For a lossy passive two-port network, $|S_{11}|$ and $|S_{21}|$ intersect at $-4\text{ dB}$.
The percentage of power dissipated in the lossy network at the intersection frequency is __.
(rounded off to two decimal places)
The question asks for the percentage of power dissipated in a lossy passive two-port network at the frequency where the magnitudes of the S-parameters, $|S_{11}|$ and $|S_{21}|$, intersect at -4 dB.
The magnitude of an S-parameter given in decibels (dB) can be converted to a power ratio using the formula:
$P_{ratio} = |S_{ij}|^2 = 10^{\frac{S_{dB}}{10}}$
where $S_{dB}$ is the S-parameter magnitude in dB.
At the intersection frequency for the lossy network, we have $|S_{11}|_{dB} = -4 \text{ dB}$ and $|S_{21}|_{dB} = -4 \text{ dB}$.
The corresponding power ratios are:
Calculating the value:
$10^{-0.4} \approx 0.3981$
Therefore, at the intersection frequency:
For a passive network, the sum of the power ratios ($|S_{11}|^2 + |S_{21}|^2$) represents the fraction of incident power that is either reflected or transmitted. The remaining power is dissipated.
The total power ratio accounted for by reflection and transmission is:
$|S_{11}|^2 + |S_{21}|^2 \approx 0.3981 + 0.3981 = 0.7962$
The fraction of power dissipated ($P_{diss\_ratio}$) is:
$P_{diss\_ratio} = 1 - (|S_{11}|^2 + |S_{21}|^2)$
$P_{diss\_ratio} \approx 1 - 0.7962 = 0.2038$
To find the percentage of power dissipated, we multiply the dissipated fraction by 100%:
$ \text{Percentage Dissipated} = P_{diss\_ratio} \times 100\% $
$ \text{Percentage Dissipated} \approx 0.2038 \times 100\% = 20.38\% $
Rounded to two decimal places, the percentage of power dissipated is 20.38%.
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