3
\$\begingroup\$

I start from the telegrapher's equation: \$-\frac{dV(z)}{dz}=(R'+j\omega L')I(z)\$, where \$V(z)\$ and \$I(z)\$ are the phasors of voltage and current respectively, in the transmission line model. \$R'\$ and \$L'\$ are resistance per unit length and inductance per unit length respectively.

The solution to the wave equation \$\frac{d^2V(z)}{dz^2}-\gamma^2V(z)=0\$ where \$\gamma=\sqrt{(R'+j\omega L')(G'+j\omega C')} \$ has the form \$V(z)=V_o^+e^{-\gamma z}+V_o^-e^{\gamma z} \$. \$G'\$ and \$C'\$ are respectively the conductance per unit length and capacitance per unit length of the transmission line.

From the telegrapher's equation we get: \$-\frac{dV(z)}{dz}=\gamma V_o^+e^{-\gamma z}-\gamma V_o^-e^{\gamma z}=\gamma (V_o^+e^{-\gamma z}-V_o^-e^{\gamma z})=(R'+j\omega L')I(z)\$

\$I(z)=\frac{\gamma}{R'+j\omega L'}(V_o^+e^{-\gamma z}-V_o^-e^{\gamma z})\$

...and I'm stuck here.

Given that characteristic impedance \$\frac{V_o^+}{I_o^+}=Z_o=\frac{V_o^-}{I_o^-}\$, how do I arrive at \$Z_o=\frac{R'+j\omega L'}{\gamma}=\sqrt{\frac{R'+j\omega L'}{G'+j\omega C'}}\$?

I'm not sure how to get \$Z_o\$ from \$\frac{V_o^+e^{-\gamma z}+V_o^-e^{\gamma z}}{I(z)}=\frac{V_o^+e^{-\gamma z}-V_o^-e^{\gamma z}}{I_o^+e^{-\gamma z}+I_o^-e^{\gamma z}}\$

\$\endgroup\$
5
\$\begingroup\$

This seems the simplest way to derive characteristic impedance. R, L, G and C represent: -

  • series resistance of cable/length
  • series inductance of cable/length
  • parallel conductance of cable/length
  • parallel capacitance of cable/length

Therefore the impedance looking into a short "lump" comprising these elements is: -

\$Z_0 = R + jwL + Z_o//\dfrac{1}{G + jwC}\$

In other words, the impedance looking in to the "lump" is the series impedance (\$R +jwL\$) plus the shunt components and the shunt components are: -

G and C plus the next "lump" which "offers" itself as another lump of \$Z_0\$.

\$Z_0 = R + jwL + \dfrac{\frac{Z_0}{G+jwC}}{Z_0 + \frac{1}{G+jwC}}\$

\$Z_0= R + jwL + \dfrac{Z_0}{1 + Z_0(G+jwC)}\$

multiplying through by \$1 + Z_0(G+jwC)\$ gives: -

\$Z_0[1 + Z_0(G+jwC)] = [R+jwL][1 + Z_0(G+jwC)]+Z_0\$

which becomes this: -

\$Z_0 + Z_0^2(G+jwC) = R+jwL + Z_0[(R+jwL)(G+jwC)]+Z_0\$

The important thing next is to recognize that \$(R+jwL)(G+jwC)\$ is insignificant as the "lump" approaches zero length and we are left with: -

\$Z_0^2(G+jwC) = R+jwL\$

hence \$Z_0 = \sqrt{\dfrac{R+jwL}{G+jwC}} \$

\$\endgroup\$
0
\$\begingroup\$

its very easy.First you put I(z)=Io+e-yz + Io-eyz.substitute Vo+=Io+*Zo and Vo-=-Io-*Zo in the equation where you got stuck .

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.