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I am practicing for my exam that I have this semester and I stumbled upon this one. How can i find inverse Fourier transform given: $$ X(j\omega) = \sum_{k=-\infty}^{\infty}\delta(\omega-2k+1) $$

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  • $\begingroup$ Have you tried using the definition of the (inverse) FT and the definition of the delta distribution? It's quite straight forward, really. $\endgroup$
    – Jazzmaniac
    Feb 6 at 19:35
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Using Duality Property, we have $X(j\omega) = \delta(\omega-\omega_{0})$. By using this and rewriting our function using $1-2k = -\omega_{0}$, we get: $$\begin{align} x(t) &= \frac{1}{2\pi}\int_{-\infty}^{\infty}\delta(\omega-\omega_{0})e^{j\omega t}d\omega \\ &= \frac{1}{2\pi}e^{j \omega_0 t} \end{align}$$ $$ \sum_{k=-\infty}^{\infty}\delta(\omega-\omega_{k})\leftrightarrow \frac{1}{2\pi}\sum_{k=-\infty}^{\infty}e^{j\omega_{k}t} $$ Hence: $$ \sum_{k=-\infty}^{\infty}\delta(\omega+1-2k)\leftrightarrow \frac{1}{2\pi}\sum_{k=-\infty}^{\infty}e^{j(2k-1)t} $$

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