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yungman
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I don't know whether this is the right place, should actually belong to EE but my experience is you really don't get a lot of help in EM over there because I notice there is only one or two people that attempted answer these.
This is regarding to finding the Q factor of a lossy tx line using as a quarter wave transformer. This is from "Field and Wave, Electromagnetics" by David Cheng 2nd edition, page 458.
In the book, it assume center frequency [itex] f_0[/itex] and frequency very close to the center frequency [itex] f_0 +\delta f [/itex]. So for the frequency very close to [itex] f_0,\;\; f= f_0 +\delta f [/itex], :
[tex] \beta l = \frac {2 \pi f}{v_p} l = \frac {2\pi ( f_0 + \delta f )} {v_p} l \;=\; \frac {n\pi}{2} + \frac {n\pi}{2} (\frac{\delta f}{f_0}) \;\;\hbox { where }\;n= 1,3,5...[/tex]
I don't agree with the book. From the link below, I proofed that the result change sign with different n. Please refer to the link below. The book gave:
[tex]cos \beta l = - sin [\frac{n\pi}{2}(\frac{\delta f}{f_0})] [/tex]
It should be:
[tex]cos \beta l = ^-_+ sin [\frac{n\pi}{2}(\frac{\delta f}{f_0})] [/tex]
Where -ve for n = 1,5,9... and +ve for n=3,7,11...
I actually posted in the calculus section to verify the trigonometry part:
https://www.physicsforums.com/showthread.php?t=475029
So my math part is correct. I cannot agree that the answer do not change sign. I have not been able to find another book I have that go so deep into working on Q factor in lossy lines. They are assume lossless line and skip this whole section. Please help.
Thanks
Alan
This is regarding to finding the Q factor of a lossy tx line using as a quarter wave transformer. This is from "Field and Wave, Electromagnetics" by David Cheng 2nd edition, page 458.
In the book, it assume center frequency [itex] f_0[/itex] and frequency very close to the center frequency [itex] f_0 +\delta f [/itex]. So for the frequency very close to [itex] f_0,\;\; f= f_0 +\delta f [/itex], :
[tex] \beta l = \frac {2 \pi f}{v_p} l = \frac {2\pi ( f_0 + \delta f )} {v_p} l \;=\; \frac {n\pi}{2} + \frac {n\pi}{2} (\frac{\delta f}{f_0}) \;\;\hbox { where }\;n= 1,3,5...[/tex]
I don't agree with the book. From the link below, I proofed that the result change sign with different n. Please refer to the link below. The book gave:
[tex]cos \beta l = - sin [\frac{n\pi}{2}(\frac{\delta f}{f_0})] [/tex]
It should be:
[tex]cos \beta l = ^-_+ sin [\frac{n\pi}{2}(\frac{\delta f}{f_0})] [/tex]
Where -ve for n = 1,5,9... and +ve for n=3,7,11...
I actually posted in the calculus section to verify the trigonometry part:
https://www.physicsforums.com/showthread.php?t=475029
So my math part is correct. I cannot agree that the answer do not change sign. I have not been able to find another book I have that go so deep into working on Q factor in lossy lines. They are assume lossless line and skip this whole section. Please help.
Thanks
Alan
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