Phase optimization

Question

$a_1$, $a_2$, and $a_3$ are three given real numbers, $b_1$, $b_2$, and $b_3$ are three given complex numbers. Solve this problem

$\underset{x} {\bf max} ~~|b_1e^{ja_1 x} + b_2e^{ja_2 x} + b_3e^{ja_3 x}|^2$

It can be further written as

\begin{align} &\underset{x}{\bf max} ~~|b_1e^{ja_1 x} + b_2e^{ja_2 x} + b_3e^{ja_3 x}|^2 \\ =&\underset{x}{\bf max} ~~|b_1|^2 + |b_2|^2 + |b_3|^2 + 2\textrm{Re}(b_1b_2^*e^{ja_1x}e^{-ja_2x}) + 2\textrm{Re}(b_2b_3^*e^{ja_2x}e^{-ja_3x}) \\ &~~~~~~~~~+ 2\textrm{Re}(b_1b_3^*e^{ja_1x}e^{-ja_3x}) \\ =&\underset{x}{\bf max} ~~|b_1||b_2|\cos((a_1-a_2)x+\angle{b_1}-\angle{b_2}) +|b_2||b_3|\cos((a_2-a_3)x+\angle{b_2}-\angle{b_3})+\\ &~~~~~~~~~~|b_1||b_3|\cos((a_1-a_3)x+\angle{b_1}-\angle{b_3}) \end{align}

If \begin{equation} \left\{ \begin{aligned} (a_1-a_2)x+\angle{b_1}-\angle{b_2} & = 2k_1\pi, k_1\in\mathbb{Z} \\ (a_1-a_3)x+\angle{b_1}-\angle{b_3} & = 2k_2\pi, k_2\in\mathbb{Z} \end{aligned} \right. \end{equation} all three $\cos()$ terms can achieve maximum value. But there are not always $k_1, k_2\in\mathbb{Z}$ that satisfy the above formula. Therefore there are some tradeoffs between three $\cos()$ terms.

Its derivative of $x$ is the addition of three $\sin()$ functions, and it is difficult to find the zero point of the derivative. I wonder if there is some way to solve this problem, even suboptimal solution. Thanks much!

Answer 1

We know the function (without the square) is at most $|b_1|+|b_2|+|b_3|$. In case of $b_1, b_2, b_3$ are all nonnegative (or nonpositive), this can be achieved for $x=0$. However, if $b_1, b_2, b_3$ don't have the same sign (and certainly things are worse when they are not even real), the solution may not even exist.

This is related to the Diophantine approximation. Consider the special case of $$f(x)=|1 - e^{(\pi i) x} - e^{(\pi i)\sqrt 2 x}|$$

Clearly, we have $f(x)\le 3$, and if $3$ can be achieved, then we must have $-1=e^{(\pi i)x}=e^{(\pi i)\sqrt 2 x}$, which implies both $x$ and $\sqrt 2x$ are odd integers, which implies $\sqrt 2 = \frac{\sqrt 2 x}{x}\in\mathbb Q$. Therefore $f(x)<3$ for all real $x$. On the other hand, thanks to a theorem of Scott as mentioned in this answer, there are infinitely many odd pairs $p, q$ such that $$|\sqrt 2 - \frac{p}{q}|\le \frac{1}{q^2} \Rightarrow|q\sqrt 2 - p| \le \frac{1}{q}$$and when $x=q$ is large, we have $$e^{(\pi i)x}=-1, e^{(\pi i) \sqrt 2 x}\approx e^{(\pi i) p}=-1$$ Hence $f(x)$ can be made to be as close to $3$ as possible.