Electrical Circuits II
(ECE233b)
Fourier Analysis Techniques
(Part 1)
Anestis Dounavis
The University of Western Ontario
Faculty of Engineering Science
Background
We have discussed the following solution methodologies
Phasor transforms
Steady state; Sinusoidal analysis
Laplace transforms
Transient and steady state; Any type of
forcing function
Fourier analysis techniques
Fourier transforms
Steady state; non-sinusoidal but
periodic signals.
The periodic signal is expresses as a summation of sinusoids
with harmonically related frequencies.
Fourier Series
A periodic function satisfies the following relationship
f(t)  f(t  nT0 )
n  1,2,3,......
Can be expressed as

f(t)  a0   Dn cos(n0 t   n )
(phasor notation)
n 1
where
 T0 is the period
 0 is the fundamental frequency (rad/s) 0=2/T0
Examples of periodic signals
sawtooth waveform
f(t)
f(t)
A
square waveform
A
T0
2T0
3T0
t
T0
2T0
3T0
t
Fourier Series
Fourier series can be expressed in the following forms
 Exponential Fourier Series
f(t)  a0 

 cn e
n  
jn ω0t

  cn e
jnω0 t
n  
n 0
 Trigonometric Fourier Series

f(t)  a0   (an cos(n0 t)  bn sin(n0 t))
n 1
where
 n is an integer
 a0, an, bn and cn are the Fourier coefficients
 20, 30, …. , n0 are the harmonic frequencies
 0 is the fundamental frequency
The two forms are similar since
e j  cos   j sin 
Fourier Series
Relationship between exponential and trigonometric form
f(t)  a0 

c e
n  
n 0

n
 a 0   cn e
n 1

jn ω0t

 a 0   cn e
n 1
jnω0 t
c e
 a0   Re(2c ne
*
n
jnω 0t
- jn ω0t
n  1
)
 a0   Re((Dn  n )e jn ω0t )
 a0   Re((an - jbn )e
n 1

  cn e
jnω0t
where
Dn  n  2c n  an  jbn
)
 a0   (an cos(n0 t)  bn sin(n0 t))
n 1
jnω0t
* = complex conjugate
n 1

n 1

jnω0 t

Fourier Series
Approximating a square wave
1
sin 2t   sin 3 * 2t 
3
1
sin 3 * 2t 
3
sin 2t 
Fourier Series
Approximating a square wave
8
1
 (2 n  1) sin ( 2n  1) * 2t 
n 1
Fourier Series
Determining the coefficients for exponential Fourier series
Any physical realizable periodic signal may

be represented over the interval t1<t<t1+T0
f(t) 
cn e jnω0 t
by the exponential Fourier series
n  

Multiplying both sides by e-jk t and integrating over the interval
t1 to t1+T0 yields
0
t 1 T 0
t 1 T0
t1
t1
- jkω0 t
f(t)e
dt 

t 1 T 0
since
e
t1
- j(n - k)ω 0t

 
jn ω0t  - jkω 0t
dt  c kT0
  cn e
e
 n  

0
dt  
T0
for n  k
for n  k
The Fourier coefficients are defined as
1
cn 
T0
t1 T 0
- jn ω0t
f(t)e
dt

t1
Fourier Series
Strategy to determine Fourier Coefficients
1. Identify T, 0 and defining equation of function
2. Determine coefficient by appropriate integration
Example 1
Determine the exponential Fourier series for the periodic voltage
waveform shown
v(t)
V
-T/4
T/4
-T/2
T/2
-V
Fourier Series
Determining the coefficients for trigonometric Fourier series

f(t)  a0   (an cos(n0 t)  bn sin(n0 t))
n 1
Using the result obtained for the exponential Fourier series
1
cn 
T0
t1 T 0
 f(t)e
- jn ω0t
-jn0 t
e
 cos(n0 t )  j sin(n0 t )
and
2c n  an  jbn
dt
t1
yields the following Fourier coefficients for a0, an and bn
2
an 
T0
t 1 T0
1
a0 
T0
 f(t)cos(n ω t)dt
0
t1
t 1 T 0
 f(t)dt
t1
2
bn 
T0
t 1 T0
 f(t)sin(nω t)dt
0
t1
Fourier Series
Trigonometric Fourier series
If a signal exhibits certain symmetrical properties, we can take
advantage of these properties to simplify the calculations of the
Fourier coefficients
There are 3 types of symmetry:
1) Even-function symmetry
2) Odd-function symmetry
3) Half-wave symmetry
Fourier Series
Even-function symmetry: A function is said to be even if
f(t)  f(-t)
Example of even functions
Even functions can only be approximated with even basis
functions eg) cos(not)
1) Therefore can only have “an” coefficients since bn=0
2) Can integrate over half cycle
2
a0 
T0
T0 / 2
 f(t)dt
0
4
an 
T0
T0 / 2
 f(t)cos(nω t)dt
0
0
Fourier Series
Odd-function symmetry: A function is said to be odd if
f(t)  f(-t)
Example of odd functions
Odd functions can only be approximated with odd basis
functions eg) sin(not)
1) Therefore can only have “bn” coefficients since an=0
2) Can integrate over half cycle
4
bn 
T0
T0 / 2
 f(t)sin(nω t)dt
0
0
Fourier Series
Half-Wave symmetry: A function is said to posses half-wave
symmetry if
 T 
f(t)  f  t - 0 
 2
This equation states that each half-cycle is an inverted version of
the adjacent half-cycle; that is, if the waveform from –T0/2 to 0 is
inverted, it is identical to the waveform from 0 to T0/2.
Example of Half-Wave symmetry
 Symmetric around horizontal axis
 Independent of where t=0
Fourier Series
Example of Half-Wave symmetry
1. The average of the function is zero, therefore a0=0
2. an=bn=0 if n is even and
4
an 
T0
4
bn 
T0
T0 / 2
 f(t)cos(nω t)dt
0
0
if n is odd
T0 / 2
 f(t)sin(nω t)dt
0
0
Also if the function is odd then an=0
and if the function is even then bn=0
Example 2
Determine the trigonometric Fourier series for the periodic
voltage waveform shown and compare results with example 1.
v(t)
V
-T/4
T/4
-T/2
T/2
-V
Example 3
Determine the trigonometric Fourier series for the periodic
voltage waveform shown.
V
-T
T
2T
Example 4
Determine the type of symmetry exhibited by the waveform.
2
1
-1
-1
-2
Time-Shifting
Time shifting of a periodic waveform f(t) is defined as
f(t) 

c e
n
time shifting f(t)
jnω0 t
n  
f(t - t 0 ) 

jn ω0 (t - t0 )
c
e
 n
n 
f(t - t 0 ) 

 (cne
jnω0t 0
)e
n 
where
jn ω0t


 kn e
jnω0 t
n  
k n  cn e jnω0 t0
Therefore, time shift in the time domain corresponds to a phase
shift in the frequency domain
To compute the phase shift in degrees
Phase shift (degrees) = 0td = (3600)td/T0
td=time delay
Example 5
If the wave form in example 1 is shifted by a quarter period as
shown below, determine the Fourier coefficients
v(t)
V
-T/4
T/4
-T/2
T/2
-V
Frequency Spectrum
Frequency Spectrum: The frequency spectrum of f(t)
expressed as a Fourier series consists of an amplitude
spectrum and a phase spectrum
Amplitude Spectrum: Plot of the amplitude of the harmonics
versus frequency.
Phase Spectrum: Plot of the phase of the harmonics versus
frequency.
Since the frequency components are discrete, the spectra is
called line spectra
The plots are based on the following equations
f(t)  a0 

 cn e
n  
jn ω0t

 a0   Re((Dn  n )e
jn ω0t
)
n 1
n 0
where
Dn  n  2c n  an  jbn
Example 6
The Fourier series for the waveform shown is given by

40
 20

v(t)    sin( n0 t )  2 2 cos( n0 t ) 
n
n 1  n

n odd
Plot the first four terms of the amplitude and phase spectra for
this signal
10
-T/2
T/2
-10
Steady-State Network Response
If a periodic signal is applied to a network the steady state
response of the circuit can be found as follows:
1. Represent the periodic forcing function by a Fourier series
If the input forcing function for a network is a voltage, the
input can be expressed as
v(t)  v 0  v1 (t)  v 2 (t)  ....
2. Use phasor analysis in the
frequency domain to determine the
network response due to each
source
3. The network response due to each
source is transformed to the time
domain
v 2 (t)
+
-
v1 (t)
+
-
v0
+
-
Network
4. Add time domain solutions due to each source (Principle of
Superposition) to obtain Fourier series for the total steadystate network response
Example 7
Determine the steady-state voltage vo(t) in for the circuit shown
if the input voltage v(t) is given by the expression

40
 20

v(t)    sin( n0 t )  2 2 cos( n0 t ) 
n
n 1  n

n odd
2
v(t)
+
-
1
1F
1
+
vo(t)
-
Average Power
Recall that average power is defined as
T
1
P   v(t)i(t)dt
T 0
If v(t) and i(t) are periodic functions, the signals can be

expressed as:
v(t)  Vdc  Vn cos(n0 t  θv n )
n 1

i(t)  Idc   I n cos(n0 t  θi n )
n 1
Note to calculate the average power involves the product of two
infinite series
However, only products at the same frequency survive integration
over one period (due to orthogonal properties of basis functions)
Average Power
Note:
T
0
0 cos(n 0 t)cos(m 0 t)dt  T/2
mn
m n
Therefore we can show that the average power can be
expressed as:

VnI n
P  Vdc Idc  
cos(θv n  θi n )
n 1 2
Example 8
For the network shown, the input voltage is
v(t)  42  16cos(377t  30 0 )  12cos(754t - 20 0 )
Compute the current i(t) and determine the average power
absorbed by the network
i(t) 100F
v(t)
+
-
20mH
16