Question 2.7: USING THE PERTURBATION METHOD TO FIND THE ATTENUATION CONSTA...
USING THE PERTURBATION METHOD TO FIND THE ATTENUATION CONSTANT Use the perturbation method to find the attenuation constant of a coaxial line having a lossy dielectric and lossy conductors.
From Example 2.1 and (2.32)\ Z_{0}=\frac{V_{o}}{I_{o}} =\frac{E_{\rho }\ln b/a}{2\pi H_{\phi }} =\frac{\eta \ln b/a}{2\pi } =\sqrt{\frac{\mu }{\epsilon } } \frac{\ln b/a}{2\pi } , the fields of the lossless coaxial line are, for \ a\lt \rho \lt b,
\ \bar{E} =\frac{V_{o}\hat{\rho } }{\rho \ln b/a}e^{-j\beta z},\ \bar{H} =\frac{V_{o}\hat{\phi } }{2\pi \rho Z_{0}}e^{-j\beta z},
where\ Z_{0}=\left(\eta /2\pi \right) \ln b/a is the characteristic impedance of the coaxial line and \ V_{o} is the voltage across the line at\ z=0.The first step is to find \ P_{o} , the power flowing on the lossless line:
\ P_{o}=\frac{1}{2} Re\int_{s}^{}{\bar{E}\times \bar{H}^{*} } \cdot d\bar{s} =\frac{\left|V_{o}\right| ^{2}}{2Z_{0}}\int_{\rho =a}^{b}{\int_{\phi =0}^{2\pi }{\frac{\rho d\rho d\phi }{2\pi \rho ^{2}\ln b/a} } } =\frac{\left|V_{o}\right|^{2} }{2Z_{0}} ,
as expected from basic circuit theory.
The loss per unit length,\ p_{l}, comes from conductor loss \ \left(P_{lc}\right)and dielectric loss\ \left(P_{ld}\right). From (1.131)\ P^{t}=\frac{R_{s}}{2} \int_{S}^{}{\left|\bar{J_{s}} \right|^{2}ds } =\frac{R_{s}}{2} \int_{S}^{}{\left|\bar{H}_{t} \right|^{2} } ds ,
the conductor loss in a 1 m length of line can be found as
\ P_{lc}=\frac{R_{s}}{2} \int_{S}^{}{\left|\bar{H}_{t} \right|^{2} } ds = \frac{R_{s}}{2}\int_{z=0}^{1}{\left\{\int_{\phi =0}^{2\pi }{\left|H_{\phi }\left(\rho =a\right) \right|^{2}a d\phi+\int_{\phi =0}^{2\pi }{\left|H_{\phi }\left(\rho =b\right) \right|^{2}b d\phi} } \right\} } dz=\frac{R_{s}\left|V_{o}\right|^{2} }{4\pi Z_{0}^{2}} \left(\frac{1}{a}+\frac{1}{b} \right)The dielectric loss in a 1 m length of line is, from (1.92),
\ P_{l}=\frac{\sigma }{2} \int_{V}^{}{\left|\bar{E} \right|^{2}dv } +\frac{\omega }{2}\int_{V}^{}{\left(\epsilon ^{\prime \prime } \left|\bar{E} \right|^{2}+\mu ^{\prime \prime }\left|\bar{H} \right|^{2} \right)dv } ,
\ P_{ld}=\frac{\omega \epsilon ^{\prime \prime }}{2}\int_{V}^{}{\left|\bar{E} \right|^{2}ds } =\frac{\omega \epsilon ^{\prime \prime } }{2} \int_{\rho =a}^{b}{\int_{\phi =0}^{2\pi }{\int_{z=0}^{1}{\left|E_{\rho }\right|^{2} \rho }\;d\rho }\;d\phi }\;dz =\frac{\pi \omega \epsilon ^{\prime \prime }}{\ln b/a} \left|V_{o}\right| ^{2},
where\ \epsilon ^{\prime \prime }is the imaginary part of the complex permittivity, \ \epsilon =\epsilon ^{\prime }-j\epsilon ^{\prime \prime } . Finally, applying (2.96) \ \alpha= \frac{P_{l}\left(z\right) }{2P\left(z\right) } =\frac{P_{l}\left(z=0\right) }{2P_{o}} . gives
\ \alpha =\frac{P_{lc}+P_{ld}}{2P_{o}} =\frac{R_{s}}{4\pi Z_{0}} \left(\frac{1}{a}+\frac{1}{b} \right) +\frac{\pi \omega \epsilon ^{\prime \prime }Z_{0}}{\ln b/a} =\frac{R_{s}}{2\eta \ln b/a} \left(\frac{1}{a}+\frac{1}{b} \right)+\frac{\omega \epsilon ^{\prime \prime }\eta }{2} ,
where\ \eta =\sqrt{\mu /\epsilon ^{\prime }} . This result is seen to agree with that of Example 2.6.