Chapter 2 – Solutions
Problem 2.1
(a)

0,



4
t<0
0 ≤ t < /2
2 t,
x(t) = 4
4

− 2 t, /2 ≤ t < 

 0,
t≥
2/
t
/2
t 6= 0 : x(t) → 0, → 0
t = 0 : x(0) → ∞, → 0
R∞
x(t)dt = 2 · 12 · 2 · 2 = 1
−∞
(b) x(t) = αe−αt u(t)
t = 0 : x(0) → ∞, α → ∞
t 6= 0 : x(t) → 0, α → ∞
R∞
R∞
x(t)dt = −∞ αe−αt u(t)dt
−∞
h −αt i∞
R∞
= α(0 −
= α 0 e−αt dt = α e−α
0
1
)
−α
=1
Problem 2.2
(a) y(t) =
=
t<0:
t>1:
R∞
−∞
R1
0
h(t − τ )x(τ )dτ
h(t − τ )dτ
R1
R1
y(t) = 0 e−|t−τ | dτ = 0 et−τ dτ
h i1
R
t 1 −τ
t e−τ
= e 0 e dτ = e −1 = et (1 − e−1 )
0
R 1 −(t−τ )
R 1 −t+τ
y(t) = 0 e
dτ = 0 e
dτ
1
R1
= e−t 0 eτ dτ = e−t [eτ ]10 = e−t (e − 1)
Rt
R1
0 ≤ t ≤ 1 : y(t) = 0 e−|t−τ | dτ + t e−|t−τ | dτ
Rt
R1
= 0 e−(t−τ ) dτ + t et−τ dτ
Rt
R1
= e−t 0 eτ dτ + et t e−τ dτ
h i1
−t τ t
t e−τ
= e [e ]0 + e −1
t
−t
= e (e − 1) + e (−e−1 + e−t )
t
t
= 1 − e−t − et−1 + 1 = 2 − e−t − et−1

t
−1

t<0
e (1 − e ),
y(t) = 2 − e−t − et−1 , 0 ≤ t ≤ 1

 −t
e (e − 1),
t>1
(b) No! h(t) 6= 0 for t < 0.
R∞
(c) H(f ) = −∞ h(t)e−jωt dt
R∞
= −∞ e−|t|−jωt dt
R0
R∞
= −∞ et(1−jω) dt + 0 e−t(1+jω) dt
h t(1−jω) i0
h −t(1+jω) i∞
= e 1−jω
+ e−(1+jω)
=
1
1−jω
=
2
1+ω 2
(d) y(t) =
2
1+ω02
+
−∞
1
1+jω
=
0
2
(1−jω)(1+jω)
ejω0 t , ω0 = 2πf0
Problem 2.3
h(t) = e−t u(t)
R∞
(a) H(f ) = −∞ h(t)e−jωt dt
R∞
= 0 e−t(1+jω) dt =
1
= 1+jω
−1
1+jω
e−t(1+jω)
∞
0
(b) x(t) = ejω0 t
y(t) = H(f0 )ejω0 t
1
= 1+jω
ejω0 t
0
= √ 1 2 ej(ω0 t−arctan ω0 ) ,
1+ω0
ω0 = 2πf0
2
(c) x(t) = cos ω0 t = R(ejω0 t )
1
j(ω
t−arctan
ω
)
0
y(t) = R √ 2 e 0
=
1+ω0
1
√ 2 cos(ω0 t
1+ω0
− arctan ω0 )
Problem 2.4
(a) Assume that x(t) = ejω0 t ; then y(t) = H(f0 )ejω0 t .
Thus,
y(t)
x(t)
H(f0 ) =
=
Z1 (f0 )
Z1 (f0 )+Z2 (f0 )
where
1
·1.25·103
jω0 10−6
1
+1.25·103
jω0 10−6
Z1 (f0 ) =
(capacitor and resistor in parallel)
5·103
4+jω0 5·10−3
=
Z2 (f0 ) = 5 · 103
Hence,
H(f0 ) =
5·103
4+jω0 5·10−3
5·103
+5·103
4+jω0 5·10−3
=
1
5(1+jω0 10−3 )
and
H(f ) =
1
5(1+jω·10−3 )
(b) Since cos ω0 t = R(ejω0 t ) we obtain
1
jω0 t
jω0 t
y(t) = R(H(f0 )e ) = R 5(1+jω0 10−3 ) e
ω0 = 103 yields
3
3
1−j
1
ej10 t = R 5(1+j)(1−j)
ej10 t
y(t) = R 5(1+j)
1−j j103 t
1
j(103 t−π/4)
√
= R 10 e
=R 5 2e
=
(c) H(f ) =
1
5(1+jω10−3 )
1
√
5 2
cos(103 t − π/4)
= 200 1031+jω
We have the Fourier transform pair (l):
e−αt u(t) ↔
1
α+jω
Hence, we have
3
h(t) = 200e−10 t u(t)
If x(t) = u(t), then
R∞
y(t) = −∞ h(τ )x(t − τ )dτ
3
R∞
3
200e−10 τ u(τ )u(t − τ )dτ
h −103 τ it
Rt
3
= 200 0 e−10 τ dτ = 200 e−103
=
−∞
0
−103 t
= 200 · 10−3 (1 − e
3
) = 15 (1 − e−10 t ),
t≥0
Problem 2.5
(a) H(f ) =
(b) y(t) =
1
(1 + jω)2
1
sin t
2
(c) In (a) we showed that H(f ) =
1
.
(1+jω)2
Hence we have
h(t) ↔ H(f ) =
We recognize
1
(1 + jω)2
1
(1+jω)2
as
1 d
1
−
j2π df 1 + jω
and in Table 2.4 we have
1
e−t u(t) ↔
1 + jω
(1)
So what is the inverse Fourier transform of
tx(t) ↔ −
dX(f )
df
? See Table 2.3.
1 dX(f )
j2π df
Inserting (1) yields
1 d
te u(t) ↔ −
j2π df
−t
1
1 + jω
=
1
(1 + jω)2
Hence,
h(t) = te−t u(t)
The output y(t) when x(t) = u(t) is obtained as
y(t) = h(t) ∗ u(t) = · · · = (1 − e−t − te−t )u(t)
4
Problem 2.8
(a) x(t) = eat u(−t),
a>0
We know that (Fourier transform pair (l))
e−αt u(t) ↔
1
α+jω
From Property 5 follows
x(−t) ↔ X(−f )
Hence, we have
eat u(−t) ↔
1
a−jω
2a
a2 +ω 2
2a
(a+jω)(a−jω)
(b) X(f ) =
=
=
1
a+jω
+
1
a−jω
Hence,
2a
a2 +ω 2
e−at u(t) + eat u(−t) ↔
or, equivalently,
2a
,
a2 +ω 2
e−a|t| ↔
a>0
Problem 2.9
(a) H(f ) =
(b) h(t) =
(c) y(t) =
=
1
jωC
1
R+ jωC
1
RC
R∞
=
1
1+jωRC
=
1
RC
1
1
+jω
RC
e−t/RC u(t)
1
−∞ RC
1
RC
e−(t−τ )/RC u(t − τ )u(τ )dτ
Rt
e−t/RC 0 eτ /RC dτ = 1 − e−t/RC ,
t>0
y(t) = (1 − e−t/RC )u(t)
1
(d) u(t) ↔ U(f ) = jω
+ 12 δ(f )
(Fourier transform pair (d))
1
1
1
U(f )H(f ) = jω
+ 12 δ(f ) RC
1
+jω
RC
=
1 1
RC jω
=
1
jω
−
=
1
jω
+ 12 δ(f ) −
1
1
+jω
RC
1
1
+jω
RC
+
1
1
δ(f ) RC
2
1
1
+jω
RC
+ 12 δ(f )
1
1
+jω
RC
Thus, we have
y(t) = u(t) − e−t/RC u(t) = (1 − e−t/RC )u(t)
5
Problem 2.10
Fourier transform pair (j)
1 − |t|, |t| < 1
↔ sinc2 (f )
0,
|t| > 0
and pair (f)
1
1
cos ω0 t ↔ δ(f − f0 ) + δ(f + f0 )
2
2
yield
1
1
x(t) ↔ sinc (f ) ∗
δ(f − f0 ) + δ(f + f0 )
2
2
1
1
= sinc2 (f − f0 ) + sinc2 (f + f0 )
2
2
2
Problem 2.11
(a) Suppose that x(t) = ejω0 t . Then we have
1
x(t)
1+jω0
y(t) =
Hence,
H(f ) =
1
1+jω
(b) From Fourier transform pair (l) we obtain
h(t) = e−t u(t)
(c) Suppose first that x(t) = ejω0 t . Then,
1
ejω0 t
y(t) = H(f0 )ejω0 t = 1+jω
0
= √ 1 2 ej(ω0 t−arctan ω0 )
1+ω0
Since cos ω0 t = R{ejω0 t }, the input cos ω0 t yields the output
1
j(ω0 t−arctan ω0 )
√
R
= √ 1 2 cos(ω0 t − arctan ω0 )
e
2
1+ω0
(d) We have
y(t) =
=
=
R∞
−∞
R∞
1+ω0
h(τ )x(t − τ )dτ
−∞
R∞
0
e−τ u(τ ) u(t − τ )dτ
|{z}
=0,τ <0
=1,τ >0
e−τ
u(t − τ )
| {z }
dτ
=0,t−τ <0⇔τ >t
=1,t−τ >0⇔τ <t;t>0
6
=
Rt
0
e−τ dτ = − [e−τ ]0 = −e−t − (−1) = 1 − e−t , t > 0
t
and
y(t) = 0, t < 0
Hence,
y(t) = (1 − e−t )u(t)
Problem 2.16
We have
x(t)
h(t)
1
0
t
1
0
1
t
The output y(t) is the convolution of x(t) and h(t), that is,
R∞
y(t) = −∞ x(t − τ )h(τ )dτ
Since h(τ ) is 0 outside the interval [0, 1] and 1 inside the interval we obtain
R1
y(t) = 0 x(t − τ )dτ
Let us draw the signal
t − τ, 0 ≤ t − τ ≤ 1
x(t − τ ) =
0,
otherwise
that is,
x(t − τ ) =
t − τ, t − 1 ≤ τ ≤ t
0,
otherwise
x(t − τ )
t−1
t
0
1
Since we shall integrate x(t − τ ) over the interval [0, 1] we get

0,
t<0

Rt


R1
(t − τ )dτ,
0≤t≤1
Ro1
y(t) = 0 x(t − τ )dτ =

(t − τ )dτ, 1 < t ≤ 2

 t−1
0,
t>2
7
Rt
0
(t − τ )dτ =
Rt
0
tdτ −
t2
2
Rt
0
τ dτ = t
Rt
0
dτ −
h 2 it
τ
2
0
t2
2
=t·t− =
h 2 i1
R1
R1
R1
R1
τ
(t
−
τ
)dτ
=
tdτ
−
τ
dτ
=
t
dτ
−
2
t−1
t−1
t−1
t−1
t−1
2
= t(1 − (t − 1)) − 12 − (t−1)
2
= 2t − t2 − 12 + 12 (t2 − 2t + 1)
= 2t − t2 − 12 + 12 t2 − t +
=t−
In summary,

0,


 2
t /2,
y(t) =
t − t2 /2,



0,
1
2
t2
2
t<0
0≤t≤1
1<t≤2
t>2
Problem 2.17
ψ(t) = rect(t/T )
(
1, |t| < T /2
rect(t/T ) =
0, |t| > T /2
ψ(t − T )
ψ(t)
−T /2
T /2
T
sin πf T
rect(t/T ) ↔ T sinc(f T ) = T
=
πf T
3T /2
t
0,
f = k/T, k integer 6= 0
6 0, otherwise
=
(a) Fourier bandwidth W = 1/T (main lobe of sinc(f T ))
1
(b) ψ(t − kτ ) = rect
(t − kτ )
T
τ = T is the smallest τ such that ψ(t) is orthogonal to ψ(t − kτ ) in every
nonzero k. Hence, the Nyquist-shift is
TN = T
8
(c) The Shannon bandwidth is
1
1
B=
=
2TN
2T
The Fourier bandwidth for rect(t/T ) is twice its Shannon bandwidth.
Problem 2.18

2

1 − |t|, |t| < T /2
ψ(t) =
T
0,
|t| > T /2
1
ψ(t)
−T /2
ψ(t − T )
T /2
According to Table 2.4, p. 57:
1 − |t|, |t| < 1
↔
0,
|t| > 1
T
3T /2
sinc2 (f )
Use Property 5 in Table 2.3 (a = 2/T ):
(
2
T
1 − |t|, |t| < T /2
↔
sinc2 (f T /2)
T
2
0,
|t| > T /2
T
2 2
2
The main lobe of
sinc (f T /2) is within the interval − ,
.
2
T T
(a) Fourier bandwidth W = 2/T
(b) Nyquist-shift TN = T
(c) Shannon bandwidth B = 1/2T
Hence, W = 4B
Problem 2.19

1
(1 + cos(2πt/T )), |t| < T /2
ψ(t) = 2
0,
|t| > T /2
9
t
1
rect(t/T )(1 + cos(2πt/T ))
2
According to Table 2.3 we have
ψ(t) =
1 + cos(2πt/T ) ↔ δ(f ) + 12 δ(f − T1 ) + 12 δ(f + T1 )
Hence,
Ψ(f ) =
=
=
T
sinc(f T ) ∗ (δ(f ) + 12 δ(f − T1 ) + 12 δ(f + T1 ))
2
T
(sinc(f T ) + 12 sinc((f − T1 )T ) + 12 sinc((f + T1 )T ))
2
T
(sinc (f T ) + 12 sinc(f T − 1) + 12 sin(f T + 1))
2
1 1
. Hence, the main lobe
The main lobe of sinc(f T ) is within the interval − ,
T T
2 2
of Ψ(f ) is within the interval − ,
.
T T
(a) Fourier bandwidth W =
2
T
(b) Nyquist-shift TN = T
(c) Shannon bandwidth B =
1
2T
Hence, W = 4B
10