Using telegrapher's equation

Consider one section of the transmission line for the derivation of the characteristic impedance. The voltage on the left would be V and on the right side would be V + dV . This figure is to be used for both the derivation methods.

The differential equations describing the dependence of the voltage and current on time and space are linear, so that a linear combination of solutions is again a solution. This means that we can consider solutions with a time dependence ${\displaystyle e^{j\omega t}}$ – doing so is functionally equivalent of solving for the Fourier coefficients for voltage and current amplitudes at some fixed angular frequency ${\displaystyle \omega }$. Doing so causes the time dependence to factor out, leaving an ordinary differential equation for the coefficients, which will be phasors, dependent on position (space) only. Moreover, the parameters can be generalized to be frequency-dependent.^[1]

Let

$${\displaystyle V(x,t)\equiv V(x)\ e^{+j\omega t}}$$

and

$${\displaystyle I(x,t)\equiv I(x)\ e^{+j\omega t}}$$

Take the positive direction for ${\displaystyle V}$ and ${\displaystyle I}$ in the loop to be clockwise.

We find that

$${\displaystyle \mathrm {d} V=-\left(R+j\omega L\right)I\ \mathrm {d} x=-Z'\ I\ \mathrm {d} x}$$

and

$${\displaystyle \mathrm {d} I=-\left(G+j\omega C\right)V\ \mathrm {d} x=-Y'\ V\ \mathrm {d} x}$$

or

$${\displaystyle {\frac {\mathrm {d} V}{\mathrm {d} x}}=-Z'\ I\qquad {\text{ and }}\qquad {\frac {\mathrm {d} I}{\mathrm {d} x}}=-Y'\ V}$$

where

$${\displaystyle Z'\equiv R+j\omega L\qquad {\text{ and }}\qquad Y'\equiv G+j\omega C~.}$$

These two first-order equations are easily uncoupled by a second differentiation, with the results:

$${\displaystyle {\frac {\mathrm {d} ^{2}V}{\mathrm {d} x^{2}}}=Z'Y'\,V}$$

and

$${\displaystyle {\frac {\mathrm {d} ^{2}I}{\mathrm {d} x^{2}}}=Z'Y'\,I}$$

Notice that both ${\displaystyle V}$ and ${\displaystyle I}$ satisfy the same equation.

Since ${\displaystyle Z'Y'}$ is independent of ${\displaystyle x}$ and ${\displaystyle t}$, it can be represented by a single constant ${\displaystyle -k^{2}}$. (The minus sign is included for later convenience.) That is:

$${\displaystyle -k^{2}\equiv Z'\,Y'}$$

so

$${\displaystyle jk=\pm {\sqrt {Z'\,Y'\,}}}$$

We can write the above equation as

$${\displaystyle k=\pm \omega {\sqrt {\left(L-j{\frac {R}{\omega }}\right)\left(C-j{\frac {G}{\omega }}\right)\,}}=\pm \omega {\sqrt {L\ C\ }}{\sqrt {\left(1-j{\frac {R}{\omega L}}\right)\left(1-j{\frac {G}{\omega C}}\right)\,}}}$$

which is correct for any transmission line in general. And for typical transmission lines, that are carefully built from wire with low loss resistance ${\displaystyle R}$ and small insulation leakage conductance ${\displaystyle G}$; further, used for high frequencies, the inductive reactance ${\displaystyle \omega L}$ and the capacitive admittance ${\displaystyle \omega C}$ will both be large, so the constant ${\displaystyle k}$ is very close to being a real number:

$${\displaystyle k\approx \pm \omega {\sqrt {LC\,}}\,.}$$

With this definition of ${\displaystyle k}$, the position- or ${\displaystyle x}$-dependent part will appear as ${\displaystyle \pm j\,k\,x}$ in the exponential solutions of the equation, similar to the time-dependent part ${\displaystyle +j\,\omega \,t}$, so the solution reads

$${\displaystyle V(x)=v_{(+)}\ e^{-jkx}+v_{(-)}e^{+jkx}}$$

where ${\displaystyle v_{(+)}}$ and ${\displaystyle v_{(-)}}$ are the constants of integration for the forward moving (+) and backward moving (−) waves, as in the prior section. When we recombine the time-dependent part we obtain the full solution:

$${\displaystyle V(x,t)~=~V(x)\ e^{+j\omega t}~=~v_{(+)}\ e^{-jkx+j\omega t}+v_{(-)}e^{+jkx+j\omega t}\,.}$$

Since the equation for ${\displaystyle I}$ is the same form, it has a solution of the same form:

$${\displaystyle I(x)=i_{(+)}\ e^{-jkx}+i_{(-)}\ e^{+jkx}\,,}$$

where ${\displaystyle i_{(+)}}$ and ${\displaystyle i_{(-)}}$ are again constants of integration.

The above equations are the wave solution for ${\displaystyle V}$ and ${\displaystyle I}$. In order to be compatible, they must still satisfy the original differential equations, one of which is

$${\displaystyle {\frac {\mathrm {d} V}{\mathrm {d} x}}=-Z'I\,.}$$

Substituting the solutions for ${\displaystyle V}$ and ${\displaystyle I}$ into the above equation, we get

$${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\left[v_{(+)}\ e^{-jkx}+v_{(-)}\ e^{+jkx}\right]=-(R+j\omega L)\left[\ i_{(+)}\ e^{-jkx}+i_{(-)}\ e^{+jkx}\right]}$$

or

$${\displaystyle -jk\ v_{(+)}\ e^{-jkx}+jk\ v_{(-)}\ e^{+jkx}=-(R+j\omega L)\ i_{(+)}\ e^{-jkx}-(R+j\omega L)\ i_{(-)}\ e^{+jkx}}$$

Isolating distinct powers of ${\displaystyle e}$ and combining identical powers, we see that in order for the above equation to hold for all possible values of ${\displaystyle x}$ we must have:

• For the co-efficients of ${\displaystyle e^{-jkx}}$:
  $${\displaystyle -j\,k\ v_{(+)}=-(R+j\omega L)\ i_{(+)}}$$

  
• For the co-efficients of ${\displaystyle e^{+jkx}}$:
  $${\displaystyle +j\,k\ v_{(-)}=-(R+j\omega L)\ i_{(-)}}$$

  

Since ${\textstyle jk={\sqrt {(R+j\omega L)(G+j\omega C)\,}}}$

$${\displaystyle {\begin{aligned}+{\frac {v_{(+)}}{i_{(+)}}}&={\frac {R+j\omega L}{jk}}={\sqrt {{\frac {R+j\omega L}{G+j\omega C}}\ }}\equiv Z_{0}\\[1ex]-{\frac {v_{(-)}}{i_{(-)}}}&={\frac {R+j\omega L}{jk}}={\sqrt {{\frac {R+j\omega L}{G+j\omega C}}\ }}\equiv Z_{0}\end{aligned}}}$$

hence, for valid solutions require

$${\displaystyle v_{(+)}=+Z_{0}\ i_{(+)}\quad {\text{ and }}\quad v_{(-)}=-Z_{0}\ i_{(-)}}$$

It can be seen that the constant ${\displaystyle Z_{0}}$, defined in the above equations has the dimensions of impedance (ratio of voltage to current) and is a function of primary constants of the line and operating frequency. It is called the “characteristic impedance” of the transmission line, and conventionally denoted by ${\displaystyle Z_{0}}$.^[2]

$${\displaystyle Z_{0}\quad =\quad {\sqrt {{\frac {R+j\omega L}{G+j\omega C}}\ }}\quad =\quad {\sqrt {\ {\frac {\ L\ }{C}}\ }}{\sqrt {{\frac {\ 1-j\left({\frac {R}{\omega L}}\right)\,}{\ 1-j\left({\frac {G}{\omega C}}\right)\,}}\,}}}$$

which holds generally, for any transmission line. For well-functioning transmission lines, with either ${\displaystyle R}$ and ${\displaystyle G}$ both very small, or with ${\displaystyle \omega }$ very high, or all of the above, we get

$${\displaystyle Z_{0}\approx {\sqrt {{\frac {L}{C}}\,}}}$$

hence the characteristic impedance is typically very close to being a real number. Manufacturers make commercial cables to approximate this condition very closely over a wide range of frequencies.

Alternative approach

We follow an approach posted by Tim Healy.^[3] The line is modeled by a series of differential segments with differential series ${\displaystyle \left(R\ \mathrm {d} x,L\ \mathrm {d} x\right)}$ and shunt ${\displaystyle \left(C\ \mathrm {d} x,G\ \mathrm {d} x\right)}$ elements (as shown in the figure above). The characteristic impedance is defined as the ratio of the input voltage to the input current of a semi-infinite length of line. We call this impedance ${\displaystyle Z_{0}\,.}$ That is, the impedance looking into the line on the left is ${\displaystyle Z_{0}}$. But, of course, if we go down the line one differential length ${\displaystyle \mathrm {d} x}$, the impedance into the line is still ${\displaystyle Z_{0}}$. Hence we can say that the impedance looking into the line on the far left is equal to ${\displaystyle Z_{0}}$ in parallel with ${\displaystyle C\,\mathrm {d} x}$ and ${\displaystyle G\,\mathrm {d} x}$, all of which is in series with ${\displaystyle R\,\mathrm {d} x}$ and ${\displaystyle L\,\mathrm {d} x}$. Hence:

$${\displaystyle {\begin{aligned}Z_{0}&=(R+j\omega L)\ \mathrm {d} x+{\frac {1}{~(G+j\omega C)\ \mathrm {d} x+{\frac {1}{\ Z_{0}\ }}\,}}\\[1ex]Z_{0}&=(R+j\omega L)\ \mathrm {d} x+{\frac {\ Z_{0}\ }{Z_{0}\ (G+j\omega C)\ \mathrm {d} x+1\,}}\\[1ex]Z_{0}+Z_{0}^{2}\ (G+j\omega C)\ \mathrm {d} x&=(R+j\omega L)\ \mathrm {d} x+Z_{0}\ (G+j\omega C)\ \mathrm {d} x\ (R+j\omega L)\ \mathrm {d} x+Z_{0}\end{aligned}}}$$

The added ${\displaystyle Z_{0}}$ terms cancel, leaving

$${\displaystyle Z_{0}^{2}\ (G+j\omega C)\ \mathrm {d} x=(R+j\omega L)\ \mathrm {d} x+Z_{0}\ (G+j\omega C)\ (R+j\omega L)\ (\mathrm {d} x)^{2}}$$

The first-power ${\displaystyle \mathrm {d} x}$ terms are the highest remaining order. Dividing out the common factor of ${\displaystyle \mathrm {d} x}$, and dividing through by the factor ${\displaystyle (G+j\omega C)}$, we get

$${\displaystyle Z_{0}^{2}={\frac {(R+j\omega L)}{\ (G+j\omega C)\ }}+Z_{0}\ (R+j\omega L)\,\mathrm {d} x\,.}$$

In comparison to the factors whose ${\displaystyle \mathrm {d} x}$ divided out, the last term, which still carries a remaining factor ${\displaystyle \mathrm {d} x}$, is infinitesimal relative to the other, now finite terms, so we can drop it. That leads to

$${\displaystyle Z_{0}=\pm {\sqrt {{\frac {\ R+j\omega L\ }{G+j\omega C}}\,}}\,.}$$

Reversing the sign ± applied to the square root has the effect of reversing the direction of the flow of current.

Sources