- #1
psie
- 108
- 10
- 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?
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?