University of California, San Diego
Spring 2014
ECE 45 Discussion 3 Solutions
Topics
• Fourier Series
• Examples
Fourier Series Introduction
Our analysis of systems up to now has been limited to sinusoidal functions. When we determined the
transfer function of a system, we could only determine its output when the input was sinusoidal.
However, with the Fourier Series, we can represent any periodic function as a sum of sinusoidal,
and due to the linearity of the system, we can calculate what the output will be. Because of certain
properties of the Fourier Series, calculations of similar waveforms can be very simple.
Say we have a periodic function that we want to send through some system H(ω), but the system behaves very differently for different frequencies. How do we know whether or not we can “recover” our
function at the output of the system? The Fourier Series allows us to see which frequencies contribute
to our signal.
Periodic Functions
Any periodic function has a Fourier Series representation. A periodic function is one which repeats
itself every period. Every periodic function has a fundamental period/frequency. ω0 = 2π/T0 . All
frequencies of components of a signal must be an integer multiple of ω0 . A function is periodic if:
f (t) = f (t + T0 ) for all t
Example: Find the fundamental frequency ω0 and the period T0 of the following functions
• f1 (t) = cos(π t + π/3)
T0 = 2 → ω0 = π
• f2 (t) = sin(2 t) + 2 cos(3 t + π/4) − cos(t/2)
T0 = 4π → ω0 = 1/2

 0 t < 0 or t > 3
P∞
t 0<t<1
• f3 (t) = n=−∞ x(t − 3 n) where x(t) =

1 1<t<3
T0 = 3 → ω0 = 2π/3
• f4 (t) = cos(t) + cos(π t)
Not periodic, since there is no ω0 such that 1 and π are integer multiples of ω0
Fourier Series
In order to represent a periodic function we need to know two things:
1) The fundamental frequency of the function
2) How that function behaves over one period
With these two pieces of information, we can decompose the function into a sum of sinusoidal components, where each component is a multiple of the fundamental frequency.
For a periodic function, f (t), its Fourier Series representation is:
f (t) =
∞
X
Fn ej n ω0 t
n=−∞
1
where Fn =
T0
Z
t0 +T0
f (t) e−j n ω0 t
t0
Note: ω0 = 2π/T0 , and t0 is ANY time. Usually it is convenient to pick t0 = 0 (depends on function).
It is helpful to think of the integral as “filtering out” any portion of the signal except the contribution of
the sinusoidal function at frequency n ω0 . Fn is how much the sinusoidal frequency at ω0 n contributes
to the signal. Thus summing over all n will yield the signal itself.
Example: Find the Fourier Series representation of the square wave described below:
We don’t have a compact expression for f (t) but we can figure out its period and model its behavior
over a period:
T0 = 2L → ω0 = π/L
since f (t + k 2L) = f (t) for all integers k
Over a period [0, 2L] :
f (t) =
H
0<t<L
−H L < t < L
This is enough for a Fourier Series representation.
Z t0 +T0
Z 2L
1
1
−j n ω0 t
f (t) e−j n π t/L
Fn =
f (t) e
=
T0 t0
2L 0
We can break up the integral of the function f (t) into two sections:
1
Fn =
2L
Z
0
L
1
H e−j n π t/L +
2L
Fn =
Z
2L
(−H) e−j n π t/L
L
H
=
2L
L
2L !
e−j n π t/L e−j n π t/L −
−j n π/L 0
−j n π/L L
H
e−jnπ − e0 − (e−jn2π − e−jnπ )
−2j n π
Since n is an integer, we can make some interesting simplifications:
e−j2πn = 1 and e−jπn = (−1)n
Fn =
H
H
(2(−1)n − 2) =
(1 − (−1)n )
−2j n π
j nπ
2H
n is odd
j nπ
Fn =
0
n is even
We can now plug this into our summation expression:
∞
X
f (t) =
n=−∞
Fn e
j n ω0 t
∞
X
2H j n π t/L
=
e
j nπ
n=−∞
n is odd
We could leave it like this, but there is another interesting simplification which utilizes Euler’s Formula:
2H
f (t) =
π
∞
X
ej n π t/L − e−j n π t/L
4H
=
jn
π
n=1,3,5...
∞
X
sin( nLπ t )
n
n=1,3,5...
Parseval’s Theorem
This is just one of the numerous useful properties of the Fourier Series. Parseval’s Theorem allows us
to determine the energy/power of a signal from either the time representation of a period or the Fourier
Series components.
With periodic functions, we generally aren’t concerned with the total power (since our function continues infinitely), rather the average power over a period.
Z
∞
X
1
|f (t)|2 dt =
|Fn |2
Pavg =
T T
n=−∞
It is often the case that we may know one but not the other or one is much easier to calculate than the
other.
Example: Determine the average power of f (t) = −16 sin(4t)
By Euler’s Formula:
f (t) = −16 sin(4t) = −16
ω0 = 4 thus F−1 =
8
j
e4t − e−4t
2j
and F1 =
−8
j
Thus Pavg = |F−1 |2 + |F1 |2 = 82 + 82 = 128
Equivalently:
2
π
Z
0
π/2
|16 sin(4t)|2 dt = 128