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Chapter 16
Fourier Series
Preview
Nonsinusoidal but periodic excitations are common.
This chapter begins the study of periodic functions and introduces the
Fourier Series view.
Preview
To evaluate the Fourier coefficients, the following integrals will be
helpful.
cos at dt sin at a
∫^ =
sin at dt cos at a
∫ =
2
t sin at dt sin at t cos at a a
∫
2
t cos at dt cos at t sin at a a
∫
Preview
Values of cosine, sine and exponential functions for integral multiples
of π.
cos 2 n π = 1
sin 2 n π = 0
cos ( 1)
n n π = −
sin n π = 0
( 1) for n even^2 cos (^2 0) for n odd
n n π ⎧ −⎪ = ⎨ ⎪⎩
21 ( 1) for n odd sin (^2 0) for n even
n n π
− ⎧ −⎪ = ⎨ ⎪⎩
2
j n
e
π
jn n
e
π
2 2 21
( 1) for n even
( 1) for n odd
n jn
n
e
j
π
−
Section 16.1 Overview
A periodic function is one that satisfies the relationship
Where n is an integer 1, 2, 3, ….
And T is the period.
Thus the function repeats with period T.
f ( ) t = f t ( ± nT )
f ( t 0 (^) ) = f t ( 0 − T ) = f ( t 0 + T )= f ( t 0 − 2 T ) = f t ( 0 + 2 T ) and so on...
The smallest value of T that satisfies the periodicity condition is the
fundamental period of f(t).
Section 16.1 Fourier Series Analysis
Fourier discovered that a periodic function can be represented by an
infinite sum of sine or cosine functions that are harmonically related.
The Fourier coefficients are av, an and b n.
0 0 1
( ) (^) v n cos (^) n sin n
f t a a n ω t b n ω t
∞
=
= + (^) ∑ +
The fundamental frequency is given by ω 0.
The harmonic frequencies of f(t) are the integer multiples of ω 0.
Section 16.1 Dirichlet’s Conditions
Keeping the mathematicians happy requires that certain conditions
exist for the convergent Fourier series.
Dirichlet’s Conditions:
1. f(t) is single-valued (at any one point).
2. f(t) has a finite number of discontinuities in one period.
3. f(t) has a finite number of maxima and minima in one period.
4. The integral exists.
0
0
( )
t T
t
f t dt
∫
Any periodic function generated by a physically realizable source
satisfies Dirichlet’s conditions.
Section 16.1 Fourier Series
We will determine f(t).
Then we will calculate the coefficients a v, an and b n.
It this course, linear lumped-parameter, time-invariant applies.
Thus so does superposition.
The procedure is straightforward and involves no new techniques of
circuit analysis.
The analysis results in the Fourier series representation of the steady-
state response of the circuit.
Section 16.2 The Fourier Coefficients
0
0
0
( ) cos
t T an (^) t f t k t dt T
ω
= ∫
0
0
t T av (^) t f t dt T
= (^) ∫
0
0
0
( ) sin
t T bn (^) t f t k t dt T
ω
= ∫
The Fourier coefficients are defined by the following integrals.
Recall that 0
T
π ω = = 2 π f
Example
Find the Fourier Series for the following signal.
The period of this signal is T = 2π.
0
T
π ω =
π
π
= = (^1) sec rad
0
T av f t dt T
∫
2
0
t dt
π
π π
= (^) ∫ ( ) 12 2 22 0
t
π
π
( )
2 2
π π
= ⎡^ −⎤
2
2
π
π
Example
0 0
( ) cos
T an f t k t dt T
= ω ∫
2
0
cos 2 2
t kt dt
π
π π
∫
2 (^2 )
cos 4
t kt dt
π
π
∫
2 (^ )
1 2 2 0
cos sin 2 k^
kt kt kt
π
π
= ⎡^ + ⎤
2 12 (^ )^12 (^ )
cos 2 2 sin 2 cos 0 0sin 0 2 k^ k
k π k π k π k k π
= ⎡^ + − + ⎤
2 2
1 1 2
k k π
⎣ ⎦ 2 [ ]
2 π
Recall the integral solution:
2
1 ∫ x^ cos^ ax dx^ =^ a (cos^ ax^ + ax^ sin^ ax )
Example
0 0
( ) sin
T bk f t k t dt T
= ω ∫
2
0
sin 2 2
t kt dt
π
π π
∫
2 (^2 )
sin 2
t kt dt
π
π
∫
2 (^ )
1 2 2 0
sin cos 2
k kt^ kt^ kt
π
π
= ⎡^ − ⎤
2 12 (^ )^12 (^ )
sin 2 2 cos 2 sin 0 0 cos 0 2 k^ k
k π k π k π k k π
= ⎡^ − − − ⎤
2 2
1 1 2
2 k^ k
k π π
k
k
π
π
π k
The integral solution that was used is
2 sin 1 (sin cos ) a ∫ x^ ax dx^ =^ ax^ − ax^ ax
Fourier Series with 10 terms Fourier Series with 100 terms
Fourier Series with 1,000 terms Section 16.3 The Effect of Symmetry on the Fourier Coefficients
Four types of symmetry may be used to simplify the task of evaluating
the Fourier coefficients.
Even function f(t) = f(-t)
Odd function f(t) = -f(-t)
Half-wave symmetry f(t) = -f(t – T/2)
Quarter-wave symmetry which has symmetry at the midpoint of both
the positive and negative half cycles.
Use symmetry at your own peril at this point. On tests or quizzes, the
use of incorrect symmetry is no excuse. You will have time to solve
the coefficients without the use of symmetry of any kind.
Section 16.3 Even Symmetry
Any even periodic function f(t) consists of cosine terms only.
A function is an even function if f(t) = f(-t).
2
f ( ) x = x
2 2
⇒ f ( − x ) = −( x )= x
1 ( ) cos(45 ) 2
f x = =
D 1 ( ) cos( 45 ) 2
⇒ f − x = − =
D
Examples of even functions.
Section 16.3 Even Symmetry
When a function is even we can use the follow abbreviated equations:
2
0
T
av f t dt
T
∫
2
0 0
( ) cos
T
an f t n t dt
T
∫
Cosine is even (and shows above) and sine is odd (and thus is not
included above).
bn = 0
Section 16.3 Odd Symmetry
Any odd periodic function f(t) consists of sine terms only.
Examples of odd functions.
3
f ( ) x = x
3 3
⇒ f ( − x ) = −( x ) = − x
1 ( ) sin(45 ) 2
f x = =
D 1 ( ) sin( 45 ) 2
⇒ f − x = − = −
D
A function is an odd function if f(t) = -f(-t).
Section 16.3 Odd Symmetry
When a function is odd we can use the follow abbreviated equations:
av = an = 0
2
0 0
( ) sin
T
bn f t n t dt
T
= (^) ∫ ω
Sine is odd (and shows above) and cosine is even (and thus is not
included above).
Also note that odd functions do not have a “dc” value.
Section 16.3 Half-wave symmetry
If a periodic signal f(t), shifted by half the period, remains unchanged
except for a sign (-) then the signal is said to have half-wave
symmetry.
T
f t = − f t −
In a signal with half-wave symmetry, all the even numbered harmonics
vanish.
Section 16.3 Quarter-wave symmetry
Quarter-wave symmetry is really reaching down into the bag just to
avoid some calculus.
Quarter-wave symmetry will not be covered and will not be tested
Read the section on your own.
Section 16.4 Compact Form of the Fourier Series
We can combine the an and b n coefficients into a single cosine (or sine
if wanted) term. The resulting series is usually called the compact
Fourier Series.
0 1
( ) (^) v n cos( (^) n ) n
f t a A n ω t θ
∞
=
= + (^) ∑ −
An cos( n ω o (^) t − θ n ) = an cos n ω o t + bn sin n ω (^) ot
2 2 1 where tan
n n n n n n
b A a b and a
θ
− ⎛^ ⎞
The dc term (ω 0 = 0) is given by av.
= an cos n ω o t + bn cos( n ω (^) ot − 90 )
D
Note that in this text, the phase term is misleading. The authors have
incorporated a minus sign in the series definition for the phase.
Example – Compact Fourier Series
Find the compact trigonometric Fourier Series for the periodic signal x(t).
In this case T 0 = π.
0 0
1 Fundamental frequency f T
=
1 Hz π
=
0 0
2
T
π ω =
2
sec
π
π
= 2 sec
rad
Therefore
0 0 1
( ) (^) v n cos (^) n sin n
f t a a n ω t b n ω t
∞
=
= + (^) ∑ + 1
v n cos^2 n sin^2 n
a a n t b n t
∞
=
= + (^) ∑ +
Section 16.5 An Application of Fourier Series
A low pass filter is excited by a square signal.
We want to find the steady-state response of the output signal.
First find the Fourier series of the input signal.
0
2
T
π ω =
Section 16.5 An Application of Fourier Series
The square wave has odd, half-wave and quarter-wave symmetry.
I will use odd and half-wave symmetry in finding the Fourier
coefficients.
av = 0 by odd symmetry
ak = 0 by odd symmetry
bk = 0 for k even by half − wave symmetry
2
0 0
( ) sin
T
bk f t k t dt for k odd by half wave symmetry T
= ω − ∫
2
0
sin
T
Vm k t dt T T
π = (^) ∫ 0 02 2
( ) cos
m T k T
V
k t T^ π
= − ω
1 1
cos cos 0 2
V m (^) k T k
k T T
π π
π =− =
Vm
k π
4 Vm
k π
Section 16.5 An Application of Fourier Series
The input square wave has the following Fourier series.
0
m sin
V
ω t π
sin 3 3
V m (^) ω t
π
1 0 1,3,5,...
sin
m g (^) n n
V
v n ω t π
∞
=
= (^) ∑
0
sin 5 .... 5
V m (^) ω t
π
The system is assumed to be linear, lumped parameter, time invariant
so superposition applies.
Now write the transfer function of the circuit and find the output
voltage.
Section 16.5 An Application of Fourier Series
The output is the voltage across the capacitor.
( )^0
g
v H j v
ω =
1 0 1,3,5,...
sin
m g n n
V
v n ω t π
∞
=
= (^) ∑
For the first term of the Fourier series (n = 1) we have
1
(^0 )
j C g j C
v v R
ω
ω
vg j ω RC
1 1 1
g o
v v j ω RC
4 0
1
V m
j RC
π
ω
D
0
4
2 2 1 0 1
1 ( ) tan ( )
V m
RC RC
π ω ω
−
D
1 (^0 ) 0
tan ( ) 1 ( )
V m RC RC
ω π ω
These are phasors.
Section 16.5 An Application of Fourier Series
In this case the phasors are referenced to the sine function rather than
the usual cosine function. Thus
Similarly the next term of the Fourier series (n = 3) is
1 0 0
(^1 ) 0
sin tan ( )
m
o
V
t RC v RC
ω ω π
ω
1 0 0
(^3 ) 0
sin 3 tan (3 ) 3
1 (3 )
m
o
V
t RC
v RC
ω ω π
ω
Section 16.5 An Application of Fourier Series
For the general k th^ case we have
So we can now write the Fourier series (steady-state) of the output
signal v o.
1 0 0
2 0
sin tan ( )
m
ok
V
k t k RC k v k RC
ω ω π
ω
− ⎡ − ∠ ⎤ ⎣ ⎦ =
1 0 0
2 1,3,5,... (^0)
4 sin^ tan^ (^ )
m o n
V n^ t^ n^ RC v n n RC
ω ω
π ω
∞ −
=
∑
Section 16.5 An Application of Fourier Series
We can now draw some circuit behavior conclusions from the Fourier
series for the output signal.
As the frequency ω = nω 0 → ∞, the magnitude of the response → 0.
1 0 0
2 1,3,5,... (^0)
4 sin^ tan^ (^ )
m o n
V n^ t^ n^ RC v n n RC
ω ω
π ω
∞ −
=
∑
The Fourier series does describe a low pass filter as expected!
1 4 sin^0 tan^ (^0 ) 0
m o
V n^ t^ n^ RC v n
ω ω
π
− ⎡ − ∠ ⎤ ⎣ ⎦ = = ∞
Section 16.5 An Application of Fourier Series
What about the effects of varying the capacitance C?
For large C
0 2 0 1,3,5,...
4 sin^90 m
n
V n^ t
RC n
ω
πω
∞
=
≈ (^) ∑
D
The output harmonics are decreasing by 1/n 2.
1 0 0
1,3,5,... 2 0
4 sin^ tan^ (^ )
m o n
V n^ t^ n^ RC v n n RC
ω ω
π ω
∞ −
=
∑
0 2 0 1,3,5,...
(^4) m cos( )
n
V n t
RC n
ω
πω
∞
=
= (^) ∑
But the input harmonics are decreasing by 1/n.
Again, we see the effects of a low pass filter.
Section 16.5 An Application of Fourier Series
Read pages 673 and 674.
You will need this information for the laboratory.
Section 16.6 Average-Power Calculations with Periodic Functions
Given the Fourier series representations of the voltage and current at a
pair of terminals.
Write v & i in accordance with the passive sign convention.
Instantaneous power p = vi.
0 1
dc n cos(^ vn ) n
v V V n ω t θ
∞
=
= + (^) ∑ −
0 1
dc n cos(^ in ) n
i I I n ω t θ
∞
=
= + (^) ∑ −
Section 16.6 Average-Power Calculations with Periodic Functions
The average power is given by
Only terms of the same frequency (of v and i) survive the integration.
Thus
0
0
1 t^ T Pavg (^) t p dt T
= (^) ∫
0 0
(^1) t T Pavg V Idc dc t (^) t for the dc components T
=
0
0
1 t T
t
vi dt T
= (^) ∫
0 0 1
n n cos(^ vn ) cos(^ in )^. n
V I n t n t dt for the harmonics T
ω θ ω θ
∞
=
Section 16.6 Average-Power Calculations with Periodic Functions
Use the following trig identity to simplify.
Then
1 1 cos α cos β= 2 cos(α − β) + 2 cos( α +β)
0 0 0 0 1
cos( ) cos( )
t T avg dc dc t n n vn in n
P V I t V I n t n t dt T T
ω θ ω θ
∞
=
= + (^) ∑ − −
0
0
0 (^1) integral = zero
[cos( ) cos(2 )] 2
n n^ t^ T dc dc (^) t vn in vn in n
V I
V I n t dt T
θ θ ω θ θ
∞ (^) +
=
= + (^) ∑ (^) ∫ − + − −
1
[cos( ) 2
n n dc dc vn in n
V I
V I T
T
θ θ
∞
=
= + (^) ∑ − 1
[cos( ) 2
n n dc dc vn in n
V I
V I θ θ
∞
=
= + (^) ∑ −
The total average power is the superposition of the average powers
associated with each harmonic.
Example
Given the Fourier series for a voltage.
Since only a finite number of Fourier terms are given, we can only
estimate the rms voltage.
v = 10 + 30cos( ω o (^) t − θ 1 ) + 20cos(2 ω o (^) t − θ 2 ) + 5cos(3ω (^) o (^) t −θ 3 ) + 2cos(5ω (^) o t −θ 5 )
2 2
1 2
n rms v n
A
V a
∞
=
∑
2 2 2 2 2 30 20 5 2 10 2 2 2 2
= 27.65 Vrms
Section 16.8 The Exponential Form of the Fourier Series
The Exponential form of the Fourier series will be covered in later
courses (EEE 180).
Read this section on your own for background information.
Section 16.9 Amplitude and Phase Spectra
Amplitude and phase spectra of the Fourier series will also be covered
in later courses (EEE 180).
Read this section on your own for background information.
End of Chapter 16
