## Appendix D Some Identities Involving Poisson Sums  
Here it is derived that  
\[\int\limits_{-\infty}^{\infty}e^{-j2\pi ft}\sum\limits_{n=-\infty}^{\infty} \delta(t-nT)dt=\frac{1}{T}\sum\limits_{m=-\infty}^{\infty}\delta\Big{(}f-\frac {m}{T}\Big{)}\] (A15)  
Alternative derivations can be found at [[https://en.wikipedia.org/wiki/Poisson_summation_formula](https://en.wikipedia.org/wiki/Poisson_summation_formula)]([https://en.wikipedia.org/wiki/Poisson_summation_formula](https://en.wikipedia.org/wiki/Poisson_summation_formula)). Before proceeding, it is worth keeping in mind that multiplying a time function by a string of Dirac deltas as found in the LHS of (A15) constitutes sampling it, which corresponds to convolving its Fourier spectrum by the string of Dirac deltas on the RHS of (A15), which constitutes a periodic extension of its Fourier spectrum. That is effectively what was done in the main text in relation to (14)-(21), but the sampling function was not just Dirac deltas, but pulses with a complex-valued \"shape\", which is a convolution of that pulse shape with a string of Dirac deltas as expressed by (17). So on the LHS is the Fourier transform of a string of Dirac deltas, which is a periodic function of time. As such, this periodic function has a Fourier series with coefficients \(\{c_{m}\}\) and can be expressed as a sum of the harmonics of the fundamental repetition frequency, \(F\), which is the reciprocal of the repetition period, \(T\). \[\sum\limits_{n=-\infty}^{\infty}\delta(t-nT)=\sum\limits_{m=-\infty}^{\infty} c_{m}e^{j2\pi nmFt}\] (A16)  
Figure A6: Direct representation of time-limited frequency-shift structure used for recovery of OAM mode. The Fourier coefficients are determined by the usual dot product of the periodic function with the complex conjugate of harmonic basis function, i.e., the integral over one period, normalized by the period length  
\[c_{m}=\frac{1}{T}\int\limits_{0}^{T}\sum\limits_{n=-\infty}^{\infty}\delta(t-nT )e^{-j2\pi\frac{\pi\pi}{T}}dt \tag{109}\]  
Only the \(n=0\) period of the string of Dirac deltas falls within the limits of the integral, so (109) becomes  
\[c_{m}=\frac{1}{T}\int\limits_{0}^{T}\delta(t)e^{-j2\pi\frac{\pi\pi}{T}}dt= \frac{1}{T}e^{-j2\pi\frac{\pi\pi}{T}0}=\frac{1}{T} \tag{110}\]  
which means the coefficients are all equal to 1/\(T\). That simplifies (108) to  
\[\sum\limits_{n=-\infty}^{\infty}\delta(t-nT)=\frac{1}{T}\sum\limits_{m=-\infty }^{\infty}e^{j2\pi mFt} \tag{111}\]  
That is now substituted in the LHS of (109) and the order of summation and integration is reversed, as each operation is linear, and really a form of addition. \[\int\limits_{-\infty}^{\infty}e^{-j2\pi ft}\frac{1}{T}\sum\limits_{m=-\infty}^ {\infty}e^{j2\pi mTt}dt=\frac{1}{T}\sum\limits_{m=-\infty}^{\infty}\int \limits_{-\infty}^{\infty}e^{-j2\pi(f-mF)t}dt \tag{112}\]  
The integral is now evaluated as a limit:  
\[ \begin{array}{l}\int\limits_{-\infty}^{\infty}e^{-j2\pi(f-mF)t}dt=\lim \limits_{\tau\rightarrow\infty}\int\limits_{-\tau/2}^{\tau/2}e^{-j2\pi(f-mF)t }dt=\cdots\\ \cdots=\lim\limits_{\tau\rightarrow\infty}\left[\frac{e^{-j2\pi(f-mF)\tau/2} -e^{j2\pi(f-mF)\tau/2}}{-j2\pi(f-mF)}\right]=\cdots\\ \cdots=\lim\limits_{\tau\rightarrow\infty}\tau\left[\frac{\sin((\tau(f-mF) \tau)\tau)}{\pi(f-mF)\tau}\right]=\delta(f-mF)\end{array}  \tag{113}\]  
The Dirac delta function in frequency arises in the limit due to the properties of the \(\sin(x)/x\) = \(\mathrm{sinc}(x/\pi)\) function: it is equal to 1 where \(x\) = 0 (as a result of another limit) and in (113) the zero-crossings collapse toward \(f\) = \(mF\) in frequency, and tend to 0 for all other \(f\).