# Z-transform of $\cos(\omega_0 n(n+1))u[n]$

I'm doing some research on Zadoff-Chu sequences and as a part of it I wanted to find the Z-transform of: $$\cos(\omega_0 n(n+1))u[n]$$ Wolfram Alpha / Mathematica couldn't help out. I couldn't find a way to do this from identities so I tried doing it from 1st principles. \begin{aligned} \mathcal{Z}\{\cos(\omega_0 n(n+1))u[n]\} &= \sum_{n = 0}^\infty \cos(\omega_0 n(n+1)) z^{-n} \\ &= \frac{1}{2}\sum_{n = 0}^\infty (e^{j\omega_0(n+1)}z^{-1})^n + \frac{1}{2}\sum_{n = 0}^\infty(e^{-j\omega_0(n+1)}z^{-1})^n \end{aligned}

This is where I'm not sure how to proceed. If the initial expression was just $$\cos(\omega_0 n)u[n]$$ then it would be straight forward to use the infinite sum of the geometric series. But here the required relation is: $$\sum_{n = 0}^\infty (ka^{(n+1)})^n = \;? \;\;,\;\; |a| = 1$$ I've tried researching it and the closest I've come is looking at Jacobi Theta functions: https://en.wikipedia.org/wiki/Theta_function

Some simple MATLAB simulations show that the sequence does converge, I'm just not sure if there's a way to find a closed-form answer to it analytically.

• Just a guess, but since there is an $n^2$ in the argument, this is a "chirp" which is directly related to a gaussian which has a known Fourier transform. Sep 4 '21 at 2:17
• Zadoff-Chu sequences are finite length sequences (or periodic sequences if you prefer) and so the chirp frequency does not increase forever. The $z$-transforms are polynomials in $z^{-1}$, not infinite series as you have written. Sep 4 '21 at 3:00