Chapter 101 Fourier Series for Periodic Functions of Period 2Π

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CHAPTER 101 FOURIER SERIES FOR PERIODIC FUNCTIONS OF PERIOD 2π

EXERCISE 362 Page 1066

1. Determine the Fourier series for the periodic function:

 −2, when −〈〈π x 0 f(x) =   +2, when 0 〈〈x π

which is periodic outside this range of period 2π.

The periodic function is shown in the diagram below.

∞ The Fourier series is given by: f(x) = a0 ++∑( anncos nx b sin nx) (1) n=1

11ππ0 10 π a0 = fx( )d x = −+2dx 2d x =[ − 2 x] +[ 2 x] 22ππ∫−−ππ{ ∫∫0 } 2 π{ −π 0 }

1 ={[(0) − (2ππ )] +[ (2 ) −= (0)]} 0 2π

11ππ0 an = f()cos x nx d x =−+ 2cosnx d x 2cos nx d x ππ∫−−ππ{ ∫∫0 }

0 π 12  2  = −+sinnx sin nx = 0 π    nn −π 0

11ππ0 bn = f( x )sin nx d x =−+ 2sinnx d x 2sin nx d x ππ∫−−ππ{ ∫∫0 } =

0 π 12  2  2 cos nx +−cos nx ={[cos0 − cosn ( −ππ )] −[ cos n − cos0]} ππ   n −π nn 0

2 When n is even, bn ={[11 −−−] [ 11]} = 0 π n 2 28 When n is odd, bn ={[1 −− 1] −[ − 11 −]} =( 4) = πn ππnn 8 8 8 Hence, b1 = , b3 = , b5 = and so on π 3π 5π Substituting onto equation (1) gives:

88 8 f(x) = 0 + 0 + sinxxx+++ sin 3 sin 5 ... ππ35 π

8 11 i.e. f(x) = sinxxx+++ sin 3 sin 5 ... π 35

π π 2. For the Fourier series in Problem 1, deduce a series for at the point where x = 4 2

π 8πππ 8385 When x = , f(x) = 2, hence 2 = sin+++ sin sin ... 2 ππ23 2 5 π 2 8 111 i.e. 2 = 1−+−+... π 357 2π 111 and =−+−+1 ... 8 357 π 111 i.e. =−+−+1 ... 4 357

3. For the waveform shown below determine (a) the Fourier series for the function and (b) the sum

of the Fourier series at the points of discontinuity.

∞ (a) The Fourier series is given by: f(x) = a0 ++∑( anncos nx b sin nx) (1) n=1

1π 1−πππ/2 /2 1π /2 1ππ 11 a0 = fx( )d x= 0dx+ 1d x + 0d x = [ x] = −− ={π} = 2π∫−π 2 π{ ∫ −−π ∫∫ ππ/2 /2 } 2ππ{ −π /2} 22 2 2 π 2

11π −π/2 ππ/2 an = f( x )cos nx d x =(0)cosnx d x ++ 1cos nx d x (0)cos nx d x ππ∫−π { ∫ −−π ∫∫ ππ/2 2 }

π /2 π /2 11 1 ππ   = cosnx d x = sin nx =sinn −− sin  n  ∫−π /2 ππ nn −π /2  22   12 When n = 1, a1 =(11 −−) = ππ 1 When n = 2, a2=(00 −===) 0 aa46 and so on for all even values of n 2π 12 When n = 3, a1 =( −−11) =− 33ππ 12 When n = 5, a5 =(11 −−) = 55ππ 2 2 Hence, a7 = − , a9 = and so on 7π 9π

11π −π/2 ππ/2 bn = f( x )sin nx d x =(0)sinnx d x ++ 1sin nx d x (0)sin nx d x ππ∫−π { ∫ −−π ∫∫ ππ/2 /2 }

π /2 π /2 11  1ππ   =sinnx d x =− cos nx  =−cosn −− cos n  ∫−π /2     ππnn −π 22  

Whatever value of n is chosen, bn = 0

Substituting onto equation (1) gives:

12 2 2 2 f(x) = +cox −cos3 x + cos5 x − cos7 x ++ ... 0 2357ππ π π

12 1 1 i.e. f(x) = +cosxxx −++ cos3 cos5 ... 2π  35

(b) The sum of the Fourier series at the points of discontinuity (i.e. at π/2, π, 3π/2, ...) is: 10+ 1 = 2 2

4. For Problem 3, draw graphs of the first three partial sums of the Fourier series and show that as

the series is added together term by term the result approximates more and more closely to the

function it represents.

2 2 2 1 In the diagram below graphs of cos x, cos 3x and cos 5x and f(x) = are shown π 3π 5π 2 2 2 2 A graph of cos x – cos 3x + cos 5x is also shown π 3π 5π 1 2 2 2 Finally, a graph of f(x) = + cos x – cos 3x + cos 5x is sketched. If further harmonics 2 π 3π 5π were added then the waveform would approach that shown in Problem 3

5. Find the term representing the third harmonic for the periodic function of period 2π given by:

 0, when−〈〈π x 0 f(x) =  1, when 0 〈〈x π

The periodic function is shown in the diagram below.

π 1 ππ1 1 sin nx an = f()cos x nx d x= 1cos nx d x = = 0 ∫∫−π 0  π ππn 0

11ππ bn = f( x )sin nx d x= sin nx d x ππ∫∫−π 0

π 1 cosnx 1 1 = − =−−=−{cosnnππ cos0} { 1cos } ππ π nn0 n

The third harmonic is when n = 3,

1 12 i.e. b3 ={1 − cos3π} ={ 1 −− 1} = 3π 33 ππ

∞ Since the Fourier series is given by: f(x) = a0 ++∑( anncos nx b sin nx) , n=1

2 the 3rd harmonic term is: sin 3x 3π

6. Determine the Fourier series for the periodic function of period 2π defined by:

  0, when−π 〈〈t 0   π f(t) =  1, when 0 〈〈t  2  π −1, when 〈〈t π  2

The periodic function is shown in the diagram below

11π 0 ππ/2 1ππ/2 a0 = ft( )d t= 0d t + 1d t + − 1d t =[ t] +−[ t] 22ππ∫−−ππ{ ∫∫∫0 π/2 } 2π{ 0 π /2}

1 ππ    = −(00)  +( −π ) −−   = 22π  2  

11π ππ/2 an = f()cosd t nt t = cosdnt t+− cos nx d x ππ∫−ππ{ ∫∫0 /2 }

ππ/2 1 sinnt sin nt 1  nππ n  =  − =sin −− 0 sinnπ − sin  ππ    n0 nnπ /2  22  

When n is even, an = 0

1ππ   12 When n = 1, a1 =sin − 0  − sinπ − sin  ={[ 1 −−− 0] [ 0 1]} = π22   ππ

13ππ  3 1 2 When n = 3, a3 =sin − 0  − sin 3π − sin  ={[ −−−−−=−1 0] [ 0 1]} 3π 2  23 ππ3

15ππ  5 1 2 When n = 5, a5 =sin − 0  − sin 5π − sin  ={[1 −−− 0] [ 0 1]} = 5π 2  25 ππ5 2 2 It follows that a7 = − , a9 = and so on 7π 9π

11π ππ/2 bn = f( t )sin nt d t = sinnt d t +− sin nt d t ππ∫−ππ{ ∫∫0 /2 } =

ππ/2 1 cosnt   cos nt  1  nππ  n  − + =−cos −+ cos 0 cosnπ − cos  ππ       nnn 0  π /2  22  

11nnππ    n π = 1− cos  +− cosnnππ cos  =−+1 2cos cos  ππnn22    2 1 When n is odd, bn ={101 −−} = 0 π n

1 42 When n is even, b2 ={1 −−+ 2( 1) 1} = = 22π ππ 1 b4 ={1 − 2(1) += 1} 0 4π 1 42 b6 ={1 −−+ 2( 1) 1} = = 6π 63ππ 2 Similarly, b8 = 0 , b10 = , and so on 5π

∞ Substituting into f(t) = a0 ++∑( anncos nt b sin nt) n=1

22 2 2 gives: f(x) = 0 + costttt−+−+ cos3 cos5 cos7 ... ππ357 π π

22 2 + sin 2tt++ sin 6 sin10 t + ... πππ35

2 11 11 i.e. f(x) = costtt− cos3 + cos5 −+ ... sin 2 tt + sin 6 + sin10 t + ... π 35 35

7. Show that the Fourier series for the periodic function of period 2π defined by:

 θ, when−〈〈 πθ 0 f(θ) =  sin,when0θ〈〈 θπ

2 1 cos2θθθ cos4 cos6 is given by: f(θ) = −−−−... π 2 (3) (3)(5) (5)(7)

The periodic function is shown in the diagram below

11ππ0 11π af0 = ()dθ θ= 0d θ + sind θθ =[ − cos θ] ={( −cosπ) −−( cos 0) } 22ππ∫−−ππ{ ∫∫0 } 22 ππ{ 0 } 1 11 = (1− cosπ ) =( 1 −− 1) = 22π ππ

1 0 π ann = 0cosθθ d+ sin θ cos n θθ d π {∫∫−π 0 }

11ππ1 = sin(θθ+nn) + sin ( θθ −) =sinθ( 1 ++ nn) sin θ( 1 −) d θ from 6, page 716 ππ∫∫0022{ } of the textbook

π 1cosθθ( 1+−nn) (1 )1cos ππ( 1 +− nn) cos( 1 ) 11 = − −cos = − − −− − 2ππ 1+n 1 − n 2 1+n 1 − n 11 +− nn  0 

11111 When n is odd, an =−−++ =0 21111π +−+−nnnn 1 cos3ππ cos()11−  11 1  1 4 2 When n = 2, a2 = − − ++ = −+−=11  − =− 2π 3−− 131233π  233 ππ 1 cos5ππ cos(3)1− 1 1 1111  When n = 4, a4 = − − ++ = −+−  2ππ 5−− 3 5 3 2 5353 

1 3535−+−  1 4  2 =  =−=−  2π (3)(5)  2 ππ (3)(5)  (3)(5) 1 cos7ππ cos(5)1− 1 1 1111  When n = 6, a6 = − − ++ = −+−  2ππ 7−− 5 7 5 2 7575 

1 5757−+−  1 4  2 =  =−=−  2π (5)(7)  2 ππ (5)(7)  (5)(7)

1 0 π bnn = 0sinθθ d+ sin θ sin n θθ d π {∫∫−π 0 }

π 1π 1 1 sinθθ (1+−nn ) sin (1 )  = −cos(θθ +−nn) cos( θθ −) =− − = 0 from 9, page ∫0  ππ2 21+−nn 1 0 716 of the textbook

∞ Substituting into f(θ) = a0 ++∑( anncos nbθθ sin n) n=1

1 22 2 gives: f(θ) = −−cos 2θθθ cos 4 − cos6 − ... + 0 π 3πππ (3)(5) (5)(7)

2 1 cos2θθθ cos4 cos6 i.e. f(θ) = −−−−... π 2 (3) (3)(5) (5)(7)