Insertion Loss in a (T) Two Port Network

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SUMMARY

The insertion loss for the two-port network was calculated using the formula 20log(N)dB, where N is defined as (R0 + R1) / (R0 - R1). With R0 set at 141.42 ohms and R1 at 100 ohms, the calculated value of N is 5.82858, resulting in an insertion loss of 15.31 dB. The calculation was confirmed as correct, with suggestions to clarify the order of operations and the origin of the R0 value derived from the filter circuit's impedance matching.

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agata78
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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[itex]_0{}[/itex]=[itex]\sqrt{R^2{}_1{}+2R_{1}R_{2}}[/itex]

I suppose by 'N' you mean I1/I2
 
Last edited:

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