Question:medium
For a lossless passive two-port network, \(|S_{11}|\) and \(|S_{21}|\) intersect at \(-3\) dB.
For a lossy passive two-port network, \(|S_{11}|\) and \(|S_{21}|\) intersect at \(-4\) dB.
The percentage of power dissipated in the lossy network at the intersection frequency is .
(rounded off to two decimal places)
For a lossless passive two-port network, \(|S_{11}|\) and \(|S_{21}|\) intersect at \(-3\) dB.
For a lossy passive two-port network, \(|S_{11}|\) and \(|S_{21}|\) intersect at \(-4\) dB.
The percentage of power dissipated in the lossy network at the intersection frequency is .
(rounded off to two decimal places)
For a lossy passive two-port network, \(|S_{11}|\) and \(|S_{21}|\) intersect at \(-4\) dB.
The percentage of power dissipated in the lossy network at the intersection frequency is .
(rounded off to two decimal places)
Show Hint
Use power conservation: fraction dissipated = 1 minus |S11|^2 minus |S21|^2, after converting the given dB value to a ratio.
Updated On: Jul 20, 2026
Show Solution
Correct Answer: 20.38
Solution and Explanation
Step 1: Set up the power balance.
At any port pair, the power that goes in either comes back as reflection, passes through as transmission, or is lost inside the network. So
\[ 1=|S_{11}|^2+|S_{21}|^2+P_{diss} \]
where $P_{diss}$ is the dissipated fraction.
Step 2: Turn the given dB numbers into plain ratios.
A value of $x$ dB means the power ratio is $10^{x/10}$. For the lossy case both curves meet at $-4$ dB, so
\[ |S_{11}|^2=|S_{21}|^2=10^{-4/10}=10^{-0.4} \]
Step 3: Work out the number.
\[ 10^{-0.4}\approx0.3981 \]
Step 4: Add the two equal fractions.
\[ |S_{11}|^2+|S_{21}|^2=2(0.3981)=0.7962 \]
Step 5: Get the dissipated share.
\[ P_{diss}=1-0.7962=0.2038 \]
As a check, the lossless curve meets at $-3$ dB, giving $2\times10^{-0.3}=2\times0.5012\approx1$, so nothing is lost there, exactly as expected for a lossless network.
Step 6: State the percentage.
\[ \boxed{20.38\%} \]
At any port pair, the power that goes in either comes back as reflection, passes through as transmission, or is lost inside the network. So
\[ 1=|S_{11}|^2+|S_{21}|^2+P_{diss} \]
where $P_{diss}$ is the dissipated fraction.
Step 2: Turn the given dB numbers into plain ratios.
A value of $x$ dB means the power ratio is $10^{x/10}$. For the lossy case both curves meet at $-4$ dB, so
\[ |S_{11}|^2=|S_{21}|^2=10^{-4/10}=10^{-0.4} \]
Step 3: Work out the number.
\[ 10^{-0.4}\approx0.3981 \]
Step 4: Add the two equal fractions.
\[ |S_{11}|^2+|S_{21}|^2=2(0.3981)=0.7962 \]
Step 5: Get the dissipated share.
\[ P_{diss}=1-0.7962=0.2038 \]
As a check, the lossless curve meets at $-3$ dB, giving $2\times10^{-0.3}=2\times0.5012\approx1$, so nothing is lost there, exactly as expected for a lossless network.
Step 6: State the percentage.
\[ \boxed{20.38\%} \]
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