July 16th, 2015, 06:39 AM  #1 
Newbie Joined: Mar 2012 Posts: 8 Thanks: 0  impedance magnitude rlc circuit parallel
Hi. I'd like to see the calculations that lead to the termination impedance formula in 1st picture >> ((a) is the right one answer). I saw another formula to calculate impedance (2nd image from left) and both are right, I've tried to insert some value for r,C,L and the result is correct in both obviously, but I'm interested in 1st one calculation, since I can't figure out how the final formula is obtained. Tvm in advance. Enrico /Italy
Last edited by skipjack; July 17th, 2015 at 05:53 AM. 
July 17th, 2015, 06:11 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,375 Thanks: 2010 
Starting with formula (a) in the first image, divide the numerator (assumed to be positive) and denominator by the numerator (so that the numerator becomes 1). If you don't see why that leads to the formula for Z in the second image, let us know your working so that we can identify any errors in it.

July 18th, 2015, 04:53 AM  #3 
Newbie Joined: Mar 2012 Posts: 8 Thanks: 0  impedance magnitude RLC circuit parallel
Tvm SkipJack. I think I solved. My question was to find calculation, steps that lead to (a) formula, that one in 1st image, but ...here it is. I think it is correct. 1st term is 1/ Y1 +Y2 +Y3 = Z=1/Y (being Y =admittance) ,then once found 2nd term I did Absolute value of that 1 (complex number absolute value = sqrt (a^2 +b^2) being complex number say x = a +jb) and I obtained 3rd term. What do you think... is it right? Tvm Skipjack for answered me Last edited by skipjack; July 18th, 2015 at 09:35 AM. 

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