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I've been going through this paper where we exploit the well known Fourier expansion of the model signal as shown in the image below.

fourier tranform of this signal

I've never come across this well known fourier expansion before. Can someone point me towards right direction to fill in the missing steps in above image and also the use of Bessel functions got me confused a bit. Any references would be of great help. Thanks.

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  • $\begingroup$ Avoid cross-posting the same question on multiple SE sites. Cross-posting is not appreciated, and should not be encouraged. You have the exact same question here, and copied the two answers below there. You seem satisfied with the answers below, so please consider deleting them there. $\endgroup$ – Gilles Sep 6 '16 at 6:00
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They're talking about the Fourier series expansion of the periodic function

$$f(\tau)=e^{-jx\sin(\tau)}=\sum_{n=-\infty}^{\infty}c_ne^{jn\tau}\tag{1}$$

which sometimes appears in the literature, e.g., when computing the spectrum of a frequency modulated sine wave. The complex Fourier coefficients in $(1)$ are given by

$$c_n=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(\tau)e^{-jn\tau}d\tau=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{j(-n\tau-x\sin(\tau))}d\tau=J_{-n}(x)\tag{2}$$

where the last equality follows from the integral representation of the Bessel function of the first kind of order $n$.

Combining $(1)$ and $(2)$ we get

$$f(\tau)=\sum_{n=-\infty}^{\infty}J_{-n}(x)e^{jn\tau}\tag{3}$$

With $x=2k_0A$ and $\tau=\omega_Dt$ you obtain the expansion used in the paper you quoted.

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    $\begingroup$ @AIaye: Sure, I hope it's clear now. $\endgroup$ – Matt L. Sep 4 '16 at 20:07
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Let the following integral be:

$$\int_0^T\sum_{n=-\infty}^\infty J_{-n}(2k_0 A)\exp\left(jn\omega_D t\right)\exp(-j\omega t)dt=B$$

\begin{align} \implies B&=\sum_{n=-\infty}^\infty J_{-n}(2k_0 A)\left(\int_0^T \exp\left(jn\omega_D t\right)\exp(-j\omega t)dt\right)\quad\\ &=\sum_{n=-\infty}^\infty J_{-n}(2k_0 A)\left(\int_0^T\exp(-j(\omega-n\omega_D) t)dt\right)\\ &=\sum_{n=-\infty}^\infty J_{-n}(2k_0 A)\cdot\underbrace{\frac{\left[\exp\left(-j(\omega-n\omega_D) T\right)\right]_0^T}{-j(\omega-n\omega_D)}}_{C}\tag{1}\\ \end{align}

The $C$ term in equation $(1)$ can be further expanded as follows:

\begin{align} \implies C&=\frac{\exp(j(n\omega_D-\omega) T)-1}{j(n\omega_D-\omega)}\\ &=\exp\left(j(n\omega_D-\omega) \frac T2\right)\frac{\left[\exp\left(j(n\omega_D-\omega) \frac T2\right)-\exp\left(-j(n\omega_D-\omega) \frac T2\right)\right]}{j(n\omega_D-\omega)}\\ &=2\exp\left(j(n\omega_D-\omega) \frac T2\right) \frac{\sin\left((n\omega_D-\omega) \frac T2\right)}{(n\omega_D-\omega)}\\ &=T\exp\left(-j(\omega-n\omega_D) \frac T2\right)\frac{\sin\left((n\omega_D-\omega) \frac T2\right)}{(n\omega_D-\omega)\frac T2}\\ &=T\exp\left(-j(\omega-n\omega_D) \frac T2\right)\mathrm{sinc}\left((\omega-n\omega_D) \frac T2\right)\tag{2} \end{align} The equation in the paper misses the $T$ term in the final summation as shown in equation $(2)$. For the Bessel functions see @Matt L.'s answer, and equation 7 in this link.

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  • $\begingroup$ You might want to use another letter instead of $A$ for the integral, because the letter $A$ is already used in the argument of the Bessel function. $\endgroup$ – Matt L. Sep 4 '16 at 13:24
  • $\begingroup$ Mysterious down-voter, a little comment would be appreciated. $\endgroup$ – Gilles Sep 8 '16 at 12:09

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