01b-Fourier Series Revision

Revision of Fourier Series

Revision of Fourier Series 2

Odd and Even FunctionsAn even function is a function that is symmetric about the value t = 0, i.e. f(t) = f(-t)

An odd function is a function that is anti- symmetric about the value t = 0, i.e. -f(t) = f(-t)

even/symmetric function

odd/anti-symmetric function


A Fourier series is a series of sine and cosine terms whose frequencies are integer multiples of a common fundamental angular frequency ω,

where and T is the fundamental period.

The numbers an and bn are called the Fourier coefficients of the series.

Fourier Series

10 n n2

n 1 n 1

f(t) a a cos(n t) b sin(n t)∞ ∞

= =

= + ω + ω∑ ∑

2Tπ

ω =


Fourier SeriesThe Fourier coefficients are given by

The constant term, , is the average or mean value of the function.

T2

0 T0

a f(t)dt= ∫

( )T

2n T

0

a f(t)cos n t dt= ω∫ ( )T

2n T

0

b f(t)sin n t dt= ω∫

102 a


Main FormulaeThe following standard integrals are useful when evaluating Fourier components.

[ ]dg dfdt dtf(t) dt f(t)g(t) g(t)dt= −∫ ∫

1sin t dt cos tωω = − ω∫ 1cos t dt sin tωω = ω∫

( )21t sin t dt sin t t cos tω

ω = ω − ω ω∫ ( )21t cos t dt cos t t sin tω

ω = ω + ω ω∫

Integration by Parts:


Main Formulae

( )32 2 21t sin t dt 2 t sin t 2cos t t cos t

ωω = ω ω + ω − ω ω∫

( )32 2 21t cos t dt t sin t 2 t cos t 2sin t

ωω = ω ω + ω ω − ω∫

( )2 2at at at1

ae sin t dt ae sin t e cos t

ω +ω = ω − ω ω∫

( )2 2at at at1

ae cos t dt ae cos t e sin t

ω +ω = ω + ω ω∫


Main Formulae

( ) ( )( )1 1 12cos( t)cos( t)dt sin t sin tω−φ ω+φ

⎡ ⎤ ⎡ ⎤ω φ = ω− φ + ω+ φ⎣ ⎦ ⎣ ⎦∫

( ) ( )( )1 1 1 12 2cos( t)sin( t)dt cos t cos tω−φ ω+φ

⎡ ⎤ ⎡ ⎤ω φ = ω−φ − ω+ φ⎣ ⎦ ⎣ ⎦∫

( ) ( )( )1 1 12sin( t)sin( t)dt sin t sin tω−φ ω+φ

⎡ ⎤ ⎡ ⎤ω φ = ω−φ − ω+φ⎣ ⎦ ⎣ ⎦∫


Before doing any examples it is wise to note some time- saving results

Odd and Even Functions

If a function f(t) is periodic and symmetric/even then it will have a Fourier series containing only cosine terms, i.e. bn = 0 if f(t) = f(-t).

If a function f(t) is periodic and antisymmetric/odd then it will have a Fourier series containing only sine terms, i.e. an = 0 if f(t) = -f(-t).


Discontinuous FunctionsWhen integrating discontinuous functions, split the integral into sections;

e.g. if

then

1

2

f (t), 0 t af(t)

f (t), a t T≤ ≤⎧

= ⎨ < <⎩

T a T

1 20 0 a

f(t)dt f (t)dt f (t)dt= +∫ ∫ ∫


Determine the Fourier series of the periodic, rectangular pulse function defined by

Example 1

A, 0 t 1f(t) , f(t 2) f(t)

0, 1 t 2≤ ≤⎧

= + =⎨ < <⎩

0 0.5 1 1.5 2 2.5 3 3.5 4


Example 1The function f(t) has period T = 2, so

The Fourier series will have the form

Since the function is neither symmetric nor anti-symmetric we have to work out both an and bn coefficients;

2Tπ

ω = = π

10 n n2

n 1 n 1

a a cos( nt) b sin( nt)∞ ∞

= =

+ π + π∑ ∑


Example 1[ ]

T 2 1 212

0 T 00 0 0 1

a f(t)dt f(t)dt A dt 0dt At A= = = + = =∫ ∫ ∫ ∫

( )T 2

2n T

0 01 2

0 11A A A

n n n0

a f(t)cos n t dt f(t)cos(n t)dt

A cos(n t)dt 0dt

sin(n t) sin(n ) sin(0) 0π π π

= ω = π

= π +

⎡ ⎤ ⎡ ⎤ ⎡ ⎤= π = π − =⎣ ⎦ ⎣ ⎦ ⎣ ⎦

∫ ∫

∫ ∫

So a0 = A

So an = 0 for n > 0


Example 1( )

T 22

n T0 0

1 21A

n 00 1

nA A An n n 2A

n

b f(t)sin n t dt f(t)sin(n t)dt

A sin(n t)dt 0dt cos(n t)

0, n evencos(n ) cos(0) ( ( 1) 1)

, n odd

π

π π ππ

= ω = π

⎡ ⎤= π + = − π⎣ ⎦

⎧⎪⎡ ⎤ ⎡ ⎤= − π − − = − − + = ⎨⎣ ⎦ ⎣ ⎦ ⎪⎩

∫ ∫

∫ ∫

So 2A 2A 2A 2A1 2 3 4 5 6 73 5 7b ,b 0,b ,b 0,b ,b 0,b ,π π π π= = = = = = =

and the Fourier Series is:

( )1 2 2 2 22 3 5 7f(t) A sin( t) sin(3 t) sin(5 t) sin(7 t)π π π π= + π + π + π + π +L


Example 1

0 0.5 1 1.5 2 2.5 3 3.5 4

1

0 0.5 1 1.5 2 2.5 3 3.5 4

1

Fundamental only

1st - 3rd harmonics

0 0.5 1 1.5 2 2.5 3 3.5 4

1

0 0.5 1 1.5 2 2.5 3 3.5 4

1

1st - 7th harmonics

1st - 15th harmonics


Example 2Determine the Fourier series of the symmetric periodic function defined by

t 0 tf(t) , f(t 2 ) f(t)

2 t t 2≤ ≤ π⎧

= + π =⎨ π − π < < π⎩

0 2 4 6 8 10 12

1

2

3


Example 2The function f(t) has period T = 2π, so

and

The coefficients an and bn can be deduced from the general formula; but since the function is symmetric all sine terms vanish, i.e. all bn = 0.

2 2 1T 2π π

ω = = =π

10 n n2

n 1 n 1

a a cos(nt) b sin(nt)∞ ∞

= =

+ +∑ ∑


Example 2Then

so a0 = π

( ) ( ) ( ) ( )( )

2 21 1

00 0

22 21 1 12 20

2 2 2 2 21 1 12 2

a f(t)dt t dt (2 t)dt

t 2 t t

0 4 2 2

π π π

π ππ

π π

π π

π

⎛ ⎞= = + π −⎜ ⎟⎜ ⎟

⎝ ⎠⎛ ⎞⎡ ⎤ ⎡ ⎤= + π −⎜ ⎟⎣ ⎦ ⎣ ⎦⎝ ⎠

= π − + π − π − π − π = π

∫ ∫ ∫


2 2

2 2

2 21 1

n0 0

221 1 1 1 1

n n nn n0

1 1 1 1n nn n

21n

a f(t)cos(nt)dt t cos(nt)dt (2 t)cos(nt)dt

cos(nt) t sin(nt) sin(nt) cos(nt) t sin(nt)

cos(n ) sin(n ) cos(0) 0sin(0)

si

π π π

π ππ

π ππ

π π

ππ

⎛ ⎞= = + π −⎜ ⎟⎜ ⎟

⎝ ⎠⎛ ⎞⎡ ⎤ ⎡ ⎤= + + − −⎜ ⎟⎣ ⎦ ⎣ ⎦⎝ ⎠

⎡ ⎤ ⎡ ⎤π + π π − +⎣ ⎦ ⎣ ⎦

= +

∫ ∫ ∫

( ) ( )

2

2

2 2 2 2 2

1 1nn

2 1 1n nn

n 2 n1 1 1 1 1 1 2n n n n n

n(2 n) cos(2 n) 2 sin(2 n)

sin(n ) cos(n ) sin(n )

( 1) ( 1) ( 1) 1

π

π π

⎛ ⎞⎜ ⎟⎜ ⎟⎡ ⎤π − π − π π⎜ ⎟⎣ ⎦⎜ ⎟

⎡ ⎤⎜ ⎟− π − π − π π⎣ ⎦⎝ ⎠

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤= − − + − − − − = − −⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

Example 2Also

24

n n na 0 if n is even, and a if n is odd−

π= =i.e.


0 2 4 6 8 10 12

1

2

3

0 2 4 6 8 10 12

1

2

3

Fundamental only

1st - 3rd harmonics

0 2 4 6 8 10 12

1

2

3

0 2 4 6 8 10 12

2

1st - 7th harmonics


Example 2

4 4 4 42 9 25 49f(t) cos(t) cos(3t) cos(5t) cos(7t)π

π π π π= − − − − −L

Hence


Determine the Fourier series of the anti-symmetric periodic function defined by

Example 3

1 12 2f(t) t t f(t 1) f(t)= − ≤ ≤ + =

1.5 1 0.5 0 0.5 1 1.5

0.5

0.5


Example 3The function f(t) has period T = 1, so

Hence

The coefficients and can be deduced from the general formula but since the function is anti-symmetric the constant part and all the cosine terms vanish, i.e. a0 = 0 and all an = 0.

2 2 2T 1π π

ω = = = π

10 n n2

n 1 n 1

f(t) a a cos(2 nt) b sin(2 nt)∞ ∞

= =

= + π + π∑ ∑


Example 3Since the function is defined over the period the Fourier integrals can be done over that interval;

1 12 2t− ≤ ≤

( )( )

2

2 2

2

12 1

21 1n 2 n(2 n) 1

212

1 1 1 14 n 4 n(2 n) (2 n)

n2(2 n)

b 2 t sin(2 nt)dt 2 sin(2 nt) t cos(2 nt)

2 sin(n ) cos(n ) sin( n ) cos( n )

2 ( 1)

ππ −−

π ππ π

−π

⎛ ⎞⎡ ⎤= π = π − π⎜ ⎟⎢ ⎥⎜ ⎟⎣ ⎦⎝ ⎠

⎡ ⎤ ⎡ ⎤= π − π − − π + − π⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

= −

∫

2n 11

n ( n)b ( 1) +

π= −So

We could evaluate similar integrals for a0 and an , but the fact that f(t) is odd guarantees that a0 = an = 0.


Example 3The Fourier series for this function is therefore:

( )21 1 1 1

4 9 16f(t) sin(2 t) sin(4 t) sin(6 t) sin(8 t)π

= π − π + π − π +L

1.5 1 0.5 0 0.5 1 1.5

0.5

0.5

1.5 1 0.5 0 0.5 1 1.5

0.5

0.5

Fundamental only

1st and 2nd harmonics

1.5 1 0.5 0 0.5 1 1.5

0.5

0.5

1.5 1 0.5 0 0.5 1 1.5

0.5

0.5

1st - 4th harmonics



Half-Range SeriesA non-periodic function f(t) may be associated with three different Fourier series. Each of these will match the function over the region , but will also repeat the same function profile periodically.

The first is the normal Full-Range Fourier Series already discussed.The second is formed by making an even extension of the function – the Half-Range Cosine Series.The third is formed by making an odd extension of the function – the Half-Range Sine Series.


For f(t) defined over 0 < t < L, the half-range cosine series has period 2L and is given by

where

and

The Half-range Cosine series of a function defined over a region is an even or symmetric extension with period 2L

Half-Range Cosine Series

( )( )0 ntn L

n 1

aC(t) a cos2

∞π

=

= +∑C(t) f(t) over 0 t L

C( t) C(t), (symmetry)C(t 2L) C(t), (periodicity)

= < <⎧⎪ − =⎨⎪ + =⎩

( )L L

nt2 20 nL L L

0 0

a f(t)dt and a f(t)cos dtπ= =∫ ∫


Half-Range Cosine Series

t

f(t)

L0

Half-range cosine series of f(t)

4L

C(t)

tL 2L 3L0–L


For f(t) defined over 0 < t < L, the half-range sine series has period 2L and is given by

where

and

The Half-range Sine series of a function defined over a region is an odd or antisymmetric extension with period 2L

Half-Range Sine Series

( )( )ntn L

n 1

S(t) b sin∞

π

=

= ∑S(t) f(t) over 0 t L

S( t) S(t), (symmetry)S(t 2L) S(t), (periodicity)

= < <⎧⎪ − = −⎨⎪ + =⎩

( )L

nt2n L L

0

b f(t)sin dtπ= ∫


Half-Range Sine Series

t

f(t)

L0

Half-range sine series of f(t)

4L

S(t)

tL 2L 3L0–L


Example 4Obtain the half-range cosine and half-range sine series for

defined over 0 < t < 1.

The half-range cosine series will have period T = 2, (since L = 1) and has the form

2f(t) 1 t= −

( )( )10 n2

n 1

C(t) a a cos n t∞

=

= + π∑


Example 4Using the general formula

and

[ ]1

12 32 1 1 40 1 3 3 30

0

a (1 t )dt 2 t t 2 1 0⎡ ⎤ ⎡ ⎤= − = − = − − =⎣ ⎦⎣ ⎦∫

( )

( ) ( ) ( ) ( ) ( ) ( ) ( )( )( ) ( ) ( ) ( ) ( ) ( ) ( )( ) [ ]

( )2 2

122

n 10

12 321 1 1 1n n n n

0

2 31 1 1 1n n n n

n4 4( n) ( n)

a (1 t )cos n t dt

2 sin n t t sin n t 2 tcos n t 2 sin n t

2 sin n sin n 2 cos n 2 sin n 0

cos n ( 1)

π π π π

π π π π

− −π π

= − π

⎡ ⎤= π − π + π − π⎢ ⎥⎣ ⎦

⎡ ⎤= π − π + π − π −⎢ ⎥⎣ ⎦= π = −

∫


Example 4So

( ) ( ) ( )( )2

1 14 92 4

3 116

cos t cos 2 t cos 3 tC(t)

cos 4 t ... π

⎛ ⎞π − π + π⎜ ⎟= +⎜ ⎟− π +⎝ ⎠


Example 4The half-range sine series will also have period T = 2, and takes the form

( )( )nn 1

S(t) b sin n t∞

=

= π∑


Example 4( )

( ) ( ) ( ) ( ) ( ) ( ) ( )( )( ) ( ) ( ) ( ) ( ) ( ) ( )( )

122

n 10

12 321 1 1 1n n n n

0

2 31 1 1 1n n n n

1n

b (1 t )sin n t dt

2 cos n t t cos n t 2 t sin n t 2 cos n t

2 cos n cos n 2 sin n 2 cos n

2

π π π π

π π π π

π

= − π

⎡ ⎤= − π − − π + π + π⎢ ⎥⎣ ⎦

⎡ ⎤= − π − − π + π + π⎢ ⎥⎣ ⎦

− − − −

∫

( )( )( )( )3

31n

4 2n( n)

0 0 2

1 cos n

π

ππ

⎡ ⎤+ +⎢ ⎥⎣ ⎦

= − π +

So 3

2n

82nn ( n)

, n evenb , n odd

π

π π

⎧⎪= ⎨ +⎪⎩


Example 4

( )( )( )

3

3

3

82 1

82 13 427

825 125

S(t) sin( t) sin(2 t)

sin(3 t) sin(4 t)

sin(5 t) ...

π ππ

π ππ

π π

= + π + π

+ + π + π

+ + π +


Example 4

2 1 0 1 2 3 4

1

2 1 0 1 2 3 4

Cosine series, (up to 32nd harmonics)

Sine series, (up to 32nd harmonics)

1 0 1 2

1

f t( ) 1 t2

0 t< 1<


Example 5Obtain the half-range sine series for a function

defined on the interval 0 < t < π.

The half-range sine series will have period T = 2π, and is given by

212 2

1 0 tf(t)

t

π

π

⎧ < <⎪= ⎨ < < π⎪⎩

( )( )nn 1

S(t) b sin nt∞

=

= ∑


Example 5

( )

2n

0

22 2 1

20

2

2 1 2 12n 2n0 2

n n2 1 2 1 2 1 2 1n 2 n 2n 2n 2

n1 2 1n 2 n n

3n

b f(t) sin(nt) dt

sin(nt) dt sin(nt) dt

cos(nt) cos(nt)

cos( ) cos(n ) cos( )

cos( ) cos(n )

if

π

π

ππ

π ππ

π π

ππ π

π ππ π π π

ππ π π

π

=

= +

⎡ ⎤ ⎡ ⎤= − + −⎣ ⎦ ⎣ ⎦

⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤= − − − + − π − −⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦= − + − π

=

∫

∫ ∫

2n

n is odd n 2,6,10,...0 n 4,8,12,...

π

⎧⎪ =⎨⎪ =⎩

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01b-Fourier Series Revision