Instructors Solutions to Assignment 3 Problem 4.6
(a) By inspection, we note that the time period T0 = 2π, which implies that the fundamental frequency ω0 = 1.
Since the CTFS coefficient a0 represents the average value of the signal, therefore, a0 = 3/2.
Using Eq. (4.31), the CTFS cosine coefficients an’s, for (n ≠ 0), are given by
[ ] [ ]
0
0 0 00 0 0
0
2 1 1
3 3
1( )cos( ) 3cos( ) 3cos( )
sin( ) sin( ) 0 0
T
n T
n n
a x t n t dt n t dt nt dt
nt n
π π
ππ
π π
π
ω ω
π
= = =
= = − =
∫ ∫ ∫
Using Eq. (4.32), the CTFS sine coefficients bn’s are given by
[ ] [ ]0
0 0 00 0
3 3 32 1
6
1( )sin( ) 3sin( ) cos( ) cos( ) cos(0) 1 ( 1)
0
Tn
n n n nT
n
b x t n t dt nt dt nt n
n odd
n even
ππ
ππ π π
π
ω π ⎡ ⎤= = = − = − + = − −⎣ ⎦
⎧ =⎪= ⎨=⎪⎩
∫ ∫
(c) By inspection, we note that the time period T0 = T, which implies that the fundamental frequency ω0 = 2π/T.
Since the CTFS coefficient a0 represents the average value of the signal, therefore, a0 = 1/2.
Since the function [x3(t) − 0.5] is odd, therefore, the CTFS cosine coefficients an = 0, for (n ≠ 0).
Using Eq. (4.32), the CTFS sine coefficients bn’s are given by
( )
π=
ω=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
ω×⎟
⎠⎞⎜
⎝⎛−+
ωω×⎟
⎠⎞⎜
⎝⎛−
ω−×−=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
ωω−×⎟
⎠⎞⎜
⎝⎛−−
ωω−×⎟
⎠⎞⎜
⎝⎛ −=
ω⎟⎠⎞⎜
⎝⎛ −= ∫
nTn
nTnTn
TnT
ntn
Tntn
Tt
T
dttnTt
Tb
T
T
n
12
)()0sin(1
)()sin(1
)(1102
)()sin(1
)()cos(12
)sin(12
0
20
20
0
0
02
0
0
0
0
00
(e) By inspection, we note that the time period T0 = 2T, which implies that the fundamental frequency ω0 = π/T.
Using Eq. (4.30), the CTFS coefficient T0 is given by
( ) ( )
[ ]
2
00 0 0 0
0
2 2 2 4
2 4 2 4 2 2 2/
1 1 1 1
11 1 1 1 1 1 1
( ) 1 0.5sin sin
cos( ) cos( ) cos(0)
T T T Tt tT T
TtT
T T T T
T T
a x t dt dt dt dtπ π
ππ π ππ
ππ −
⎡ ⎤= = − = −⎣ ⎦
= + × = + − = − =⎡ ⎤⎣ ⎦
∫ ∫ ∫ ∫
Assignment 2 CSE3451: Signals and Systems 2
Using Eq. (4.31), the CTFS cosine coefficients an’s, for (n ≠ 0), are given by
( ) ( )0 0 00 0 0
2 22 1 11 0.5sin cos( ) cos( ) sin cos( )T T T
t tn T T
A B
TT Ta n t dt n t dt n t dtπ πω ω ω
= =
⎡ ⎤= − = −⎣ ⎦∫ ∫ ∫1 4 4 2 4 43 1 4 4 44 2 4 4 4 43
where Integrals A and B are simplified as
[ ] [ ] [ ]0 0001 1 1sin( ) sin( ) 0 sin( ) 0 0T
n nn TA n t n T nπ πω ω ω π= = − = − =
and
( ) ( ) ( ) ( )
[ ][ ]
00 0 0
0 0
2 2 4
4 4( 1)/ ( 1)/
4 ( 1)
1 1 1
1 1 1 1
1 1
sin cos( ) sin cos sin ( 1) sin ( 1)
cos ( 1) cos ( 1) for 1
1 cos ( 1)
T T Tt t n t t tT T T T T
T Tt tT T
T T T
T Tn T n T
n
B n t dt dt n n dt
n n n
n
π π π π π
π ππ π
π
ω
π+ −
+
−
⎡ ⎤= = = + − −⎣ ⎦
= × + + × − ≠⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦
= − + −
∫ ∫ ∫
[ ]
2
4 ( 1)
4 ( 1) 4 ( 1) ( 1)12 2
1 cos ( 1)
0 10 1n
n n n
n
n oddn odd
n evenn even
π
π π π
π−
+ − −
− −
≠ =≠ = ⎧⎧⎪ ⎪= =⎨ ⎨− =− =⎪ ⎪⎩ ⎩
For [ ]2 20
04 4 82 /1 1 1 11, sin cos 1 cos2 0T
Tt tT TT T Tn B dtπ π
ππ π−= = = × = − =⎡ ⎤⎣ ⎦∫ .
In other words, 2( 1)1
0
n
n oddB
n evenπ −
=⎧⎪= ⎨− =⎪⎩
which implies that
2( 1)1
0n
n
n odda A B
n evenπ −
=⎧⎪= − = ⎨ =⎪⎩
Using Eq. (4.32), the CTFS sine coefficients bn’s are given by
bn =22T 1− 0.5sin π t
T( )⎡⎣ ⎤⎦sin(nω0t)dt0
T
∫ = 1T sin(nω0t)dt0
T
∫=C
− 12T sin π t
T( )sin(nω0t)dt0
T
∫=D
where Integrals C and D are simplified as
[ ] [ ] [ ]0 0001 1 1
2
0cos( ) cos( ) cos(0) 1 cos( )T
n nn Tn
n evenC n t n T n
n oddπ πωπ
ω ω π=⎧⎪= − = − + = − = ⎨=⎪⎩
and
Assignment 2 CSE3451: Signals and Systems
3
( ) ( ) ( )
[ ][ ]
0 0
0 0
2 4
4 4( 1)/ ( 1)/
4 ( 1) 4 ( 1)
1 1
1 1 1 1
1 1
sin sin cos ( 1) cos ( 1)
sin ( 1) sin ( 1) for 1
sin ( 1) sin(0) sin ( 1) sin
T Tt n t t tT T T T
T Tt tT T
T T
T Tn T n T
n n
D dt n n dt
n n n
n n
π π π π
π ππ π
π ππ π− +
− +
⎡ ⎤= = − − +⎣ ⎦
= × − − × + ≠⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦
= − − − + −
∫ ∫
[ ][ ]
(0)
0 for 1n= ≠
For (n = 1),
D = 12T sin2 π t
T( )dt0
T
∫ = 14T 1− cos 2π t
T( )⎡⎣ ⎤⎦dt0
T
∫ = 14 −
14T×2π /T sin
2π tT⎡⎣ ⎤⎦0
T
=0
⎛
⎝⎜⎜
⎞
⎠⎟⎟= 14 .
In other words, 14 1
0 1
nD
n
=⎧⎪= ⎨>⎪⎩
.
Therefore, 4
2 1
2
0
1
1 .n
n
n even
b C D n
n oddπ
π
=⎧⎪⎪= − = − =⎨⎪
≠ =⎪⎩
▌
Problem 4.11
(a) By inspection, we note that the time period T0 = 2π, which implies that the fundamental frequency ω0 = 1. Using Eq. (4.44), the DTFS coefficients Dn’s are given by
( )⎪⎩
⎪⎨⎧
≠−
==
π== π−
π
π−
−
ω− ∫∫ .01
0,3
21)(1
23
23
0
2
20
0
0
0
ne
ndtedtetx
TD jn
nj
jntT
T
tjnn
or, ( )⎪⎪⎩
⎪⎪⎨
⎧
≠
=
=−−π
=
π n
nn
n
njD
jn
nn
odd
.0,even 0
0
)1(123
3
23
The magnitude and phase spectra are given by
Magnitude Spectrum: ⎪⎩
⎪⎨
⎧
≠=
=
π . odd,.0,even ,0
0,
3
23
nnn
nD
n
n
Phase Spectrum: ⎪⎩
⎪⎨
⎧
<>−=
π
π
.0, odd,0, odd,
even ,0
2
2nnnnn
Dn
The magnitude and phase spectra are shown in row 1 of the subplots included in Fig. S4.11.
(c) By inspection, we note that the time period T0 = Τ, which implies that the fundamental frequency ω0 = 2π/T. Using Eq. (4.44), the DTFS coefficients Dn’s are given by
Assignment 2 CSE3451: Signals and Systems 4
( )( )⎪
⎪
⎩
⎪⎪
⎨
⎧
≠−
==×
=−==
∫∫∫
ω−
ω−ω−
.011
0,11)(1
0
21
21
00 0
00
ndteT
ndte
Tdtetx
TD T
tjnTt
TTT
tjnTt
Ttjn
n
For (n ≠ 0), the DTFS coefficients are given by
( ) ( ) ( )Ttjn
T
tjn
Tt
Ttjn
Tt
n jne
jnedte
TD
02
0
1
00 )()(111 00
0
⎥⎥⎦
⎤
⎢⎢⎣
⎡
ω−−−
ω−−=−=
ω−ω−ω−∫ ,
which reduces to
π
=⎥⎥⎦
⎤
⎢⎢⎣
⎡
ω−−
ω−+
ω−−=
ω−
njjnjne
jnD
T
T
Tjn
TTn 21
)(1
)()(10
02
0
12
0
1
0
10
.
Combining the two cases, we get
⎪⎩
⎪⎨⎧
≠
==
π .0,
0,
2121
n
nD
njn
The magnitude and phase spectra are given by
Magnitude Spectrum: ⎪⎩
⎪⎨⎧
≠
==
π.0,
0,
2121
n
nD
nn
Phase Spectrum: ⎪⎩
⎪⎨⎧
>π−<π=
=.0,5.00,5.00,0
nnn
Dn
The magnitude and phase spectra are shown in row 3 of the subplots included in Fig. S4.11.
(e) By inspection, we note that the time period T0 = 2Τ, which implies that the fundamental frequency ω0 = π/T. For (n = 0), the exponential DTFS coefficients is given by
( ) ( )
[ ]
2
00 0 0 0
0
2 2 2 4
2 4 2 4 2 2/
1 1 1 1
1 1 1 1 1 1 1
( ) 1 0.5sin sin
cos( ) cos( ) cos(0)
T T T Tt tT T
TtT
T T T T
T T
D x t dt dt dt dtπ π
ππ ππ π
⎡ ⎤= = − = −⎣ ⎦
= + × = + − = −⎡ ⎤⎣ ⎦
∫ ∫ ∫ ∫
For (n = 0), the exponential DTFS coefficients is given by
Dn =12T 1− 0.5sin π t
T( )⎡⎣ ⎤⎦e− jnω0t dt =
0
T
∫ 12T e− jnω0t dt
0
T
∫=A
− 14T sin π t
T( )e− jnω0t dt0
T
∫=B
.
Solving for Integrals A and B, we get
Assignment 2 CSE3451: Signals and Systems
5
0 0
000
2 2 2 21 1 1 11 1 ( 1)T
Tjn t jn t jn nT j n T j n j nA e dt e eω ω π
ω π π− − −
− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤= = = − = − −⎣ ⎦ ⎣ ⎦⎣ ⎦∫
and
( )
( )
( 1) ( 1)
( 1) ( 1)
0 0 0
( 1) ( 1)0
( 1)1 1( 1)
4 8 8
8
8
1 1 1
1
1
sin
for 1
1
jn t j t j t jn t j n t j n tT T T T T T
j n t j n tT T
T T TtT
TT T
j n j n
j nn
T j T j T
j T
B e dt e e e dt e e dt
e e n
e
π π π π π π
π π
π
π π
ππ
− − − − − − +
− − − +
− − +
− −−
⎡ ⎤ ⎡ ⎤= = − = −⎣ ⎦ ⎣ ⎦
⎡ ⎤= + ≠ ±⎣ ⎦
= − −
∫ ∫ ∫
( )2
2
( 1)( 1)
( 1) ( 1)1 1 1( 1) ( 1) 4 ( 1)
14 ( 1)
81
1
( 1) 1 1 ( 1)
1 ( 1)
j nn
n nn n n
nn
e ππ
π
π
π
− ++
− −−− + −
−−
⎡ ⎤−⎣ ⎦⎡ ⎤ ⎡ ⎤ ⎡ ⎤= − − − = − −⎣ ⎦ ⎣ ⎦⎣ ⎦
⎡ ⎤= + −⎣ ⎦
For n = ± 1, Integral B reduces to
2 2
200
8 8 8 81 1 1For 1, 1 = =
j t j tT T
T TTjj T j T j T j
Tn B e dt t eπ π
π
− −⎡ ⎤ ⎡ ⎤= = − = +⎣ ⎦ ⎣ ⎦∫
and 2 2
200
8 8 8 81 1 1For 1, 1 = =
j t j tT T
T TTjj T j T j T j
Tn B e dt e tπ π
π
− − −⎡ ⎤ ⎡ ⎤= − = − = − − −⎣ ⎦ ⎣ ⎦∫ .
In other words, 2
18
14 ( 1)
1
1 ( 1) j
nn
nB
otherwiseπ
−−
± = ±⎧⎪= ⎨ ⎡ ⎤+ −⎪ ⎣ ⎦⎩
Combining, the above cases, the CTFS coefficients can be expressed as
2
1 12 2
18
14 ( 1)
12
2
2
1
1
0
1 ( 1) 1
1 ( 1) 1 ( 1)
1
nn j
n nn
j n
j n
n
D n
otherwise
π
π
π
π −
⎧ − =⎪⎪ ⎡ ⎤= − − = ±⎨ ⎣ ⎦⎪ ⎡ ⎤ ⎡ ⎤− − + + −⎪ ⎣ ⎦ ⎣ ⎦⎩
−
=
m
( )( )
( )2
1
18
14 ( 1)
1 12
2
1
1
0
1
1 ( 1) 1 ( 1)
1
n nn j n
n
j n
otherwise
π
π
π
π
π−
⎧ =⎪⎪ − = ±⎨⎪ ⎡ ⎤ ⎡ ⎤+ − + − −⎪ ⎣ ⎦ ⎣ ⎦⎩
−
=
m
( )2
18
12 ( 1)
1
1
0
1
0
1n
jn
n
j n
n even
n oddπ
π
π
−
⎧ =⎪
− = ±⎪⎪⎨
≠ =⎪⎪ ± ≠ =⎪⎩
m
The expressions for the magnitude and phase spectra are given by
Assignment 2 CSE3451: Signals and Systems 6
Magnitude Spectrum:
( )
22
1 12
18
0.1592112 ( 1)
1
0.31831
0.3408 01 0
0.1933 1 1
0 0
1 1
nnn
nn
nn
nnD
n evenn even
n oddn odd
π
π
π
π
−−
≈ =⎧ ⎧− =⎪ ⎪
≈ = ±− = ±⎪ ⎪⎪ ⎪= =⎨ ⎨≈ ≠ =≠ =⎪ ⎪⎪ ⎪≈ ± ≠ =± ≠ = ⎪⎪ ⎩⎩
Phase Spectrum: 2
12
0 0
( ) 1 , 0
( ) 1 , 0n
jn
n even n even
D j n n odd n
n odd n odd n
π
π
⎧ = =⎧⎪ ⎪⎪ ⎪= = ± = − = >⎨ ⎨⎪ ⎪
± ≠ = = <⎪⎪ ⎩⎩
R R m
R
The magnitude and phase spectra are shown in in row 5 of the subplots included in Fig. S4.11.
▌
Problem 4.13
In each case, we show that the exponential CTFS coefficients obtained directly from Eq. (4.44) are identical to those obtained from the trigonometric CTFS coefficients.
(a) From the solution of Problem P4.6(a), we know that
10 2a = , 0na = , and
6
0n
n n oddb
n evenπ⎧ =⎪= ⎨
=⎪⎩.
Using Eq. (4.45), the exponential CTFS coefficients for x1(t) are given by
( )( )
[ ]0 0
32
3
3
1 12 21 12 2
0 0 0 0 0
0 0
00
=odd, 0
n n n n n n
n n n
n
n
a n a nD a jb n jb n a a
a jb n jb n
nn even
j n nj
π
π
−
− − −
= =⎧ ⎧⎪ ⎪= − > = − > = =⎨ ⎨⎪ ⎪+ < <⎩⎩
==
=− >−
Q
32
3
0 0
=odd, 0 jn
nn evenn odd
n n π
⎧ ⎧ =⎪ ⎪⎪ = =⎨ ⎨⎪ ⎪ =⎩⎪ <⎩
(c) From the solution of Problem P4.6(c), we know that
π=== nnn baa 121
0 and,0, .
Using Eq. (4.45), the exponential CTFS coefficients for x3(t) are given by
( )( )
[ ]0 0
12
12
12
1 12 21 12 2
0 0 0 0 0
0 0
0 0
n n n n n n
n n n
n
n
a n a nD a jb n jb n a a
a jb n jb n
nj nj
π
π
−
− − −
= =⎧ ⎧⎪ ⎪= − > = − > = =⎨ ⎨⎪ ⎪+ < <⎩⎩
== − >
−
Q
12
2
0
0 0
jn
nn
n π−
⎧=⎧⎪ =⎨ ⎨ ≠⎩⎪ <⎩
(e) From the solution of Problem P4.6(e), we know that
Assignment 2 CSE3451: Signals and Systems
7
0 2 21 1a π= − ,
2( 1)1
0n
n
n odda
n evenπ −
=⎧⎪= ⎨ =⎪⎩, and
42 1
2
0
1
1n n
n
n even
b n
n oddπ
π
=⎧⎪⎪= − =⎨⎪
≠ =⎪⎩
Using Eq. (4.45), the exponential CTFS coefficients for x5(t) are given by
(n = 0): 0 2 21 1D π= −
(n = 1): ( ) ( ) ( )1 1 112 2 4 8
2 1 1 1jD a jb jπ π= − = − − = −
(n = −1): ( ) ( ) ( )1 1 112 2 4 8
2 1 1 1jD a jb jπ π− = + = − = − −
(n > 1): ( )22
1
112 ( 1)2 ( 1)
12
jjnn
n n nnn
n oddn oddD a jb
n evenn evenππ
ππ −−
⎧ =⎧− =⎪ ⎪= − = =⎨ ⎨ ==⎪ ⎪⎩⎩
(n < −1): ( )( )
22
122
112 ( 1)2 ( 1)
12
jjnn
n n nnn
n oddn oddD a jb
n evenn evenππ
ππ
−− −
−−
⎧ =⎧=⎪ ⎪= + = =⎨ ⎨ ==⎪ ⎪⎩⎩
Combining the above results, we obtain
( )( )2
1 12
18
12 ( 1)
1
1
1 0
1
0
1 .
n
n
jn
n
j nD
n even
n odd
π
π
π
π
−
⎧ − =⎪± − = ±⎪⎪= ⎨
≠ =⎪⎪ ± ≠ =⎪⎩
Problem 5.2
(a) By definition,
[ ] [ ] [ ]
[ ] [ ] [ ]).2/(sinc3
66)2/sin(2
133)(
2/
2/)2/sin(
/212/)2/sin(2/2/3
0
2/2/2/330)(1
ωπ=
×==ωπ−−=
−−=−−===ω
ωπ−
ωπωπ
πωπ−
ωωπωπ−ωπ−
ω
πωπωπ−ωπ−
ωωπ−
ωπ
ω−ω−∫
ω−
j
jjjj
jjjj
jjj
etj
e
eeje
eeeedteXtj
(b) By definition,
Assignment 2 CSE3451: Signals and Systems 8
[ ] [ ]
0.5 1.5 0.5 1.5
2 ( ) ( )0.5 0.50.5 0.5
0.5 0.5 1.5 0.50.5 1
0.5 1
sin0.50.5
( ) 0.5 0.5
2 sin(0.5 ) 2 sin(0.5 )
j t j tT T T Tj t j t e e
j jT TT T
j T j T j T j Tj j
j Tj j
TT
X e dt e dt
e e e e
j T e j T
ω ωω ωω ω
ω ω ω ωω ω
ωω ω
ω
ω ω
− −− −− −−
−
− − −
−
⎡ ⎤ ⎡ ⎤= + = +⎣ ⎦ ⎣ ⎦
⎡ ⎤ ⎡ ⎤= − − − −⎣ ⎦ ⎣ ⎦
= − − − −
= ×
∫ ∫
( ) ( )
(0.5 ) sin(0.5 )0.50.5
0.5 T 0.5 T
2
0.5 sinc sinc .
T Tj T TT
j T
e
T Te
ω ωωω ω
ωω ωπ π
−
−
⎡ ⎤ ⎡ ⎤+ ×⎣ ⎦ ⎣ ⎦= +
(c) By definition,
( ) ( ) ( )
( ) ( )
( ).1
0
11)(
222
22
2
11111
)(11
)(1
)(1
0)(1
)(0
3
TjTjTj
TjT
jTjje
T
T
je
Tje
Tt
Ttj
Tt
ee
dteX
Tj
tjtj
ω−ωωωω
ω−ω
ω−ω−ω−
ω−ω−ω−
−+=++−=
⎥⎦⎤
⎢⎣⎡ −−−⎥⎦
⎤⎢⎣⎡ −−=
⎥⎦⎤
⎢⎣⎡ −−−=−=ω
ω−
ω−ω−
∫
For ω = 0, ( ) ( ) 220
22
03 011)( TT
T
TtT
T
Tt dtX =+=⎥⎦
⎤⎢⎣⎡ −−=−=ω ∫ .
(d) By definition,
( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
[ ] .)(sinc)cos(1
00
11
11)(
5.02)5.0()5.0(sin
)5.0/(14)5.0(sin222
)(11
)(1
)(1
)(1
)(11
)(1
0)(1
)(
0
)(1
)(
0
0
4
2
2
22
2
2
2222
22
πω
ωω
ωω×
ω
ω−ω−ω−ω−ω−ω−
ω−ω−−ω−ω−
ω−
−
ω−
=×==ω−=
⎥⎦⎤
⎢⎣⎡ −+−−−+⎥⎦
⎤⎢⎣⎡ +−−=
⎥⎦⎤
⎢⎣⎡ −−−+⎥⎦
⎤⎢⎣⎡ −+=
−++=ω
ω−ω
ω−ω−ω−ω−
∫∫
TTT
TTT
T
jTjje
Tje
TjTj
T
je
Tje
Tt
Tje
Tje
Tt
Ttj
Tt
T
tjTt
TT
dtedteX
TjTj
tjtjtjtj
(e) By definition,
X 5(ω ) = 1− 0.5sin π tT( )⎡⎣ ⎤⎦e
− jωt dt =0
T
∫ e− jωt dt0
T
∫=A
− 0.5 sin π t
T( )e− jωt dt0
T
∫=B
We consider different cases for the above integral.
Assignment 2 CSE3451: Signals and Systems
9
Case I: ( ω = 0)
( ) ( )
[ ]
50 0
0 2/0.5 1
(0) ( ) 1 0.5sin 0.5 sin
cos( ) cos( ) cos(0) (1 )
T Tt tT T
TtTT
T T
X x t dt dt dt dt
T T T T
π π
ππ πππ π
−∞ ∞
∞ −∞
⎡ ⎤= = − = −⎣ ⎦
= + = + − = − = −⎡ ⎤⎣ ⎦
∫ ∫ ∫ ∫
Case II: ( ω ≠ 0, ω ≠ π/T):
[ ]0
0
1 1 11 1 0T
Tj t j t j T j Tj j jA e dt e e eω ω ω ωω ω ω ω− − − −
− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤= = = − = − ≠⎣ ⎦ ⎣ ⎦ ⎣ ⎦∫
B = 0.5 e− jωtπ2
T2−ω 2
− jω sin π tT( )− π
T cosπ tT( ){ }⎡
⎣⎢
⎤
⎦⎥0
T
for ω ≠ 0,± πT
= 0.5T 2
π 2−ω 2T 2e− jωt jω sin π t
T( )=0 at t=0,T
+ πT cos
π tT( )
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥0
T
= 0.5T 2
π 2−ω 2T 2− πT e
− jωT − πT
⎡⎣ ⎤⎦ =0.5πT
π 2−ω 2T 21+ e− jωT⎡⎣ ⎤⎦
Case III: ( ω = π/T):
( )
2
2
2
( ) ( )0.5 0.52 2
0 0 0
0.52
0
0.52
0
0.5 sin
1
,
1
t tT T T T
tT
j tT
tT
T T Tj j j t j tj t j tt
T j j
Tj
j
Tj
j
T
T
B e dt e e e dt e e dt
e dt
As e is periodic with period T e
e dt
π π π π
π
π
π
ω ωω ωπ
π
π
ω
ω
− − − − +− −
−
± ±
⎡ ⎤ ⎡ ⎤= = − = −⎣ ⎦ ⎣ ⎦
⎧ ⎡ ⎤− =⎪ ⎣ ⎦⎪= ⎨⎪ ⎡ ⎤− − = −⎪ ⎣ ⎦⎩
∫ ∫ ∫
∫
∫
[ ]
2
0
0.5 0.52 20
0j tT
T
T Tj j
dt
t
π⎡ ⎤=⎢ ⎥
⎣ ⎦
= ± = ±
∫
Combining, the above results, the CTFT can be expressed as
2 2 2
1
0.55 2
0.5
1
1
(1 ) 0
( ) 1
1 1
j T Tj T
j T j TTT
j
j
T
X e
e e
π
ω π
ω ωππ ω
ω
ω
ω
ω ω−
− −−
− =
⎡ ⎤= − = ±⎣ ⎦⎡ ⎤ ⎡ ⎤− − +⎣ ⎦ ⎣ ⎦
m
2 2 2
1
4
0.5
2
1
(1 ) 0
1 1
Tj T
j T j TTT
j
j
T
otherwise
T
e e
π
π
ω ωππ ω
π
ω
ω
ω− −
−
⎧⎪⎪⎨⎪⎪⎩
− =
= ± = ±
⎡ ⎤ ⎡ ⎤− − +⎣ ⎦ ⎣ ⎦
m
otherwise
⎧⎪⎪⎨⎪⎪⎩
▌
Problem 5.4
(a) The partial fraction expansion is given by
Assignment 2 CSE3451: Signals and Systems 10
)3(
2)2(
1)3)(2(
)1()(1 ω++
ω+−≡
ω+ω+ω+=ω
jjjjjX
Calculating the inverse CTFT, we obtain
)(2)()( 321 tuetuetx tt −− +−= .
(b) The partial fraction expansion is given by
)3(
5.0)2(
1)1(
5.0)3)(2)(1(
1)(2 ω++
ω+−+
ω+≡
ω+ω+ω+=ω
jjjjjjX
Calculating the inverse CTFT, we obtain
)(5.0)()(5.0)( 322 tuetuetuetx ttt −−− +−= .
(c) The partial fraction expansion is given by
)3(5.0
)2(1
)2(0
)1(5.0
)3()2)(1(1)( 223 ω+
−+ω+
−+ω+
+ω+
≡ω+ω+ω+
=ωjjjjjjj
X
Calculating the inverse CTFT, we obtain
)(5.0)()(5.0)( 323 tuetutetuetx ttt −−− +−= . ▌
Problem 5.9
(a) Applying the linearity property,
( ) { } {} { } { })()3sin(7)10cos(315)()3sin(7)10cos(35 221 tutettutetX tt −− ℑ−ℑ+ℑ=−+ℑ=ω .
By selecting the appropriate CTFT pairs from Table 5.2, we get
( ) {}2213)2(
21)10(3)10(31)(10+ω+
−−ωπδ+−ωπδ+ℑωδ=ωj
X .
(b) Entry (8) of Table 5.2 provides the CTFT pair
ω⎯⎯ →← jt 2CTFT)sgn( .
Using the duality property, )sgn(2CTFT2 ω−π⎯⎯ →←jt ,
or, )sgn(CTFT1 ω−⎯⎯ →←π jt .
(c) Entry (7) of Table 5.2 provides the CTFT pair
ω+− ⎯⎯ →← jte 4
8CTFT4 .
Using the time shifting property, ω−ω+
−− ⎯⎯ →← 548CTFT54 jj
t ee .
Using the frequency differentiation property,
{ }ω+ω−
ω−− ⎯⎯ →← j
jddt ejet 4
852CTFT5422
2)(
Assignment 2 CSE3451: Signals and Systems
11
or, 3)4(15
415CTFT542 16200
ω+ω−
ω+ω−−− +⎯⎯ →←
jj
jjt eeet .
(d) Entry (17) of Table 5.2 provides the CTFT pair
( )πωππ ⎯⎯ →←= 6
CTFT3
)3sin( rect3)3(sinc3 ttt
and ( )πωππ ⎯⎯ →←= 10
CTFT5
)5sin( rect5)5(sinc5 ttt
Using the multiplication property
( ) ( )[ ]πω
πω
ππ
ππ
ππ ∗⎯⎯ →←××π 1062
CTFT)5sin()3sin(2 rectrect2
tt
tt
or, ( ) ( )[ ]πω
πωπππ ∗⎯⎯ →← 1062
CTFT)5sin()3sin( rectrect2ttt ,
or, ( ) ( )2sin(3 )sin(5 ) CTFT 5
2 6 105 rect rectt tt
π π π ω ωπ π⎡ ⎤←⎯⎯→ ∗⎣ ⎦ ,
where * is the convolution operation.
(e) Entry (17) of Table 5.2 provides the CTFT pair
( )πωππ ⎯⎯ →←= 6
CTFT3
)3sin( rect3)3(sinc3 ttt
and ( )πωππ ⎯⎯ →←= 8
CTFT4
)4sin( rect4)4(sinc4 ttt .
Using the time differentiation property,
( )πωππ ω⎯⎯ →← 8
CTFT)4sin(1 rect)( jtt
dtd .
Using the convolution property
( ) ( )[ ]πω
πω
πππ
πππ ω×⎯⎯ →←∗×π 862
CTFT)4sin(1)3sin(2 rectrect2
jtt
dtd
tt
or, ( ) ( )[ ]πω
πωπππ ω×⎯⎯ →←∗ 862
CTFT)4sin()3sin( rectrect jtt
dtd
tt ,
or, ( )πωππ π⎯⎯ →←∗ 6CTFT)4sin()3sin( rect24 jt
tdtd
tt . ▌
Problem 5.15
(a) Using the time scaling property, ( ) ( )221CTFT2 ω⎯⎯⎯ →← Xtx .
Using the frequency shifting property, ( ) ( )25
21CTFT5 2 +ω− ⎯⎯ →← Xtxe tj .
Substituting the value of X(ω), we obtain
Assignment 2 CSE3451: Signals and Systems 12
{ }5
5 1 32
1112112
1 5 3(2 )0 elsewhere
11 55 1
0 elsewhere.
j te x tω
ω
ω
ω
ωω
+−
+
−
⎧ − + ≤⎪ℑ = ⎨⎪⎩
− ≤ ≤ −⎧⎪= − ≤ ≤ −⎨⎪⎩
(b) Using the frequency differentiation property,
( ) 2
2CTFT2)(ω
⎯⎯ →←dXdtxjt ,
or, ( ) 2
2CTFT2ω
−⎯⎯ →←dXdtxt .
The CTFT of t2 x(t) is given by
{ } ( )[ ] ( )[ ] [ ] [ ])3()3()3()3(rect)( 332
2
2+ωδ−−ωδ=−ωδ−+ωδ−=−=Δ−= ω
ωω
ω dd
ddtxtF .
(c) Express ( ) dtdx
dtdx
dtdx tt 55 +=+ .
Using the time differentiation property, the CTFT of dxdt is given by
)(CTFT ωω⎯⎯ →← Xjdtdx .
Applying the frequency differentiation property to the above CTFT pair, gives
[ ] ωω ω−ω−=ωω⎯⎯⎯ →← ddX
dd
dtdx XXjjt )()(CTFT .
The CTFT of ( )5 dxdtt + is given by
( ){ } )(5)(5 ωω+ω−ω−=+ℑ ω XjXt ddX
dtdx .
Substituting the value of X(ω), we obtain
( ){ }( ) ( )( ) ( )
⎪⎩
⎪⎨
⎧≤ω≤−+−+ω≤ω≤−−−ω
=+ℑ ωω
ωω
elsewhere.00311530115
5 32
3
32
3jj
t dtdx
(d) Using the time multiplication property,
( ) ( ) ( ) ( )[ ]ω∗ω⎯⎯ →←⋅ π XXtxtx 21CTFT ,
which implies that
{ } ( ) ( )[ ]3321)()( ωωπ Δ∗Δ=⋅ txtxF .
(e) Using the time convolution property,
Error! Objects cannot be created from editing field codes.,
which reduces to
Assignment 2 CSE3451: Signals and Systems
13
{ }⎪⎩
⎪⎨⎧ ≤ω−+
=
⎪⎪⎩
⎪⎪⎨
⎧≤ω⎥⎦
⎤⎢⎣⎡ −
=∗ωω
ω
elsewhere.0
31
elsewhere0
31)()( 3
29
2
32
txtxF
(f) Using the time multiplication property,
( ) ( ) ( ) ( ) ( )021
021CTFT
0cos ω+ωπδ∗ω+ω−ωπδ∗ω⎯⎯ →←ω⋅ ππ XXttx ,
or, ( ) ( ) ( )021
021CTFT
0cos ω+ω+ω−ω⎯⎯ →←ω⋅ XXttx
Case I: For ω0 = 3/2, we obtain
( ) ( ) ( )2321
23
21CTFT)2/3cos( +ω+−ω⎯⎯ →←⋅ XXttx .
The two replicas overlap over (−3/2 ≤ ω < 3/2), therefore,
{ }⎪⎪⎩
⎪⎪⎨
⎧
≤ω≤+≤ω≤−−≤ω≤−+
= −ω
+ω
elsewhere.0
1)2/3cos()(
29
23
62/3
21
23
23
23
29
62/3
21
ttxF
Case II: For ω0 = 3, we obtain
( ) ( ) ( )333cos 21
21CTFT +ω+−ω⎯⎯ →←⋅ XXttx .
Since there is no overlap between the two shifted replicas,
{ }⎪⎪⎩
⎪⎪⎨
⎧
≤−ω−
≤+ω−
= −ω
+ω
elsewhere.0
331
331
3cos)( 3333
21ttxF
or, { }⎪⎪⎩
⎪⎪⎨
⎧
<ω≤−
<ω≤−−
= −ω
+ω
elsewhere.0
60
06
3cos)( 63
21
63
21
ttxF
Case III: For ω0 = 6, we obtain
( ) ( ) ( )666cos 21
21CTFT +ω+−ω⎯⎯ →←⋅ XXttx .
Since there is no overlap between the two shifted replicas,
{ }⎪⎪⎩
⎪⎪⎨
⎧
≤−ω−
≤+ω−
= −ω
+ω
elsewhere.0
361
361
3cos)( 3636
21ttxF
Assignment 2 CSE3451: Signals and Systems 14
or, { }⎪⎪⎩
⎪⎪⎨
⎧
<ω≤−
−<ω≤−−
= −ω
+ω
elsewhere.0
93
39
3cos)( 66
21
66
21
ttxF ▌
Problem 5.20
(a) Calculating the CTFT of both sides and applying the time differentiation property, yields
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )ω=ω+ωω+ωω+ωω XYYjYjYj 6115 23 ,
or, ( ) ( ) ( )( ) ( ) ( )ω=ω+ω+ω+ω XYjjj 6115 23 ,
or, ( ) ( )( ) ( ) ( ) ( ) 6115
123 +ω+ω+ω
=ωω=ω
jjjXYH .
The impulse response h(t) can be obtained by calculating the inverse CTFT of H(ω), which can be expressed as
)3(
5.0)2(
1)1(
5.0)3)(2)(1(
1)(ω+
+ω+
−+ω+
≡ω+ω+ω+
=ωjjjjjj
H
Calculating the inverse CTFT, we obtain
)(5.0)()(5.0)( 32 tuetuetueth ttt −−− +−= . (b) Calculating the CTFT of both sides and applying the time differential property, yields
( ) ( ) ( ) ( ) ( ) ( )ω=ω+ωω+ωω XYYjYj 232 ,
or, ( ) ( )( ) ( ) ( )ω=ω+ω+ω XYjj 232 ,
or, ( ) ( )( ) ( ) ( ) 23
12 +ω+ω
=ωω=ω
jjXYH .
The impulse response h(t) can be obtained by calculating the inverse CTFT of H(ω), which can be expressed as
( ) ( ) )2(
1)1(
123
1)( 2 ω+−
ω+≡
+ω+ω=ω
jjjjH
Calculating the inverse CTFT, we obtain
)()()( 2 tuetueth tt −− −= .
(c) Calculating the CTFT of both sides and applying the time differentiation property, yields
( ) ( ) ( ) ( ) ( ) ( )ω=ω+ωω+ωω XYYjYj 22 ,
or, ( ) ( )( ) ( ) ( )ω=ω+ω+ω XYjj 112 ,
or, ( ) ( )( ) ( ) ( ) 12
12 +ω+ω
=ωω=ω
jjXYH .
Assignment 2 CSE3451: Signals and Systems
15
The impulse response h(t) can be obtained by calculating the inverse CTFT of H(ω), which can be expressed as
( )211)(ω+
=ωj
H
Calculating the inverse CTFT, we obtain
)()( tuteth t−= . (d) Calculating the CTFT of both sides and applying the time differentiation property, yields
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )ω+ωω=ω+ωω+ωω XXjYYjYj 4862 ,
or, ( ) ( )( ) ( ) ( )( ) ( )ω+ω=ω+ω+ω XjYjj 4862 ,
or, ( ) ( )( )
( )( ) ( ) ω+
=+ω+ω
+ω=ωω=ω
jjjj
XYH
21
864
2.
The impulse response h(t) can be obtained by calculating the inverse CTFT of H(ω), which is given by
)()( 2 tueth t−= .
(e) Calculating the CTFT of both sides and applying the time differential property, yields
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )ω=ω+ωω+ωω+ωω XYYjYjYj 12198 23 ,
or, ( ) ( ) ( )( ) ( ) ( )ω=ω+ω+ω+ω XYjjj 12198 23 ,
or, ( ) ( )( ) ( ) ( ) ( ) 12198
123 +ω+ω+ω
=ωω=ω
jjjXYH .
The impulse response h(t) can be obtained by calculating the inverse CTFT of H(ω), which can be expressed as
( )( ) ( ) ( ) )4(
3/1)3(2/1
)1(6/1
12198123 ω+
+ω+
−+ω+
≡+ω+ω+ω
=ωjjjjjj
H
Calculating the inverse CTFT, we obtain
)()()()( 4313
21
61 tuetuetueth ttt −−− +−= . ▌
Problem 5.29
(a) In Example 3.6, it was shown that
[ ] ).()()()( 22 tueetuetuety tttt −−−− −=∗=
(b) From Table 5.2, the CTFT of ( )x t and ( )h t are obtained as
ω+=ω jX 11)( , and
ω+=ω jH 21)( .
The CTFT of the output is then given by
Assignment 2 CSE3451: Signals and Systems 16
( ) ( )1 1 1 1
1 21 2( ) ( ) ( ) j jj jY H X ω ωω ωω ω ω + ++ += = = − .
Calculating the inverse CTFT results in the output signal
[ ] ).()( 2 tueety tt −− −=
(c) As ω+ω
ω ==ω jXYH 2
1)()()( , the Fourier-domain input-output relationship can be expressed as
)()(2)( ω=ω+ωω XYYj .
Calculating the inverse CTFT of both sides results in the following differential equation
)()(2 txtydtdy =+ .
The output can be obtained by solving the differential equation with input ( ) ( )tx t e u t−=
and zero initial conditions y(0−) = 0.
Zero-input Response: Due to zero initial condition, the zero-input response is yzi(t) = 0.
Zero-state Response: The characteristics equation is given by (s + 2) = 0 resulting in a single pole at s = −2. The homogenous component of the zero-state response is given by
.)( 2thzs Aety −=
Since the input x(t) = exp(−t) u(t), the particular solution is of the form
( )p tzsy t Ke−= for 0t ≥ .
Inserting the particular solution in the differential equation results in K = 1. Therefore,
( ) ( )p tzsy t e u t−= .
The overall zero-state response is, therefore, given by
ttzs eAety −− += 2)(
for t ≥ 0. To determine the value of A, we insert the initial condition y(0−) = 0 giving
1 0 1A A+ = ⇒ = −
or, A = −1. The zero state response is given by
( )2( ) ( )t tzsy t e e u t− −= −
.
Total Response: By adding the zero-input and zero-state responses, the overall output is given by
{ ( )20
( ) ( ) ( ) ( ).t tzi zsy t y t y t e e u t− −
=
= + = −
It is observed that Methods (a) – (c) yield the same result. ▌
Problem 5.31
(a) The magnitude spectra of the two systems are calculated below
Assignment 2 CSE3451: Signals and Systems
17
( ) 12
2
400400
1 ==ωω+ω+H
( )⎩⎨⎧ ≥ω=ω
elsewhere.0201
2H
The magnitude spectra are plotted in Fig. S5.31). From Fig. S5.31(a), we observe that the magnitude |H1(ω)| is 1 at all frequencies. Therefore, System H1(ω) is an all pass filter.
From Fig. S5.31(b), we observe that the magnitude |H2(ω)| is zero at frequencies below 20 radians/s. At frequencies above 20 radians/s, the magnitude is 1. Therefore, System H2(ω) is a highpass filter.
ω0
1( )ω1H
cωcω−ω
0
1( )ω1H
cωcω− ω
0
1( )ω2H
2020−ω
0
1( )ω2H
2020− (a) (b)
Fig. S5.31: Magnitude Spectra for Problem 5.31.
(b) Calculating the inverse CTFT, the impulse response of the two systems is given by
{ } { } { } )()(401)( 20120401
2020401
1 ttueth tjj
j δ−=ℑ−ℑ=ℑ= −−ω+
−ω+ω−−− .
{ } { } { }1 1 1 20 202 40 40( ) 1 rect( ) 1 rect( ) ( ) sinc( )th t tω ω
π πδ− − −=ℑ − =ℑ −ℑ = − . ▌
Problem 5.32
The transfer functions for the three LTIC systems are given by
System (a): 21 )1(2)(ω+
=ωj
H .
System (b): ω
+ωπδ=ωj
H 1)()(2 .
System (c): ω+ω−=
ω++−=ω
jj
jH
221
252)(3
The following Matlab code generates the magnitude and phase spectra of the three LTIC systems.
%MATLAB Program for Problem P5.32
%System (a)
clear; % clear the MATLAB environment
num_coeff = [2]; % NUM coeffs. in decreasing powers of s
denom_coeff = [1 2 1]; % DEN coeffs. in decreasing powers of s
Assignment 2 CSE3451: Signals and Systems 18
sys = tf(num_coeff,denom_coeff); % specify the transfer function
figure(1)
bode(sys,{0.02,100}); grid; % sketch the Bode plots
title('Bode Plot for System-1')
%System (b)
clear; % clear the MATLAB environment
num_coeff = [1]; % NUM coeffs. in decreasing powers of s
denom_coeff = [1 0]; % DEN coeffs. in decreasing powers of s
sys = tf(num_coeff,denom_coeff); % specify the transfer function
figure(2)
bode(sys,{0.02,100}); grid; % sketch the Bode plots
title('Bode Plot for System-2')
%System (vc)
clear; % clear the MATLAB environment
num_coeff = [-2 1]; % NUM coeffs. in decreasing powers of s
denom_coeff = [1 2]; % DEN coeffs. in decreasing powers of s
sys = tf(num_coeff,denom_coeff); % specify the transfer function
figure(3)
bode(sys,{0.02,100}); grid; % sketch the Bode plots
title('Bode Plot for System-3')
The resulting Bode plots are shown in Fig. S5.32.
Calculating Output:
System (a): Using the modulation property, the output of system (a) is given by
[ ] ( )[ ]
2 21 1
1 2 (1 1) (1 1)
2( ) ( 1) ( 1) 2 ( 1) ( 1)(1 )
( 1) ( 1) .
j jY
jj
ω π δ ω δ ω π δ ω δ ωω
π δ ω δ ω
+ −= × − + + = − + +
+= − − − +
Calculating the inverse CTFT, we obtain
tty sin)(1 = .
System (b): Using the modulation property, the output of system (a) is given by
[ ]
[ ]
1 12
1( ) ( ) ( 1) ( 1) ( 1) ( 1)
( 1) ( 1) .
j jYj
j
ω πδ ω π δ ω δ ω π δ ω δ ωω
π δ ω δ ω
−⎡ ⎤ ⎡ ⎤= + × − + + = − + +⎢ ⎥ ⎣ ⎦⎣ ⎦
= − − − +
Calculating the inverse CTFT, we obtain
tty sin)(2 = .
Assignment 2 CSE3451: Signals and Systems
19
-80
-60
-40
-20
0
20
Mag
nitu
de (d
B)
10-1 100 101 102-180
-135
-90
-45
0
Phas
e (d
eg)
Bode Plot for System-1
Frequency (rad/sec) (a)
-40
-20
0
20
40
Mag
nitu
de (d
B)
10-1 100 101 102-91
-90.5
-90
-89.5
-89
Phas
e (d
eg)
Bode Plot for System-2
Frequency (rad/sec)
-10
-5
0
5
10
Mag
nitu
de (d
B)
10-1 100 101 102180
225
270
315
360
Phas
e (d
eg)
Bode Plot for System-3
Frequency (rad/sec) (b) (c)
Figure S5.32. Magnitude and phase spectra for systems in Problem 5.32.
System (c): Using the modulation property, the output of system (a) is given by
[ ]
[ ]
2
1 2 1 22 2
1 2( ) ( 1) ( 1)2
( 1) ( 1)
( 1) ( 1) .
j jj j
jYj
j
ωω π δ ω δ ωω
π δ ω δ ω
π δ ω δ ω
− ++ −
−= × − + ++
⎡ ⎤= − + +⎣ ⎦= − − − +
Calculating the inverse CTFT, we obtain
tty sin)(3 = .
To explain why the three systems produce the same output for input x(t) = cost, consider Eq. (5.77), which for ω0 = 1 is given by
( ) ( ))1(cos)1(cos 0)( SymmetricHermitian HtHt H <+ω⎯⎯⎯⎯⎯⎯⎯⎯⎯ →⎯ ω .
Assignment 2 CSE3451: Signals and Systems 20
In other words, the output for x(t) = cos(t) depends only on the magnitude and phase of the system at ω = 1. For the three systems, we note that
1)1()1()1( 321 === HHH
and
2321 )1()1()1( π−===< HHH .
Since the magnitudes and phases of the three systems at ω = 1 are the same, the three systems produce the same output ( ) siny t t= for x(t) = cos(t). ▌