# Communication Systems/Wired Transmission

This page will discuss the topic of signal propagation through physical
mediums, such as wires.

## Transmission Line Equation

Many kinds of communication systems require signals at some point to be
conveyed over copper wires.

The following analysis requires two assumptions:

:   

    :   • A transmission line can be decomposed into small, distributed
        passive electrical elements
    :   • These elements are independent of frequency (i.e. although
        *reactance* is a function of frequency, resistance, capacitance
        and inductance are not)

These two assumptions limit the following analysis to frequencies up to
the low MHz region. The second assumption is particularly difficult to
defend since it is well known that the resistance of a wire increases
with frequency because the conduction cross-section decreases. This
phenomenon is known as the skin effect and is not easy to evaluate.

:   

    :   ![](transmission_line.gif "fig:transmission_line.gif")

The purpose behind the following mathematical manipulation is to obtain
an expression that defines the voltage (or current) at any time (*t*)
along any portion (*x*) of the transmission line. Later, this analysis
will be extended to include the frequency domain.

Recall the characteristic equations for inductors and capacitors:

$$v = L\frac{{\partial i}}{{\partial t}}$$ and
$i = C\frac{{\partial v}}{{\partial t}}$

### Kirchoff\'s Voltage Law

Kirchoff\'s voltage law (KVL) simply states that the sum of all voltage
potentials around a closed loop equal zero. Or in other words, if you
walked up a hill and back down, the net altitude change would be zero.

:   

    :   Applying KVL in the above circuit, we obtain:

$$v\left( {x,t} \right) = R\Delta xi\left( {x,t} \right) + L\Delta x\frac{{\partial i}}{{\partial t}}\left( {x,t} \right) + v\left( {x + \Delta x,t} \right)$$

:   

    :   Rearranging:

$$v\left( {x,t} \right) - v\left( {x + \Delta x,t} \right) = R\Delta xi\left( {x,t} \right) + L\Delta x\frac{{\partial i}}{{\partial t}}\left( {x,t} \right)$$

:   

    :   But the LHS (left hand side) of the above equation, represents
        the voltage drop across the cable element $\Delta v$, therefor:

$$\Delta v = R\Delta xi\left( {x,t} \right) + L\Delta x\frac{{\partial i}}{{\partial t}}\left( {x,t} \right)$$

:   

    :   Dividing through by $\Delta x$, we obtain:

$$\frac{{\Delta v}}{{\Delta x}} = Ri\left( {x,t} \right) + L\frac{{\partial i}}{{\partial t}}\left( {x,t} \right)$$

:   

    :   The LHS is easily recognized as a derivative. Simplifying the
        notation:

$$\frac{{\partial v}}{{\partial x}} = Ri + L\frac{{\partial i}}{{\partial t}}$$

This expression has both current and voltage in it. It would be
convenient to write the equation in terms of current or voltage as a
function of distance or time.

### Simplifying the Equation (trust me)

The first step in separating voltage and current is to take the
derivative with respect to the position *x* (**Equation 1**):

$$\frac{{\partial ^2 v}}{{\partial x^2 }} = R\frac{{\partial i}}{{\partial x}} + L\frac{{\partial ^2 i}}{{\partial x\partial t}}$$

:   

    :   The next step is to eliminate the current terms, leaving an
        expression with voltage only. The change in current along the
        line is equal to the current being shunted across the line
        through the capacitance *C* and conductance *G*. By applying KCL
        in the circuit, we obtain the necessary information (**Equation
        2**):

$$\frac{{\partial i}}{{\partial x}} = Gv + C\frac{{\partial v}}{{\partial t}}$$

:   

    :   Taking the derivative with respect to time, we obtain
        (**Equation 3**):

$$\frac{{\partial ^2 i}}{{\partial x\partial t}} = G\frac{{\partial v}}{{\partial t}} + C\frac{{\partial ^2 v}}{{\partial t^2 }}$$

:   

    :   Substituting (**Equation 2**) and (**Equation 3**) into
        (**Equation 1**), we obtain the desired simplification:

$$\frac{{\partial ^2 v}}{{\partial x^2 }} = R\left[ {Gv + C\frac{{\partial v}}{{\partial t}}} \right] + L\left[ {G\frac{{\partial v}}{{\partial t}} + C\frac{{\partial ^2 v}}{{\partial t^2 }}} \right]$$

:   

    :   Collecting the terms, we obtain:

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:   

    :   **The Transmission Line Equation for Voltage**

$$\frac{{\partial ^2 v}}{{\partial x^2 }} = RGv + \left( {RC + LG} \right)\frac{{\partial v}}{{\partial t}} + LC\frac{{\partial ^2 v}}{{\partial t^2 }}$$

This equation is known as the transmission line equation. Note that it
has voltage at any particular location *x* as a function of time *t*.

:   

    :   Similarly for current, we obtain:

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:   

    :   **The Transmission Line Equation for Current**

$$\frac{{\partial ^2 i}}{{\partial x^2 }} = RGi + \left( {RC + LG} \right)\frac{{\partial i}}{{\partial t}} + LC\frac{{\partial ^2 i}}{{\partial t^2 }}$$

But we\'re not quite done yet.

### Solving the Transmission Line Equation

Historically, a mathematician would solve the transmission line equation
for *v* by assuming a solution for *v*, substituting it into the
equation, and observing whether the result made any sense. An engineer
would follow a similar procedure by making an "educated guess" based on
some laboratory experiments, as to what the solution might be. Today
there are more sophisticated techniques used to find solutions. In this
respect, the engineer may lag behind the mathematician by several
centuries in finding applications for mathematical tools.

To solve the transmission line equation, we shall guess that the
solution for the voltage function is of the form:

$$v\left( t \right) = e^{j\omega t} e^{ - \gamma x}$$

The first term represents a unity vector rotating at an angular velocity
of $\omega$ radians per second, in other words, a sine wave of some
frequency. The second term denotes the sinusoid being modified by the
transmission line, namely its amplitude decaying exponentially with
distance. If we let $\gamma$ be a complex quantity, we can also include
any phase changes which occur as the signal travels down the line.

:   The sine wave is used as a signal source because it is easy to
    generate, and manipulate mathematically. Euler's Identity shows the
    relationship between exponential notation and trigonometric
    functions:

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:   

    :   **Euler\'s Identity**

$$e^{j\omega t}  = \cos \left( {\omega t} \right) + j\sin \left( {\omega t} \right)$$

:   

    :   Going back to our educated guess, we will let
        $\gamma  = \alpha  + j\beta$, therefore:

$$v\left( t \right) = e^{j\omega t} e^{ - \left( {\alpha  + j\beta } \right)x}  = e^{ - \alpha x} e^{\left( {\omega t - \beta x} \right)j}$$

:   

    :   The term $e^{ - \alpha x}$ represents the exponential amplitude
        decay as this signal travels down the line. $\alpha$ is known as
        the attenuation coefficient and is expressed in Nepers per
        meter.

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:   

    :   The term $e^{\left( {\omega t - \beta x} \right)j}$ represents
        the frequency of the signal at any point along the line. $\beta$
        component is known as the phase shift coefficient, and is
        expressed in radians per meter.

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:   

    :   Substituting our *educated guess*

$$v\left( t \right) = e^{j\omega t} e^{ - \left( {\alpha  + j\beta } \right)x}$$

:   

    :   into the transmission line equation for voltage, we obtain:

$$\frac{{\partial ^2 }}{{\partial x^2 }}\left[ {e^{j\omega t} e^{ - \left( {\alpha  + j\beta } \right)x} } \right] = RG\left[ {e^{j\omega t} e^{ - \left( {\alpha  + j\beta } \right)x} } \right] + \left( {RC + LG} \right)\frac{\partial }{{\partial t}}\left[ {e^{j\omega t} e^{ - \left( {\alpha  + j\beta } \right)x} } \right] + LC\frac{{\partial ^2 }}{{\partial t^2 }}\left[ {e^{j\omega t} e^{ - \left( {\alpha  + j\beta } \right)x} } \right]$$

This looks pretty intimidating, but if you can do basic differentials
and algebra, you can do this!

#### Simplifying the Equation (trust me)

The idea now is to work through the math to see if we come up with a
reasonable solution. If we arrive at a contradiction or an unreasonable
result, it means that our educated guess was wrong and we have to do
more experimenting and come up with a better guess as to how voltage and
current travel down a transmission line.

Let\'s look at this equation one term at a time:

:   

    :   LHS = RHS Term 1 + RHS Term 2 + RHS Term 3

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:   

    :   Starting with the left hand side (LHS) we get the following
        simplification:

$$\frac{{\partial ^2 }}{{\partial x^2 }}\left[ {e^{j\omega t} e^{ - \left( {\alpha  + j\beta } \right)x} } \right] = \frac{\partial }{{\partial x}}\left[ { - \left( {\alpha  + j\beta } \right)e^{j\omega t} e^{ - \left( {\alpha  + j\beta } \right)x} } \right] = \left( {\alpha  + j\beta } \right)^2 e^{j\omega t} e^{ - \left( {\alpha  + j\beta } \right)x}$$

:   

    :   Believe it or not, the RHS Term 1 does not need simplifying.

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:   

    :   Simplifying the RHS Term 2, we obtain:

$$\left( {RC + LG} \right)\frac{\partial }{{\partial t}}\left[ {e^{j\omega t} e^{ - \left( {\alpha  + j\beta } \right)x} } \right] = \left( {RC + LG} \right)j\omega \left( {e^{j\omega t} e^{ - \left( {\alpha  + j\beta } \right)x} } \right)$$

:   

    :   Simplifying the RHS Term 3, we obtain:

$$LC\frac{{\partial ^2 }}{{\partial t^2 }}\left[ {e^{j\omega t} e^{ - \left( {\alpha  + j\beta } \right)x} } \right] = LC\frac{\partial }{{\partial t}}\left[ {j\omega e^{j\omega t} e^{ - \left( {\alpha  + j\beta } \right)x} } \right] =  - LC\omega ^2 e^{j\omega t} e^{ - \left( {\alpha  + j\beta } \right)x}$$

:   

    :   Let\'s put it all back together:

$$\left( {\alpha  + j\beta } \right)^2 e^{j\omega t} e^{ - \left( {\alpha  + j\beta } \right)x}  = RG\left( {e^{j\omega t} e^{ - \left( {\alpha  + j\beta } \right)x} } \right) + \left( {RC + LG} \right)j\omega \left( {e^{j\omega t} e^{ - \left( {\alpha  + j\beta } \right)x} } \right) - LC\omega ^2 \left( {e^{j\omega t} e^{ - \left( {\alpha  + j\beta } \right)x} } \right)$$

:   

    :   Note that each of the four terms contain the expression
        $e^{j\omega t} e^{ - \left( {\alpha  + j\beta } \right)x}$.

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:   

    :   Therefore we end up with:

$$\left( {\alpha  + j\beta } \right)^2  = RG + \left( {RC + LG} \right)j\omega  - LC\omega ^2$$

:   

    :   And this can be further simplified to:

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:   

    :   **Attenuation and Phase Shift Coefficients**

$$\alpha  + j\beta  = \gamma  = \sqrt {\left( {R + j\omega L} \right)\left( {G + j\omega C} \right)}$$

This result is not self contradictory or unreasonable. Therefore we
conclude that our educated guess was right and we have successfully
found an expression for attenuation and phase shift on a transmission
line as a function of its distributed electrical components and
frequency.

### Lossless Transmission Line

Signal loss occurs by two basic mechanisms: signal power can be
dissipated in a resistor \[or conductance\] or signal currents may be
shunted to an AC ground via a reactance. In transmission line theory, a
lossless transmission line does not dissipate power. Signals, will still
gradually diminish however, as shunt reactances return the current to
the source via the ground path. For the power loss to equal zero, R = G
= 0. This condition occurs when the transmission line is very short. An
oscilloscope probe is an example of a very short transmission line. The
transmission line equation reduces to the voltage equation:

$$\frac{{\partial ^2 v}}{{\partial x^2 }} = LC\frac{{\partial ^2 v}}{{\partial t^2 }}$$

:   

    :   and the current equation:

$$\frac{{\partial ^2 i}}{{\partial x^2 }} = LC\frac{{\partial ^2 i}}{{\partial t^2 }}$$

To determine how sinusoidal signals are affected by this type of line,
we simply substitute a sinusoidal voltage or current into the above
expressions and solve as before, or we could take a much simpler
approach. We could start with the solution for the general case:

$$\alpha  + j\beta  = \gamma  = \sqrt {\left( {R + j\omega L} \right)\left( {G + j\omega C} \right)}$$

:   

    :   Let *R* = *G* = 0, and simplify:

$$\alpha  + j\beta  = \sqrt {\left( {j\omega L} \right)\left( {j\omega C} \right)}  = \omega \left( {LC} \right)^2 j$$

:   

    :   Equating the real and imaginary parts:

$$\alpha  = 0$$

$$\beta  = \omega \sqrt {LC}$$

This expression tells us that **a signal travelling down a lossless
transmission line, experiences a phase shift directly proportional to
its frequency.**

#### Phase Velocity

A new parameter, known as phase velocity, can be extracted from these
variables:

$$V_p  = \frac{1}{{\sqrt {LC} }} = \frac{\omega }{\beta }$$ meters per
second

Phase velocity is the speed at which a fixed point on a wavefront,
appears to move. In the case of wire transmission lines, it is also the
velocity of propagation., typically: 0.24c \< Vp \< 0.9c .

The distance between two identical points on a wavefront is its
wavelength ($\lambda$) and since one cycle is defined as 2$\pi$ radians:

$$\lambda  = \frac{{2\pi }}{\beta }$$ and $\omega  = 2\pi f$

:   

    :   therefore:

$$V_p  = \lambda f$$

In free space, the phase velocity is 3 x 10^8^ meters/sec, the speed of
light. In a cable, the phase velocity is somewhat lower because the
signal is carried by electrons. In a waveguide transmission line, the
phase velocity exceeds the speed of light.

### Distortionless Transmission Line

A distortionless line does not distort the signal phase, but does
introduce a signal loss. Since common transmission lines are not super
conductors, the signal will decrease in amplitude but retain the same
shape as the input. This characteristic is essential for long cable
systems.

Phase distortion does not occur if the phase velocity *V~p~* is constant
at all frequencies.

By definition, a phase shift of 2$\pi$ radians occurs over one
wavelength $\lambda$.

:   

    :   Since

$$V_p  = \lambda f\quad \quad \lambda  = \frac{{2\pi }}{\beta }\quad \quad f = \frac{\omega }{{2\pi }}$$

:   

    :   Then:

$$V_p  = \frac{{2\pi }}{\beta } \times \frac{\omega }{{2\pi }} = \frac{\omega }{\beta }$$

This tells us that in order for phase velocity *V~p~* to be constant,
the phase shift coefficient $\beta$, must vary directly with frequency
$\omega$.

:   

    :   Recall

$$\gamma  = \sqrt {\left( {R + j\omega L} \right)\left( {G + j\omega C} \right)}  = \alpha  + j\beta$$

The problem now is to find $\beta$. This can be done as follows:

$$\gamma  = \sqrt {\left( {\frac{{R + j\omega L}}{{j\omega L}}} \right)\left( {j\omega L} \right)\left( {\frac{{G + j\omega C}}{{j\omega C}}} \right)\left( {j\omega C} \right)}  = j\omega \sqrt {LC} \sqrt {1 + \frac{R}{{j\omega L}}} \sqrt {1 + \frac{G}{{j\omega C}}}$$

It may seem that we have lost $\beta$, but do not give up. The 2nd and
3rd roots can be expanded by means of the Binomial Expansion.

:   

    :   Recall:

$$\left( {1 + x} \right)^n  = 1 + nx + \frac{{n\left( {n - 1} \right)}}{{2!}}x^2  + \frac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}x^3  +  \cdots$$

:   

    :   In this instance *n* = 1/2. Since the contribution of successive
        terms diminishes rapidly, $\gamma$ is expanded to only 3 terms:

$$\gamma  \approx j\omega \sqrt {LC} \left( {1 + \frac{1}{2}\frac{R}{{j\omega L}} - \frac{1}{8}\left( {\frac{R}{{j\omega L}}} \right)^2 } \right)\left( {1 + \frac{1}{2}\frac{G}{{j\omega C}} - \frac{1}{8}\left( {\frac{G}{{j\omega C}}} \right)^2 } \right)$$

This may seem complex, but remember it is only algebra and it will
reduce down to simple elegance. Expanding the terms we obtain:

$$\gamma  \approx j\omega \sqrt {LC} \left\{ \begin{array}{l}
 1 + \frac{1}{2}\frac{G}{{j\omega C}} - \frac{1}{8}\left( {\frac{G}{{j\omega C}}} \right)^2  + \frac{1}{2}\frac{R}{{j\omega L}} - \frac{1}{4}\frac{{RG}}{{\omega ^2 LC}} \\ 
  - \frac{1}{{16}}\frac{R}{{j\omega L}}\left( {\frac{G}{{j\omega C}}} \right)^2  - \frac{1}{8}\left( {\frac{R}{{j\omega L}}} \right)^2  \\ 
  - \frac{1}{{16}}\left( {\frac{R}{{j\omega L}}} \right)^2 \frac{G}{{j\omega C}} + \frac{1}{{64}}\left( {\frac{R}{{j\omega L}}} \right)^2 \left( {\frac{G}{{j\omega C}}} \right)^2  \\ 
 \end{array} \right\}$$

:   

    :   Since $\gamma  = \alpha  + j\beta$, we merely have to equate the
        real and imaginary terms to find $\beta$.

$$\beta  \approx \omega \sqrt {LC} \left\{ {1 + \underbrace {\frac{1}{8}\left( {\frac{G}{{\omega C}}} \right)^2  - \frac{1}{4}\frac{{RG}}{{\omega ^2 LC}} + \frac{1}{8}\left( {\frac{R}{{\omega L}}} \right)^2 }_{{\rm{Difference}}\;{\rm{of}}\;{\rm{squares}}} + \underbrace {\frac{1}{{64}}\left( {\frac{R}{{\omega L}}} \right)^2 \left( {\frac{G}{{\omega C}}} \right)^2 }_{{\rm{Very}}\;{\rm{small}}}} \right\}$$

:   

    :   Or

$$\beta  \approx \omega \sqrt {LC} \left\{ {1 + \frac{1}{8}\left( {\frac{R}{{\omega L}} - \frac{G}{{\omega C}}} \right)^2 } \right\}$$

:   

    :   Note that if $\frac{R}{{\omega L}} = \frac{G}{{\omega C}}$ then
        $\beta  \approx \omega \sqrt {LC}$

From this we observe that $\beta$ is directly proportional to $\omega$.

:   

    :   Therefore the **requirement for distortionless transmission
        is:**

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:   

    :   *RC = LG*


:   

    :   This is one of the essential design characteristics for a
        broadband coax cable network.

If we equate the real terms, we obtain:

$$\alpha  \approx \sqrt {RG}$$

So there is a reason to study algebra after all!

## The Frequency Domain

Signal analysis is often performed in the frequency domain. This tells
us how the transmission line affects the spectral content of the signals
they are carrying.

To determine this, it is necessary to find the Fourier Transform of the
transmission line equation. Recall:

$$\frac{{\partial ^2 v}}{{\partial x^2 }} = RGv + \left( {RC + LG} \right)\frac{{\partial v}}{{\partial t}} + LC\frac{{\partial ^2 v}}{{\partial t^2 }}$$

and recall (hopefully) the Fourier Transform (which converts the time
domain to the frequency domain):

$$\mathbb{F}\left\{ {f\left( t \right)} \right\} = F\left( \omega  \right) = \int\limits_{ - \infty }^\infty  {e^{ - j\omega t} f\left( t \right)dt}$$

To prevent this analysis from 'blowing up', we must put a stipulation on
the voltage function namely, that it vanishes to zero at an infinite
distance down the line. This comprises a basic boundary condition.

$${\text{let}}\quad v \to 0\quad {\text{as}}\quad x \to \infty$$

This stipulation is in agreement with actual laboratory experiments. It
is well known that the signal magnitude diminishes as the path
lengthens.

Likewise, a time boundary condition, that the signal was zero at some
time in the distant past and will be zero at some time in the distant
future, must be imposed.

$${\text{let}}\quad v \to 0\quad {\text{as}}\quad t \to \infty$$

Although engineers have no difficulty imposing these restrictions,
mathematical purists, are somewhat offended. For this and other reasons,
other less restrictive transforms have been developed. The most notable
in this context, is the Laplace transform, which does not have the same
boundary conditions.

Having made the necessary concessions in order to continue our analysis,
we must find the Fourier Transform corresponding to the following terms:

$$\mathbb{F}\left\{ v \right\}\quad \quad \quad \mathbb{F}\left\{ {\frac{{\partial v}}
{{\partial t}}} \right\}\quad \quad \quad \mathbb{F}\left\{ {\frac{{\partial ^2 v}}
{{\partial t^2 }}} \right\}$$

$${\text{Let:}}\quad \quad \quad \mathbb{F}\left\{ v \right\} = V$$

:   

    :   Then applying the transform on the derivative, we obtain:

$$\mathbb{F}\left\{ {\frac{{\partial v}}
{{\partial t}}} \right\} = \int\limits_{ - \infty }^\infty  {e^{ - j\omega t} \frac{{\partial v}}
{{\partial t}}dt}$$

This equation can be solved by using integration by parts:

$$\int {u\,dv}  = uv - \int {v\,du}$$

$${\text{let}}\quad u = e^{ - j\omega t} \quad \quad \quad \therefore du =  - j\omega e^{ - j\omega t}$$

$${\text{and}}\quad dv = \frac{{\partial v}}{{\partial t}}\quad \quad \quad \therefore v = v$$

$$\therefore \mathbb{F}\left\{ {\frac{{\partial v}}
{{\partial t}}} \right\} = e^{ - j\omega t} \left. v \right|_{ - \infty }^\infty   - \int\limits_{ - \infty }^\infty  {v\left( { - j\omega e^{ - j\omega t} } \right)dt}$$

Applying the boundary conditions when t goes to infinity makes the 1st
term disappear.

$$\therefore \mathbb{F}\left\{ {\frac{{\partial v}}
{{\partial t}}} \right\} = j\omega \int\limits_{ - \infty }^\infty  {e^{ - j\omega t} v\,dt}$$

Note that the resulting integral is simply the Fourier Transform. In
other words:

$$\mathbb{F}\left\{ {\frac{{\partial v}}
{{\partial t}}} \right\} = j\omega \mathbb{F}\left\{ v \right\} = j\omega V$$

:   

    :   similarly:

$$\mathbb{F}\left\{ {\frac{{\partial ^2 v}}
{{\partial t^2 }}} \right\} = \left( {j\omega } \right)^2 \mathbb{F}\left\{ v \right\} = \left( {j\omega } \right)^2 V$$

We can now write the transmission line equation in the frequency domain:

$$\frac{{\partial ^2 V}}
{{\partial x^2 }} = RGV + \left( {RC + LG} \right)j\omega V + LC\left( {j\omega } \right)^2 V$$

:   

    :   where:

$$V = V\left( \omega  \right) = \mathbb{F}\left\{ {v\left( t \right)} \right\}$$

:   

    :   Rearranging the terms, we obtain:

$$\frac{{\partial ^2 V}}
{{\partial x^2 }} = \left[ {RG + \left( {RC + LG} \right)j\omega  + \left( {j\omega L} \right)\left( {j\omega C} \right)} \right]V$$

:   

    :   or

$$\frac{{\partial ^2 V}}
{{\partial x^2 }} = \left[ {\left( {R + j\omega L} \right)\left( {G + j\omega C} \right)} \right]V$$

:   

    :   since:

$$\sqrt {\left( {R + j\omega L} \right)\left( {G + j\omega C} \right)}  = \alpha  + j\beta  = \gamma$$

:   

    :   then

$$\frac{{\partial ^2 V}}{{\partial x^2 }} = \gamma ^2 V$$

:   

    :   or

$$\frac{{\partial ^2 V}}{{\partial x^2 }} - \gamma ^2 V = 0$$

This represents the most general form of the transmission line equation
in the frequency domain. This equation must now be solved for V to
observe how voltage (or current) varies with distance and frequency.
This can be done by assuming a solution of the form:

$$V = \underbrace {Ae^{ - \gamma x} }_{{\text{forward}}\;{\text{wave}}} + \underbrace {Be^{\gamma x} }_{{\text{reverse}}\;{\text{wave}}}$$

These terms represent an exponential decay as the signal travels down
the transmission line. If we ignore any reflections, assuming that the
cable is infinitely long or properly terminated, this simplifies to:

$$V = V_0 e^{ - \gamma x}$$

To verify whether this assumption is correct, substitute it into the
equation, and see if a contradiction occurs. If there is no
contradiction, then our assumption constitutes a valid solution.

$$\frac{{\partial ^2 }}{{\partial x^2 }}V_0 e^{ - \gamma x}  - \gamma ^2 V_0 e^{ - \gamma x}  = 0$$

$$\frac{\partial }{{\partial x}}\left( { - \gamma ^2 V_0 e^{ - \gamma x} } \right) - \gamma ^2 V_0 e^{ - \gamma x}  = 0$$

$$\gamma ^2 V_0 e^{ - \gamma x}  - \gamma ^2 V_0 e^{ - \gamma x}  = 0$$

$$0 = 0$$

Thus we validate the assumed solution. This tells us that in the
frequency domain, the voltage or current on a transmission line decays
exponentially:

$$V = V_0 e^{ - \gamma x}$$

:   

    :   where:

$$\gamma  = \sqrt {\left( {R + j\omega } \right)\left( {G + j\omega } \right)}  = \left| \gamma  \right|\angle \varphi  = \alpha  + j\beta$$

$$\gamma  = {\text{ propagation}}\;{\text{constant}}$$

$$\alpha  = {\text{ attenuation}}\;{\text{coeficient}}$$

$$\beta  = {\text{phase}}\;{\text{coefficient}}$$

In exponential notation, a sinusoid may be represented by a rotating
unity vector, of some frequency:

$$e^{j\omega t}  = \cos \omega t + j\sin \omega t$$

Note that the magnitude of this function is 1, but the phase angle is
changing as a function of *t*.

:   

    :   If we let: $V_0  = e^{j\omega t}$

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:   

    :   Then:
        $V_0  = e^{j\omega t} e^{ - \gamma x}  = e^{j\omega t} e^{ - \left( {\alpha  + j\beta } \right)x}  = \underbrace {e^{ - \alpha x} }_{{\text{attenuation}}\;vs.\;x}\overbrace {e^{j\left( {\omega t - \beta x} \right)} }^{{\text{phase}}\;vs{\text{.}}\;t\;{\text{and}}\;x}$

This result is quite interesting because it is the same solution for the
transmission line equation in the time domain. The term
${e^{ - \alpha x} }$ represents an exponential decay. The signal is
attenuated as length *x* increases. The amount of attenuation is defined
as:

:   

    :   Attenuation in Nepers:
        $N = \left| {\ln e^{ - \alpha x} } \right| = \alpha x$

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:   

    :   Attenuation in dB:
        $= 20\log e^{ - \alpha x}  \approx 8.68589\alpha x$

This allows us to determine the attenuation at any frequency at any
point in a transmission line, if we are given the basic line parameters
of*R, L, G, & C*.

The term ${e^{j\left( {\omega t - \beta x} \right)} }$ represents a
rotating unity vector since:

$$e^{j\left( {\omega t - \beta x} \right)}  = \cos \left( {\omega t - \beta x} \right) + j\sin \left( {\omega t - \beta x} \right)$$

The phase angle of this vector is ${\beta x}$ radians.

## Characteristic Impedance

The characteristic impedance of a transmission line is also known as its
surge impedance, and should not be confused with its resistance. If a
line is infinitely long, electrical signals will still propagate down
it, even though the resistance approaches infinity. The characteristic
impedance is determined from its AC attributes, not its DC ones.

Recall from our earlier analysis:
