condition for minimum attenuation

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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

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

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

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