I have a complex number s of the form s=[1/sqrt(NN\$_{t})] e^{j\phi_{k}}\$ S\$_{o}\$ is another complex number. It is given that |s-s\$ _{o}\$|<= \$\epsilon\$ where 0< \$\epsilon\$< 2
It is stated in the literature that this constrain can be written as
$$ \phi_{k}=\arg s \in\left[\gamma , \gamma + \delta \right] $$
where \$ \gamma \$ and \$ \delta \$ are given by \$ \gamma \$ =arg S\$ _{o} \$ – arccos(1- \$\epsilon ^2/2)\$
and \$\delta \$ =2arccos(1- \$ \epsilon^2/2)\$
Can anyone explain how is that possible?
Best Answer
This is only possible if \$|s|=|s_0|=1\$. Squaring the original inequality gives
$$|s-s_0|^2=|s|^2+|s_0|^2-2\cos(\Delta\phi)\tag{1}\le\epsilon^2$$
where \$\Delta\phi=\arg\{s\}-\arg\{s_0\}\$ is the phase difference between \$s\$ and \$s_0\$. If \$|s|=|s_0|=1\$ is satisfied we get from (1)
$$2(1-\cos(\Delta\phi))\le\epsilon^2$$
or, equivalently,
$$\cos(\Delta\phi)\ge1-\frac{\epsilon^2}{2}\tag{2}$$
From this inequality it follows that
$$|\Delta\phi|\le\arccos\left(1-\epsilon^2/2\right)$$
which is equivalent to the condition in your question.