# Electrical – Fourier series coefficient for sin(wt+theta)

fouriermath

Am trying to find the Fourier series coefficient ck for the following function
$$x\left(t\right)=\sin\left(wt+\theta \right)$$
Here is my work
$$c_k=\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}\:x\left(t\right)e^{-jkwt}dt$$
Now x(t) can be written as
$$x\left(t\right)=\frac{\left(e^{j\left(wt+\theta \right)}-e^{-j\left(wt+\theta \:\right)}\right)}{2j}$$
However:
$$c_k=\frac{1}{T}\int _{-\frac{T}{2}}^{\frac{T}{2}}\:\frac{\left(e^{j\left(wt+\theta \:\right)}-e^{-j\left(wt+\theta \:\:\right)}\right)}{2j}e^{-jkwt}dt$$
would look very messy and am ending up with 2 sinc functions for my answer that I don't even know what do with. I can't put all the steps here but if somebody would steer me in the right direction I would really appreciate it.

1. pull the $$\\frac1{2j}\$$ out of your integral, it's just a constant.
2. The integral is a linear operation – use that to split your sum, so you have two integrals; one over $$\e^{j\omega t + j\theta}e^{-jk\omega t}\$$ and one over $$\e^{-j\omega t - j\theta}e^{-jk\omega t}\$$.
3. $$\e^{j\omega t + \theta}=e^{j\omega t}\cdot e^{j\theta}\$$; this applies to both integral. $$\e^{\pm j\theta}\$$ is a constant and can be pulled out of your integral.
4. You end up with an integrated product: $$\e^{\pm j\omega t}\cdot e^{-jk\omega t}\$$. Combine that into a single exponential.
5. Notice how that integral is over a length of $$\T\$$, which is one period. It can only be non-zero for a single value.