Fourier Series - Recap
0 ,( ) nn
in tf t F e
0
Given a function ( ), of period ,fundamental frequency 2
= . The complex Fourier series expansion is
f t T
T
0
0
where 1
( )T
nin tF f t dt
Te
0 0
0 0
0 0 01
2
For a real-valued ( ), the sine-cosine Fourier series is
1 1( )cos
( ) cos( ) sin( ),
( ) ,
( )s
whe
in(
re
) .
T T
n
n
n
n nf t a a n t
a f t n t dt b f t n
f t
t dtT
n
T
b t
Interpretation of Fourier series: expansion in an orthonormal basis.
Analogy: orthonormal basis in 3-dimensional space:
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1 2 3Vectors , , are orthonormal when they:
are mutually orthogonal: for
have unit length: 1 for 1,2,3.
m n
n
m n
n
e e e
e e
e
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0 if This can be expressed as , ,
1 if where , denotes the
(or dot product) of vecto
i
rs
nner
, .
Recall, , = cos( )
product
m n
m n
m ne e
u v
u v
u v u v
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v
Expanding a vector in an orthonormal basis.
1 1 2 2 3 3
1 2 3
,
where the scalar coefficients
, , can be expressed as
, for 1,2,3.nn
v v v
v v v
v n
v e e e
v e
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v
3v
2v
1v
1 1 2 2 3 3
2 2 2 21 2 3
Additionally, we have the relationship
, ,
and in particular the vector length satisfies
, + .
u v u v u v
v v v
u v
v v v
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Analogy between vector and Fourier expansions
3
1,
,
Vectors:
n n
nn
n
v
v
v
v e
e
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0
The analogy can be made precise by defining an inner product 1
between periodic functions: ( ), ( ) ( ) ( )T
T f t g t f t g t dtT
0
0
0 ,
1( )
( )
Fourier Series:
T
n
n
in t
n
in t
F f t dtT
f t F
e
e
0
,
Then, the basis functions ( ) play the role of th
( .
e ,
an ( )d we have ( )) ( ) ,
n n
nn n nn
in t
F f tf t F t
t
t
e e
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0 PREVIOUS
LECTURE
0
0
0 0
0( )
For a complete analogy, we check the orthonormality
of the basis functions ( ) .
1( ), ( )
0 if 1
1 if
nT
T
in t
im t in tm n
i m n t
t
t t dtT
m ndt
m nT
e
e e
e
2
0 0
So the functions ( ) are mutually orthogonal in this abstract
sense, and of unit "size". What is this notion of size?
1 1( ), ( ) ( ) ( ) ( )
This is called the root-mean-square (RMS
n
T T
t
f f t f t f t f t dt f t dtT T
) value of the periodic
function ( ).f t
Example: “sawtooth” function, of period T = 2.
11 32 2
0 1 1
RMS value of ( ) is
1 1 1( )
2 6 3
T
f t
tf f t dt t dt
T
f
t0
1
1 2 3-1-2
2f
0
1
1 2 3-1-2
2f
t
2
0
1( ) Interpretation for the RMS value
T
f t dtT
f It is one of many possible “norms” (i.e. notions of size) of a function. Why this choice?
• Mathematically, it has nice properties, like vector length.• Physical interpretation based on power:
Example: f(t)= I(t), current going through a unit resistor R=1. 2
2 2
RMS0
Instantaneously, the dissipated power is ( ) . If ( ) is periodic,1
the mean power dissipated over one period is ( ) .T
I t I t
I t dt IT
RMS
Equals the power burned by the constant (DC) current ( ) . I t I
0
2 00 0RMS
0 0
Special case (familiar from AC circuits): ( ) cos( ). Here
1 1 sin(2 )cos ( )
2 4 2
TT
I t I t
t t II I t dt I
T T
One more ingredient in the analogy
2 2 2 21 2 3
3
1
,
,
+
Vectors:
n
n
n
n
n
v
v
v
v
v
v e
v
v e
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������������� � 0 0
0
0
( )
,
1( ),
( )
Fourier Series:
T
nin t in t
nn
in t
F f t dt f tT
e
f F
e
t e
2 2 2
RMS0
1( )
T
nn
f f t dt FT
PARSEVAL’S RELATION
2
0 0 0
2
0
0
0
(1 1 1
( ) )
1
Proof:
( ) ( )
( )
n
T T
nn
n
T
T
n
in t
in t
n n
F
f t dt dt dtT T T
dt F
f t F
FT
f t f t
f t
e
e
2 2
0
1( )PARSEVAL'S RELATION:
T
nn
f t dt FT
2
0 0
2
0
2 20
0
1
1
2Also, for a real-valued ( ) cos( ) sin( ),
we have the Parseval re1
( )lation 2T
n n
nn n
n
f t a a n t b n t
f t dt a a bT
2 2 2
0 0
0
0
0
,Note that if we isolate from the expansion
the harmonic term ,its mean square value is equal to
1 1
So we get the following interpretation fo
( )
nT T
n n n
nn
in t
in t
in t
dt dt FT T
f t F
F
F F
e
e
e
r Parseval: the mean square value (i.e. the mean power) of ( ) is equal to the sum of the mean powers of its harmonics.
f t
Example: Square Wave:
01
01
odd
2( ) sin( )
4sin( )
nn
nn
f t b n t
n tn
2
22
0
22
1 1 1odd odd
1 2 8 ( ) 2 2
1T
nn n n
n n
f t dtT n
bn
2
21
odd
1 1Therefore, we conclude that 1
8 9 25
Not an obvious sum!
1
nn
n
2Since ( ) 1, the left-hand side is easily found be to be 1.f t
f
tT 2T02T
1
-1
Application of Fourier series: power conversion.
AC
Rectifier
DC
0 0( ) sin( )x t V t ( )y t
• A rectifier is a circuit that converts the power to DC (used by electronic equipment such as computers, audio, ...). Ideally, y(t) should be a perfectly flat, constant DC voltage. • In practice, one gets an approximation to DC, with some remaining oscillations (“AC component”).
+
0
0 00
1
The signal ( ) is a pure sinusoid (AC power) as in the voltage you find in the wall outlet. Its Fourier expansion is immediate:
( ) , so = , and 0 for 12 2 n
i t i t
x t
x t V X X ni i
Ve e
A simple rectifier: ( ) ( )y t x t
Nonlinear, time invariant and memoryless system.Can be approximately implemented by a diode circuit.
0 0
0 0
( ) sin( )
Period is ;2
2
y
y
y t V tT
T
0 0( ) sin( ). Period s
i .x t V t
T
Not quite flat, but let’s see how much of y is DC.
0( )Represent ( )by a Fourier Series yn
n
in tYty t y e
0 000
0 0
0 0
1( ) sin( )
for 0 we have 0 sin( ) 0.2
y yn
y y
y y
y yT Tin t in tV
Y y t dt t dtT TT
T t T t t
e e
0
0
00
0
0
0 0
0 0(1 2 (1
22
0
22
0
2
0
) 2 )
2Therefore sin( )
2
2
n
T
in t
Ti t i t
in t
T
i n t i n t
VY t dt
T
Vdt
T i
Vdt
Ti
e
e e e
e e
0 0
0
0
0 0
0 0
0 0
(1 2 (1
2 2
22
0
(1 2 ) (1 2 )
) )
(1 2 ) (1 2 )
n
T T
T
i n t i n t
i n t i n t
i n i n
VY dt
Ti
V
Ti
e e
e e
02
0
1 1
(1 2 ) (1 2 )
(1 2 ) (1 2 )
1 1i n i n
n n
V
Ti
e e
0 02(1 2 ) (1 2 ) (1 4 )
2 1 1 1 2 1 2
2 n n n
V V n n
02(1 4 )
2n
n
VY
0
2
00
DCAC COMPONENTSCOMPONENT
0 0
0 0
2
(1 4 )
2( ) y yn
n n
in t in tV
n
VYt Yy e e
0
0
2
0
00
COMP
1
DC AC COMPONENTS
1
2
4
(1 4 )
Re , 0 ( ) cos( )
2cos( )
Alternatively, in sine-cosine form we get
n n n n n y
y
n
n
a
V
n
Y Y b t a n t
Vn t
a y
0 0
0
So we obtain a desired DC component, plus harmonics of the fundamental frequency 2 .In the power supply, 2 60 . The harmonics will have frequencies 2 120 , 2 240 ,...
y
HzHz Hz
2 22 20 0
0 0
1The total power is ( ) sin ( )
2
y yT T
y y
V Vy t dt t dt
T T
22 0
0 2
4The DC power is 81% of the total power.
The AC power is the remaining 19%. Not too bad, not too good as a DC source.
VY
02V
0
DCAC
0
0
2( ) yn
n
in tVYty e
From the graph, the ACpart looks significant.
2
0
2 2
0
DC POWER AC POWERTOTAL
POWER
0
1( )
T
nn
y t dtT
Y Y
Power analysis by Parseval: