Insertion Loss in a (T) Two Port Network

  • Thread starter agata78
  • Start date
  • #1
agata78
139
0

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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Answers and Replies

  • #2
gneill
Mentor
20,945
2,886

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.
 
  • #3
shaltera
90
0
INSLOSS.jpg


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

I suppose by 'N' you mean I1/I2
 
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