condition for minimum attenuation

The attenuation constant \alpha of a transmission line  is

\alpha =\sqrt{\frac{(RG-\omega ^{2}LC)+\sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}}{2}}-----EQN(1)

\alpha depends up on the  4 primary constants  along with frequency \omega. In order to achieve the minimum attenuation , these primary constants should be varied in turn.

case 1:-

consider the  case where the attenuation (\alpha )  variation depends only on Inductance(L) .

i.e by considering R, G , C and \omega as constants 

from EQN(1), 2\alpha ^{2} =(RG-\omega ^{2}LC)+\sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}

differentiating the above equation w.r. to L

\therefore 4\alpha \frac{d\alpha }{dL} = -\omega ^{2}C + \frac{2L(\omega ^{2}G^{2}+\omega ^{4}C^{2})}{2 \sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}}.

4\alpha \frac{d\alpha }{dL} =\frac{-\omega ^{2}C \sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}+\omega ^{2}L(G^{2}+\omega ^{2}C^{2})}{\sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}}.

\frac{d\alpha }{dL} =\frac{-\omega ^{2}C \sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}+\omega ^{2}L(G^{2}+\omega ^{2}C^{2})}{4\alpha\sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}}.

now      \frac{d\alpha }{dL}=0.

-\omega ^{2}C \sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}+\omega ^{2}L(G^{2}+\omega ^{2}C^{2})=0.

\omega ^{2}C \sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}=\omega ^{2}L(G^{2}+\omega ^{2}C^{2}).

C \sqrt{(R^{2}+\omega ^{2}L^{2})}=L\sqrt{(G^{2}+\omega ^{2}C^{2})}.

C (R^{2}+\omega ^{2}L^{2})=L(G^{2}+\omega ^{2}C^{2}).

\Rightarrow LG=RC.

\therefore L=\frac{RC}{G}  H/m.

The condition for L to get minimum attenuation is   \frac{RC}{G}  H/m but in general L should be maintained at  less than this value.

case 2:-

consider the  case where the attenuation (\alpha )  variation depends only on Capacitance(C) .

i.e by considering R, L, G and \omega as constants 

from EQN(1), 2\alpha ^{2} =(RG-\omega ^{2}LC)+\sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}

differentiating the above equation w.r. to L

\therefore 4\alpha \frac{d\alpha }{dL} = -\omega ^{2}L + \frac{2C(\omega ^{2}R^{2}+\omega ^{4}L^{2})}{2 \sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}}.

4\alpha \frac{d\alpha }{dL} =\frac{-\omega ^{2}L \sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}+\omega ^{2}C(R^{2}+\omega ^{2}L^{2})}{\sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}}.

\frac{d\alpha }{dL} =\frac{-\omega ^{2}L \sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}+\omega ^{2}C(R^{2}+\omega ^{2}L^{2})}{4\alpha\sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}}.

now      \frac{d\alpha }{dL}=0.

-\omega ^{2}L \sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}+\omega ^{2}C(R^{2}+\omega ^{2}L^{2})=0.

\omega ^{2}L \sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}=\omega ^{2}C(R^{2}+\omega ^{2}L^{2}).

C \sqrt{(R^{2}+\omega ^{2}L^{2})}=L\sqrt{(G^{2}+\omega ^{2}C^{2})}.

C (R^{2}+\omega ^{2}L^{2})=L(G^{2}+\omega ^{2}C^{2}).

\Rightarrow LG=RC.

\therefore C=\frac{LG}{R} F/m.

The condition for C to get minimum attenuation is  \frac{LG}{R}  F/m but in general C should be maintained at  more  than this value.

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Author: Lakshmi Prasanna Ponnala

Completed M.Tech in Digital Electronics and Communication Systems and currently working as a faculty.