Insertion Loss in a (T) Two Port Network

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The discussion focuses on calculating the insertion loss for a two-port network using the formula N = (R0 + R1) / (R0 - R1). The calculated value of N is approximately 5.83, leading to an insertion loss of 15.31 dB. Participants confirm the accuracy of the calculations but suggest clarifying the derivation of R0 and using parentheses for better readability in the math. Additionally, there is a mention that 'N' may refer to the current ratio I1/I2. Overall, the calculations and methodology are validated, with minor recommendations for improvement.
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Homework Statement



Calculate the insertion loss for the network in the diagram?

Homework Equations



20log(N)dB

N = R0 + R1 / R0 - R1

The Attempt at a Solution




N = R0 + R1 / R0 - R1

N = 141.42 + 100 / 141.42 - 100

N = 241.42 / 41.42

N = 5.82858

20log(5.82858) = 15.311

Insertion Loss = 15.31 dB

Can someone please confirm I am correct or incorrect?
 

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agata78 said:

Homework Statement



Calculate the insertion loss for the network in the diagram?

Homework Equations



20log(N)dB

N = (R0 + R1) / (R0 - R1)

The Attempt at a Solution




N = (R0 + R1) / (R0 - R1)

N = (141.42 + 100) / (141.42 - 100)

N = 241.42 / 41.42

N = 5.82858

20log(5.82858) = 15.311

Insertion Loss = 15.31 dB

Can someone please confirm I am correct or incorrect?


Looks good. You might use some parentheses to make the order of operations on your math clear :wink:

You probably should have explained where the R0 value came from, although it can be derived from the filter circuit itself with the assumption that it is impedance matched to the source and sink impedances.
 
INSLOSS.jpg


R_0{}=\sqrt{R^2{}_1{}+2R_{1}R_{2}}

I suppose by 'N' you mean I1/I2
 
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