Fourier transform of $\sin(\frac{\pi}{4} n)$

Question

$x(n)=\sin(\frac{\pi}{4}n)$

$X(v)=\frac{1}{2j}\delta(v-\frac{1}{8})-\frac{1}{2j}\delta(v+\frac{1}{8})$ ?

Is this result correct?

Answer 1

We know that Fourier Transform of $\dfrac{1}{2\pi}$ is $\sum\limits_{k=-\infty}^\infty\delta(v-2\pi k)$, i.e. \begin{align} \mathscr{F}\left(\dfrac{1}{2\pi}\right)&=\sum_{k=-\infty}^\infty\delta(v-2\pi k)\\ \implies\mathscr{F}(1)&=\pi\sum_{k=-\infty}^\infty\delta(v-2\pi k)\\ \implies\mathscr{F}\left(1\cdot e^{jv_0n}\right)&=2\pi\sum_{k=-\infty}^\infty\delta(v-v_0-2\pi k)\\ \implies\mathscr{F}\left(e^{jv_0n}\right)&=2\pi\sum_{k=-\infty}^\infty\delta(v-v_0-2\pi k)\\ \end{align} Now, \begin{align} \mathscr{F}\left[\sin\left(\frac{\pi}{4} n\right)\right]&=\dfrac{1}{2j}\mathscr{F}\left[e^{j\frac{\pi}{4}n}-e^{-j\frac{\pi}{4}n}\right]\\ &=\dfrac{1}{2j}\left[2\pi\sum_{k=-\infty}^\infty\delta(v-\frac{\pi}{4}-2\pi k)-2\pi\sum_{k=-\infty}^\infty\delta(v+\frac{\pi}{4}-2\pi k)\right]\\ &=\dfrac{\pi}{j}\sum_{k=-\infty}^\infty\left[\delta(v-\frac{\pi}{4}-2\pi k)-\delta(v+\frac{\pi}{4}-2\pi k)\right] \end{align}