# Fourier transform of a sum of chirps

A chirp signal is defined as follow: $$x(t)=\sin(\omega t^2)$$ I have the following modified chirp: $$y(t,N)=\sum_{k=1}^{N}\sin(\omega t+k\beta t^2)$$ My problem is to find the Fourier transform $Y(\Omega,N)$ of $y(t,N)$. Thanks in advance.

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This is somewhat standard being a Gaussian integral. You will have $$\int_{-\infty}^\infty e^{-i\Omega t}\sum_{k=1}^N\sin(\omega t+k\beta t^2)dt.$$ This reduces to the evaluation of the two integrals $$\int_{-\infty}^\infty e^{-i\Omega t\pm i(\omega t+k\beta t^2)}dt.$$ Now, complete the square in the exponent as $\left[\pm\sqrt{k}\beta t+\frac{(\Omega\pm\omega)}{2\sqrt{k}\beta}\right]^2-\frac{(\Omega\pm\omega)^2}{4k\beta^2}$ and you are left with $$e^{i\frac{(\Omega\pm\omega)^2}{4k\beta^2}}\int_{-\infty}^\infty e^{-i\left[\pm\sqrt{k}\beta t+\frac{(\Omega\pm\omega)}{2\sqrt{k}\beta}\right]^2}dt.$$ Finally, $$e^{i\frac{(\Omega\pm\omega)^2}{4k\beta^2}}\frac{1}{\sqrt{k}\beta}\int_{-\infty}^\infty e^{-i\tau^2}d\tau=e^{i\frac{(\Omega\pm\omega)^2}{4k\beta^2}}\frac{1}{\sqrt{k}\beta}(1-i)\sqrt{\frac{\pi}{2}}.$$ Turning back to your sum, you are done.