# Upper bound on "time-continuous Zadoff–Chu signals"

I'm trying to find a tight (as tight as possible) upper bound for $$\lvert x_u(t) \rvert$$, where $$x_u(t)$$ is the $$T$$-periodic signal defined by the Fourier sum $$x_u(t) = \frac{1}{N}\sum_{k=-N_0}^{N_0}X_u[k] e^{i2\pi kt/T}.$$ Here, $$N$$ is any odd prime, $$N_0 = (N-1)/2$$, and $$X_u[k]$$ are the DFT coefficients of the Zadoff–Chu sequence $$x_u[n]$$, namely \begin{align} X_u[k] &= \sum_{n=0}^{N-1}x_u[n] e^{-i 2 \pi nk/N} & x_u[n] &= e^{-i \pi u n (n+1)/N} \end{align} for any $$u=1,2,\dots,N-1$$. The upper bound does not need to depend on $$u$$ - in other words, an upper bound for $$\max_{u,t} \lvert x_u(t) \rvert$$ is sufficient.

I've obviously tried the triangle inequality and the Cauchy–Schwarz inequality, both of which result in $$\lvert x_u(t) \rvert \le \sqrt{N}$$ (it can be proven that $$\lvert X_u[k]\rvert = \sqrt{N}$$ independently of $$k$$ and $$u$$), but the bound is too loose.

The next thing I tried was to expand the $$X_u[k]$$ coefficients and write $$x_u(t)=\sum_{n=0}^{N-1}x_u[n]D_N\Bigl(\frac{t}{T} - \frac{n}{N}\Bigr)$$ where we introduce the Dirichlet kernel $$D_N(y) = \frac{1}{N}\sum_{k=-N_0}^{N_0}e^{i2\pi k y} =\frac{\sin(N\pi y)}{N\sin(\pi y)} =\frac{1}{N}+\frac{2}{N}\sum_{k=1}^{N_0-1}\cos(2\pi k y).$$ Then, by the triangle inequality and some symmetry arguments that allow fixing $$t=T/(2N)$$, one obtains $$\lvert x_u(t) \rvert \le \sum_{n=0}^{N-1}\Bigl\lvert D_N\Bigl(\frac{t}{T} - \frac{n}{N}\Bigr)\Bigr\rvert \le \sum_{n=0}^{N-1}\Bigl\lvert D_N\Bigl(\frac{2n-1}{2N}\Bigr)\Bigr\rvert \le \frac{4}{\pi}\sum_{n=1}^{2N-1}\frac{1}{n} - \frac{2}{\pi}\sum_{n=1}^{N-1}\frac{1}{n}$$ after using $$y(1-y/\pi) \le \sin y \le (4/\pi)y(1-y/\pi)$$ to bound the kernel. By well-known results about harmonic numbers, one can find the large-$$N$$ bound $$\lvert x_u(t) \rvert \lesssim \frac{2}{\pi} \ln N + \frac{4}{\pi} \ln 2 + \frac{2}{\pi} \gamma\tag{1}\label{eq:d_bound}$$ where $$\gamma$$ is the Euler—Mascheroni constant. This is a much better bound than the previous one, but graphing the function shows that it is still too pessimistic (case $$N=7$$ and case $$N=53$$ - in both cases, parameter $$u$$ is set to provide the highest possible peak).

At this point, I've started to suspect that one cannot rely only on the fact that $$\lvert x_u[n]\rvert = 1$$ and $$\lvert X_u[k]\rvert = \sqrt{N}$$ to find a meaningful bound, and the actual value of those coefficients must be taken into account. With this in mind, I tried to apply the approach of Upper bound for sum of the nth roots of unity to $$\lvert x_u(t)\rvert^2 = \frac{1}{N}\biggl\lvert \sum_{k=-N_0}^{N_0} e^{i \pi u u^{-1} k(u^{-1}k + 1)/N}e^{i2\pi k t/T}\biggr\rvert^2 =\frac{1}{N} \sum_{k=-N_0}^{N_0} e^{i \pi u u^{-1} k(u^{-1}k + 1)/N}e^{i2\pi k t/T} \sum_{l=-N_0}^{N_0} e^{-i \pi u u^{-1} l(u^{-1}l + 1)/N}e^{-i2\pi k t/T}.$$ The first identity follows from the fact that $$X_u[k] = X_u[0] x_u^*[u^{-1}k]$$, where $$u^{-1}$$ is the inverse of $$u$$ modulo $$N$$ (recall that $$N$$ is prime, so $$u^{-1}$$ is well defined) and where $$(\cdot)^*$$ denotes complex conjugate. Unfortunately, because of the quadratic term in the exponent (i.e., $$u^{-1} k(u^{-1}k + 1)$$), I don't see the way to express the double sum in terms of $$k-l$$ only and, in turn, to move forward.

Another similar approach consists in manipulating the above expression into $$\lvert x_u(t)\rvert^2 = 1 + \frac{2}{N}\sum_{m=1}^{N-1} \operatorname{Re}\Bigl(a_m e^{-i 2\pi m t/T}\Bigr)\tag{2}\label{eq:last}$$ where $$a_m = \sum_{k=-N_0}^{N_0-m}x_u^*[u^{-1}k]x_u[u^{-1}(k+m)] =x_u[u^{-1}m]\sum_{k=-N_0}^{N_0-m}e^{-i 2\pi u^{-1} km/N} =x_u[u^{-1}m]e^{i \pi u^{-1} m^2/N}\frac{\sin(\pi u^{-1}m(N-m)/N)}{\sin(\pi u^{-1}m/N)}.$$ This time, again with a similar strategy as the one in Upper bound for sum of the nth roots of unity, I may be able to estimate $$\lvert a_m \rvert$$ (although it'll probably require approximating the $$\cos$$ with a polynomial of degree 8, instead of 4) but I'm not sure it's worth the effort since I wouldn't know how to use them - some numerical experiments show that, if I just apply $$\operatorname{Re}(a_m e^{-i 2\pi m t/T}) \le \lvert a_m \rvert$$ inside of $$\eqref{eq:last}$$, I end up with a bound that is slightly worse than $$\eqref{eq:d_bound}$$.

Attacking the problem from a somehow different angle, one can also think about a more "analysis-oriented" solution. For instance, looking to \eqref{eq:last}, we have $$\lvert x_u(t) \rvert^2 = f(e^{i2\pi t/T})$$ where $$f(z) = \operatorname{Re}\Bigl(1 + \frac{2}{N}\sum_{m=1}^{N-1}a_m^* z^m\Bigr)$$ is a function that is harmonic on the whole complex plane, since it's the real part of an entire function. Then I was thinking about Harnack's inequality, but I couldn't really see how to apply it, since I would have to estimate a lower bound on $$f(z)$$ for $$\lvert z \rvert \le R$$, $$R > 1$$, in order to shift the function and ensure its nonnegativity on the ball of radius $$R$$, but this sounds as hard as the original problem.

To conclude, let me add a few thoughts and minor results that may be useful, although I couldn't find an application.

• Since $$x_u(nT/N) = x_u[n]$$, we readily see that $$\lvert x_u(nT/N)\rvert = 1$$, for all $$n \in \mathbb Z$$.
• Symmetry arguments, together with the previous point, suggest that the maximum is located at a point of the form $$t=(2m+1)/(2N)$$ for some $$m =0,1,\dots, N-1$$. Actually, the plots suggest that $$t=(2N-1)/(2N)$$ is the maximum point (I couldn't prove either conjecture).
• Besides being $$N$$-periodic, the Zadoff–Chu sequence and its DFT exhibit the symmetries \begin{align} x_u[N-n-1]&=x_u[n] & X_u[-k] &= X_u[k]e^{-i2\pi k/N}. \end{align}
• Using the second identity above, it is straightforward to show that shifting $$x_u(t)$$ towards the right by $$T/(2N)$$ results in an even function, that is $$x_u(t-T/(2N))=x_u(-t-T/(2N))$$.