Digital Signal Processing Reference
In-Depth Information
20 cm
Signal
Conductor
e
r
=
4.5
Reference
Conductor
100 ps
100 ps
(a)
0.498
nH
0.498
nH
0.498
nH
0.198
pF
0.198
pF
0.198
pF
1
2
142
(b)
Figure 3-10
Equivalent circuit of a transmission line.
SOLUTION Since the digital waveform has rise and fall times of 100 ps (
t
r
=
t
f
=
100 ps), equations (3-22) and (3-23) give the required values for the model:
10
8
)
100 ps
(
3
.
0
×
10
−
3
m
z
=
10
√
4
.
5
=
1
.
41
×
l
z
=
0
.
2
N
s
=
=
141
.
8 segments
10
−
3
1
.
41
×
Since it is inconvenient to build an equivalent-circuit model with 141.8 segments,
N
s
is rounded up to 142. The inductance and capacitance values for each segment
are calculated with equations (3-24):
10
−
10
)
(
0
.
2
)(
1
.
41
×
10
−
13
F
C
z
=
=
1
.
98
×
142
10
−
7
)
(
0
.
2
)(
3
.
54
×
10
−
10
H
L
z
=
=
4
.
98
×
142
Figure 3-10b shows the equivalent circuit for this transmission line.
3.2.4 Wave Equation in Terms of
LC
The wave equations, which were used as the basis for analyzing propagating
electromagnetic fields, were derived in Section 2.3.1. To analyze transmission
lines in terms of circuit parameters, we rederive the wave equation from the
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