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

AI Thread Summary
The discussion focuses on calculating the insertion loss of a T-network using given resistor values. The calculated insertion loss is approximately 3.015 dB, but the user expresses uncertainty about the simplicity of the result. To verify the calculation, alternative methods such as KCL (Kirchhoff's Current Law) analysis are suggested, though the user struggles with the lack of voltage values for proper analysis. The conversation emphasizes the need for establishing node voltages to proceed with KCL equations effectively. Ultimately, the user is encouraged to calculate power delivered to the load with and without the network components to confirm the insertion loss.
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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