Solution to Homework Assignment 1
Solution to Problem 1: test
(a)
G(f) =
{
e−j2pift0 f ∈ [−f0, f0]
0 elsewhere
.
By using the definition of inverse FT:
f(t)=
∫ f0
−f0
e−j2pift0ej2piftdf =
∫ f0
−f0
ej2pif(t−t0)df =
1
j2pi(t− t0)e
j2pif(t−t0)|f0−f0
=
sin[2pif0(t− t0)]
pi(t− t0) = 2f0sinc[2f0(t− t0)].
(b) G(f) = rect
(
f
2f0
)
. Thus g(t) = 2f0sinc(2f0t).
Or from the definition of inverse FT:
f(t)=
∫ f0
−f0
ej2piftdf =
1
j2pit
ej2pift
∣∣f0
−f0 =
sin(2pif0t)
pit
= 2f0sinc(2f0t).
Solution to Problem 2: From the plot,
g1(t) = sin t[u(t)− u(t− pi)].
By definition:
G1(f) =
∫ pi
0
sin(t)e−j2piftdt
= −
∫ pi
0
e−j2piftd cos(t)
= − cos(t)e−j2pift∣∣pi
0
+
∫ pi
0
cos(t)de−j2pift
= 1 + e−j2pi
2f + (−j2pif)
∫ pi
0
cos(t)e−j2piftdt
= 1 + e−j2pi
2f + (−j2pif)
∫ pi
0
e−j2piftd sin(t)
= 1 + e−j2pi
2f + (−j2pif)
[
sin(t)e−j2pift
∣∣pi
0
−
∫ pi
0
sin(t)de−j2pift
]
= 1 + e−j2pi
2f − (−j2pif)2
∫ pi
0
sin(t)e−j2piftdt
= 1 + e−j2pi
2f + 4pi2f 2G1(f)
After move the 4pi2f 2G1(f) to the left of the equation, we get
G1(f)− 4pi2f 2G1(f) = 1 + e−j2pi2f .
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ECE 380 Introduction to Communication Systems
Hence,
G1(f) =
1 + e−j2pi
2f
1− 4pi2f 2
By properties: Notice that
g1(t) = sin t u(t) + sin(t− pi)u(t− pi).
sin t u(t)⇐⇒ pi
2j
[
δ
(
f − 1
2pi
)
− δ
(
f +
1
2pi
)]
+
1
1− (2pif)2 .
By using the time-shifting property,
sin(t− pi) u(t− pi)⇐⇒
{
pi
2j
[
δ
(
f − 1
2pi
)
− δ
(
f +
1
2pi
)]
+
1
1− (2pif)2
}
e−j2pi
2f .
By linearity of FT, we have
G(f)=
{
pi
2j
[
δ
(
f − 1
2pi
)
− δ
(
f +
1
2pi
)]
+
1
1− (2pif)2
}
(1 + e−j2pi
2f ).
As δ
(
f ± 1
2pi
)
(1 + e−j2pi
2f ) = δ
(
f ± 1
2pi
)
(1 + e±jpi) = 0, we have
F (ω) =
1 + e−j2pi
2f
1− 4pi2f 2 .
For the second signal,
g2(t) = e
−at[u(t)− u(t− T )].
By definition,
G2(f) =
∫ T
0
e−ate−j2piftdt = − 1
a+ j2pif
e−(a+j2pif)t
∣∣∣∣T
0
=
1− e−(a+j2pif)T
a+ j2pif
Notice that
g2(t) = e
−atu(t)− e−aT e−a(t−T )u(t− T ).
e−atu(t)⇐⇒ 1
a+ j2pif
.
By using the time-shifting property,
e−a(t−T )u(t− T )⇐⇒ 1
a+ j2pif
e−j2pifT .
By the linearity of FT, we have
F (ω)=
1− e−(a+j2pif)T
a+ j2pif
.
Solution to Problem 3: test.
(a) Since
g(t) sin(2pifct) =
1
2j
g(t)
[
ej2pifct − e−j2pifct] ,
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ECE 380 Introduction to Communication Systems
from frequency-shifting property and linearity,
g(t) sin(2pifct)

1
2j
[G(f − fc)−G(f + fc)].
(b) Since 2+ cos(2pif0t)
2δ(f) + 12δ(f − f0) + 12δ(f + f0), by using the property in (a), we have
S(f) =
1
2j
[
2δ(f − 100) + 1
2
δ(f − 100− f0) + 1
2
δ(f − 100 + f0)
−2δ(f + 100)− 1
2
δ(f + 100− f0)− 1
2
δ(f + 100 + f0)
]
.
The spectrum is as in Figure 1.
Figure 1: Spectra for Problem 3.
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ECE 380 Introduction to Communication Systems
Solution to Problem 4: test.
(a) The signals u(t), ue(t), uo(t) are represented by the following figure.
Figure 2: Signals for Problem 4.
(b)We have
ue(t) =
1
2

1
2
δ(f).
uo(t) =
1
2
sgn (t)
1
j2pif
.
u(t)
1
2
δ(f) +
1
j2pif
.
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