# How to find inverse Fourier transform of summ of delta functions?

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: $$X(j\omega) = \sum_{k=-\infty}^{\infty}\delta(\omega-2k+1)$$

• Have you tried using the definition of the (inverse) FT and the definition of the delta distribution? It's quite straight forward, really. Feb 6 at 19:35

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: $$\sum_{k=-\infty}^{\infty}\delta(\omega+1-2k)\leftrightarrow \frac{1}{2\pi}\sum_{k=-\infty}^{\infty}e^{j(2k-1)t}$$