0
\$\begingroup\$

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?

\$\endgroup\$
1
  • 1
    \$\begingroup\$ I adjusted your display equation to try to make it say what I think you wanted to say. If I got it wrong, feel free to revert my edits. \$\endgroup\$
    – The Photon
    Commented Oct 8, 2014 at 19:22

1 Answer 1

1
\$\begingroup\$

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.

\$\endgroup\$
3
  • \$\begingroup\$ I got the lower limit for \$ \phi _{k} \$ from your equation but I couldn't get upper limit \$\endgroup\$ Commented Oct 9, 2014 at 4:22
  • \$\begingroup\$ The upper limit is just \$\arg\{s_0\}+\arccos(1-\epsilon^2)\$, and the lower limit is \$\arg\{s_0\}-\arccos(1-\epsilon^2)\$, so the difference between the two arguments is never greater than \$\arccos(1-\epsilon^2)\$. \$\endgroup\$
    – Matt L.
    Commented Oct 9, 2014 at 7:18
  • \$\begingroup\$ Does it mean they have assumed maximum magnitude of arg{S \$ _{o}\$} is arccos(1- \$ \epsilon ^2/2)\$ \$\endgroup\$ Commented Oct 9, 2014 at 9:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.