Page 473 - Advanced engineering mathematics

13.5 Phase Angle Form 453

Now suppose f has fundamental period p. Its Fourier series on [−p/2, p/2], with L = p/2,
is
∞
1
a 0 + (a n cos(2nπx/p) + b n sin(2nπx/p)),
2
n=1
where
2 p/2
a n = f (x)cos(2nπx/p)dx
p −p/2
and
2 p/2
b n = f (x)sin(2nπx/p)dx.
p −p/2
It is sometimes convenient to write this series in a different way. Let
2π
ω 0 = .
p
The Fourier series of f (x) on [−p/2, p/2] is
∞
1
a 0 + (a n cos(nω 0 x) + b n sin(nω 0 x)),
2
n=1
where
2 p/2
a n = f (x)cos(nω 0 x)dx
p −p/2
and
2 p/2
b n = f (x)sin(nω 0 x)dx.
p −p/2
Now look for numbers c n and δ n so that
a n cos(nω 0 x) + b n sin(nω 0 x) = c n cos(nω 0 x + δ n ).
To solve for these constants, use a trigonometric identity to write this equation as

a n cos(nω 0 x) + b n sin(nω 0 x) = c n cos(nω 0 x)cos(δ n ) − c n sin(nω 0 x)sin(δ n ).
One way to satisfy this equation is to put
c n cos(δ n ) = a n and c n sin(δ n ) =−b n .
If we square both sides of these equations and add the results, we obtain
2
2
2
c = a + b ,
n n n
so

2
c n = a + b .
2
n n
Next, divide to obtain
c n sin(δ n ) b n
= tan(δ n ) =− ,
c n cos(δ n ) a n
assuming that a n = 0. Then

b n
δ n =−arctan .
a n

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October 14, 2010 14:57 THM/NEIL Page-453 27410_13_ch13_p425-464