1) For the function:
f(t) = {0 for h < t < 2p
= {1 for 0 < t < h
Over 0 < t < 2 p h > 0
Give a simplified expression for the associated Fourier coefficient if: c n = 1/2 p ò 2p 0 f(t) exp (-int) dt
= 1/ 2p ò 2p h e - int dt
Solution:
c n = 1/ 2p ò 2p h e - int dt = 1/ 2p (1/ -in) e - int ] 2p h
= 1/ -2p in [e - i2n p - e - inh ]
c n = i/ 2pn [1 - e - inh ]
2) F+ (k) = Ö (2/ p ) ò ¥ 0 f(x) exp (ikx) dx
Find F+ (k) for: x N e -x
(N an integer)
Solution:
F+ (k) = Ö (2/ p ) ò ¥ 0 x N e -x e ikx dx
By Euler's formula: e ikx = cos kx + i sin kx, so we can substitute the imaginary part, i.e. Im (e ikx ) = i sin kx into the integral to obtain:
F+ (k) = Ö (2/ p ) i ò ¥ 0 x N e -x sin kx dx
From CRC Mathematics Tables we find:
ò ¥ 0 x n e -ax sin (bx) dx =
i Ö (2/ p ) n! [(a +ib) n+1 + (a - ib) n+1]/ 2i (a 2 + b 2) n+1
Then with a = 1, b = k, n = N:
i Ö (2/ p ) ò ¥ 0 x N e -x sin kx dx =
i Ö (2/ p ) N! [(1 +ik) N+1 + (1 - ik) N+1]/ 2i (1 + k 2) N+1
F+ (k) = (2/ p ) { n! [(1 +ik) n+1 + (1 - ik) n+1]/ 2i (1 + k 2) n+1
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