CHAPTER 104 FOURIER SERIES OVER ANY RANGE
CHAPTER 104 FOURIER SERIES OVER ANY RANGE EXERCISE 366 Page 1083
1. The voltage from a square wave generator is of the form: 0, − 10 〈 t 〈 0 v(t) = 5, 0 〈 t 〈 10 and is periodic of period 20. Show that the Fourier series for the function is given by: v(t) =
5 10 π t 1 3π t 1 5π t + sin + sin + sin + ... 2 π 10 3 10 5 10
The square wave is shown below
∞
The Fourier series is of the form:
v(t) = a 0 +
n =1
{∫
2π nt 2π nt cos + bn sin L L
∑ a
n
a0 =
1 1 L /2 1 10 v(t ) d t = v(t ) d t = ∫ ∫ 20 −10 L − L /2 20
an =
2 L /2 2 10 2π nt 2π nt = v t t ( ) cos d v(t ) cos dt ∫ ∫ L /2 − − 10 L 20 20 L
=
0 −10
10
}
0 d t + ∫ 5d t = 0
1 5 10 [5t ] 0 = 20 2
10 1 0 π nt π nt ∫ −10 0 cos d t + ∫ 0 5cos dt 10 10 10
10
π nt 5sin 1 10 = 5 [sin π n − sin 0] = 0 for n = 1, 2, 3, … = πn 10 π n 10 0
bn =
2 L /2 2 10 2π nt 2π nt v(t ) sin v(t ) sin dt dt = ∫ ∫ L /2 − − 10 L 20 20 L
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© 2014, John Bird
=
10 1 0 π nt π nt ∫ −10 0sin d t + ∫ 0 5sin dt 10 10 10
10
π nt −5cos −5 1 10 = = [cos π n − cos 0] πn 10 πn 10 0
When n is even, b n = 0 When n is odd, b 1 =
−5
π
(–1 –1) =
10
π
, b3 =
10 10 −5 , and so on (−1 − 1) = , b 5 = 5π 3π 3π
Thus the Fourier series for the function v(t) is given by: v(t) =
5 10 π t 1 3π t 1 5π t + sin + sin + sin + ... 2 π 10 3 10 5 10
2. Find the Fourier series for f(x) = x in the range x = 0 to x = 5
The periodic function is shown in the diagram below
The Fourier series is given by:
∞ 2π nx 2π nx f(x) = a0 + ∑ an cos + bn sin where L = 5 L L n =1
5
1 L 1 5 1 x2 1 52 5 = = = = = a0 f ( x ) d x x d x L ∫0 5 ∫0 5 2 0 5 2 2 5
an
2 L L ∫0
2π nx 2π nx x sin cos 2 5 2 2π nx 2π nx 5 + 5 by parts f ( x)= cos d x x = cos d x 2 5 ∫0 5 2π n L 5 2π n 5 0 5
2 5sin 2π n cos 2π n 1 = 0 = + − 0+ 2 2 2 π n 5 2π n 2π n 5 5 5 1539
© 2014, John Bird
5
2π nx 2π nx sin x cos 2 L 2 5 2 5 5 2π nx 2π nx by parts bn = f ( x) sin x sin − + d x = d x = 2 5 ∫0 5 L ∫0 2π n L L 2π nx 5 5 0
2 5cos 2π n sin 2π n 2 5cos 2π n 5 5 = − = = − + − + − − 0 0 cos 2π n = ( ) 2 5 2π n 2π n 5 2π n πn πn 5 5 5 5 5 5 5 Hence, b1 = − , b2 = − , b3 = − , b4 = − , and so on π 2π 3π 4π
Thus,
∞ 2π nx 2π nx f(x) = a0 + ∑ an cos + bn sin L L n =1
i.e.
f(x) =
5 5 2π x 5 4π x 5 6π x − sin sin sin − − − ... 2 π 5 2π 5 3π 5
i.e.
f(x) =
5 5 2π x 1 4π x 1 6π x − sin + sin + sin + ... 2 π 5 2 5 3 5
3. A periodic function of period 2π is defined by: − 3, − 2 〈 x 〈 0 f(x) = + 3, 0 〈 x 〈 2 Sketch the function and obtain the Fourier series for the function.
The periodic function is shown in the diagram below
The function is odd since it is symmetrical about the origin, i.e. an = 0 Thus,
a0 =
∞ 2π nx f(x) = a0 + ∑ bn sin L n =1
1 L /2 1 f ( x) d x = ∫ L − L /2 4
{∫
0 −2
2
where L = 4
} 14 {[−3x]
−3d x + ∫ 3d x = 0
1540
0 −2
}
+ [3 x ] 0 = 2
1 {( 0 ) − ( 6 ) + ( 6 ) − ( 0 )} = 0 4 © 2014, John Bird
2 2 L /2 2 0 2π nx 2π nx 2π nx −3sin bn = f ( x) sin dx= d x + ∫ 3sin d x ∫ ∫ 0 L − L /2 4 −2 L 4 4
0 2 π nx π nx 3cos 3cos 1 3cos 0 2 1 3cos 0 3cos(−π n) 3cos π n 2 + − = = − + − −− 2 π n π n 2 π n πn πn π n 2 − 2 2 2 0 2 2 2
6 1 6 6 = − cos π n = (1 − cos π n ) 2 π n π n πn 2 2
When n is even, bn = 0 Hence, = b1 Thus,
6
π
12
1) (1 − −=
π
b3 =
,
6 12 6 12 1) , = b5 1) , and so on (1 − −= (1 − −= 3π 3π 5π 5π
∞ 12 π x 12 3π x 12 5π x 2π nx f(x) = a0 + ∑ bn sin = 0 + π sin 2 + 3π sin 2 + 5π sin 2 + ... L n =1
i.e.
12 π x 1 3π x 1 5π x sin + sin + sin + ... π 2 3 2 5 2
f(x) =
4. Determine the Fourier series for the half-wave rectified sinusoidal voltage V sin t defined by: V sin t , 0 〈 t 〈 π f(t) = 0, π 〈 t 〈 2π which is periodic of period 2π.
The periodic function is shown in the diagram below
The Fourier series is given by: a= 0
1 L 1 f (t ) d= t ∫ 0 2π L
∫
π 0
∞ 2π nt 2π nt f(t) = a0 + ∑ an cos + bn sin where L = 2π L L n =1
V sin t d= t
V V V V π [ − cos t ]= [(− cos π ) − (− cos 0)=] [1 + 1=] 0 2π 2π 2π π © 2014, John Bird 1541
an
2 L 2 π V 2π nt 2π nt dt dt = = f (t ) cos V sin t cos ∫ ∫ 2π 0 L 0 π L 2π =
= a1
π 0
sin t cos nt d t
(1)
π
1 [sin(n + 1)t + sin(1 − n)t ] ∫ 2 π
V
0
V = 2π =
∫
V 2π
π
cos(n + 1)t cos(1 − n)t − (n + 1) − (1 − n) 0
cos(n + 1)π cos(1 − n)π 1 1 − − − −− (n + 1) (1 − n) (n + 1) (1 − n)
V 1 1 1 1 − − − − − = 0 = a3 = a5 and so on 2π 2 0 2 0
V 1 1 1 1 V a2 = + − − − = 2π 3 −1 3 −1 2π
2 V −2 = 3 2π
2V 4 − =− 3π 3
V 1 1 1 1 V 2 2 V a4 = + − − − = − = 2π 5 −3 5 −3 2π 5 3 2π V 1 1 1 1 V a6 = + − − − = 2π 7 −5 7 −5 2π
6 − 10 V = (3)(5) 2π
4 2V − =− π (3)(5) (3)(5)
2 2 V 10 − 14 V 7 − 5 =2π (5)(7) =2π
4 2V − =− π (5)(7) (5)(7) and so on
bn
V 2 L 2 π 2π nt 2π nt = = f t x V sin t sin ( ) sin d dt ∫ ∫ 0 0 L π 2π L 2π
∫
π 0
sin t sin nt d t
It may be shown that bn = 0 except when n = 1 V
π∫
When n= = 1, b1
π 0
sin t= sin t d t
V
π∫
=
0
V 2π
= sin 2 t d t
V
π∫
π 0
1 (1 − cos 2t ) d t 2
π
V V sin 2t t π 0 0 − = − − = ( ) ( ) 2 0 2π 2
∞ 2π nt 2π nt f(t) = a0 + ∑ an cos + bn sin L L n =1
Thus,
i.e. f(x) =
i.e.
π
V
π
−
f(t) =
2V 2V 2V V 2π (2)t 2π (4)t 2π (6)t 2π (1)t cos cos cos − − − ... + sin 3π 2 2π π (3)(5) 2π π (5)(7) 2π 2π V
π
+
V 2V sin t − 2 π
cos 2t cos 4t cos 6t + + + ... (1)(3) (3)(5) (5)(7)
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© 2014, John Bird
EXERCISE 367 Page 1085 1. Determine the half-range Fourier cosine series for the function f(x) = x in the range 0 ≤ x ≤ 3. Sketch the function within and outside of the given range.
The periodic function is shown in the diagram below. Since a half-range cosine series is required, the function is symmetrical about the f(x) axis and
= a0
1 L
{∫
L 0
}
f= ( x) d x
∞ nπ x f ( x= ) a0 + ∑ an cos L n =1
3 1 3 1 x 2 3 = x d x = 3 ∫0 3 2 0 2
2 L nπ x 2 3 nπ x = x cos ∫ 0 f ( x) cos d x d x ∫ L L 3 0 3
an
3 n x n x π π x sin cos 2 1 3 3 2 3sin nπ cos nπ = + = + − 0+ 2 2 2 3 nπ nπ 3 nπ nπ nπ 3 3 3 3 3 0
by parts
2 2 cos nπ 1 2 3 6 = 0 + nπ − 1} −= = ( cos nπ − 1) {cos 2 2 n 2π 2 3 nπ nπ 3 nπ 3 3
When n is even, an = 0 and
Thus,
i.e.
6 12 6 12 ( −2 ) =− 2 , a3 = 2 2 ( −2 ) =− 2 2 , 2 π (1) π π (3) π (3)
a1 =
2
a5 = −
12 , and so on π (5) 2 2
∞ 12 12 nπ x 3 12 πx 3π x 5π x f ( x) = a0 + ∑ an cos cos cos cos − − − = − ... L 2 π2 3 π 2 (3) 2 3 π 2 (5) 2 3 n =1
3 12 π x 1 3π x 1 5π x f ( x) = − cos + 2 cos + 2 cos + ... 2 2 π 3 3 3 5 3
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© 2014, John Bird
2. Find the half-range Fourier sine series for the function f(x) = x in the range 0 ≤ x ≤ 3. Sketch the function within and outside of the given range.
The periodic function is shown in the diagram below. Since a half-range sine series is required, the function is symmetrical about the origin and
∞ nπ x f ( x) = ∑ bn sin L n =1
3
nπ x nπ x x cos sin 2 L 2 3 3 3 nπ x nπ x 2 by parts − + bn = f ( x) sin x sin d x = d x = 2 L ∫0 3 ∫0 nπ L 3 3 nπ 3 3 0
2 3cos nπ sin nπ 2 cos nπ 6 = cos nπ − + − ( 0 + 0 ) = − = − 2 3 nπ nπ nπ nπ 3 3 3 6 6 6 6 b1 == , b2 − , b3 = , b4 = − , and so on 2π 3π 4π π ∞ 6 πx 6 2π x 6 3π x 6 4π x nπ x sin sin sin Thus, f ( x) = ∑ bn sin − + − + ... = sin 3 2π 3 3π 3 4π 3 L π n =1
i.e.
f(x) =
6 πx 1 2π x 1 3π x 1 4π x sin − sin + sin − sin + ... π 3 2 3 3 3 4 3
3. Determine the half-range Fourier sine series for the function defined by: t, 0 〈 t 〈 1 f(t) = (2 − t ), 1 〈 t 〈 2
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© 2014, John Bird
The periodic function is shown in the diagram below. Since a half-range sine series is required, the function is symmetrical about the origin and
= bn
∞ nπ t f (t ) = ∑ bn sin L n =1
2 2 L 2 1 nπ t nπ t nπ t f (t ) = sin dt t sin d t + ∫ (2 − t ) sin dt ∫ ∫ 1 L 0 2 0 L 2 2 1 2 n t n t n t n t n t π π π π π t cos sin 2 cos t cos sin 2 2 2 2 2 = − + + − + − 2 2 n π n π n π n π nπ 2 2 2 2 2 1 0
− 2 cos nπ + 2 cos nπ − sin nπ nπ nπ 2 nπ cos nπ sin nπ 2 2 2 2 2 = − + − (0 + 0) + 2 nπ π n nπ nπ 2 cos cos 2 2 2 + 2 −− nπ nπ 2 2
by parts
nπ sin − 2 2 nπ 2
nπ 2sin 8 nπ 2 = = sin 2 n 2π 2 2 nπ 2 When n is even, bn = 0
8 8 8 b1 = , b3 = − , b5 = , and so on 2 2 2 π (3) π (5) 2 π 2 ∞ 8 8 8 nπ t πt 3π t 5π t Thus, f (t ) = ∑ bn sin = 2 sin − 2 2 sin + 2 2 sin − ... L π 2 (3) π 2 (5) π 2 n =1
i.e.
f(t) =
8 πt 1 3π t 1 5π t sin − sin + sin − ... π 2 2 32 2 52 2
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© 2014, John Bird
4. Show that the half-range Fourier cosine series for the function f(θ) = θ 2 in the range 0 to 4 is given by: f(θ) =
16 64 πθ 1 2πθ 1 3πθ − cos − 2 cos + 2 cos − ... 2 3 π 4 2 4 3 4
Sketch the function within and outside of the given range.
The periodic function is shown in the diagram below. Since a half-range cosine series is required, the function is symmetrical about the f(θ) axis and
1 = a0 L an
{∫
L 0
∞ nπθ f (θ= ) a0 + ∑ an cos L n =1
}
4 1 4 2 1 θ 3 1 64 16 = θ dθ f (θ= ) dθ = = 4 ∫0 4 3 0 4 3 3
2 L nπθ 2 4 2 nπθ = θ cos ∫ 0 f (θ ) cos dθ ∫ L L 4 0 4 θ 2 sin nπθ 1 4 = 2 nπ 4
nπθ 2θ cos + 4 2 nπ 4
nπθ 2sin − 4 3 nπ 4
1 16sin nπ 8cos nπ 2sin nπ = + − 2 3 2 nπ nπ nπ 4 4 4
dθ 4 by parts 0
64 1 8cos nπ 1 8(16) = cos nπ = − ( 0 ) = cos nπ 2 2 2 n 2π 2 2 nπ 2 n π 4
64 64 64 64 64 64 , a2 = ( −1) =−= (1) 2 2 , a3 = 2 2 (−1) =− 2 2 , and so on 2 2 2 2 π (3) π (3) π (1) π π (2) π (2)
and a1 =
2
Thus, ∞ nπθ f (θ ) = a0 + ∑ an cos L n =1
16 64 πθ − cos = 3 π2 4
64 2πθ + 2 2 cos π (2) 4 1546
64 3πθ − 2 2 cos π (3) 4
+ ...
© 2014, John Bird
i.e.
16 64 πθ − f (θ ) = cos 3 π2 4
1 2πθ − 2 cos 2 4
1 3πθ + 2 cos 3 4
1547
− ...
© 2014, John Bird
1. The voltage from a square wave generator is of the form: 0, − 10 〈 t 〈 0 v(t) = 5, 0 〈 t 〈 10 and is periodic of period 20. Show that the Fourier series for the function is given by: v(t) =
5 10 π t 1 3π t 1 5π t + sin + sin + sin + ... 2 π 10 3 10 5 10
The square wave is shown below
∞
The Fourier series is of the form:
v(t) = a 0 +
n =1
{∫
2π nt 2π nt cos + bn sin L L
∑ a
n
a0 =
1 1 L /2 1 10 v(t ) d t = v(t ) d t = ∫ ∫ 20 −10 L − L /2 20
an =
2 L /2 2 10 2π nt 2π nt = v t t ( ) cos d v(t ) cos dt ∫ ∫ L /2 − − 10 L 20 20 L
=
0 −10
10
}
0 d t + ∫ 5d t = 0
1 5 10 [5t ] 0 = 20 2
10 1 0 π nt π nt ∫ −10 0 cos d t + ∫ 0 5cos dt 10 10 10
10
π nt 5sin 1 10 = 5 [sin π n − sin 0] = 0 for n = 1, 2, 3, … = πn 10 π n 10 0
bn =
2 L /2 2 10 2π nt 2π nt v(t ) sin v(t ) sin dt dt = ∫ ∫ L /2 − − 10 L 20 20 L
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© 2014, John Bird
=
10 1 0 π nt π nt ∫ −10 0sin d t + ∫ 0 5sin dt 10 10 10
10
π nt −5cos −5 1 10 = = [cos π n − cos 0] πn 10 πn 10 0
When n is even, b n = 0 When n is odd, b 1 =
−5
π
(–1 –1) =
10
π
, b3 =
10 10 −5 , and so on (−1 − 1) = , b 5 = 5π 3π 3π
Thus the Fourier series for the function v(t) is given by: v(t) =
5 10 π t 1 3π t 1 5π t + sin + sin + sin + ... 2 π 10 3 10 5 10
2. Find the Fourier series for f(x) = x in the range x = 0 to x = 5
The periodic function is shown in the diagram below
The Fourier series is given by:
∞ 2π nx 2π nx f(x) = a0 + ∑ an cos + bn sin where L = 5 L L n =1
5
1 L 1 5 1 x2 1 52 5 = = = = = a0 f ( x ) d x x d x L ∫0 5 ∫0 5 2 0 5 2 2 5
an
2 L L ∫0
2π nx 2π nx x sin cos 2 5 2 2π nx 2π nx 5 + 5 by parts f ( x)= cos d x x = cos d x 2 5 ∫0 5 2π n L 5 2π n 5 0 5
2 5sin 2π n cos 2π n 1 = 0 = + − 0+ 2 2 2 π n 5 2π n 2π n 5 5 5 1539
© 2014, John Bird
5
2π nx 2π nx sin x cos 2 L 2 5 2 5 5 2π nx 2π nx by parts bn = f ( x) sin x sin − + d x = d x = 2 5 ∫0 5 L ∫0 2π n L L 2π nx 5 5 0
2 5cos 2π n sin 2π n 2 5cos 2π n 5 5 = − = = − + − + − − 0 0 cos 2π n = ( ) 2 5 2π n 2π n 5 2π n πn πn 5 5 5 5 5 5 5 Hence, b1 = − , b2 = − , b3 = − , b4 = − , and so on π 2π 3π 4π
Thus,
∞ 2π nx 2π nx f(x) = a0 + ∑ an cos + bn sin L L n =1
i.e.
f(x) =
5 5 2π x 5 4π x 5 6π x − sin sin sin − − − ... 2 π 5 2π 5 3π 5
i.e.
f(x) =
5 5 2π x 1 4π x 1 6π x − sin + sin + sin + ... 2 π 5 2 5 3 5
3. A periodic function of period 2π is defined by: − 3, − 2 〈 x 〈 0 f(x) = + 3, 0 〈 x 〈 2 Sketch the function and obtain the Fourier series for the function.
The periodic function is shown in the diagram below
The function is odd since it is symmetrical about the origin, i.e. an = 0 Thus,
a0 =
∞ 2π nx f(x) = a0 + ∑ bn sin L n =1
1 L /2 1 f ( x) d x = ∫ L − L /2 4
{∫
0 −2
2
where L = 4
} 14 {[−3x]
−3d x + ∫ 3d x = 0
1540
0 −2
}
+ [3 x ] 0 = 2
1 {( 0 ) − ( 6 ) + ( 6 ) − ( 0 )} = 0 4 © 2014, John Bird
2 2 L /2 2 0 2π nx 2π nx 2π nx −3sin bn = f ( x) sin dx= d x + ∫ 3sin d x ∫ ∫ 0 L − L /2 4 −2 L 4 4
0 2 π nx π nx 3cos 3cos 1 3cos 0 2 1 3cos 0 3cos(−π n) 3cos π n 2 + − = = − + − −− 2 π n π n 2 π n πn πn π n 2 − 2 2 2 0 2 2 2
6 1 6 6 = − cos π n = (1 − cos π n ) 2 π n π n πn 2 2
When n is even, bn = 0 Hence, = b1 Thus,
6
π
12
1) (1 − −=
π
b3 =
,
6 12 6 12 1) , = b5 1) , and so on (1 − −= (1 − −= 3π 3π 5π 5π
∞ 12 π x 12 3π x 12 5π x 2π nx f(x) = a0 + ∑ bn sin = 0 + π sin 2 + 3π sin 2 + 5π sin 2 + ... L n =1
i.e.
12 π x 1 3π x 1 5π x sin + sin + sin + ... π 2 3 2 5 2
f(x) =
4. Determine the Fourier series for the half-wave rectified sinusoidal voltage V sin t defined by: V sin t , 0 〈 t 〈 π f(t) = 0, π 〈 t 〈 2π which is periodic of period 2π.
The periodic function is shown in the diagram below
The Fourier series is given by: a= 0
1 L 1 f (t ) d= t ∫ 0 2π L
∫
π 0
∞ 2π nt 2π nt f(t) = a0 + ∑ an cos + bn sin where L = 2π L L n =1
V sin t d= t
V V V V π [ − cos t ]= [(− cos π ) − (− cos 0)=] [1 + 1=] 0 2π 2π 2π π © 2014, John Bird 1541
an
2 L 2 π V 2π nt 2π nt dt dt = = f (t ) cos V sin t cos ∫ ∫ 2π 0 L 0 π L 2π =
= a1
π 0
sin t cos nt d t
(1)
π
1 [sin(n + 1)t + sin(1 − n)t ] ∫ 2 π
V
0
V = 2π =
∫
V 2π
π
cos(n + 1)t cos(1 − n)t − (n + 1) − (1 − n) 0
cos(n + 1)π cos(1 − n)π 1 1 − − − −− (n + 1) (1 − n) (n + 1) (1 − n)
V 1 1 1 1 − − − − − = 0 = a3 = a5 and so on 2π 2 0 2 0
V 1 1 1 1 V a2 = + − − − = 2π 3 −1 3 −1 2π
2 V −2 = 3 2π
2V 4 − =− 3π 3
V 1 1 1 1 V 2 2 V a4 = + − − − = − = 2π 5 −3 5 −3 2π 5 3 2π V 1 1 1 1 V a6 = + − − − = 2π 7 −5 7 −5 2π
6 − 10 V = (3)(5) 2π
4 2V − =− π (3)(5) (3)(5)
2 2 V 10 − 14 V 7 − 5 =2π (5)(7) =2π
4 2V − =− π (5)(7) (5)(7) and so on
bn
V 2 L 2 π 2π nt 2π nt = = f t x V sin t sin ( ) sin d dt ∫ ∫ 0 0 L π 2π L 2π
∫
π 0
sin t sin nt d t
It may be shown that bn = 0 except when n = 1 V
π∫
When n= = 1, b1
π 0
sin t= sin t d t
V
π∫
=
0
V 2π
= sin 2 t d t
V
π∫
π 0
1 (1 − cos 2t ) d t 2
π
V V sin 2t t π 0 0 − = − − = ( ) ( ) 2 0 2π 2
∞ 2π nt 2π nt f(t) = a0 + ∑ an cos + bn sin L L n =1
Thus,
i.e. f(x) =
i.e.
π
V
π
−
f(t) =
2V 2V 2V V 2π (2)t 2π (4)t 2π (6)t 2π (1)t cos cos cos − − − ... + sin 3π 2 2π π (3)(5) 2π π (5)(7) 2π 2π V
π
+
V 2V sin t − 2 π
cos 2t cos 4t cos 6t + + + ... (1)(3) (3)(5) (5)(7)
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© 2014, John Bird
EXERCISE 367 Page 1085 1. Determine the half-range Fourier cosine series for the function f(x) = x in the range 0 ≤ x ≤ 3. Sketch the function within and outside of the given range.
The periodic function is shown in the diagram below. Since a half-range cosine series is required, the function is symmetrical about the f(x) axis and
= a0
1 L
{∫
L 0
}
f= ( x) d x
∞ nπ x f ( x= ) a0 + ∑ an cos L n =1
3 1 3 1 x 2 3 = x d x = 3 ∫0 3 2 0 2
2 L nπ x 2 3 nπ x = x cos ∫ 0 f ( x) cos d x d x ∫ L L 3 0 3
an
3 n x n x π π x sin cos 2 1 3 3 2 3sin nπ cos nπ = + = + − 0+ 2 2 2 3 nπ nπ 3 nπ nπ nπ 3 3 3 3 3 0
by parts
2 2 cos nπ 1 2 3 6 = 0 + nπ − 1} −= = ( cos nπ − 1) {cos 2 2 n 2π 2 3 nπ nπ 3 nπ 3 3
When n is even, an = 0 and
Thus,
i.e.
6 12 6 12 ( −2 ) =− 2 , a3 = 2 2 ( −2 ) =− 2 2 , 2 π (1) π π (3) π (3)
a1 =
2
a5 = −
12 , and so on π (5) 2 2
∞ 12 12 nπ x 3 12 πx 3π x 5π x f ( x) = a0 + ∑ an cos cos cos cos − − − = − ... L 2 π2 3 π 2 (3) 2 3 π 2 (5) 2 3 n =1
3 12 π x 1 3π x 1 5π x f ( x) = − cos + 2 cos + 2 cos + ... 2 2 π 3 3 3 5 3
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© 2014, John Bird
2. Find the half-range Fourier sine series for the function f(x) = x in the range 0 ≤ x ≤ 3. Sketch the function within and outside of the given range.
The periodic function is shown in the diagram below. Since a half-range sine series is required, the function is symmetrical about the origin and
∞ nπ x f ( x) = ∑ bn sin L n =1
3
nπ x nπ x x cos sin 2 L 2 3 3 3 nπ x nπ x 2 by parts − + bn = f ( x) sin x sin d x = d x = 2 L ∫0 3 ∫0 nπ L 3 3 nπ 3 3 0
2 3cos nπ sin nπ 2 cos nπ 6 = cos nπ − + − ( 0 + 0 ) = − = − 2 3 nπ nπ nπ nπ 3 3 3 6 6 6 6 b1 == , b2 − , b3 = , b4 = − , and so on 2π 3π 4π π ∞ 6 πx 6 2π x 6 3π x 6 4π x nπ x sin sin sin Thus, f ( x) = ∑ bn sin − + − + ... = sin 3 2π 3 3π 3 4π 3 L π n =1
i.e.
f(x) =
6 πx 1 2π x 1 3π x 1 4π x sin − sin + sin − sin + ... π 3 2 3 3 3 4 3
3. Determine the half-range Fourier sine series for the function defined by: t, 0 〈 t 〈 1 f(t) = (2 − t ), 1 〈 t 〈 2
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© 2014, John Bird
The periodic function is shown in the diagram below. Since a half-range sine series is required, the function is symmetrical about the origin and
= bn
∞ nπ t f (t ) = ∑ bn sin L n =1
2 2 L 2 1 nπ t nπ t nπ t f (t ) = sin dt t sin d t + ∫ (2 − t ) sin dt ∫ ∫ 1 L 0 2 0 L 2 2 1 2 n t n t n t n t n t π π π π π t cos sin 2 cos t cos sin 2 2 2 2 2 = − + + − + − 2 2 n π n π n π n π nπ 2 2 2 2 2 1 0
− 2 cos nπ + 2 cos nπ − sin nπ nπ nπ 2 nπ cos nπ sin nπ 2 2 2 2 2 = − + − (0 + 0) + 2 nπ π n nπ nπ 2 cos cos 2 2 2 + 2 −− nπ nπ 2 2
by parts
nπ sin − 2 2 nπ 2
nπ 2sin 8 nπ 2 = = sin 2 n 2π 2 2 nπ 2 When n is even, bn = 0
8 8 8 b1 = , b3 = − , b5 = , and so on 2 2 2 π (3) π (5) 2 π 2 ∞ 8 8 8 nπ t πt 3π t 5π t Thus, f (t ) = ∑ bn sin = 2 sin − 2 2 sin + 2 2 sin − ... L π 2 (3) π 2 (5) π 2 n =1
i.e.
f(t) =
8 πt 1 3π t 1 5π t sin − sin + sin − ... π 2 2 32 2 52 2
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© 2014, John Bird
4. Show that the half-range Fourier cosine series for the function f(θ) = θ 2 in the range 0 to 4 is given by: f(θ) =
16 64 πθ 1 2πθ 1 3πθ − cos − 2 cos + 2 cos − ... 2 3 π 4 2 4 3 4
Sketch the function within and outside of the given range.
The periodic function is shown in the diagram below. Since a half-range cosine series is required, the function is symmetrical about the f(θ) axis and
1 = a0 L an
{∫
L 0
∞ nπθ f (θ= ) a0 + ∑ an cos L n =1
}
4 1 4 2 1 θ 3 1 64 16 = θ dθ f (θ= ) dθ = = 4 ∫0 4 3 0 4 3 3
2 L nπθ 2 4 2 nπθ = θ cos ∫ 0 f (θ ) cos dθ ∫ L L 4 0 4 θ 2 sin nπθ 1 4 = 2 nπ 4
nπθ 2θ cos + 4 2 nπ 4
nπθ 2sin − 4 3 nπ 4
1 16sin nπ 8cos nπ 2sin nπ = + − 2 3 2 nπ nπ nπ 4 4 4
dθ 4 by parts 0
64 1 8cos nπ 1 8(16) = cos nπ = − ( 0 ) = cos nπ 2 2 2 n 2π 2 2 nπ 2 n π 4
64 64 64 64 64 64 , a2 = ( −1) =−= (1) 2 2 , a3 = 2 2 (−1) =− 2 2 , and so on 2 2 2 2 π (3) π (3) π (1) π π (2) π (2)
and a1 =
2
Thus, ∞ nπθ f (θ ) = a0 + ∑ an cos L n =1
16 64 πθ − cos = 3 π2 4
64 2πθ + 2 2 cos π (2) 4 1546
64 3πθ − 2 2 cos π (3) 4
+ ...
© 2014, John Bird
i.e.
16 64 πθ − f (θ ) = cos 3 π2 4
1 2πθ − 2 cos 2 4
1 3πθ + 2 cos 3 4
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− ...
© 2014, John Bird
