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

������������� �

������������� � 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: