Using the telegrapher's equation 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 {{nowrap|e^{j \omega t}.}} Doing so allows to factor out the time dependence, 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. Consider a
steady-state problem such that the voltage and current can be written as: \begin{align} v(x,t) &= V(x) e^{j \omega t}\\[.5ex] i(x,t) &= I(x) e^{j \omega t} \end{align} Take the positive direction for and in the loop to be clockwise. Substitution in the telegraph equations and factoring out the time dependence e^{j \omega t} now gives: \begin{align} \frac{\mathrm{d}V}{\mathrm{d}x} &= -\left( R + j \omega L \right) I = -Z I,\\[.5ex] \frac{\mathrm{d}I}{\mathrm{d}x} &= -\left( G + j \omega C \right) V = -Y V, \end{align} with impedance and
admittance . Derivation and substitution of these two
first-order differential equations results in two uncoupled second-order differential equations: \begin{align} \frac{\mathrm{d}^2 V}{\mathrm{d}x^2} &= k^2 V,\\[.5ex] \frac{\mathrm{d}^2 I}{\mathrm{d}x^2} &= k^2 I, \end{align} with and called the
propagation constant. The solution to these types of equations can be written as: \begin{align} V(x) &= A e^{-k x} + B e^{k x}\\[.5ex] I(x) &= A_{1} e^{-k x} + B_{1} e^{k x} \end{align} with , , and the
constants of integration. Substituting these constants in the first-order system gives: \begin{align} A_1 &= \hphantom{-}A \frac{k}{R+ j \omega L}\\[.5ex] B_1 &= -B \frac{k}{R + j \omega L} \end{align} where \frac{A}{A_1} = -\frac{B}{B_1} = \frac{R + j \omega L}{k} = \sqrt{\frac{R+ j\omega L}{G + j \omega C}} = \sqrt{\frac{Z}{Y}} = Z_{0}. It can be seen that the constant , 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 '''' of the transmission line. Taking the positive root, this equation simplifies to: \ Z_\mathrm{IT} = \sqrt{ \frac{\ Z\ }{ Y }\ }\
Derivation Using this insight, many similar derivations exist in several books Here, we follow an approach posted by Tim Healy. The line is modeled by a series of differential segments with differential series elements () and shunt elements () (as shown in the figure at the beginning of the article). 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 . That is, the impedance looking into the line on the left is . But, of course, if we go down the line one differential length , the impedance into the line is still . Hence we can say that the impedance looking into the line on the far left is equal to in parallel with and , all of which is in series with and . Hence: \begin{align} &Z_0 = (R + j\omega L)\operatorname{d}\!x + \frac{ 1 }{(G + j\omega C)\operatorname{d}\!x + \frac{1}{Z_0}} \\ &Z_0 = (R + j\omega L)\operatorname{d}\!x + \frac{Z_0}{Z_0(G + j\omega C)\operatorname{d}\!x + 1} \\ &Z_0 + Z_0^2(G + j\omega C)\operatorname{d}\!x = (R + j\omega L)\operatorname{d}\!x + Z_0(G + j\omega C)\operatorname{d}\!x \,(R + j\omega L)\operatorname{d}\!x + Z_0 \end{align} The added terms cancel, leaving Z_0^2(G + j\omega C)\operatorname{d}\!x = (R + j\omega L)\operatorname{d}\!x + Z_0(G + j\omega C) (R + j\omega L) (\mathop{}\!\operatorname{d}\!x)^2 The first-power terms are the highest remaining order. Dividing out the common factor of , and dividing through by the factor , we get Z_0^2 = \frac{(R + j\omega L)}{(G + j\omega C)} + Z_0(R + j\omega L)\operatorname{d}\!x. In comparison to the factors whose divided out, the last term, which still carries a remaining factor , is infinitesimal relative to the other, now finite terms, so we can drop it. That leads to 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. == Lossless line ==