Hi Gang:
I am just starting to learn F.T and I'm stuck on a question.
How do i obtain the fourier series of this function:

s(t) = (Ac)[1+ Mcos((Wm)(t))]*cos((Wc)(t))

where c & m are subscripts, and * is multiply.

I tried using product of sums and differences (trig identity) on cos(Wmt) and cos(Wct) (as it was advised), plus I converted the sinusoids into Euler's form. But I still have no idea where that takes me. Can someone help? thnx

Hi Milkymoo,

The Fourierseries consists of terms that are either cosines or sines. So the challenge is to write everything in there as either a cosine or a sine.

In this case, you would want to use the formula:

cos(x)*cos(y) = (1/2)*cos(x-y)+ (1/2)*cos(x+y)

This would lead to:

s(t)
= (Ac)*cos((Wc)(t)) + (Ac*M)*cos((Wm)(t))*cos((Wc)(t))
= (Ac)*cos((Wc)(t)) + (Ac*M/2)*cos(Wm(t) - Wc(t)) + (Ac*M/2)*cos(Wm(t)+Wc(t))

To continue, you would need to know what Wc(t) and Wm(t) are and perform the integrals to know the coefficients in the fourier decomposition.

Bye!

Crisp

Hi Milkymoo,

I recognize your function. As a Broadcast Engineer, I had better know it! It's the amplitude modulation equation. Here, (written in Mathematica script) is an example of how to find the Fourier series coefficients for this function:

fc=Sin[10*2*Pi*t]; (* 10Hz Carrier, with t=time *)
fm=Sin[1 *2*Pi*t]; (* 1Hz Modulation *)
f=(1+fm)*fc; (* 100% Modulated AM Signal *)
an=(2/1)*Integrate[f*Cos[n*2*Pi*t],{t,0.75,1.75}];
bn=(2/1)*Integrate[f*Sin[n*2*Pi*t],{t,0.75,1.75}];

The fourth line gives the cosine coefficients and the fifth line gives the sine coefficients for each value of n (each multiple of the fundamental frequency). In this case, the lower limit of integration is 0.75 seconds, while the upper limit is 1.75 second. The "period" here is 1 second. The factor to the left of "Integrate" is 2 divided by the period, or in this case; 2 divided by 1 second.

The only values of n in these equations that (in this case) do not return zero, are n=9, n=10, and n=11. This is what you would expect given a carrier of 10Hz and a 1Hz modulating signal. The resulting modulated signal will contain a lower sideband of 9Hz, an upper sideband of 11Hz, as well as the original carrier of 10Hz.

The cosine coefficient given by the equation on the forth line for n=9 is 0.5, and for n=11 it's a negative 0.5. The sine coefficient given by the equation on the fifth line for n=10 is 1.0. These three weighted sin and cos functions when added together will give a signal level vs. time plot identical to the original modulation function (try it);

f=0.5*Sin[9*2*Pi*t]-0.5*Sin[ 11*2*Pi*t]+1*Cos[10*2*Pi*t]

Fourier's equations rely on the property of orthogonal functions whereby; Integrate[sn(x)*sm(x), {x,a,b}]=0 if n does not equal m (analogously, the "inner product" of orthogonal vectors is zero.) The Fourier series are one of many such orthogonal functions (i.e., Legendre, Lagurre, Hermite, Tchybecheff's polynomials as well as Bessel functions).

The analysis of arbitrary functions in terms of a set of orthogonal functions is one of the most important techniques used in physics and engineering. It's also an incredible idea. I'm so fascinated by orthogonal polynomials that I took "orthogonal" as my Sciforums name. Lord Kelvin shared my fascination; he described Fourier's Series as a "mathematical poem."

Good luck,
Michael

Thnx guys! That clarified things alot.
If I need help now I know where to go. :)