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How to calculate REAL and IMAGINARY impedance from reflection coeffcient?

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john fiction

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Hello everyone,

I've got a return loss value of -12.616 dB and a phase of -1.1339. And I need to calculate the real and imaginary part of the impedance. How do I do this?

Thanks in advance!
 
Last edited:

simple imaginary to polar co ordinates

a+jb=(sqrt(a^2+b^2))phase(b/a)

tan(theta)=b/a, a=r*cos(theta), b= r*sin(theta)

Hope that helps

Regards
Elchiquito
 
Thanks for your response elchiquito. What are the values a and b in your formula?
 

Don't bother I think I know what you're doing there. But which values are a and b in your equation?
 

ok. Let me put it straight for you.

reflection coefficient=(Zload-Z0)/(Zload+Z0);=(Zn-1)/(Zn+1)

Where Zload is the load impedance Z0=characteristic impedance

reflection coefficient is a quantity with both magnitude and phase; perform the algebra using the equations and solve for the magnitude and phase.

S11=reflection coefficient= r*cos(theta)+Jr*sin(theta)

and then rest is algebra. r=-12.616 theta=-1.139

Hope that helps

Regards
Elchiquito
 
Alright and if I calculated the values of a and b, I can then paste these in the Eq. 06 and 07 in the link:
**broken link removed**

a = -12.616*cos(-1.139) = -12.614
b = -12.616*sin(-1.139) = 0,251
Z0 = 50


Zin(REAL) = 50 * ( (1-(-12.61)^2 - (0,251)^2) / ((1--12.61)^2 + (0,251)^2) = -42.65 ohm
Why is this a negative?

Other than that, thank you very much for helping me, I really appreciate it.
 

sorry there is a small mistake, the values that you have presented in the post is in dB convert them to normal values and then perform the algebra and you should be able to get the correct value of zin

Regards
Elchiquito
 
It still gives a negative Zin(Real) => -80 Ohm.. Perhaps the return loss (s11(db)) should be a positive value instead of a negative one..?

edit: is it -20 or 20?
 

instead of -20 it's 20 you should be getting .227. now you should be able to get right value.
 
I meant -20 or 20 in the formula you gave me: s11(db)=20log(s11). But I guess it's 20. So it's 10^(-12.616/20) = 0,234
I get 80 ohm that's good.
 

Yeah now I see what you meant by that. Thanks for helping me out man. I couldnt' have done it without you!
 

no worries brother. it is always wise to handle this problems using smith chart.smith chart simplifies it to such an extent you wouldn't have to do any algebra whatsoever.

All the best brother

Regards
Elchiquito
 

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