Derivation of Fourier Coefficients

Derivation of Fourier Coefficients

Derivation of Fourier Coefficients
Swapnil Sunil Jain
December 28, 2006

Derivation of Fourier Coefficients

As you know, any periodic function

f

⁢

(

t

)

can be written as a

Fourier series

like the following

f

⁢

(

t

)

=

c

0

+

∑

n

=

1

∞

a

n

⁢

cos

⁡

(

ω

n

⁢

t

)

+

b

n

⁢

sin

⁡

(

ω

n

⁢

t

)

(1)

where

ω

n

=

n

⁢

ω

0

and

ω

0

=

2

⁢

π

T

In the process to find an explicit

expression

for the coefficients

c

0

,

a

n

,

b

n

in terms of

f

⁢

(

t

)

, we write (1) in a slightly different way as the following

f

⁢

(

t

)

=

⁢

c

0

+

a

1

⁢

cos

⁡

(

ω

1

⁢

t

)

+

a

2

⁢

cos

⁡

(

ω

2

⁢

t

)

+

…

+

a

k

⁢

cos

⁡

(

ω

k

⁢

t

)

+

…

+

b

1

⁢

sin

⁡

(

ω

1

⁢

t

)

+

b

2

⁢

sin

⁡

(

ω

2

⁢

t

)

+

…

+

b

k

⁢

sin

⁡

(

ω

k

⁢

t

)

+

…

(2)

where

k

is a

positive integer

.

In order to derive the coefficient

c

0

, we take the integral of both sides of (2) over one period.

∫

τ

f

⁢

(

t

)

⁢

𝑑

t

=

⁢

∫

τ

c

0

⁢

𝑑

t

+

∫

τ

a

1

⁢

cos

⁡

(

ω

1

⁢

t

)

⁢

𝑑

t

+

∫

τ

a

2

⁢

cos

⁡

(

ω

2

⁢

t

)

⁢

𝑑

t

+

…

+

∫

τ

a

k

⁢

cos

⁡

(

ω

k

⁢

t

)

⁢

𝑑

t

+

…

+

∫

τ

b

1

⁢

sin

⁡

(

ω

1

⁢

t

)

⁢

𝑑

t

+

∫

τ

b

2

⁢

sin

⁡

(

ω

2

⁢

t

)

⁢

𝑑

t

+

…

+

∫

τ

b

k

⁢

sin

⁡

(

ω

k

⁢

t

)

⁢

𝑑

t

+

…

where

τ

=

[

t

0

,

t

0

+

T

]

. After evaluating the above equation, all the integrals on the right side with a sine or a cosine term drop out (since the integral of a sine or cosine over one period is zero) and we get

∫

τ

f

⁢

(

t

)

⁢

𝑑

t

=

⁢

c

0

⁢

∫

τ

𝑑

t

⇒

∫

τ

f

⁢

(

t

)

⁢

𝑑

t

=

⁢

c

0

⁢

(

T

)

⇒

c

0

=

⁢

1

T

⁢

∫

τ

f

⁢

(

t

)

⁢

𝑑

t

=

1

T

⁢

∫

t

0

t

0

+

T

f

⁢

(

t

)

⁢

𝑑

t

Now, in order to find

a

k

, we multiply both sides of (2) by

cos

⁡

(

ω

k

⁢

t

)

and we arrive at

f

⁢

(

t

)

⁢

cos

⁡

(

ω

k

⁢

t

)

=

⁢

c

0

⁢

cos

⁡

(

ω

k

⁢

t

)

+

a

1

⁢

cos

⁡

(

ω

1

⁢

t

)

⁢

cos

⁡

(

ω

k

⁢

t

)

+

a

2

⁢

cos

⁡

(

ω

2

⁢

t

)

⁢

cos

⁡

(

ω

k

⁢

t

)

+

…

+

a

k

⁢

cos

2

⁡

(

ω

k

⁢

t

)

+

…

+

b

1

⁢

sin

⁡

(

ω

1

⁢

t

)

⁢

cos

⁡

(

ω

k

⁢

t

)

+

b

2

⁢

sin

⁡

(

ω

2

⁢

t

)

⁢

cos

⁡

(

ω

k

⁢

t

)

+

…

+

b

k

⁢

sin

⁡

(

ω

k

⁢

t

)

⁢

cos

⁡

(

ω

k

⁢

t

)

+

…

Then we take the integral of both sides of the above equation over one period and we get

∫

τ

f

⁢

(

t

)

⁢

cos

⁡

(

ω

k

⁢

t

)

⁢

𝑑

t

=

⁢

∫

τ

c

0

⁢

cos

⁡

(

ω

k

⁢

t

)

⁢

𝑑

t

+

∫

τ

a

1

⁢

cos

⁡

(

ω

1

⁢

t

)

⁢

cos

⁡

(

ω

k

⁢

t

)

⁢

𝑑

t

+

∫

τ

a

2

⁢

cos

⁡

(

ω

2

⁢

t

)

⁢

cos

⁡

(

ω

k

⁢

t

)

⁢

𝑑

t

+

…

+

∫

τ

a

k

⁢

cos

2

⁡

(

ω

k

⁢

t

)

⁢

𝑑

t

+

…

+

∫

τ

b

1

⁢

sin

⁡

(

ω

1

⁢

t

)

⁢

cos

⁡

(

ω

k

⁢

t

)

⁢

𝑑

t

+

∫

τ

b

2

⁢

sin

⁡

(

ω

2

⁢

t

)

⁢

cos

⁡

(

ω

k

⁢

t

)

⁢

𝑑

t

+

…

+

∫

τ

b

k

⁢

sin

⁡

(

ω

k

⁢

t

)

⁢

cos

⁡

(

ω

k

⁢

t

)

⁢

𝑑

t

+

…

By using

orthogonality

relationships or by literally evaluating the above integrals, we get the following

∫

τ

f

⁢

(

t

)

⁢

cos

⁡

(

ω

k

⁢

t

)

⁢

𝑑

t

=

⁢

∫

τ

a

k

⁢

cos

2

⁡

(

ω

k

⁢

t

)

⁢

𝑑

t

⇒

∫

τ

f

⁢

(

t

)

⁢

cos

⁡

(

ω

k

⁢

t

)

⁢

𝑑

t

=

⁢

a

k

⁢

(

T

2

)

⇒

a

k

=

⁢

2

T

⁢

∫

τ

f

⁢

(

t

)

⁢

cos

⁡

(

ω

k

⁢

t

)

⁢

𝑑

t

=

2

T

⁢

∫

t

0

t

0

+

T

f

⁢

(

t

)

⁢

cos

⁡

(

ω

k

⁢

t

)

⁢

𝑑

t

Now, the process of finding

b

k

is

similar

. We multiply both sides of (2) by

sin

⁡

(

ω

k

⁢

t

)

and we get

f

⁢

(

t

)

⁢

sin

⁡

(

ω

k

⁢

t

)

=

⁢

c

0

⁢

cos

⁡

(

ω

k

⁢

t

)

+

a

1

⁢

cos

⁡

(

ω

1

⁢

t

)

⁢

sin

⁡

(

ω

k

⁢

t

)

+

a

2

⁢

cos

⁡

(

ω

2

⁢

t

)

⁢

sin

⁡

(

ω

k

⁢

t

)

+

…

+

a

k

⁢

cos

⁡

(

ω

k

⁢

t

)

⁢

sin

⁡

(

ω

k

⁢

t

)

+

…

+

b

1

⁢

sin

⁡

(

ω

1

⁢

t

)

⁢

sin

⁡

(

ω

k

⁢

t

)

+

b

2

⁢

sin

⁡

(

ω

2

⁢

t

)

⁢

sin

⁡

(

ω

k

⁢

t

)

+

…

+

b

k

⁢

sin

2

⁡

(

ω

k

⁢

t

)

+

…

Then we take the integral of both sides of the above equation over one period and we arrive at

∫

τ

f

⁢

(

t

)

⁢

sin

⁡

(

ω

k

⁢

t

)

⁢

𝑑

t

=

⁢

∫

τ

c

0

⁢

cos

⁡

(

ω

k

⁢

t

)

⁢

𝑑

t

+

∫

τ

a

1

⁢

cos

⁡

(

ω

1

⁢

t

)

⁢

cos

⁡

(

ω

k

⁢

t

)

⁢

𝑑

t

+

∫

τ

a

2

⁢

cos

⁡

(

ω

2

⁢

t

)

⁢

cos

⁡

(

ω

k

⁢

t

)

⁢

𝑑

t

+

…

+

∫

τ

a

k

⁢

cos

⁡

(

ω

k

⁢

t

)

⁢

sin

⁡

(

ω

k

⁢

t

)

⁢

𝑑

t

+

…

+

∫

τ

b

1

⁢

sin

⁡

(

ω

1

⁢

t

)

⁢

cos

⁡

(

ω

k

⁢

t

)

⁢

𝑑

t

+

∫

τ

b

2

⁢

sin

⁡

(

ω

2

⁢

t

)

⁢

cos

⁡

(

ω

k

⁢

t

)

⁢

𝑑

t

+

…

+

∫

τ

b

k

⁢

sin

2

⁡

(

ω

k

⁢

t

)

⁢

𝑑

t

+

…

By using orthogonality relationships or by literally evaluating the above integrals, we get the following

∫

τ

f

⁢

(

t

)

⁢

sin

⁡

(

ω

k

⁢

t

)

⁢

𝑑

t

=

⁢

∫

τ

b

k

⁢

sin

2

⁡

(

ω

k

⁢

t

)

⁢

𝑑

t

⇒

∫

τ

f

⁢

(

t

)

⁢

sin

⁡

(

ω

k

⁢

t

)

⁢

𝑑

t

=

⁢

b

k

⁢

(

T

2

)

⇒

b

k

=

⁢

2

T

⁢

∫

τ

f

⁢

(

t

)

⁢

sin

⁡

(

ω

k

⁢

t

)

⁢

𝑑

t

=

2

T

⁢

∫

t

0

t

0

+

T

f

⁢

(

t

)

⁢

sin

⁡

(

ω

k

⁢

t

)

⁢

𝑑

t

Title

Derivation of Fourier Coefficients

Canonical

name

DerivationOfFourierCoefficients1

Date of creation

2013-03-11 19:30:41

Last modified on

2013-03-11 19:30:41

Owner

swapnizzle (13346)

Last modified by

(0)

Numerical id

1

Author

swapnizzle (0)

Entry type

Definition

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