
+
3
] − u n
)
.
E x
=
∞
∑ n
= −∞
0.9
2 n
( u
[ n
+
3
] − u
[ ] )
2 =
−
1 ∑ n
= −
3
0.9
4 n =
0.9
−
12 +
0.9
−
8 +
0.9
−
4 =
7.3879
If one period of x
( )
is described by x t
= rect 2 t
−
4 ,
−
1
< t
<
1 , find the numerical average signal power of x
( )
.
P x
=
1
T
∫
T x
( ) 2 dt
=
1
2
∫
1
−
1

( ) even
−
4
2 dt
=
1
∫
0
⎡⎣ rect 2
( ) +
16
− ( ) ⎤⎦ dt
P x
=
1/ 4
0
∫ dt
+
16
1
0
∫ dt
−
8
1/ 4
0
∫ dt
=
1 / 4
+
16
−
2
=
57 / 4
=
14.25

3.
2.
Solution of EECS 315 Test 1 F10
1. e
(
15
+ j 4
) f
Find the numerical magnitude and phase in radians of this function. (If the phase is undefined just write "undefined").
at f
= −
0.2
0.36788
∠ −
0.6
Find the numerical fundamental periods of these signals.
(a) x
[ ]
=
13
δ
48
[ ]
−
11
δ
60
[ ]
Fundamental period is LCM of 48 and 60 which is 240
(c) x
[ ] =
4 sin 18
π n / 6
) −
3cos 10
π n / 4
) x
[ ]
=
(
π n
(
9 / 6
) )
−
(
π n
(
5 / 4
) )
=
(
π n
(
3 / 2
) )
−
(
π n
(
5 / 4
) )
Fundamental period is LCM of 2 and 4 which is 4.
A continuous-time function is defined by x
( ) = t
−
2 u
( )
and y
( )
is the generalized derivative of x
( )
.
(a) (1 pt) Is x
( )
even, odd or neither?
Neither
(b) Is y
( )
even, odd or neither?
Even

4.
6.
5.
If x
( ) =
13
δ
(
7
( t
−
2
) )
, find the numerical value of
∫
8
−
3 x t dt .
∫
8
−
3 x
( ) dt
=
8
−
3
∫
13
δ
(
7
( t
−
2
) ) dt
8
= (
13 / 7
) δ ( t
−
2
) dt
−
3
∫ =
13 / 7
≅
1.8571
If x
[ ]
=
8
δ
9
⎡⎣
3
( n
−
2
) ⎤⎦ , find the numerical value of
8 ∑ n
= −
3 x
[ ]
. n
8 ∑
= −
3 x
[ ] = n
8 ∑
= −
3
8
δ
9
⎡⎣ 3
( n
−
2
)
⎤⎦
8
=
8
δ n
∑
= −
3
9
⎡⎣ 3
( n
−
2
)
⎤⎦ = ( +
0
+
1
+
0
+
0
+
1
+
0
+
0
+
1
+
0
+
0
+
1
) =
32
The amplitude scaling factor is A and the time scaling and shifting between the original signal and the scaled and shifted signal is in the form t
→ t − t w
0 . Find numerical values of A , t
0
and w .
Original Scaled and Shifted
2
1
-1
0
5
4
3
-2
-3
-4
-5
2
1
0
-1
5
4
3
-2
-3
-4
-5
A = 0.25 , t
0
= -4 , w = -3

9.
7.
8.
If x
( ) = −
2 t 2 +
11 t
+
1 and x e
( )
is the even part of x
( )
, find the numerical value of x e
( )
. x e
( ) = − 2 t 2 + 1 ⇒ x e
( ) = − 49
Find the numerical signal energy of x
[ ] =
0.7
2 n
( u
[ n
+
4
] − u
[ n
+
1
] )
.
E x
=
∞
∑ n
= −∞
0.7
2 n
( u
[ n
+
4
] − u
[ n
+
1
] )
2 =
−
2 ∑ n
= −
4
0.7
4 n =
0.7
−
16 +
0.7
−
12 +
0.7
−
8 =
390.5
If one period of x
( )
is described by x t
= rect t
−
1 ,
−
1
< t
<
1 , find the numerical average signal power of x
( )
.
P x
=
1
T
∫
T x
( ) 2 dt
=
1
2
∫
1
−
1 rect
( ) even
−
1
2 dt
=
1
∫
0
⎡⎣ rect 2
( ) +
1
−
2 rect
( ) ⎤⎦ dt
P x
=
1/2
0
∫ dt
+
1
0
∫ dt
−
2
1/2
0
∫ dt
=
1 / 2
+
1
−
1
=
1 / 2