Vrftsjtryk

From Schuam’s Outlines, Digital Signal Processing, Second Edition, 2012, page 44:

Book claims that solving this system of equations:

$$left[begin{matrix}1&1\e^{i pi/3}&e^{-i pi/3}\end{matrix}right]left[begin{matrix}A\B\end{matrix}right]=left[begin{matrix}0.5\0.75\end{matrix}right]$$

Yields:

$$left[begin{matrix}A\B\end{matrix}right]=ifrac{sqrt3}{3}left[begin{matrix}frac{1}{2}e^{-i pi/3}-frac{3}{4}\-frac{1}{2}e^{i pi/3}+frac{3}{4}\end{matrix}right]$$

I’m curious about the technique used to deal with the “complex conjugates” such that it yeilds that result of the book. When I try it, I get stuck when it comes to similfying the numerator and denominator down to book’s solution.

For Example:

$$A=0.5 – B$$
$${A e}^{i pi/3} + B e^{-i pi/3}=0.75$$
$${(0.5-B) e}^{i pi/3} + B e^{-i pi/3}=0.75$$
$$(0.5)(e^{i pi/3})-B (e^{i pi/3}) + B e^{-i pi/3}=0.75$$
$$(0.5)(e^{i pi/3})+(B)left(- e^{i pi/3} + e^{-i pi/3} right)=0.75$$
$$Bleft( e^{-i pi/3}- e^{i pi/3}right)=0.75 - (0.5)(e^{i pi/3})$$
$$B=frac{0.75 -(0.5)(e^{i pi/3})}{e^{-i pi/3}- e^{i pi/3}}$$

(At this point, I’m thinking that I need to multiply the numerator and denominator by the complex conjugate of the denominator to remove the complex portion from the denominator… however, I’m not really sure…is there some short cut you an take to handle this all in polar form?)

How can I convert my result for B into the result the book claims:

$$B=ifrac{sqrt3}{3}left(-frac{1}{2}e^{jpi/3}+frac{3}{4}right)$$

Picking $frac{1}{e^{-ipi /3}-e^{ipi /3}}$ and multiply and divide it by the complex conjugate of the denominator. This gives

$$frac{e^{ipi /3}-e^{-ipi /3}}{(e^{-ipi /3}-e^{ipi /3})*(e^{ipi /3}-e^{-ipi /3})}=frac{e^{ipi /3}-e^{-ipi /3}}{1-e^{-2ipi /3}-e^{2ipi /3}+1}=frac{e^{ipi /3}-e^{-ipi /3}}{1-(e^{-2ipi /3}+e^{2ipi /3})+1}$$

Using Moivre's relationship we know that $cos(theta)=frac{e^{itheta}+e^{-itheta}}2$ and hence:
$$e^{-2ipi /3}+e^{2ipi /3}=2cos(pi/3)=-1$$

Hence we have the last expression equal to: $$frac{e^{ipi /3}-e^{-ipi /3}}{3}$$
Using the Moivre's relationship again we have: $sin(theta)=frac{e^{itheta}-e^{-itheta}}{2i}$ and hence
$e^{ipi /3}-e^{-ipi /3}=2isin(pi/3)$ which lead us to the final result:
$$frac{2i sin(pi/3)}{3}=frac{isqrt 3}{3}$$