Deriving the elliptical polarization equation from Kraus

[yungman]

This is not a home work, it is part of the textbook on elliptical polarization. Attached is a page in Kraus Antenna book, I cannot verify the equation on the last line. Here is my work

$$E_y=E_2(\sin{\omega} t \cos \delta \;+\; \cos \omega {t} \sin \delta)$$ , $$\sin\omega {t} =\frac {E_x}{E_1}\;,\; \cos \omega {t} =\sqrt{1-(\frac{E_x}{E_1})^2}$$

$$\Rightarrow\; E_y=\frac {E_2 E_x\cos \delta}{E_1}\;+\;E_2\sqrt{1-(\frac {E_x}{E_1})^2} \;\sin\delta$$

$$\Rightarrow\; \sin \delta \;=\;\frac {E_y}{E_2\sqrt{1-(\frac{E_x}{E_1})^2}}\;-\; \frac{E_x\cos\delta}{E_1 \sqrt{1-(\frac{E_x}{E_1})^2}}$$

$$\Rightarrow\; \sin^2\delta\;=\;\frac{E^2_y}{E_2^2\;(1\;-\;(\frac{E_x}{E_1})^2)}\;-\;\frac{2E_y\;E_x\;\cos\delta}{E_1\;E_2\;(1\;-\;(\frac{E_x}{E_1})^2)}\;+\;\frac {E_x^2\;\cos^2\;\delta}{E_1^2\;(1\;-\;(\frac{E_x}{E_1})^2)}$$

Compare to the last line in the book, I just cannot get the last equation of the book. I checked it a few times and I just cannot see anything wrong with my derivation. Please take a look and see what I did wrong.

Thanks

Alan

 

[kuruman]

There are times when one has to declare the book wrong. I am saying this because
1. I cannot find anything wrong with your derivation and neither can you.
2. In order to reconcile your expression with the book's it must be true that $$1-\left(\frac{E_x}{E_1}\right)^2=1 ~\rightarrow~\frac{E_x}{E_1}=0$$
That last result is highly unlikely, therefore I think it's a typo which is more likely.