The voltage between the two conductors and the current flowing
through them will depend on distance
along the transmission line as
well as time. We express this dependence as
and
.
When we place a sinusoidal source at one end of the transmissionline, these voltages and currents will also be sinusoidal because the
transmission line model consists of linear circuit elements. Asis customary in analyzing linear circuits, we express voltages
and currents as the real part of complex exponential signals,and write circuit variables as a complex amplitude—here
dependent on distance—times a complex exponential:
and
.
Using the transmission line circuit model, we find from KCL,KVL, and v-i relations the equations governing the complex
amplitudes.
Kcl at center node
V-i relation for rl series
Rearranging and taking the limit
yields the so-called
transmission line equations .
By combining these equations, we can obtain a single equation
that governs how the voltage's or the current's complexamplitude changes with position along the transmission line.
Taking the derivative of the second equation and plugging thefirst equation into the result yields the equation governing the
voltage.
This equation's solution is
Calculating its second derivative and comparing the result with
our equation for the voltage can check this solution.
Our solution works so long as the quantity
satisfies
Thus,
depends on
frequency, and we express it in terms of real and imaginaryparts as indicated. The quantities
and
are constants determined by the source and physical
considerations. For example, let the spatial origin be themiddle of the transmission line model
[link] .
Because the circuit model contains simple circuit elements,physically possible solutions for voltage amplitude cannot
increase with distance along the transmission line. Expressing
in terms of its real and
imaginary parts in our solution shows that such increases are a(mathematical) possibility.
The voltage cannot increase without limit; because
is always positive, we must segregate the solution for negative
and positive
. The first term
will increase exponentially for
unless
in this region; a similar result applies to
for
.
These physical constraints give us a cleaner solution.
This solution suggests that voltages (and currents too) will
decrease
exponentially along a transmission
line. The
space constant , also known as the
attenuation constant , is the distance over which
the voltage decreases by a factor of
.
It equals the reciprocal of
,
which depends on frequency, and is expressed by manufacturers inunits of dB/m.
The presence of the imaginary part of
,
,
also provides insight into how transmission lines work. Becausethe solution for
is proportional to
,
we know that the voltage's complex amplitude will
vary
sinusoidally in space . The complete solution for the
voltage has the form
The complex exponential portion has the form of a
propagating wave . If we could take a snapshot of
the voltage (take its picture at
),
we would see a sinusoidally varying waveform along thetransmission line. One period of this variation, known as the
wavelength , equals
.
If we were to take a second picture at some later time
,
we would also see a sinusoidal voltage. Because
the second waveform appears to be the first one, but
delayed—shifted to the right—in space. Thus, thevoltage appeared to move to the right with a speed equal to
(assuming
). We denote this
propagation speed by
, and it equals
In the high-frequency region where
and
,
the quantity under the radical simplifies to
,
and we find the propagation speed to be
For typical coaxial cable, this propagation speed is a fraction
(one-third to two-thirds)of the speed of light.
