UNIT – I FOURIER SERIES PROBLEM 1
[Pages:63]UNIT ? I FOURIER SERIES
PROBLEM 1:
The turning moment T lb feet on the crankshaft of steam engine is given for a series of values of the crank angle degrees
0
T0
30 60
90
120 150 180
5224 8097 7850 5499 2626 0
Expand T in a series of sines.
Solution:
Let T = b1 sin + b2 sin2 + b3 sin3 + ........., Since the first and last values of T are repeated neglect last one
T
T sin T sin2
0
0
30 5224
60 8097
90 7850
120 5499
150 2624
Total
0 2612 7012.2 7850 4762.27 1313 23549.47
0 4524.11 7012.2
0 - 4762.27 - 2274.18 4499.86
T sin3
0 5224
0 -7850
0 2626
0
b1 = 2 [mean value of T sin ] = 2[ 23549.47 ] = 7849.8 6
b2 = 2 [mean value of T sin2 ] = 2[ 4499.86 ] = 1499.95 6
b3 = 2 [mean value of T sin2 ] = 0
f ( x ) = 7849.8 sin + 1499.95 sin2
PROBLEM 2:
Analyze harmonically the given below and express y in Fourier series upto the third harmonic.
x0
2
4 5
2
3
3
3
3
y 1.0 1.4 1.9 1.7 1.5 1.2
1
Solution:
Since the last value of y is a repetition of the first, only the six values will be used. The length of the interval is 2.
Let y =
a0 2
+ (a1 cosx + b1 sinx) + (a2 cos2x + b2 sin2x) + (a3 cos3x + b3 sin3x) + ...
(1)
x
y
cosx
0 1.0
1
sinx cos2x sin2x cos3x sin3x
0
1
0
1
0
1.4
0.5
0.866 - 0.5 0.866 -1
0
3
2 1.9
- 0.5
0.866 - 0.5 - 0.866 1
0
3
1.7
-1
0
1
0
-1
0
4 1.5
- 0.5
- 0.866 - 0.5 0.866 1
0
3
5 1.2
0.5
- 0.866 - 0.5 - 0.866 -1
0
3
a0 = 2 [mean value of y] = 2 (8.7) = 2.9 6
a1 = 2 [mean value of y cosx] = 2 [1 (1.0) + 0.5 (1.4) ? 0.5 (1.9) ? 1 (1.7) ? 0.5 (1.5) + 0.5 (1.2)] 6 = - 0.37
b1 = 2 [mean value of y sinx] = 2 [0.866 (1.4 + 1.9 -1.5 ? 1.2)] 6 = 0.17
a2 = 2 [mean value of y cos2x]
= 2 [1(1.0 + 1.7) ? 0.5 (1.4 + 1.9 + 1.5 + 1.2)] 6
= - 0.1 b2 = 2 [mean value of y sin2x]
= 2 [0.866(1.4 ? 1.9 + 1.5 ? 1.2)] 6
= - 0.06 a3 = 2 [mean value of y cos3x]
= 2 [1.0 ? 1.4 + 1.9 ? 1.7 + 1.5 ? 1.2] 6
= 0.03 b3 = 2 [mean value of y sin3x]
= 0
Hence y = 1.45 + (-0.37 cosx + 0.17 sinx) ? (0.1 cos2x + 0.06 sin2x) + 0.03 cos3x.
PROBLEM 3:
Find the Fourier series expansion for the function f(x) = x sinx in 0 < x < 2 and deduce
that 1 1 1 ......... 2
1.3 3.5 5.7
4
SOLUTION:
The Fourier series is
f (x)
a0 2
(an
n 1
cos nx
bn
sin
nx)..................(1)
Given f(x) = x sinx
a0
=
1
2
f (x)dx
0
=
1
2
x sin xdx
0
=
1
[x (- cosx) - (1)(- sinx)]0 2
= 1 (-2) = -2
an
=
1
2
f (x) cos nxdx
0
=
1
2
x sin x cos nxdx
0
=
1 2
2
[x sin(1 n)x sin(1 n)x]dx
0
[ 2SinACosB Sin(A B) Sin(A B)]
1 2
x(
cos(1
1 n
n)x
cos(1 n)x 1 n
)
(
sin(1 n)x ( 1 n)2
sin(1 n)x (1 n)2
2 ) 0
= 1 [2( cos 2(1 n) cos 2(1 n))] (
2
1 n
1 n
[ 1 1 ] 1 n 1 n
[1
n 1 1-n2
n
]
=
an
2 n2 1
where n 1
a1
=
1
2
f (x) cos xdx
0
=
1
2
x sin x cos xdx
0
=
1 2
2
x sin 2xdx
0
=
1 2
x(
cos 2
2x
)
(1)(
sin 2x 4
)
0
2
= 1 [2 (- cos 4 ) = - 1
2
2
2
1 [2( cos 4)
2
2
bn
=
1
2
0
f
(x)
sin
nxdx
=
1
2
0
x
sin
x
sin
nxdx
=
1 2
2
0
x[cos(1
n)x
cos(1
n
)x]dx
[ 2SinASinB Cos(A B) Cos(A B)]
=
1 2
x(
sin(1 n)x 1 n
sin(1 n)x 1 n
)
(
cos(1 (1
n)x n)2
cos(1 (1
n
n)x )2
)
2 0
=
1 cos 2(1 n)
2
(1 n)2
cos 2(1 n) (1 n)2
cos 0 (1 n)2
cos 0
(1
n)2
=
1 1
2
(1
n)2
1 (1 n)2
1 (1 n)2
1
(1
n
)2
= 0 where n 1
b1
=
1
2
0
f
(x)
sin
xdx
=
1
2
0
x
sin
x
sin
xdx
=
1
2
0
x
sin
2
xdx
=
1
2
0
x(1
cos 2
2x
)dx
=
1 2
2
0
x(1
cos
2x)dx
1 2
x(x
sin 2x ) 2
(1)( x2 2
cos 2x 4
) 0
2
= 1 [22 1 1]
2
44
=
Substitute in (1), we get
f
(x)
=
1
1 2
cos x
n2
n
2 2
1
cos
nx
sin
x
.......(2) +
...................(2)
Deduction Part:
Put x [x is a point of continuity] [x= is a point of continuity]
2
2
f ( ) = f ( ) sin =
2
2 2 22 2
(sin = 1) 2
(2)
(2)
2
1
0
n2
2 n2 1
cos
n 2
= -1-0+
2
1
2
n2
1 n2
1
cos
n 2
-
+1
=
- 1 2
1 cos n +1 =
2
n2 (n 1)(n 1) 2
- 1
1
cos n =
4 2 n2 (n 1)(n 1) 2
=
1
cos n 2
n2 (n 1)(n 1) 2
4
1 (1) 1 (0) 1 (1) ........ 2
1.3
2.4 3.5
4
(1)[ 1 1 1 ........] [ 2]
1.3 3.5 5.7
4
1 1 1 ........ [ 2]
1.3 3.5 5.7
4
Problem:4
Find the Fourier series for f(x) = sin x in - < x <
Solution:
Given f(x) = sin x
This is an even function. bn = 0
f(x) =
a0 2
a nCosnx.....................................(1) n 1
a0
=
2
0
f
(x)dx
=
2
0
Sinx
dx
=
2
0
sin
xdx
=
2
[
cos
x]0
= - 2 (cos cos 0)
= - 2 [11] [-1-1]
= 4
an
=
2
0
f
(x)
cos
nxdx
=
2
0
Sinx
cos nxdx
=
2
0
sin
x
cos
nxdx
=
2
1 2
0
[sin(n
1)x
sin(n
1)x]dx
=
1 [ cos(n 1)x n 1
cos(n 1)x n 1
]0
= 1 [ cos(n 1) cos(n 1) 1 1 ]
n 1
n 1 n 1 n 1
= 1 [ (1)n1 (1)n1 1 1 ] n 1 n 1 1 n n 1
= 1 [(1)n (1)n 1 1 ] n 1 n 1 n 1 n 1
= 1 [(1)n 1 (1)n 1]
n 1
n 1
= (1)n 1[ 1 1 ] n 1 n 1
=
(1)n
1[n
1 n n2 1
1]
=
(1)n
1[
2 n2
] 1
=
2[(1)n 1] (n 2 1)
if n 1.
a1
=
2
0
f
(x)
cos
xdx
=
2
0
Sinx
cos xdx
=
2
0
sin
x
cos
xdx
=
2
1 2
0
sin
2xdx
=
1
[
cos 2x 2
]0
= 1 [ 1 1] = 0 22
Substitute in equation (1), we get
f(x) =
2
0
n2
2[(1)n (n 2 1)
1]
cos
nx
................
................
In order to avoid copyright disputes, this page is only a partial summary.
To fulfill the demand for quickly locating and searching documents.
It is intelligent file search solution for home and business.
Related download
- phƯƠng trÌnh lƯỢng giÁc
- basic trigonometric identities common angles
- ipostedavideocoveringthishandouthere
- math 202 jerry l kazdan
- trigonometry identities ii double angles
- trigonometric identities miami
- double angle identity practice weebly
- cos x cos x cos x x cos x x
- fourier series college of the redwoods
- trigonometria disequazioni trigonometriche
Related searches
- series i bonds redemption chart
- series i savings bond rates
- current interest rate on series i bonds
- series i savings bonds
- series i savings bonds value
- series i savings bond
- series i bond redemption tables
- series i savings bond value calculator
- current series i bond rates
- are series i bonds worth it
- series i bonds current value
- treasury bond series i rate