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I'm just beginning to learn about Fourier series and I'm trying to figure out how to find the Fourier series coefficients for $$x(t) = e^{j100\pi t}$$

I know that $$x(t) = \sum_{-\infty}^{\infty} a_{k} e^{jk(2\pi/T)t}$$

How do I go about finding the coefficient(s)? I know that I can get T = 1/50, but beyond that I don't even know where to begin. I think I'm supposed to be able to do this just by looking at it without having to solve the integral equation for $a_k$, but I don't know what I'm supposed to be looking for.

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The answer is right there in the question. You have expressed the signal as linear sum of scaled complex exponentials.

Now the question is fourier series coefficients of a complex exponential$$e^{j100\pi t}$$

So you can see from the summation of x(t) that you have mentioned in the question that there is only one fourier series coefficient $a_1$ with a fundamental frequency $\omega_0$ because there is only one complex exponential in the input.The value of $\omega_0$ is $$ 100\pi$$. So $f_0$ = $(100\pi)/(2\pi)$ = 50

The value of $a_1$ is 1.

In other words you can express $$x(t)=(1).\ e^{(100 \pi)(1)t}$$ to see that k=1 and $a_1$=1.

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The answer depends on what period you assume / assign the signal to have (i.e. what sampling rate you use ). As long as the ground frequency is divisible by $50$ you will get exactly one non-zero Fourier coefficient, the position being dependent on the quotient between 50 and the ground frequency.

The special case Karan brings up above is the Fourier series with highest possible ground frequency still able to represent the function.

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    $\begingroup$ The period of a periodic function $f(t)$ is defined as the smallest value $T$ satisfying $f(t)=f(t+T)$. So for the function $f(t)=e^{j100\pi t}$ it makes a lot of sense to say that its fundamental frequency is $50$ Hz, and not $25$ Hz or $10$ Hz or whatever. Also, I'm not sure what you mean by "what sampling rate you use", because no mention has been made about discrete-time signals; the question exclusively deals with continuous-time signals. $\endgroup$
    – Matt L.
    Oct 30, 2015 at 23:00
  • $\begingroup$ I agree with you on the definition of periodic function and the most "natural" way to choose frequency may be as you say, but there exist many applications where you may want to choose another base frequency. Maybe you plan to measure other frequencies later on and then it would be convenient to plan ahead to have some place to put them or you may want to do operations on the signal in the Fourier domain afterwards and then the most compact representation would give less freedom for what you could do. $\endgroup$
    – mathreadler
    Oct 31, 2015 at 10:40

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