# Minimizing maximum instantaneous magnitude by applying phase shift

Suppose I have the following sum, where all the input variables are positive real numbers while the output is complex-valued data

$$a e^{i \left( x + y \right)} + a e^{i \left( x - y \right)} + a b e^{i \left( x + z \right)} \tag{1}$$

Here I'm assuming $$x$$ is running from zero upwards while all the other variables stay constant.

Now, if $$b$$ was $$0$$, then the value of $$z$$ would have no effect. In that case, the instantaneous magnitude of the resulting sum would go as high as $$2$$ and as low as $$0$$ because then the sum would actually be equal to

$$a e^{i x} \cdot 2 \cos \left( y \right) \tag{2}$$

Therefore, if I divided the lowest resulting magnitude (i.e. absolute value) by the highest (let's call them $$m_{\textrm{l}}$$ and $$m_{\textrm{h}}$$), I would get $$\frac{0}{2} = 0$$. Let's call the result $$q$$ (meaning quotient).

If $$b$$ was equal to $$1$$ and at the same time $$z$$ was equal to a multiple of $$\pi$$, then I would again get $$q = \frac{m_{\textrm{l}}}{m_{\textrm{h}}} = 0$$ because then I would have $$m_{\textrm{l}} = 0$$ and $$m_{\textrm{h}}$$ = 3.

However, if $$b = 1$$ and at the same time $$z$$ is an odd multiple of $$\pi/2$$, then $$m_{\textrm{l}}$$ never goes all the way down to $$0$$ and also $$m_{\textrm{h}}$$ never goes all the way up to $$3$$. So in this case, $$q$$ is no longer $$0$$.

My question is: "What values should I choose for $$b$$ and $$z$$ if I want to minimize $$m_{\textrm{h}}$$ and at the same time maximize $$q$$?

• You stated all the other variables remain constant, so in that case what is $y$? Your formula $ae^{jx}2\cos(y)$ could always be 0 if $y=\pi/2$. Based on your further description I am guessing that besides sweeping $x$ you also sweep $y$, is that correct? What is this for? Mar 3 at 12:49
• Oops, my fault. Of course, you were right, I totally forgot about that. The y was supposed to run from 0 upwards as well, only at a much slower rate than the x. Which means that I’ve accidentally specified some absolutely incorrect conditions here. Not sure if I should actually edit my post. My idea was something like taking a sum of 3 complex exponentials of different frequencies, let's say 40 Hz and 45 Hz and 50 Hz, with the 45 Hz one having a different starting phase from the other two (where its individual amplitude coefficient is given by b and the phase offset is given by z). Mar 3 at 23:27
• Ok please don’t edit it such that it is a different question since my answer I gave now matches the question as it was originally asked. If my answer seems correct to you, it would be better to accept that as you did and then post another question if what you want to ask is different- thanks! Mar 4 at 0:23
• But feel free to post your new question, however I think what you want is $ae^{j(\omega_1-y)t} + ae^{j(\omega_1+y)t} + abe^{j(\omega_1t+z)}$, right? (In which case $\omega_1 = 2\pi 45$ rad/sec, $y = 2\pi 5$ to get your 40, 45 and 50 Hz and to have $z$ as a phase offset rather than a frequency offset). Mar 4 at 2:52

Assuming I understood the formulation properly, I don't see that there is a solution for a specific $$b$$ and $$z$$ that would simultaneously minimize $$m_H$$ and maximize $$q$$. However, for any given $$b$$, we maximize $$q$$ while minimizing $$m_H$$ when $$z= \pm\pi/2$$ for $$z \in [0, 2\pi)$$.

Here is how I proceeded:

Starting with the OP's formula:

$$m = ae^{j(x+y)} + ae^{j(x-y)} + abe^{j(x+z)}$$

I factored out the common $$ae^{jx}$$ to get:

$$m = ae^{jx}(e^{jy}+e^{-jy} + be^{jz}) = ae^{jx}(2\cos(y) + be^{jz})$$

With:

\begin{align} m_H &= |m|_{max},\\ m_L &= |m|_{min},\\ q &= m_L/m_H \end{align}

Note that $$ae^{jx}$$ has a constant magnitude $$|ae^{jx}|=a$$, so the phasor $$e^{jx}$$ has no effect on $$m_H$$, $$m_L$$ or $$q$$. $$a$$ itself is just a constant scaling, so as long as $$a>0$$, if we maximize the inner term $$2\cos(y)+be^{jz}$$ we also maximize $$m$$. $$ae^{jx}$$ also factors out completely when we calculate $$q$$. Thus I will simplify the OP's formula to work on:

$$n = 2\cos(y) + be^{jz}$$

And from that ask if we can minimize $$|n|_{max}$$ while maximizing $$q=|n|_{min}/|n|_{max}$$?

The OP did not say this specifically but I assume that $$y$$ is not restricted such that $$y \in [0, 2\pi)$$. A vector diagram on a complex plane will be helpful towards possible solutions. Below first shows what would occur if we were to choose $$z=0$$. For all values of $$|b|\le2$$, the sum will still pass through the origin and thus $$q=0$$. Thus $$|b|>2$$ when $$z=0$$, but we also note that when $$b=2$$, $$|n|_{max}=4$$ and increases from there as $$b+2$$.

For any other angle $$z$$, the sum will not go through the origin (such that $$|n|_{min}>0$$), and the maximum as the Cartesian distance from the origin will be smaller for any given $$b$$ with $$|b|>0$$. We minimize that Cartesian distance while maximizing the minimum magnitude by choosing $$z=\pi/2$$ (or similarly $$z=-\pi/2$$):

Below is a plot of the results as we sweep $$b$$ using the optimum value for $$z$$ as $$z=\pi/2$$:

Where we see as we can with the math that $$q \rightarrow 1$$ as $$b \rightarrow \infty$$, but $$|n|_{max}\rightarrow \infty$$ as $$b \rightarrow \infty$$, so there is no local maxima of $$q$$ with a minimum $$|n|_{max}$$, so we cannot choose a value of $$b$$ and $$z$$ to do this. However, for any given $$b$$, we can maximize $$q$$ while minimizing $$|n|_{max}$$ when $$z=\pi/2$$ as demonstrated here if that is what the OP intended to seek, and more generally with $$z=\pm \pi/2$$ for $$z \in [0, 2\pi)$$.