# Fourier series coefficient for sin(wt+theta)

Am trying to find the Fourier series coefficient ck for the following function $$\begin{equation} x\left(t\right)=\sin\left(wt+\theta \right) \end{equation}$$ Here is my work $$\begin{equation} c_k=\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}\:x\left(t\right)e^{-jkwt}dt \end{equation}$$ Now x(t) can be written as $$\begin{equation} x\left(t\right)=\frac{\left(e^{j\left(wt+\theta \right)}-e^{-j\left(wt+\theta \:\right)}\right)}{2j} \end{equation}$$ However: $$\begin{equation} c_k=\frac{1}{T}\int _{-\frac{T}{2}}^{\frac{T}{2}}\:\frac{\left(e^{j\left(wt+\theta \:\right)}-e^{-j\left(wt+\theta \:\:\right)}\right)}{2j}e^{-jkwt}dt \end{equation}$$ 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.

## 2 Answers

Just pointers:

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.
• thank you very much for your answer. I am going through the integration and using the fact that \begin{equation} w_0\:=\:\frac{2\pi }{T_0} \end{equation} I am ending up with a zero!! can you elaborate more please – Raykh Feb 10 '19 at 19:15
• exactly! That's the beauty: Only for the exactly right frequency, you end up with a constant $\ne0$ that you integrate from $-\frac T2$ to $\frac T2$. perfect. That shows you're doing it right; it's what we call orthogonal frequencies. – Marcus Müller Feb 10 '19 at 19:47
• yeah but the whole thing is being like \begin{equation} c_k\:=\:0 \end{equation} That is, the entire integral is converging to zero. Would you please elaborate that in the answer. I really appreciate your help but am kinda lost hahaha – Raykh Feb 10 '19 at 19:59

From the double angle formulas, multiplying any harmonically related sine and/or cosine terms will result in zero when integrated over a period of the fundamental, except for $$\sin^2 k\theta\$$ and $$\cos^2 k\theta\$$.