Expert: Josh - 9/1/2007

Question
sir can u plz help in fourier series..

i just wanna know how to recognize even and odd terms in the given f(x)
i am also confused in determining An and Bn when the limits are 0<x<2pi. In some books after finding An they say to find An when n>1 and then for n=1. I am not able to understand the difference.

plz help me ...
thanks

Answer
Dear Harsh Rathore

Sorry about the delay. I actually answered this question three days ago, but for some reason my message did not get through to the system after I clicked reply.

Let f(t) = A0/2+ SUM_{n=1 to n=infinity} A(n)cos(nwot)+ B(n)sin(nwot).

A(n)= (2/T)*[Integral_{t=-T/2,t=T/2} f(t)cos(nwot) dt]
B(n)= (2/T)*[Integral_{t=-T/2,t=T/2} f(t)sin(nwot) dt]

where wo=2*pi/T is the fundamental frequency of the signal, T is the periodicity of the function.

Observe that taking a linear combination of cosine harmonics "SUM_{n=1 to n=infinity} A(n)cos(nwot)" always produces an even function. Thus, if f(t) is an odd function, it does not require any weighted contributions from any cos(nwot). Hence, A(n)=0 for all n>=1. Similarly, sin(nwot) for any n>=1 is odd. If f(t) is even, the Fourier series does not require any contribution from these sinusoidal terms, hence B(n)=0 for all n>=1. In general, if f(t) is neither odd or even, we cannot assume A(n)=0 or B(n)=0 for n>=1.

The Fourier coefficients A(n) and B(n) are calculated using the integral equations given earlier. This is essentially an inner product in the continuous-time domain (Hilbert space) evaluated over one period. It does not matter whether the limits go from t=-T/2 to t=T/2, or from t=0 to t=T. If the signal has a fundamental period of 2*pi, then T=2*pi.

Note that the formulae for A(n) and B(n) do not always yield solutions which are defined for n=1.

e.g., Consider the half-wave rectified sine function defined by
f(t) = { sin(pi*t) for 2k<=t<=2k+1
    = { 0         for 2k+1<=t<=2k+2 where k is an integer.

This question is not conceptually difficult, but there are many algebraic steps involved, starting with the observation that sin(wt) can be written as [exp(jwt)-exp(-jwt)]/(2j), where j=sqrt(-1). I am not going to go through the mechanics since integrals are difficult to type here, but I will illustrate one point.

Draw a sketch yourself to see that the function has period T=2. It can be shown that the Fourier series is given by
f(t)=(1/pi) + 0.5sin(pi*t) + {SUM_{n=2,n=inf} [1-cos((n+1)pi)]*cos(n*pi*t)/[(1-n^2)pi]}  _____(**)

In particular, B(n)= (2/T)*[Integral_{t=-1,t=1} f(t)sin(nwot) dt] = sin((n+1)pi)/[(n^2-1)pi]....intermediate steps are omitted

Thus, B(n)=0 for n>1, but it is not defined when n=1 (cannot have zero in denominator).
So, we need to compute B(n) separately when n=1.
We can easily show that B(1) = [Integral_{t=-1,t=1} [sin(pi*t)]^2 dt] = 1/2. This accounts for the second term in the Fourier series expansion (**).

You'll find there are similar considerations when A(n) is computed.