Fourier series of translated function

[psie]

Homework Statement

Find the Fourier series of $h(t)=e^{3it}f(t-4)$, when $f$ has period $2\pi$ and satisfies $f(t)=1$ for $|t|<2$, $f(t)=0$ for $2<|t|<\pi$.

Relevant Equations

Previously I worked an exercise where I showed that if $f$ has Fourier coefficients $(c_n)$, then the function $t\mapsto e^{iat}f(t)$ has Fourier coefficients $(c_{n-a})$ for $a\in\mathbb Z$. And similarly, the function $t\mapsto f(t-b)$ has Fourier coefficients $(e^{-inb}c_n)$ for $b\in\mathbb R$.

So here is my attempt. The result doesn't look very nice, so maybe there's a cleaner solution:

From the relevant equations, the coefficients of $h(t)$ should be $(e^{-i(n-3)4}c_{n-3})$, so I need to find $(c_n)$. They are given by, assuming $n\neq0$, \begin{align}\frac1{2\pi}\int_{-\pi}^\pi f(t)e^{-int}dt&=\frac1{2\pi}\int_{-2}^2 e^{-int}dt \nonumber \\ &=\frac1{2\pi}\left[-\frac{e^{-int}}{in}\right]_{-2}^2 \nonumber \\ &=\frac1{2\pi}\left(\frac{e^{i2n}}{in}-\frac{e^{-i2n}}{in}\right) \nonumber \\ &=\frac{\sin(2n)}{\pi n}.\nonumber\end{align} For $n=0$, we get simply $\frac2{\pi}$.

Recall the coefficient of $h(t)$ should be $(e^{-i(n-3)4}c_{n-3})$, so they are $$e^{-i(n-3)4}\frac{\sin (2(n-3))}{\pi(n-3)}\text{ for }n\neq 3,\quad \frac{2}{\pi} \text{ for }n=3 .$$ Therefor the (complex) Fourier series of $h(t)$ must be $$h(t)\sim\frac{2}{\pi}e^{i3t}+\sum_{\substack{k\in\mathbb Z \\ k\neq 3}}e^{-i(n-3)4}\frac{\sin (2(n-3))}{\pi(n-3)}e^{int}.$$

Unfortunately my book does not provide any answer to this exercise, so hence the post. Is this going in the right direction?

 

[anuttarasammyak]

You can check your result by performing inverse Fourier transform and seeing if it comes back.