University of California, San Diego
Spring 2014
ECE 45 Discussion 6 Solutions
Topics
• Fourier Transform
• Properties of F.T.
Fourier Transform
The Fourier transform is a way of representing a function in terms of its frequency components. We can
think of it as a Fourier series with an infinite period (i.e. it never repeats itself), so it can have frequency
contributions from any frequency.
Z ∞
1
F (jω)ejωt dω
f (t) =
2π −∞
Z ∞
f (t)e−jωt dt
F (jω) =
−∞
Fourier Transform as an Input to an LTI System:
Since a Fourier transform is an integral (sum with infinitesimally small intervals) of sinusoidal components, we know how to analyze the output of a system with a periodic function as its input. For any
input x(t).
Z ∞
Z ∞
1
1
jωt
x(t) =
X(jω)e dω −→ H(ω) −→ y(t) =
|H(ω)|X(jω) ej(ωt+∠H(ω)) dω
2π −∞
2π −∞
We can represent y(t) as a Fourier Transform:
Y (jω) = X(jω) H(ω)
The Fourier transform is a one-to-one mapping, so if we know either the time function or the frequency
function, we know the other as well.
Parseval’s Theorem:
Power is the same whether we look at our signal in the time or frequency domain
Z ∞
Z ∞
1
2
|x(t)| dt =
|X(jω)|2 dω
2π
−∞
−∞
Time Shifting:
f (t − t0 ) ↔ F (jω) e−jωt0
Frequency Shifting:
f (t) ejω0 t ↔ F (j(ω − ω0 ))
Linearity:
if f1 (t) ↔ F1 (jω) and f2 (t) ↔ F2 (jω)
then a f1 (t) + b f2 (t) ↔ a F1 (jω) + b F2 (jω)
Conjugation:
x∗ (t) ↔ X ∗ (−jω)
Time Scaling:
1
f (c t) ↔ X
|c|
jω
c
Time Derivative:
df (t)
↔ jωF (jω)
dt
Very useful exercise to derive the properties using the definitions:
Z ∞
1
f (t) =
F (jω)ejωt dω
2π −∞
Z ∞
F (jω) =
f (t)e−jωt dt
−∞
Unit Step Function (Heaviside step function)
Useful for representing a signal “turning on” or starting at a certain value
0 t<0
u(t) =
1 t≥0
u(t − t0 ) =
0 t < t0
1 t ≥ t0
for any function f (t)
0
t < t0
f (t) u(t) =
f (t) t ≥ t0
Example 1: Determine the Fourier transform of f (t) = u(t) e−a t where a > 0
Z ∞
Z ∞
1
−jωt
e−a t e−jωt dt =
f (t)e
dt =
F (jω) =
a + jω
0
−∞
Example 2: Determine the Fourier transform of g(t) = u(−t) ea t where a > 0
g(t) = f (−t)so we can use the time scaling property with c = −1
G(jω) =
1
1
F (jω/(−1)) = F (−jω) =
| − 1|
a − jω
Example 3: Determine the Fourier transform of h(t) = ea |t|
We calculated this last week, but we can use the linearity property to confirm our result, since h(t) =
f (t) + g(t)
H(jω) = F (jω) + G(jω) =
1
1
2a
+
= 2
a + jω a − jω
a + ω2
Example 4: Given the input to an LTI system is x(t) = u(t) e−t , determine the output y(t) when the
e−jω
transfer function of the is H(ω) = 1+jω
X(ω) =
1
1 + jω
Y (ω) = X(ω)H(ω) =
1
y(t) =
2π
Z
e−jω
(1 + jω)2
∞
Y (jω)ejωt dω
−∞
Let G(jω) =
1
(1+jω)2
Note: Y (jω) = G(jω) e−jω
From the table of Fourier transform pairs:
G(jω) =
1
↔ g(t) = t e−t u(t)
(1 + jω)2
Using the time shifting property:
y(t) = g(t − 1) = (t − 1) e−(t−1) u(t − 1)
Example 5: Determine the inverse Fourier transform (time domain representation) of
F (ω) =
19 − jω
10 + ω 2 + 3jω
Use partial fractions to break up F (ω)
F (ω) =
A
B
19 − jω
=
+
(5 − jω)(2 + jω)
5 − jω 2 + jω
A(2 + jω) + B(5 − jω) = 19 − jω
2A + 5B = 19 and A − B = −1 → A = 2 and B = 3
F (ω) =
2
3
+
= 2u(−t) e5t + 3u(t) e−2t
5 − jω 2 + jω
Example 6: What is the power of the signal f (t) = π5 sinc(5t)
Z ∞
Z
25 ∞
2
|f (t)| dt = 2
P =
|sinc(5t)|2 dt =???
π
−∞
−∞
Recall the rectangle function:
0
) is a rectangular pulse with height A, width W , and centered at t0 .
A rect( t−t
W
A (t0 − W/2) ≤ t ≤ (t0 + W/2)
t−t0
A rect( W ) =
0 else
Use Parseval’s to simplify calculation! From the transform pairs table (and from last week)
1 |ω| ≤ 5
F (jω) = rect(ω/10) =
0 else
1
P =
2π
Z
∞
1
|F (jω)| dω =
2π
−∞
2
Z
5
1 dω =
−5
5
π
Example 7: Determine the Fourier transform of
t 0≤t≤2
g(t) =
0 else
We could do the integration, but we’d have to use integration by parts, so let’s use the properties instead.
1 −1 ≤ t ≤ 1
Let f (t) = rect(t/2)
→ F (jω) = 2 sinc(ω)
0 else
Note f (t) =
d
g(t
dt
+ 1) so F (jω) = jω ejω G(jω)
Thus G(jω) =
F (jω)
2 e−jω sinc(ω)
=
jω ejω
jω