condition for minimum attenuation

The attenuation constant $\alpha$ of a transmission line  is

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

$\alpha$ 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), $2\alpha&space;^{2}&space;=(RG-\omega&space;^{2}LC)+\sqrt{(R^{2}+\omega&space;^{2}L^{2})(G^{2}+\omega&space;^{2}C^{2})}$

differentiating the above equation w.r. to L

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

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

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

now      $\frac{d\alpha&space;}{dL}=0$.

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

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

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

$C&space;(R^{2}+\omega&space;^{2}L^{2})=L(G^{2}+\omega&space;^{2}C^{2})$.

$\Rightarrow&space;LG=RC$.

$\therefore&space;L=\frac{RC}{G}$  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), $2\alpha&space;^{2}&space;=(RG-\omega&space;^{2}LC)+\sqrt{(R^{2}+\omega&space;^{2}L^{2})(G^{2}+\omega&space;^{2}C^{2})}$

differentiating the above equation w.r. to L

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

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

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

now      $\frac{d\alpha&space;}{dL}=0$.

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

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

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

$C&space;(R^{2}+\omega&space;^{2}L^{2})=L(G^{2}+\omega&space;^{2}C^{2})$.

$\Rightarrow&space;LG=RC$.

$\therefore&space;C=\frac{LG}{R}$ 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.