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

fouriermath

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.

Best Answer

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.
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