How can I obtain de inverse discrete Fourier Transform of the following equation:
\begin{equation} X(j\omega)=\sum_{k=-\infty}^{\infty} (-1)^k\delta(\omega-\frac{k\pi}{2}) \end{equation}
I have tried by expanding its Fourier series and then doing the inverse and by tables, but I end up obtaining different results, and I am unsure which is the correct one. A step-by-step solution would be very helpful.
Solution 1 (by Fourier Series)
The signal could be written by its Fourier series in this way
\begin{equation}
X(j\omega)=\frac{1}{\pi}\sum_{k=-\infty}^{\infty} (1-e^{-jk\pi})e^{jk2\omega}
\end{equation}
Doing the inverse by definition
\begin{equation}
x(n)=\frac{1}{2\pi}\int_{-\pi}^{\pi} X(j\omega)e^{j\omega n}d\omega
\end{equation}
Obtaining
\begin{equation}
x(n)=\frac{1}{\pi}\sum_{k=-\infty}^{\infty} (\delta(n+2k)-e^{-jk\pi}\delta(n+2k))
\end{equation}
Solution 2 (by table) \begin{equation} x(n)=\frac{1}{2\pi}\sum_{k=0}^{N-1} (-1)^k e^{jkn\frac{\pi}{2}} \end{equation}
Why don’t we just take the inverse Fourier Transform directly:
$$ \begin{align} x(n)&=\frac{1}{2\pi}\int_{-\infty}^{\infty}{X\left(j\omega\right)\cdot e^{jwn}\phantom{x}d\omega}\\ \\ &=\frac{1}{2\pi} \int_{-\infty}^{\infty}{\sum_{k=-\infty}^{\infty}{(-1)^{k}\cdot\delta\left(\omega-\frac{k\pi}{2}\right)}\cdot e^{jwn}\phantom{x}d\omega}\\ \\ &=\frac{1}{2\pi}\sum_{k=-\infty}^{\infty}{(-1)^{k}\cdot e^{jkn\frac{\pi}{2}}} \end{align} $$
This aligns with the table so now we know your first method is incorrect or can be further simplified.