FOURIER SERIES Jean-Baptiste Fourier (France, 1768 - 1830) proved that almost any period

function can be represented as the sum of sinusoids with integrally related frequencies.

The Fourier series is one example of an orthogonal set of basis functions, as a very important example for engineers.

Trigonometric form of Fourier Series Let us map the functions 1, and by the following :

The purpose of this nothing deeper than to map the conventional Fourier series onto the notation we have derived for orthogonal functions

T

mt2sin T

mt2cos

0/2sin

01

0/2cos

mTmt

m

mTmt

tm

The Fourier series is a special case of the more general theory of orthogonal functions.

Now calculate the value of m from

ie

The value of m for is simply the power in a sinewave (cosine wave)

The value of 0 is the power in a DC signal of unit amplitude.

Now we can derive immediately the Euler formula from equation

by substituting in the values of and m from the above equations then

nm

nmdttt

t

t knntt 01 *

12

2

1

T

T

T

m

t

tmdt

T

t

tmdt

T

t

tmdt

T

Tmt

Tmt

0

0

0

0

0

0

05.01

011

05.01

/2sin

1

/2cos

2

2

2

0m

dttt

ttx

T n

T

nna

*0

0

1

tm

In the Fourier series, instead of using the term a-m as the coefficient of the cosine terms in the Fourier expansion we usually use the term am with bm reserved for the sine terms.

The important point to realize here is that the Fourier series expansion is only a special case of an expansion in terms of orthogonal functions

There are many other function (e.g. Walsh function), so using the Fourier series as an example, try and understand the more general orthogonal function approach

When we write a periodic function using a Fourier series expansion in terms of a DC term and sine and cosine terms the problem which remains is to determine the coefficients a0 , am and bm

T

T

T

m

t

tmdtTmttx

T

t

tmdttx

T

t

tmdtTmttx

T

a0

0

0

0

0

0

0/2sin2

01

0/2cos2

10

2sin

2cos

mmm T

mt

T

mttx baa

Solving a problem using Fourier series Consider a sawtooth wave which rises from -2V to 2V in a second. It passes

through a linear time invariant communication channel which does not pass frequencies greater than 5.5 Hz.

What is the power lost in the channel ? Assume the output and input impedance are the same. (Use sine or cosine Fourier series).

Remove all

frequencies > 5.5 Hz

u(t) y(t)

t

2V

t=0

1s

The first step is set up the problem mathematically.

The time origin has not been specified in the problem.

Since the system is time invariant it doesn't matter when t = 0 is located since it is will not change the form of the output.

Choose time t = 0 at the center of the rise of the sawtooth because it makes the function which we now call u(t), into an odd function.

Since the system is specified in terms of it frequency response, i.e. what it will do if a sinwave of a given frequency is input, it makes a lot of sense to express as a sum of either sines and cosines or complex exponentials since as we know what happens to these functions.

If it's a sinewave or cosine wave and has a frequency less than 5.5 Hz it is transmitted, otherwise it is eliminated.

The situation with complex exponential is a little trickier, if its in the range [-5.5,5.5] Hz then will be transmitted otherwise will be eliminated.

It would do no good to find the response to each sinewave individually, because we could not then add up these individual response to form the total output, because that would require superposition to hold.

Let us calculate the Fourier series for a sawtooth wave or arbitrary period and

amplitude. Now with the choice of t = 0, we can write the input as mathematically

within the period as

We don't need to worry that outside the range [-T/2 < t < T/2] the above formula is

incorrect since all the calculation are done within the range [-T/2 < t < T/2].

As we strict to the range given the mathematical description is identical to the

sawtooth and all will be well.

We want the input to written in the form

22

2 Tt

T

T

Attu

10

2sin

2cos

nnn T

ntb

T

ntaatu

Go back to Euler formula for Fourier series which have derived earlier from the general orthogonality conditions

Now for the DC value of the sawtooth

For an the coefficients of the cosine terms

dtT

nt

T

At

Tdt

T

nttu

Ta

T

T

T

T

n

2/

2/

2/

2/

2cos

222cos

2

02

2

211

2/

2/

2

20

2/

2/

2/

2/

0

T

T

T

T

T

T

t

T

Aa

dtT

At

Tdttu

Ta

dtT

nttu

Tb

dtT

nttu

Ta

dttuT

a

T

T

n

T

T

n

T

T

2/

2/

2/

2/

2/

2/

0

2sin

2

2cos

2

1

The next step is use integration by parts, i.e.

Therefore

gdt

dfgf

dt

dgf

dt

dgfg

dt

dfgf

dt

d

.

.

dtT

nt

n

T

T

nt

n

tT

T

Aa

dtT

nt

n

T

T

nt

n

tT

T

Aa

T

T

T

T

n

T

T

T

T

n

2/

2/

2/

2/

2/

2/

2/

2/

2sin

2

2sin

2

4

2sin

2

2sin

2

4

0

coscos2

sin2

4

2cos

2sin

2sin

22

4

2sin

2

2sin

2

4

222

2

2/

2/

2

2

2/

2/

2/

2/2

n

n

T

T

n

T

T

T

T

n

a

nnn

Tn

n

T

T

Aa

T

nt

n

Tn

Tn

T

n

T

T

Aa

dtT

nt

n

T

T

nt

n

tT

T

Aa

This is a lot of work for 0, when it is fairly obvious that the integral of the product of an odd function (sawtooth) and an even function (the cosine term) is always zero when we integrate from [-v,v] whatever value of v, as shown below

The procedure for calculating for is almost identical, the final answer is

Now we know a0 , an and bn , we can write down the Fourier series representation for u(t) after substituting T = 1 and A = 2

The series has all the sine terms present, i.e.bm is never zero and there are no cosine terms.

3

6sin

2

4sin2sin

4 ttttu

n

nbn

cos4

X =

How can be check that our calculations for u(t) is correct ? We can calculate the power in the signal by

and also by Parseval's theorem when applied to sine and cosine functions

Where the 0.5 came from ?

Remember that is m which has the basis function was equal to sin(t) and the power in the sinewave, m = 0.5 .

Having calculated the Fourier series and having checked it using the Parseval's theorem it only remains to calculate the power in the first 5 harmonics, i.e. those with a frequency less than 5.5 Hz

3

4

42/

2/

22/

2/

2

P

dttdttuPT

T

T

T

3

45.0

6

4

5.04

1

3

1

2

11

4

22

2222

b

b

P

P

Thus the power transmitted by the channel is

The power loss is therefore 1.3333 - 1.1863 = 0.1470, and the power gain in dB is thus -0.51 dB.

The final answer is that the channel attenuates the signal by 0.51 dB.

1863.1

2

1

5

1

4

1

3

1

2

11

422222

t

t

P

P