- #1

yungman

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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

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

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

[tex]\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}}[/tex]

[tex]\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)}[/tex]

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

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

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

[tex]\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}}[/tex]

[tex]\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)}[/tex]

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

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