University of California, San Diego
Spring 2014
ECE 45 Homework 3 Solutions
3.1 The unit-triangle function ∆(t) is defined as
1 − 2|t| |t| ≤ 1/2
∆(t) =
0
else
Calculate the Fourier transform D(jω) of the unit-triangle function.
1 − 2t 0 ≤ t ≤ 1/2
Let f (t) =
0
else
1 + 2t −1/2 ≤ t ≤ 0
Then f (−t) =
0
else
∆(t) = f (t) + f (−t) ←→ D(jω) = F (jω) + F (−jω)
Z
∞
−jωt
f (t) e
F (jω) =
(1 − 2t) e
dt =
−∞
0
F (jω) =
−jωt
1/2
e−jωt (jω (2t − 1) + 2) dt =
(jω)2
t=0
2 + jω − 2ejω/2
2 − jω − 2e−jω/2
−→
F
(−jω)
=
ω2
ω2
D(jω) =
Note:
1/2
Z
4 − 2ejω/2 − 2e−jω/2
4 − 4 cos(ω/2)
=
2
ω
ω2
1 − cos(2t)
= sin2 (t) −→ 4(1 − cos(ω/2)) = 8 sin2 (ω/4)
2
2
8 sin2 (ω/4)
1 4 sin(ω/4)
sinc2 (ω/4)
D(jω) =
=
=
ω2
2
ω
2
3.2 Use the results from number 3.1 and the properties of the Fourier transform to determine the Fourier
transform G(jω) of the general triangle function:
t − t0
A − A τ2 |t − t0 | if |t − t0 | ≤ τ2
=
A∆
0
else
τ
Note: A convenient way to think about the triangle function is it is a triangle of height A, centered at
t0 , with width τ
0
g(t) = A ∆ t−t
τ
By the amplitude scaling, time scaling, and time shift properties:
G(jω) = A
1
| τ1 |
D
jω
1
τ
−jωt0
e
−jωt0
= A |τ | D (jωτ ) e
A |τ |
2 τω
=
sinc
e−jωt0
2
4
3.3 Use the general triangle function and the properties of the Fourier transform to determine the Fourier
transform of the function:

if 1 ≤ t ≤ 2
 3
−3 if 2 < t ≤ 3
f (t) =

0
else
Let g(t) = 3∆
f (t) =
t−2
2
←→ G(jω) = 3 sinc2 (ω/2) e−2jω
dg(t)
←→ F (jω) = jωG(jω)
dt
Alternatively:
f (t) = 3 rect(t − 3/2) − 3 rect(t − 5/2) ←→ F (jω) = 3/2 sinc(ω/2) e−jω3/2 − 3/2 sinc(ω/2) e−jω5/2
We could obtain either from the other result by manipulation using trig properties
3.4 The signal g1 (t) + g2 (t) is sent into an LTI system. Using the Fourier transforms of g1 and g2 and
the transfer function H(ω), determine the output y(t) in terms of g1 (t) and g2 (t).
H(ω) =
2 e−jω −3 ≤ ω ≤ 3
0
else
Y (jω) = H(ω)[G1 (jω) + G2 (jω)] = H(ω)G1 (jω) + H(ω)G2 (jω) = 2 e−jω G2 (jω)
Y (jω) = 2 e−jω G2 (jω) ←→ y(t) = 2 g2 (t − 1)
3.5 Given that f (t) = 5∆2 (t/5), evaluate the Fourier transform F (jω) at ω = 0
Z ∞
Z ∞
Z ∞
−jωt
f (t)e
dt|ω=0 =
f (t) dt =
5∆2 (t/5) dt
F (0) = F (jω)|ω=0 =
−∞
−∞
−∞

 (1 + 2t/5)2 −5/2 ≤ t ≤ 0
(1 − 2t/5)2 0 < t ≤ 5/2
∆2 (t/5) =

0
else
Z
0
Z
2
−5/2
Z
5 (1 − 2t/5)2 dt
0
0
1 + 4t/5 + 4t2 /25 dt +
=5
5/2
5 (1 + 2t/5) dt +
F (0) =
−5/2
Z
!
5/2
1 − 4t/5 + 4t2 /25 dt
0
=
25
3
3.6 Determine the Fourier transform G(ω) of the signal g(t) shown below. What is the phase of the
Fourier transform?
g(t) = ∆(t/2) − rect(t − 3/2) − rect(t + 3/2)
G(jω) =
2
sinc2 (ω 2/4) − sinc(ω/2) ejω3/2 + e−jω3/2
2
G(jω) = sinc2 (ω/2) − 2 sinc(ω/2) cos(3ω/2) = sinc(ω/2)[sinc(ω/2) − 2 cos(3ω/2)]
∠G(jω) = 0
3.7 Show that for real-value signals x(t), the Fourier transform X(jω) satisfies the property X(−jω) =
X ∗ (jω). What does this tell us about X(0) for real signals?
For a complex number C = a + jb, C ∗ = a − jb. If C is purely real, C = C ∗ , so x(t) = x∗ (t)
We can write both x(t) and x∗ (t) in terms of their Fourier transforms
1
x(t) =
2π
Z
1
x (t) =
2π
Z
∗
∞
X(jω) ejωt dω
−∞
∞
X ∗ (jω) e−jωt dω
−∞
Let ω 0 = −ω, dω 0 = −dω
1
x (t) = −
2π
∗
Z
−∞
0
jω 0 t
X (−jω ) e
∞
1
x(t) = x (t) →
2π
∗
∗
Z
Z
1
dω =
2π
Z
∞
jωt
X(jω) e
−∞
∞
1
dω =
2π
0
0
X ∗ (−jω 0 ) ejω t dω 0
−∞
∞
0
X ∗ (−jω 0 ) ejω t dω 0
−∞
−→ X(−jω) = X ∗ (jω)
−→ X(0) = X ∗ (0) −→ X(0) is purely real
3.8 Determine the output y(t) to an LTI system when the input is x(t) = u(t + 1) − 2u(t − 1) + u(t − 3)
1
and the transfer function is H(ω) = jω

−1 ≤ t < 1
 1
−1 1 ≤ t < 3
x(t) =

0
else
X(jω)
d
→ X(jω) = jωY (jω) → x(t) = y(t)
jω
dt
 Rt
Z t
−1 ≤ t < 1
 R−1 1 dτ R
1
t
y(t) =
x(τ ) dτ =
1 dτ − 1 1 dτ
1≤t<3
 −1
−∞
0
else

 1 + t −1 ≤ t < 1
t−1
3−t
1 ≤ t < 3 = 2∆
=

4
0
else
Y (jω) = H(jω)X(jω) =
3.9 Determine the input to the system
1
jω + 3
−3t
−4t
which produces the output y(t) = e u(t) − e u(t)
H(ω) =
Y (jω) =
1
1
1
1
−
=
= X(jω)H(jω) = X(jω)
3 + jω 4 + jω
(3 + jω)(4 + jω)
3 + jω
X(jω) =
3 + jω
1
=
←→ x(t) = u(t) e−4t
(3 + jω)(4 + jω)
4 + jω
3.10 Suppose that a signal x(t) has the Fourier transform X(jω). Consider a signal g(t) whose shape
is the same as the shape of X(jω), i.e. g(t) = X(jt).
Show the Fourier transform G(jω) of g(t) has the same shape as 2πx(−t), i.e. G(jω) = 2πx(−ω)
Z
∞
−jωt
G(jω) =
g(t) e
−∞
Z
∞
dt =
X(jt) e
−jωt
dt = 2π
−∞
1
Since x(−a) =
2π
Z
1
2π
Z
∞
X(jt) e
−jωt
dt = 2πx(−ω)
−∞
∞
X(jb) e−jba db
−∞
Note: This is why the Fourier transform of a rectangular pulse is shaped like a sinc function, and the
Fourier transform of a sinc function is shaped like a rectangular pulse
3.11 Given the following differential equation, determine the output y(t) when x(t) = u(t) e−3t
d2 y(t) dy(t)
+
− 2y(t) = x(t)
dt2
dt
(jω)2 Y (jω) + jω Y (jω) − 2 Y (jω) = X(jω) =
Y (jω) =
1
3 + jω
X(jω)
1
A
B
C
=
=
+
+
2
(jω) + jω − 2
(3 + jω)(jω − 1)(jω + 2)
3 + jω jω − 1 2 + jω
A(jω − 1)(2 + jω) + B(3 + jω)(2 + jω) + C(3 + jω)(jω − 1) = 1
A(−ω 2 + jω − 2) + B(−ω 2 + 5jω + 6) + C(−ω 2 + 2jω − 3) = 1
A+B+C =0
A + 5B + 2C = 0
−2A + 6B − 3C = 1
A=
Y (jω) =
1
1
−1
, B=
, C=
4
12
3
1/4
1/12
1/3
1
1
1
−
−
←→ y(t) = u(t) e−3t −
u(−t) et − u(t) e−2t
3 + jω 1 − jω 2 + jω
4
12
3