Question

derivation for attenuation constant and phase constant for propagation constant of plane wave in lossy media

Answer 1

i'd done some equating stuffs and got
\[\alpha ^{2}-\beta^{2}+j2\alpha \beta=-w^{2}\mu \epsilon'+jw^{2}\mu \epsilon''\]
now i just need to equate the real and imaginary parts which will be\[\alpha^{2}-\beta^{2}=-w^2 \mu \epsilon'\]and\[2 \alpha \beta=w^2 \mu \epsilon''\]

Answer 2

but i cant reach the final derived expression,\[\alpha=w \sqrt{\frac{\mu \epsilon'}{2}[\sqrt{1+(\frac{\epsilon''}{\epsilon'})^{2}}-1]}\]\[\beta=w \sqrt{\frac{\mu \epsilon'}{2}[\sqrt{1+(\frac{\epsilon''}{\epsilon'})^{2}}+1]}\]

Answer 3

its like after equating the real and imaginary parts, the alpha and beta cant be factorized ompltely

Answer 4

use wolframalpha and check

Answer 5

wolfrom allows so many unknowns?

Answer 6

it can

Answer 7

how to make it interpret correctly? i tried
solve a^2-b^2+j(2ab)= -(omega)^2 (mu)(epsilon)+j(omega)^2(mu)(sigma/omega)
but it doesnt return the answer i need

Answer 8

isolate b from 2ab = ...
put it in a^2-b^2 = ....
you will get something in b^2 and b^4
assume b^2 =x
so you will get quadratic in x
use quadratic formula to solve
then plug back x = b^2 and take square root to find b
(cumbersome process, but you should get that)