1.1.5
Characteristic Impedance
For a lossless line, the first of the two telegrapher’s equations is
−
∂v
∂i
v=L
∂z
∂t
(1.27)
Let us consider just the forward traveling wave,
∂ +
∂
v (z − ut) = L i+ (z − ut)
∂z
∂t
(1.28)
−v + (z − ut) = L (−u)i+ (z − ut)
(1.29)
−
Using the chain rule this becomes
Integrating both side with respect to x = z − ut gives
− v + (x) dx = −u L
i+ (x) dx
(1.30)
−v + = −uLi+ + constant
(1.31)
Since i+ = 0 if v + = 0, the constant is zero, and we have
v+
L
=
=uL= √
i+
LC
L
C
(1.32)
The constant on the right-hand side has units of Ohms, and we call this the ”characteristic impedance” of
the line:
L
Zo =
(1.33)
C
If we repeat this derivation for the reverse traveling wave, we get
L
v−
= −u L = −
= −Zo
i−
C
(1.34)
The total current can then be related to the total voltage using
v(z, t) = v + (z − ut) + v − (z + ut)
v + (z − ut) v − (z + ut)
i(z, t) =
−
Zo
Zo
(1.35)
(1.36)
This allows us to find the current on a transmission line if we know the forward and reverse voltage waveforms.
The minus sign in the second equation is important. The characteristic impedance Zo is not a simple resistance produced by the conductors in the line, because even a transmission line constructed from perfect
conductors has a finite, nonzero characteristic impedance. An impedance is the ratio of total voltage to total
current, whereas the characteristic impedance is the ratio of the forward voltage waveform to the associated
current and the negative of the ratio of the reverse voltage waveform to the associated current. If Eq. (1.36)
had a positive sign instead of a negative sign, then Zo would be a regular impedance. Can you understand
this minus sign physically?
6
1.2
Transients on Transmission Lines
Consider the transmission line circuit in Fig. 1.3.
T-line
Rg
t=0
Vg
Z0
RL
Z
Z=l
Z=0
Figure 1.3: Transmission line circuit.
When the switch closes a step voltage appears across the generator end of the line. What is the value of
that voltage? The characteristic impedance is not the total impedance of the line but rather the relationship
between the forward and reverse traveling voltages and currents. But when the switch first closes there is
no reverse wave, because the forward step has not had time to travel down to the end and back. Therefore,
the characteristic impedance of the line is the total impedance of the line imediately after the switch closes.
This leads to the equivalent circuit shown in Fig. 1.4 when the switch is first closed. From the equivalent
Rg
Z0
Figure 1.4: Equivalent circuit at t = 0.
circuit we can use a voltage divider relationship to calculate the magnitude of the step voltage,
v1+ (z = 0, t = 0) =
Zo
Vg
Zo + Rg
(1.37)
This initial pulse then starts to travel down the line at speed u. At time t = /(2u), for example the step is
halfway down the line. At t = /u, the step arrives at the load.
7
1.2.1
Reflection Coefficient
What happens when the step hits the load? The pulse will reflect, and the v − term in the wave equation
solution will no longer be zero. We need to find the amplitude of the reflected wave. This is easy to do using
boundary conditions at the load end of the transmission line. The boundary conditions are
v(, T ) = vL (T )
(1.38)
i(, T ) = iL (T )
(1.39)
where T = /u and vL (T ) and iL (T ) are the voltage across and the current through the load resistor. vL
and iL are related by Ohm’s law: vL = iL RL . Putting Ohm’s law together with the boundary conditions at
the load end of the line, we obtain
(1.40)
v(, T ) = RL i(, T )
Using Eqs. (1.35) and (1.36), this can be rewritten as
v1+ + v1− = RL
v1+ v1−
−
Zo
Zo
(1.41)
where the line voltages and currents are all evaluated at z = and t = T . Solving this for v1− gives
v1− =
RL − Zo +
v
RL + Zo 1
(1.42)
We call the constant in this expression the load reflection coefficient:
ΓL =
RL − Zo
RL + Zo
(Load reflection coefficient)
(1.43)
From this we can find the magnitude of the reflected wave:
v1− = ΓL v1+ =
RL − Zo Zo
Vg
RL + Zo Zo + Rg
(1.44)
The voltage waveform on the transmission line at a time that is after the reflection from the load, and before
the reflected pulse arrives at the generator, is shown in Fig. 1.5.
V (z, 3T/2)
L
+
+
V1
V1
(1+
l
l /2
Z
Figure 1.5: Voltage waveform at time t = 3T /2.
8
) V1+
L
What happens when the reflected wave gets back to the generator? Using the same idea as at the load end,
we can show that
Rg − Zo −
v2+ =
v = Γg v1− = Γg ΓL v1+
(1.45)
Rg + Zo 1
Can you derive this expression on your own?
Figure 1.6 shows the voltage waveform after the reflected pulse has reflected again from the generator end
of the transmission line. The total voltage at a point on the transmission line is the sum of all the reflected
and forward steps (v1+ , v1− , v2+ ) that have occured up until the current time.
g
V1
(1+
l
l /4
0
) V1+
L
Z
Figure 1.6: Voltage waveform at time t = 9T /4.
If we look at the results we have found so far, we can see a pattern:
v1+
v1−
v2+
v2−
v3+
..
.
=
=
=
=
=
Zo
Zo +Rg
ΓL v1+
Vg
Γg ΓL v1+
ΓL Γg ΓL v1+
Γg ΓL Γg ΓL v1+
(Voltage divider at t = 0)
(First reflection at the load)
(Reflection at generator)
(1.46)
We can write out the total voltage at some point on the line at t = ∞ as an infinite series:
v(z, t = ∞) = v1+ + ΓL v1+ + Γg ΓL v1+ + ΓL Γg ΓL v1+ + Γg ΓL Γg ΓL v1+ + · · ·
= v1+ (1 + ΓL ) + (1 + ΓL )ΓL Γg + (1 + ΓL )Γ2L Γ2g + · · ·
= v1+ (1 + ΓL ) 1 + ΓL Γg + Γ2L Γ2g + · · ·
Geometric series
1
= v1+ (1 + ΓL )
1 − ΓL Γg
(1.47)
(1.48)
(1.49)
If we plug in the definitions of ΓL and Γg , this reduces to
v(z, t = ∞) =
RL
Vg
Rg + R L
This is the steady state voltage on the transmission line. Does this result make sense?
9
(1.50)