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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?

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    \$\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 Oct 8 '14 at 19:22
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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.

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  • \$\begingroup\$ I got the lower limit for \$ \phi _{k} \$ from your equation but I couldn't get upper limit \$\endgroup\$ – Aashish Sharma Oct 9 '14 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. Oct 9 '14 at 7:18
  • \$\begingroup\$ Does it mean they have assumed maximum magnitude of arg{S \$ _{o}\$} is arccos(1- \$ \epsilon ^2/2)\$ \$\endgroup\$ – Aashish Sharma Oct 9 '14 at 9:07

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