Digital Signal Processing Reference
In-Depth Information
50
Ω
20 mA
20 mA
1V
50
Ω
VIR
=×
R
= 50
Ω
l
=
20 mA50
1V
×
Ω
=
2V
(a)
50
Ω
2
0 mA
20 mA
1V
VIR
=×
50
Ω
20 mA0
0
=
×
Ω
=
2V
(b)
50
Ω
0 mA
20 mA
1V
50
Ω
V
R
1
V
∞
I
=
=
= 0
2V
(c)
Figure 11.3
Launched voltage and current remains in the proper ratio (a) when the load resistance
equals the transmission line resistance, but not when the load is (b) a short circuit or (c) an open
circuit.
voltage pulse. Equation (11.1) is still satisfied when the pulses reach the 50
Ω
load
resistor, but this is not the case when the load has any other value.
For instance, changing the load to a short circuit as is shown in Figure 11.3(b)
requires the load voltage to be zero. However, the energy represented by the inci-
dent voltage and current pulses cannot simply vanish when the waves reach the
short-circuited load. A similar problem shown in Figure 11.3(c) occurs when the
load is infinite, representing an open circuit. There the current must become 0 at
the load, even though 20 mA has been launched down the line.
The solution to these dilemmas centers on conservation of energy. The incident
energy wave launched by the generator is perfectly absorbed by the load when
R
l
matches the transmission line impedance, but this is not so for any other value of
load resistance. For instance, the launched energy cannot be absorbed by the open
circuit, and because the line is lossless the energy is not dissipated there. Instead,
when the energy wave encounters the infinite impedance of the open circuit, it re-
bounds and travels back up the line toward the generator. The rebound is a reflec-
tion, and wave theory is used to explain its behavior [1-4].
