Frequency Domain Representation

1. Chapter 3
Fourier Series and Fourier Transform

2. The Fourier Series
Joseph Fourier
1768 to 1830
Fourier studied the mathematical theory of heat
conduction. He established the partial differential
equation governing heat diffusion and solved it by
using infinite series of trigonometric functions.

3. The Fourier Series
Fourier proposed in 1807
A periodic waveform f(t) could be broken down into an infinite
series of simple sinusoids which, when added together, would
construct the exact form of the original waveform.
Consider the periodic function
f (t)  f (t  nT) ; n  1,2, 3,
T = Period, the smallest value

4. The Fourier Series
The expression for a Fourier Series is
N N
f (t)  a0  an cosnω0t bn sinnω0t
n1 n1
C0  a0 and Cn are the Complex Coefficients
Fourier Series = a finite sum of harmonically related sinusoids
N
f (t)  C0  Cn cos(nω0t θn)
n1
a0, an , and bn are real and are called
Fourier Trigonometric Coefficients
Or, alternative form
and 0
ω 
2π
T

5. The Fourier Series
Definition
A Fourier Series
N  
is an accurate representation of a
periodic signal and consists of the sum of sinusoids at
the fundamental and harmonic frequencies.
The waveform f(t) depends on the amplitude and phase
of every harmonic components, and we can generate any
non-sinusoidal waveform by an appropriate combination
of sinusoidal functions.
http://archives.math.utk.edu/topics/fourierAnalysis.html

6. The Fourier Series
To be described by the Fourier Series the waveform f(t)
must satisfy the following mathematical properties:
1. The integral
f(t) has a finite number of discontinuities within
the period T.
3. f(t) has a finite number of maxima and minima
within the period T.
f (t) dt   for any t0.
0
t0 T
t

7. The Fourier Series
The Fourier Trigonometric Coefficients can be obtained
from
0
t0 T
0
t0 T
t0 T
0
0
0
0
f (t)dt
f (t)cosnω t dt
f (t)sinnω t dt n > 0
t
n t
n
t
T
a 
1
T
a 
2
T
b 
2

average value over one period
n > 0

8. 0
t0 T
0
T / 2 T / 4
T / 2 T /4
f (t)dt
T
1
T
a 
1
1
T
1
2
f (t)dt  1dt 
t
  
0
2
a 
1
Example 1 determine Fourier Series and plot for N = 7
average or DC value

9. Example 1(cont.)
An even function exhibits symmetry around the vertical axis
at t = 0 so that f(t) = f(-t).
0
t0 T
0
T /4
0
f (t)sinnω t dt
1 sinn
T /4
ω t dt 0
n t
T
 b 
2

2
T 
T /4
0
0
0
1 cosn
2
sinnωt |
T /4
T /4
n ω t dt
T /4
T
a 
2
Tωn


Determine only an

10. Example 15.3-1(cont.)
2
n
a 
1 
sinπn   sin πn
 πn

 2   
    
an  0 when n  2, 4, 6, 
and 2(1)q
when n  1, 3, 5, 
an 
π
n
where q 
(n 1)
cos
nω0t
n1,odd
1
2
2(1)q
πn
f (t) 
N
2
 
1 3 5 7
π 3π 5π
a 
2
,a 
2
,a 
2
,a 
2

11. Symmetry of the Function
Four types
1. Even-function symmetry
2. Odd-function symmetry
3. Half-wave symmetry
4. Quarter-wave symmetry
Even function
f (t)  f (t) All bn = 0
T /2
0
0
f (t)cosn
n ω t dt
T
a 
4


12. Symmetry of the Function
Odd function
f (t)   f (t)
0
All an = 0
T /2
0
f (t)sinn
n ω t dt
T
b 
4

Half-wave symmetry
f (t)   f (t 
T
)
2
an and bn = 0 for even values of n and a0 = 0

13. Symmetry of the Function
Quarter-wave symmetry
Odd & Quarter-wave
All an = 0 and bn = 0 for even values of n and a0 = 0
T /4
0
0
f (t)sinn ; for odd n
n ω t dt
T
b 
8


14. Symmetry of the Function
For Even & Quarter-wave
All bn = 0 and an = 0 for even values of n and a0 = 0
T /4
0
0
f (t)cosn ; for odd n
n ω t dt
T
a 
8

Table 15.4-1 gives a summary of Fourier coefficients
and symmetry.

15. Example 15.4-1 determine Fourier Series and N = ?
2
m
f  4 and T 
π
s
0
2 T
T 
π
 ω 
2π
 4 rad/s
To obtain the most advantages form of symmetry,
we choose t1 = 0 s  Odd & Quarter-wave
All an = 0 and bn = 0 for even values of n and a0 = 0
0
T /4
0
0
f (t)sinn ; for odd n
n ω t dt
T
b 
8


16. Example 15.4-1(cont.)
f (t)  ; 0  t  T / 4
T /4 T
t 
4 fm
t
fm
π
4
 f (t) 
32
t
2
; 0  t  T / 4
0
0
π
T /4
0
tsinn
32
sin
nπ
π2
n2
; for odd n
2
n ω t dt
T
n ω nω
0
512 sin nω t t cosnω t 
T /4

π
2
π
b 
8  32 
 
 

0
 0
2 2 
 0



17. Example 15.4-1(cont.)
The Fourier Series is
0
2
2
sin sinn
1
n
π
ω t ; for odd n
N
f (t)  3.24
n1 n
32
π2
The first 4 terms (upto and including N = 7)
9 25 49
f (t)  3.24(sin 4t 
1
sin12t 
1
sin 20t 
1
sin 28t)
Next harmonic is for N = 9 which has magnitude
3.24/81 = 0.04 < 2 % of b1 ( = 3.24)
Therefore the first 4 terms (including N = 7) is enough for
the desired approximation

18. Exponential Form of the Fourier Series
n1
C0 is the average (or DC) value of f(t) and
N
f (t)  C0  Cn cos(nω0t θn)
(an  jbn )
2
n n n
 C θ
C 
2
a2
 b2
 n n
n n
C  C
and
where
n
tan1  bn 
; if a  0
n
n
n
; if a  0
 a 
 n 
θ  



 b 
180  tan1

 a 

 n 


19. Exponential Form of the Fourier Series
or
an  2Cn cosθn and bn  2Cn sinθn
Writing cos(nω0t θn )in exponential form using
Euler’s identity with N  
0 0
0
f (t)  C  jnω t
n n
jnω t
 
n n
n0
 
C e  C e
where the complex coefficients are defined as
0
0
t0 T
n
n n
 jnω t jθ
f (t)e dt  C e
t
T
C 
1
; the coefficients for negative n are the
complex conjugates of the coefficients for positive n
n n
AndC  C*

20. Example 15.5-1 determine complex Fourier Series
Even function
The average value of f(t) is zero  C0  0
t0 T
f (t)e jnω0t
dt
0
n
t
T
C 
1

We select and define
0
2
t  
T
0
jnω  m

21.  
f (t)e jnω0t
dt
T /2
T /2
T / 4 T /4 T /2
T / 2 T / 4 T / 4
1
T
1
T
|
mt T /2
T / 2 T / 4 T /4
mt T /4
emt
|T /4
e | e
0
2
n
T
1
T
mt mt mt
Ae dt  Ae dt  Ae dt
A
C 
1
mT

A
2ejnπ /2
 2e jnπ /2
 e jnπ
 ejnπ

jnω0T
A
 4sin
2πn 
n
π



 

  
; for even n
sinn
 2sin(nπ)  
2A
; for odd n
π
2
 A
sin x
2
where x 
nπ


nπ
x

Example 15.5-1(cont.)

22. Example 15.5-1(cont.)
Since f(t) is even function, all Cn are real and = 0 for n even
1 1
C 
Asinπ / 2

2A
 C
π /2
π
For n = 1
For n = 2
2
π 2
C  A
sinπ  0  C
For n = 3
C 
Asin(3π / 2)

2A
 C
3 3
3π / 2
3
π

23. Example 15.5-1(cont.)
The complex Fourier Series is
0
n1
nodd
f (t) 
3π π π
3π
π
3
π

2A
e j3ω0t

2A
e jω0t

2A
ejω0t

2A
ej3ω0t
cosnω t
q
n
π

4A
cosω t 
4A
cos3ω t 
π 0
3π 0

4A
 (1)



2A
ejω0t
 e jω0t

2A
ej3ω0t
 e j3ω0t

where q 
n 1
2
For real f(t)  Cn  Cn
ejx
ejx
 e jx
 2cos x
 e jx
 2 jsin x

24. Example 15.5-2 determine complex Fourier Series
Even function
Use jnω0  m
T /4
1emt
dt
T /4
emt
|T /4
T /4
1
n
T
C 
1
mT

1
emT / 4
 emT / 4

mT



25. 
(n1) / 2
1
 jn2π

0 ; n even, n  0
(1) ; n odd
n e jnπ / 2
 e jnπ / 2

C 


Example 15.5-2(cont.)
To find C0
0
0
T /4
T /4
f (t)dt
T
1
T
C 
1
1
2
1dt 
T




26. The Fourier Spectrum
The complex Fourier coefficients
Cn  Cn θn
Cn
ω
Amplitude spectrum
ω
θn
Phase spectrum

27. The Fourier Spectrum
The Fourier Spectrum is a graphical display of the
amplitude and phase of the complex Fourier coeff.
at the fundamental and harmonic frequencies.
Example
A periodic sequence of pulses each of width δ

28. The Fourier Spectrum
The Fourier coefficients are
T /2
Ae jnω0t
dt
T /2
n
T
C 
1

For n  0
 
0 0
jnω δ / 2
jnω0T
0
0
2A
sin
2
n
δ /2
e jnω0t
dt
δ /2
T
A
e jnω δ / 2
 e
nω δ
nω T

 
  
 

C 
A

29. 0
0
sin(n
Aδ ω δ / 2)
T (nω δ / 2)

Aδ sin x
T x
where x  nω0δ /2
n
C 
The Fourier Spectrum
0
T T
δ /2

For n  0 C 
1 δ /2
Adt 
Aδ

30. The Fourier Spectrum
L'Hôpital's rule
x
sin x
1 for x  0
sin(nπ
)
nπ
 0 ; n  1, 2, 3,
ω  5ω0
ω  10ω0
ω  0

31. 0
n N
jnω t
f (t)   S (t)
 C e
The Truncated Fourier Series
A practical calculation of the Fourier series requires that
we truncate the series to a finite number of terms.
N
n N
The error for N terms is
ε(t)  f (t)  SN (t)
We use the mean-square error (MSE) defined as
2
0
1
T
ε (t)dt
T
MSE  
MSE is minimum when Cn = Fourier series’ coefficients

32. The Truncated Fourier Series
overshoot 10%

33. Circuits and Fourier Series
Example 15.3-1
An RC circuit excited by a periodic voltage vS(t).
It is often desired to determine the response of a circuit
excited by a periodic signal vS(t).
Example 15.8-1 An RC Circuit vO(t) = ?
R  1 , C  2 F, T  π sec

34. Circuits and Fourier Series
An equivalent circuit.
Each voltage source
is a term of the
Fourier series of vs(f).

35. Each
input
is a
Sinusoid.
Using
phasors
to find
steady-state
responses
to the
sinusoids.
Example 15.8-1
(cont.)

36. cos
nω0t
1
2
2(1)q
πn
N

n1,odd
s
v (t)  
Example 15.8-1 (cont.)
where q 
(n 1)
2
The first 4 terms of vS(t) is
1
2 π 3π
vs0
(t) vs1
(t) vs3
(t)
5π
vs5
(t)

2
cos2t 
2
cos6t 
2
cos10t
s
v (t) 
0
ω  2 rad/s
The steady state response vO(t) can then be found using
superposition. vo (t)  vo0(t)  vo1(t)  vo3(t)  vo5(t)

37. Example 15.8-1 (cont.)
The impedance of the capacitor is
0
1
; for n  0,1, 3, 5,
C
jnω C
Z 
jnω0C
1
jnω0C
Vsn
1 jnω0CR
1
4
on sn
V ; for n  0,1, 3,5,
R 
V 

We can find

38. Example 15.8-1 (cont.)
The steady-state response can be written as
von (t)  Von cos(nω0t  Von
 tan1
4n)
0
116n2
Vsn
sn
cos(nω t  V

s0

1
2

2
for n 1, 3,5
sn
n
π
Vsn  0 for n  0,1, 3,5
V
V
In this example we have

39. Example 15.8-1 (cont.)
2
o0
v (t) 
1
2
on
v (t)  cos(n2t  tan1
4n) ; for n 1,3,5
nπ 116n2
vo1(t)  0.154 cos(2t  76)
vo3 (t)  0.018cos(6t  85)
vo5(t)  0.006cos(10t  87)
o
 v (t) 
1
 0.154cos(2t  76)  0.018cos(6t  85)
2
 0.006cos(10t  87)

40. Summary
The Fourier Series
Symmetry of the Function
Exponential Form of the Fourier Series
The Fourier Spectrum
The Truncated Fourier Series
Circuits and Fourier Series