Fourier Series of Continuous Function
Fourier Series of Continuous Function
Fourier Series of Continuous Function
Dr.R.Subasri
Professor, EIE
Kongu Engineering College
𝐴 cos(𝑛𝜃+𝜃) cos(𝑛𝜃−𝜃) 𝜋
== [− − ]
2𝜋 (𝑛+1) (𝑛−1) 0
𝐴
= [1 + −1𝑛 ]
𝜋(1−𝑛2 )
2𝐴
𝑎𝑛 = ; n is even ; for n is odd 𝑎𝑛 = 0
𝜋(1−𝑛2 )
2 𝜋
𝑏𝑛 = ∫ 𝐴 sin 𝜃 sin 𝑛𝜃 𝑑𝜃
2𝜋 0
𝜋
Integral Identity : ∫0 𝑠𝑖𝑛 𝑚𝜃 sin 𝑛𝜃 𝑑𝜃 = 0
2 𝜋 1 𝜋 𝐴
𝑏1 = ∫ 𝐴 sin 𝜃 sin 𝜃 𝑑𝜃 = ∫ 𝐴𝑠𝑖𝑛2 𝜃 𝑑𝜃 =
2𝜋 0 𝜋 0 2
𝐴 2𝐴 𝐴
f(t) = + ∑∞
𝑛=2,4,6… cos 𝑛𝜔0 𝑡 + sin 𝜔0 𝑡
𝜋 𝜋(1−𝑛2 ) 2
Full wave rectified sinusoid:
𝜋
f(t) = A sin t for 0 < t < ; 𝜔 = ;
𝑇
f(t) = 𝑎0 + ∑∞ ∞
𝑛=0 𝑎𝑛 cos 𝑛𝜔0 𝑡 + ∑𝑛=0 𝑏𝑛 sin 𝑛𝜔0 𝑡
1 𝜋 𝜃 2𝐴
𝑎0 = ∫ 𝐴 sin 2 𝑑𝜃 =
𝜋 0 𝜋
𝜋
2
𝑎𝑛 = ∫ 𝐴 sin 𝜃 cos 2𝑛𝜃 𝑑𝜃
𝜋 0
𝐴 𝜋
= ∫0 [sin(2𝑛𝜃 + 𝜃) + sin(𝜃 − 2𝑛𝜃)]𝑑𝜃
𝜋
𝐴 cos(𝜃+2𝑛𝜃) cos(𝜃−2𝑛𝜃) 𝜋
= = [− − ]
𝜋 (1+2𝑛) (1−2𝑛) 0
2𝐴
= [1 + −1𝑛 ]
𝜋(1−4𝑛2 )
4𝐴
𝑎𝑛 = ;
𝜋(1−4𝑛2 )
2 𝜋
𝑏𝑛 = ∫ 𝐴 sin 𝜃 sin 2𝑛𝜃 𝑑𝜃
2𝜋 0
𝜋
Integral Identity : ∫0 𝑠𝑖𝑛 𝑚𝜃 sin 𝑛𝜃 𝑑𝜃 = 0
2𝐴 4𝐴
f(t) = + ∑∞
𝑛=2,4,6… cos 2𝑛𝜔0 𝑡
𝜋 𝜋(1−4𝑛2 )
(or)
2𝐴 4𝐴
f(t) = + ∑∞
𝑛=1 cos 𝑛𝜔0 𝑡
𝜋 𝜋(1−4𝑛2 )
𝒙(𝒕) = ∑ 𝑪𝒏 𝒆𝒋𝝎𝟎 𝒏𝒕
𝒏=−∞
𝟏 𝑻
𝑪𝒏 = ∫ 𝒙(𝒕)𝒆−𝒋𝝎𝟎 𝒏𝒕 𝒅𝒕
𝑻 𝟎
Half wave rectified sinewave:
𝟏 𝑻
𝑪𝒏 = ∫ 𝑨 𝒔𝒊𝒏𝝎𝟎 𝒕 𝒆−𝒋𝝎𝟎 𝒏𝒕 𝒅𝒕
𝑻 𝟎
𝐴 𝜋
𝐶𝑛 = ∫ 𝑠𝑖𝑛 𝜃 𝑒 −𝑗𝑛𝜃 𝑑𝜃
2𝜋 0
𝜋
𝐴 𝑒 −𝑗𝑛𝜃 (−𝑗𝑛 𝑠𝑖𝑛𝜃 − 𝑐𝑜𝑠 𝜃)
= [ ]
2𝜋 (−𝑗𝑛)2 + 1 0
𝐴
= 2𝜋(1−𝑛2) [1 + (−1)𝑛 ]
𝐴
Cn = for all even values of n
𝜋(1−𝑛2 )
𝐴 𝑇
C1 = ∫ 𝑠𝑖𝑛 𝜃 𝑒 −𝑗𝜃 𝑑𝜃
2𝜋 0
𝐴 𝜋 𝑒 𝑗𝜃 −𝑒 −𝑗𝜃
= ∫ [ ] 𝑒 −𝑗𝜃 𝑑𝜃
2𝜋 0 2𝑗
𝐴
=
𝑗4
𝐴 𝐴 𝐴
𝒇(𝒕) = ∑∞
𝒏=−∞,≠±𝟏 𝒆𝒋𝝎𝟎 𝒏𝒕 + 𝒆𝒋𝝎𝟎 𝒕 - 𝒆−𝒋𝝎𝟎 𝒕
𝜋(𝑛2 −1) 𝑗4 𝑗4
𝐴
𝒇(𝒕) = ∑∞
𝒏=−∞ 𝒆𝒋𝟐𝝎𝟎 𝒏𝒕
𝜋(1−4𝑛2)
Parseval’s Theorem: Periodical Function
Power in time domain = power in frequency domain
2
1 1
T
T
0
f (t ) dt = a02 +
2
ak2 + bk2
1 𝐴 2 1
𝑎02 + ∑ 𝑎𝑘2 + 𝑏𝑘2 = ( ) + [𝑎22 + 𝑎42 + ⋯ + 𝑏12 ]
2 𝜋 2
𝐴 2 1 2𝐴 2 2𝐴 2 𝐴 2
= ( ) + [( ) + ( ) + ⋯ + (2) ]
𝜋 2 −3𝜋 −15𝜋
𝐴 2 4 4 1 𝐴 2
= ( ) [1 + + ]+ ( )
𝜋 2×9 225 × 2 2 2
𝐴 2 𝐴 2 1.23 1
= ( ) (1.23) + ( ) = 𝐴2 ( 2 + )
𝜋 2 𝜋 8
= 0.249 A2
Power in Both domain is same
1 2𝐴 2 1
𝑎02 + ∑ 𝑎𝑘2 + 𝑏𝑘2 = ( ) + [𝑎22 + 𝑎42 + ⋯ ]
2 𝜋 2
2𝐴 2 1 16𝐴2 16𝐴2
=( ) + [ 2+ + ⋯]
𝜋 2 15𝜋 63𝜋2
𝐴 2 16 16
= ( ) [4 + + + ⋯]
𝜋 15 × 2 63 × 2
= 0.472 A2
Power in Both domain is same.
Parseval’s Theorem: Aperiodical Function
Energy in time domain = Energy in frequency domain
∞ ∞
1 ∞
∫ |𝑓(𝑡)|2 𝑑𝑡 = ∫ |𝑋(𝑓)|2 𝑑𝑓 = ∫ |𝑋(𝜔)|2 𝑑𝜔
−∞ −∞ 2𝜋 −∞
𝟏 𝟏 𝝎 ∞ 𝟏
= 𝒕𝒂𝒏−𝟏 ( ) = (𝒕𝒂𝒏−𝟏 ∞ − 𝒕𝒂𝒏−𝟏 (−∞))
𝟐𝝅 𝜶 𝜶 −∞ 𝟐𝝅𝜶
𝟏 𝟏
= 𝝅=
𝟐𝝅𝜶 𝟐𝜶
For a discrete periodical signal with time period with N samples, the discrete Fourier
series of a discrete signal x(n) is expressed as:
𝑘−1
𝑥(𝑛) = ∑ 𝐶𝑘 𝑒 𝑗2𝜋𝑘𝑛/𝑁
𝑘=0
Where n is 0,1,2,…..N-1
The Fourier coefficients Ck is given as
𝑁−1
1
𝐶𝑘 = ∑ 𝑥(𝑛) 𝑒 −𝑗2𝜋𝑘𝑛/𝑁
𝑁
𝑛=0
Where k is 0,1,2,…..N-1
Frequency Spectrum is Discrete.
3
1 1 1
𝐶0 = ∑ 𝑥(𝑛)𝑒 −0 = [𝑥(0) + 𝑥(1) + 𝑥(2) + 𝑥(3)] =
4 4 4
𝑛=0
3
1
𝐶1 = ∑ 𝑥(𝑛)𝑒 −𝑗2𝜋𝑛/4
4
𝑛=0
1
= [𝑥(0)𝑒 −0 + 𝑥(1)𝑒 −𝑗𝜋/2 + 𝑥(2)𝑒 −𝑗𝜋2/2 + 𝑥(3)𝑒 −𝑗𝜋3/2 ]
4
1 1 𝜋 𝜋 𝑗
= (𝑒 −𝑗𝜋/2 ) = (cos − 𝑗 sin ) = −
4 4 2 2 4
3
1 𝑗4𝜋𝑛 1
𝐶2 = ∑ 𝑥(𝑛)𝑒 − 4 = [𝑥(0)𝑒 −0 + 𝑥(1)𝑒 −𝑗𝜋 + 𝑥(2)𝑒 −𝑗2𝜋 + 𝑥(3)𝑒 −𝑗3𝜋 ]
4 4
𝑛=0
1 −𝑗𝜋 1 1
= (𝑒 ) = (cos 𝜋 − 𝑗 sin 𝜋) = −
4 4 4
3
1 𝑗6𝜋𝑛 1
𝐶3 = ∑ 𝑥(𝑛)𝑒 − 4 = [𝑥(0)𝑒 −0 + 𝑥(1)𝑒 −𝑗3𝜋/2 + 𝑥(2)𝑒 −𝑗3𝜋 + 𝑥(3)𝑒 −𝑗9𝜋/2 ]
4 4
𝑛=0
1 1 𝑗
= (𝑒 −𝑗3𝜋/2 ) = (cos 3𝜋/2 − 𝑗 sin 3𝜋 /2) =
4 4 4
1 𝑗 1 𝑗
Ck={ , − , , }
4 4 4 4
1 1 1 1
|𝐶𝑘 | = { , , , }
4 4 4 4
Ck = {0.-90, 0,90}={0,-/2,0, /2}
Using Twiddle matrix:
Ck={0.6035,-0.25,0.1035,-0.25}
|𝐶𝑘 | = 𝐶𝑘
Ck = {0, 0, 0, 0}
Parseval’s Theorem:
Power in discrete time domain = power in frequency domain
1 𝑁−1
𝑃= ∑𝑁−1 2
𝑛=0 |𝑥(𝑛)| = ∑𝑘=0 |𝐶𝑘 |
2
𝑁
Half wave rectified Sinusoid:
1 1 1
∑𝑁−1 2
𝑛=0 |𝑥(𝑛)| = (0 + 12 + 0 + 0) =
𝑁 4 4
1 2 1 2 1 2 1 2 1
∑𝑁−1 2
𝑘=0 |𝐶𝑘 | = ([ ] + [ ] + [ ] + [ ] ) =
4 4 4 4 4