How Do You Calculate Insertion Loss in a T-Network?

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Discussion Overview

The discussion revolves around calculating the insertion loss of a T-network given specific resistor values. Participants explore different methods to verify the calculation, including KCL (Kirchhoff's Current Law) analysis and power delivery comparisons.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • The initial calculation of insertion loss using a transmission matrix is presented, yielding a result of 3.015 dB.
  • Some participants suggest verifying the insertion loss by using KCL equations, questioning whether the same result is obtained through that method.
  • There is a discussion about the challenges of performing KCL analysis without known voltage or current values, leading to difficulties in solving the equations.
  • One participant proposes labeling arbitrary voltages at each node to facilitate KCL analysis, but expresses difficulty in proceeding without a voltage source value.
  • Another approach is suggested, where participants calculate power delivered to the load with and without certain resistors in place to determine the insertion loss through power ratios.

Areas of Agreement / Disagreement

Participants generally agree on the need to verify the insertion loss calculation through alternative methods, but there is no consensus on the best approach or resolution of the challenges faced in the KCL analysis.

Contextual Notes

Participants express uncertainty regarding the necessity of known voltage values for KCL analysis and the implications of using different methods to calculate insertion loss. The discussion remains open-ended with various proposed methods but no definitive resolution.

Jason-Li
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Homework Statement



Calculate the insertion loss of the T-network given the values:
Rs = 75Ω
Rl = 100Ω
Ra = 13Ω
Rb = 13Ω
Rc = 213Ω
The transmission matrix and Network are below:

Q.png


2. Homework Equations

q1.png


The Attempt at a Solution



Constructing the matrix:
A = 1+Ra/Rc = 1+13/213 = 1.061
B = Ra+Rb+(Ra*Rb)/Rc = 13+13+(13*13)/213 = 26.7934
C = 1/Rc = 1/213 = 0.004694
D = 1+Rb/Rc = 1+13/213 = 1.061

[ A B ] = [ 1.061 26.7934 ]
[ C D ] = [ 0.004694 1.061 ]

Insertion Loss Ail = 20log [ (ARl+B+(C*Rl+D)*Rs) / ( Rs + Rl ) ]
= 20log [ (1.061*100+26.7934+(0.00469*100+1.061)*75) / (75+100) ]
=20log [ 247.6434 / 175 ]
=20log [ 1.415 ]
=3.015 db

Would someone be able to advise if this is correct, seems far too simple to be correct. Any help is appreciated and thanks in advance.
 

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You can check your answer by writing the KCL equations and solving them. Do you get the same insertion loss using that method? Can you show that work? :smile:
 
berkeman said:
You can check your answer by writing the KCL equations and solving them. Do you get the same insertion loss using that method? Can you show that work? :smile:

Hi Berkeman, I'm trying to do the KCL using loop analysis to check but as I don't have values for voltage or current I can't seem to be able to work it out? It's the value of V I am struggling to cancel out.
 
KCL analysis uses node equations, not loops. Just label arbitrary voltages at each node, and calculate Vo/Vi...
 
berkeman said:
KCL analysis uses node equations, not loops. Just label arbitrary voltages at each node, and calculate Vo/Vi...

I'm trying to do this but can't seem to work it without a value for the voltage source.

So i made the three nodes at the top V1, V2 & V3 and made the following equations

V2/13 - V3/13 - V3/100 = 0
V2/13 - V1/13 - V1/75 = 0
V1/13 + V3/13 - V2/213 - 2*V2/13 = 0

Struggling to move on from here without at least one of the V Values. Even if I do V3 / V1 so that
( V2/13 - V3/13 - V3/100 ) / ( V2/13 - V1/13 - V1/75 ) = 0
Which I got to V3 = (100*V2) / 113
 
Let the circle just below be a voltage source of Vs. Calculate the power delivered to the load without Ra, Rb and Rc in place; with just a direct connection from the output of Rs to RL. Then calculate the power delivered to RL WITH Ra, Rb and Rc in place. The ratio of the two powers will give the insertion loss.
 

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