User:Constant314/Telegrapher's equations in the frequency domain

The telegrapher's equations are a set of two coupled, linear equations that predict the voltage and current distributions on a linear electrical transmission line. The equations are important because they allow transmission lines to be analyzed using circuit theory. The equations and their solutions are applicable from 0 Hz to frequencies at which the transmission line structure can support higher order waveguide modes. The equations can be expressed in both the time domain and the frequency domain. In the time domain approach the dynamical variables are functions of time and distance. The resulting time domain equations are partial differential equations of both time and distance. In the frequency domain approach the dynamical variables are functions of frequency, $$ \omega $$, or complex frequency, $$ s $$, and distance $$ x $$. The frequency domain variables can be taken as the Laplace transform or Fourier transform of the time domain variables or they can be taken to be phasors. The resulting frequency domain equations are ordinary differential equations of distance. An advantage of the frequency domain approach is that differential operators in the time domain become algebraic operations in frequency domain.

The Telegrapher's Equations are developed in similar forms in the following references: Kraus, Hayt, Marshall, Sadiku, Harrington, Karakash, Metzger.

Model
A transmission line of overall length $$ d $$ is modeled as a succession of short segments having a length of $$ \Delta x $$. Each segment has a series resister, $$ R \Delta x $$, a series inductor, $$ L \Delta x $$, a shunt capacitor, $$ C \Delta x $$, and a shunt conductance, $$ G \Delta x $$. The value of each element is proportional to the length of the segment. For example, if $$ R $$ was 1 Ω per meter and the segment length, $$ \Delta x $$, was 10 meters, then the resister for that segment would have a value of 10 Ώ. This model will be accurate, so long as the segment length is much shorter than the shortest wavelength of interest. Mathematically it is possible to reduce the length of each segment to zero and determine equations for the liming values.

The circuit to the right represents a signal originating from a source on the left and propagating toward a load on the right. $$ x $$ is the distance from the left side of the transmission line. Each segment represents a length $$ \Delta x $$ of the transmission line. Using circuit theory, the following equations can be derived:
 * eqn. (1):    $$ V(x+\Delta x)  =  V(x)-(R\Delta x + sL\Delta x)I(x+\Delta x)    $$
 * eqn. (2):    $$ I(x+\Delta x)  =  I(x)-(G\Delta x + sC\Delta x)V(x)   $$

Dividing all quantities by $$ \Delta x$$ and allowing $$ \Delta x \rightarrow 0$$ produces two first order ordinary linear differential equations.

Move $$V(x)$$ and $$I(x)$$ to left side
 * $$ V(x+\Delta x) -V(x) = -(R\Delta x + sL\Delta x)I(x+\Delta x)    $$
 * $$ I(x+\Delta x) -I(x) = -(G\Delta x + sC\Delta x)V(x)   $$

Divide by $$ \Delta x $$
 * $$ \frac {V(x+\Delta x) -V(x)} {\Delta x} = -(R + sL)I(x+\Delta x)    $$
 * $$ \frac {I(x+\Delta x) -I(x)} {\Delta x} = -(G + sC)V(x)   $$

Let $$ \Delta x \rightarrow 0 $$


 * eqn. (3):    $$ \frac{d}{dx} V(x) = -(R + sL)I(x)  $$
 * eqn. (4):    $$ \frac{d}{dx} I(x) = -(G + sC)V(x)  $$

The first equation means that $$V_x$$,the propagating voltage at point $$x$$, is decreased by the voltage drop produced by $$I_x$$, the current at that point, passing through the series impedance $$(R + sL)$$. The second equation means that $$I_x$$, the propagating current at point $$x$$, is decreased by the shunt current produced by $$V_x$$, the voltage at that point, appearing across the shunt admittance $$(G + sC)$$. The two equations can be combined to produce two almost identical second order ordinary differential equation that have two natural solutions each. One natural solution is a forward (left to right) propagating wave and the other solution is a reverse (right to left) propagating waver. The complete solution is the a weighted sum of the forward solution and the reverse solution. The weights are determined by the conditions at the ends of the transmission line.

Differentiate equ. (3) and eqn. (4)
 * $$ \frac{d^2}{dx^2} V(x) = -(R + sL)\frac{d}{dx} I(x) $$
 * $$ \frac{d^2}{dx^2} I(x) = -(G + sC)\frac{d}{dx} V(x) $$

Substitute equ. (4) and eqn. (3) for $$ \frac{d}{dx} I(x) $$ and $$ \frac{d}{dx} V(x)  $$
 * $$ \frac{d^2} {dx^2} V(x) = (R + sL)(G + sC)V(x) $$
 * $$ \frac{d^2} {dx^2} I(x) = (R + sL)(G + sC)I(x) $$


 * eqn. (5):    $$ \frac{d^2} {dx^2} V(x) = \gamma^2 V(x)  $$ where $$ \gamma \equiv \sqrt { (R + sL)(G + sC)}  $$.
 * eqn. (6):    $$ \frac{d^2} {dx^2} I(x) = \gamma^2 I(x)  $$

The solutions for eqn. (5) are
 * eqn. (7):    $$ V_F(x) = V_F(0) e^{-\gamma x}$$  (forward voltage wave at point x).
 * eqn. (8):    $$ V_R(x) = V_R(d) e^{-\gamma (d-x)}$$  (reverse voltage wave at point x).
 * eqn. (9):    $$ V(x)=V_F(x)+V_R(x)$$  (total voltage at point x).

The solutions for eqn. (6) are
 * eqn. (10):    $$ I_F(x) = I_F(0) e^{-\gamma x} $$  (forward current wave at point x).
 * eqn. (11):    $$ I_R(x) = I_R(d) e^{-\gamma (d-x)} $$  (reverse current wave at point x).
 * eqn. (12):    $$ I(x)=I_F(x)-I_R(x)$$  (total current). The negative sign indicates that the current in the reverse wave is traveling in the opposite direction.

where
 * $$ \gamma = \alpha + j \beta = \sqrt{(R + s L)(G + s C)} $$.  $$  \alpha $$ is called the attenuation constant and $$  \beta $$ is called the phase constant.


 * $$ v = \frac \omega \beta $$ = the propagation velocity.
 * $$ \lambda = \frac {2 \pi} \beta $$ = wavelength.
 * $$ x $$ = distance from the left side of the transmission line.
 * $$ d $$ = length of the transmission line.
 * $$ V_F(0), I_F(0)   $$ = a phasors or functions that depends on conditions at $$  x=0   $$.
 * $$ V_R(d), I_R(d)   $$ = a phasors or functions that depends on conditions at $$  x=d   $$.
 * $$ I_F(0)   $$ = a phasor or function that depends on conditions at $$  x=0   $$.
 * $$ I_R(d)   $$ = a phasor or function that depends on conditions at $$  x=d   $$.

Solutions with $$ e^{-\gamma x}$$ are forward propagating solutions that traveling from left to right. Solutions with $$ e^{-\gamma (d-x)}$$ are reverse propagating solutions that traveling from right to left. The minus sign in the second reverse solution recognizes that the current in the reverse solution is going the opposite direction from the current reference direction which is taken to be the forward direction (left to right).

Characteristic impedance
Characteristic impedance is defined as
 * $$ Z_c \equiv \frac {V_F(x)}{I_F(x)} =\frac {V_R(x)}{I_R(x)} $$

It can be shown that

Substituting forward voltage and forward current into into eqn. (3) yields
 * $$ \frac{d}{dx} V_F(x) = -(R + sL)I_F(x) $$

Substituting eqn. (8) into that yields
 * $$ \frac{d}{dx} V_F(0) e^{-\gamma x} = -(R + sL)I_F(x) $$

Computing the derivative yields
 * $$ -\gamma V_F(0) e^{-\gamma x} = -(R + sL)I_F(x) $$

Noting that
 * $$ \gamma V_F(0) e^{-\gamma x} = \gamma V_F(x) $$

yields
 * $$ -\gamma V_F(x) = -(R + sL)I_F(x) $$

or
 * $$ \frac {V_F(x)} {I_F(x)} = \frac {(R + sL)} {\gamma} = \frac {(R + sL)} {\sqrt { (R + sL)(G + sC)}} =\sqrt{\frac {(R + s L)} {(G + s C)}}  $$

Therefore


 * eqn. (13):$$ Z_c = \sqrt{\frac {(R + s L)} {(G + s C)}}   $$.  = the characteristic impedance.

Input impedance of a semi-infinite transmission line


If a current, $$ I_{in} $$ is forced into the left end of an semi-infinite transmission line, there will be a voltage, $$ V_{in} $$. Whatever voltage is produced, the ratio of that voltage to the current that produced it will be an impedance. That impedance is called the input impedance. $$ Z_{in} \equiv \frac {V_{in}}{I_{in}} $$.

A transmission line is modeled as a succession of short segments having a length of $$\Delta x$$, as shown in sub-figure A. If the semi-infinite line is cut at any point, the line to the right of the cut is still a semi-infinite line. The ratio of voltage at that point to the current entering the remaining semi-infinite line is still $$ Z_{in} $$. If the cut is to the right of the first segment, all the segments to the right of the cut may be replaced with an impedance of $$ Z_{in} $$, as shown in sub-figure B. The impedance looking into the first segment terminated on the right by $$ Z_{in} $$ must be $$ Z_{in} $$, as shown in sub-figure C.  The mathematical equivalent of sub-figure C is:
 * $$ Z_{in}= R \Delta x + s L \Delta x + \frac 1 {G \Delta x + s C \Delta x + \frac 1 {Z_{in}}}  $$

This can be solved for

Multiplying the rightmost fraction by $$ \frac {Z_{in}} {Z_{in}} $$ and gathering similar terms yields
 * $$ Z_{in}= (R + s L) \Delta x + \frac {Z_{in}} {(G + s C )\Delta xZ_{in} + 1}   $$

Multiplying both sides of the equation by $$(G + s C )\Delta xZ_{in} + 1$$ yields:
 * $$ Z_{in}((G + s C )\Delta x Z_{in} + 1)= ((G  + s C )\Delta xZ_{in} + 1)(R + s L) \Delta x +  Z_{in}  $$

Subtracting $$ Z_{in} $$ from both sides of the equation
 * $$ Z_{in}((G + s C )\Delta x Z_{in} )= ((G  + s C )\Delta xZ_{in} + 1)(R + s L) \Delta x   $$

Dividing both sides of the equation by $$ \Delta x $$
 * $$ Z_{in}((G + s C )Z_{in} )= ((G  + s C )\Delta xZ_{in} + 1)(R + s L)    $$

The equation must work for all values of $$ \Delta x $$ so set $$ \Delta x =0$$.
 * $$ Z_{in}^2(G + s C )= (R + s L)    $$

therefore


 * $$ Z_{in} = \sqrt { \frac {( R + s L )}  {(G   + s C  )} } \equiv Z_c$$

Thus the input impedance of a semi-infinite transmission line is the same as the characteristic impedance.

$$ Z_c, \gamma, \alpha, \beta, \lambda,$$ and $$v$$ are secondary parameters which means that they can be expressed in terms of the primary parameters $$  R, G, L,$$ and $$C  $$.

General one-way solution


Johnson gives the following solution,

$$

\frac {V_L} {V_S} = {[(\frac {H^{-1} +H} 2)(1+\frac {Z_S}{Z_L})+(\frac {H^{-1} - H} 2)(\frac {Z_S}{Z_C} + \frac{Z_C}{Z_L}) ]}^{-1}

= \frac {Z_L Z_C} {Z_C(Z_L+Z_S)\cosh {\gamma x}+( {Z_L Z_S} + {Z_C}^2)\sinh {\gamma x} }

$$

with

$$ H = e^{-\gamma x} $$

Interpretation of the natural solutions
Substituting $$  \alpha + j \beta$$ for $$  \gamma $$ yields the following forward solution:
 * $$ V_F(x) = V_{(F,0)} e^{-(\alpha + j \beta) x} = V_{(F,0)}e^{-\alpha  x} e^{- j \beta x}$$

The term $$ e^{-\alpha x} $$ implies that the magnitude of the voltage decreases exponentially with distance $$ x $$ an extinction coefficient of $$ \alpha   $$ as the wave propagates in the forward direction. This means, for example, if the magnitude decreases by 1 dB in 100 m, then it decrease 2 dB in 200 m and 3dB in 300m. The units of $$ \alpha  $$ are nepers per unit length ( 1 neper ≈ 8.686 dB). The term $$ e^{-j\beta x} $$ implies that the phase lag of the voltage increases linearly with distance $$ x $$. This means that if the phase lags 30° after the first 100 m, then it lags 60° after 200 m and 90° after 300 m. The forward wave can also be considered to be a currnet wave
 * $$ I_F(x) = \frac {V_{(F,0)}} {Z_c} e^{-(\alpha + j \beta) x} $$

Reformulation using parameter ratios
The formulas of characteristic impedance and propagation constant can be reformulated into terms of simple parameter ratios by factoring.
 * $$ Z_c = \sqrt{\frac {(R_\omega + j \omega L_\omega)} {(G_\omega + j \omega C_\omega)}} = \sqrt { \frac {L_\omega} {C_\omega} } \sqrt{\frac {(1 - j r_\omega)} {(1 - j g_\omega)}}  $$



\gamma = \alpha + j \beta = \sqrt{(R_\omega + j \omega L_\omega)(G_\omega + j \omega C_\omega)} = j \omega \sqrt {L_\omega C_\omega} \sqrt{(\frac {R_\omega} { j \omega L_\omega} + 1)(\frac {G_\omega} { j \omega C_\omega} +1 )} = j \omega \sqrt {L_\omega C_\omega} \sqrt{(1 - j r_\omega)(1 - j g_\omega)} $$.


 * where $$ r_\omega=\frac {R_\omega} { \omega L_\omega},  g_\omega=\frac {G_\omega} {  \omega C_\omega}  $$  Note, $$ g_\omega $$ is also called dielectric loss tangent.

Where $$ \alpha $$ is called the attenuation constant and $$  \beta $$ is called the phase constant.

In conventional transmission lines, $$ C_\omega $$ and $$ L_\omega $$ are relatively constant compared to $$ r_\omega $$ and $$ g_\omega $$. Behavior of a transmission line over many orders of frequency is mainly determined by $$ r_\omega $$ and $$ g_\omega $$, each of which can be characterized as either being much less than unity, about equal to unity, much greater than unity, or infinite (at 0 Hz). Including 0 Hz, there are ten possible frequency regimes although in practice only six of them occur.

For round coaxial cable with a round center conductor, $$ R_\omega $$ and $$ L_\omega $$ can be computed from the formula for the impedance of round wire. $$ C_\omega $$ depends on the cable geometry and the dielectric constant of the insulation. $$ G_\omega $$ depends on the cable geometry, the dielectric loss tangent of the insulation, and the conductivity of the insulation. $$ C_\omega $$ and $$ G_\omega $$ have the same dependence on geometry, so the ratio $$ g_\omega=\frac {G_\omega} { \omega C_\omega}  $$ depends only on the parameters of the insulation.

These parameterizations were used to produce the following charts:


 * $$ C_\omega = C_0 $$ a constant independent of frequency.