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Prof. D. R. Wilton
Note 2 Note 2 Transmission LinesTransmission Lines
(Time Domain)(Time Domain)
ECE 3317
Disclaimer:
Transmission lines is the subject of Chapter 6 in the book. However, the subject of wave propagation on transmission line in the time domain is not treated very thoroughly there, appearing only in the latter half of section 6.5.
Therefore, the material of this Note is roughly independent of the book.
Note about Notes 2
Approach:
Transmission line theory can be developed starting from either circuit theory or from Maxwell’s equations directly. We’ll use the former approach because it is simpler, though it doesn’t reveal the approximations or limitations of the approach.
A transmission line is a two-conductor system that is used to transmit a signal from one point to another point.
Transmission Lines
Two common examples:
coaxial cable twin line
r a
bz
A transmission line is normally used in the balanced mode, meaning equal and opposite currents (and charges) on the two conductors.
Transmission Lines (cont.)
coaxial cable
Here’s what they look like in real-life.
twin line
Another common example (for printed circuit boards):
microstrip line
w
hr
Transmission Lines (cont.)
microstrip line
Transmission Lines (cont.)
Transmission Lines (cont.)
Some practical notes:
Coaxial cable is a perfectly shielded system (no interference).
Twin line is not a shielded system – more susceptible to noise and interference.
Twin line may be improved by using a form known as “twisted pair.”
l
-l
+
++
+
++
-
---
- -
E
coax twin line
E
Transmission Lines (cont.)
Transmission line theory must be used instead of circuit theory for any two-conductor system if the travel time across the line, TL, (at the speed of light in the medium) is a significant fraction of a signal’s period T or rise time for periodic or pulse signals, respectively.
LoadLV
LI
GV
GI
LT
L T
speed of light
3[m] 30[cm] 300[ m] 3[ m]
10[nS]* 100[pS]* 1[pS]* 10[f S]*
60 [Hz] 16.7[mS]
1 [kHz] 1[mS]
1 [MHz] 1[ S]
1 [GHz] 1[nS]
1 [THz] 1[pS]
L L L LT T T T
L L L L
Frequency Period
X X
X X X
Transmission line theory
Circuit theory
X * 1.0r r assumed
4 parameters
symbol:
z
Note: We use this schematic to represent a general transmission line, no matter what the actual shape of the conductors.
Transmission Lines (cont.)
4 parameters
C = capacitance/length [F/m]
L = inductance/length [H/m]
R = resistance/length [/m]
G = conductance/length [S/m]
Four fundamental parameters that characterize any transmission line:
z These are “per unit length” parameters.
capacitance between the two wires
inductance due to stored magnetic energy
resistance due to the conductors
conductance due to the filling material between the wires
Transmission Lines (cont.)
Circuit Model
Circuit Model:
Rz Lz
Gz Cz
z
z z
z
Coaxial Cable
Example (coaxial cable)
r a
bz
0
0
2F/m
ln
ln H/m2
rCba
bL
a
2S/m
ln
1 1 1/m
2 2
d
m
Gba
Ra b
2
m
(skin depth of metal)
d = conductivity of dielectric [S/m].
m = conductivity of metal [S/m].
Coaxial Cable (cont.)
Overview of derivation: capacitance per unit length
l
-l
+
++
+
++
-
---
- -
E
1QC
V z V
0 0
ln2 2
b b
r ra a
bV E d d
a
02
ln
rCba
Coaxial Cable (cont.)
Overview of derivation: inductance per unit length
y
E
x
Js
1L
I z
0 0
0
2
ln2
b b
a a
Iz H d z d
I bz
a
0 ln2
bL
a
Coaxial Cable (cont.)
Overview of derivation: conductance per unit length
2
ln
dGba
d
C G
d
RC Analogy:
2
lnC
ba
Coaxial Cable (cont.)
Relation between L and C:
0 02F/m ln H/m
2ln
r bC L
b aa
0 0 rLC
Speed of light in dielectric medium: 1
dc
2
1
d
LCc
Hence: This is true for ALL two-conductortransmission lines.
Telegrapher’s Equations
Apply KVL and KCL laws to a small slice of line:
+ V (z,t)-
Rz Lz
Gz Cz
I (z,t)
+ V (z+z,t)-
I (z+z,t)
zz z+z
( , )KVL : ( , ) ( , ) ( , )
( , )KCL : ( , ) ( , ) ( , )
I z tV z t V z z t I z t R z L z
tV z z t
I z t I z z t V z z t G z C zt
Hence
( , ) ( , ) ( , )( , )
( , ) ( , ) ( , )( , )
V z z t V z t I z tRI z t L
z tI z z t I z t V z z t
GV z z t Cz t
Now let z 0:
V IRI L
z tI V
GV Cz t
“Telegrapher’s Equations (TE)”
Telegrapher’s Equations (cont.)
To combine these, take the derivative of the first one
with respect to z:
2
2
2
2
V I IR L
z z z t
I IR L
z t z
V V VR GV C L G C
t t t
Take the derivative of the first TE with respect to z.
Substitute in from the second TE.
Telegrapher’s Equations (cont.)
2 2
2 2( ) 0
V V VRG V RC LG LC
z t t
The same equation also holds for i.
Hence, we have:
2 2
2 2
V V V VR GV C L G C
z t t t
There is no exact solution to this differential equation, except for the lossless case.
Telegrapher’s Equations (cont.)
2 2
2 2( ) 0
V V VRG V RC LG LC
z t t
The same equation also holds for i.
Lossless case:
2 2
2 20
V VLC
z t
0R G
Note: The current satisfies the same differential equation:
Telegrapher’s Equations (cont.)
2 2
2 20
I ILC
z t
The same equation also holds for i.
2 2
2 2 2
10,
d
V V
z c t
Hence we have
Solution:
,
,.
d d
d
V z t V z c t V z c t
V V
z c t I z t
where and are
func
tions of (Similarly for !)
arbitrary
Solution to Telegrapher's Equations
This is called the D’Alembert solution to the Telegrapher's Equations (the solution is in the form of traveling waves).
2 2
2 2 2
10
d
I I
z c t
Traveling Waves
2 2
2 2 2
1
d
V V
z c t
,d d
V z t V z c t V z c t
Proof of solution:
2
2
22 2
2
,
,
d d
d d d d
V z tV z c t V z c t
zV z t
c V z c t c V z c tt
It is seen that the differential equation is satisfied by the general solution.
General solution:
Example:
z
z0
z
t = 0 t = t1 > 0 t = t2 > t1
z0 + cd t1z0 z0 + cd t2
V(z,t)
Traveling Waves
,d
V z t V z c t
( )V z
Example:
zz0
z
t = 0t = t1 > 0t = t2 > t1
z0 - cd t1 z0z0 - cd t2
V(z,t)
Traveling Waves
,d
V z t V z c t
( )V z
Loss causes an attenuation in the signal level, and it also causes distortion (the pulse changes shape and usually becomes broader).
z
t = 0 t = t1 > 0t = t2 > t1
z0 + cd t1z0 z0 + cd t2
V(z,t)
Traveling Waves (cont.)
(These effects can be studied numerically.)
CurrentV I
RI Lz t
V IL
z t
lossless
,d d
V z t V z c t V z c t
,d d
V z tV z c t V z c t
z
,d d
I z t I z c t I z c t
,d d d d
I z tc I z c t c I z c t
t
(first TE)
Our goal is to now solve for the current on the line.
Assume the following forms:
The derivatives are:
Current (cont.)
d d d d d dV z c t V z c t L c I z c t c I z c t
V IL
z t
d d dV z c t L c I z c t
d d dV z c t L c I z c t
1, , Constant
d
I z t V z tLc
1, , Constant
d
I z t V z tLc
This becomes
Equating terms with the same space and time variation, we have
Hence we have
Constants represent independent DC voltages or currents on the line. Assuming no initial line voltage or Current we conclude Constants =0
1 1d
LLc L L
CLC
Then
0
0
1, ,
1, ,
I z t V z tZ
I z t V z tZ
Define characteristic impedance Z0:
0
LZ
C The units of Z0 are Ohms.
Current (cont.)
Observation about term:
or
0
0
, ,
, ,
V z t Z I z t
V z t Z I z t
General solution:
0
,
1,
d d
d d
V z t V z c t V z c t
I z t V z c t V z c tZ
For a forward wave, the current waveform is the same as the voltage, but reduced in amplitude by a factor of Z0.
For a backward traveling wave, there is a minus sign as well.
Note that without this minus sign, the ratio of voltage to current would be constant rather than varying from point-to-point and over time along the line as is generally the case!
Current (cont.)
Current (cont.)
z
Picture for a forward-traveling wave:
0
,
1,
d
d
V z t V z c t
I z t V z c tZ
+
- ,V z t
,I z t
forward-traveling wave
z
Physical interpretation of minus sign for the backward-traveling wave:
0
,
1,
d
d
V z t V z c t
I z t V z c tZ
+
- ,V z t
,I z t
backward-traveling wave
,I z t
The minus sign arises from the reference direction for the current.
Current (cont.)
Coaxial Cable
Example: Find the characteristic impedance of a coax.
0
0
2F/m
ln
ln H/m2
rCba
bL
a
0
00
ln2
2
ln
r
bL a
ZC
ba
00
0
1 1ln
2r
bZ
a
r a
bz
Coaxial Cable (cont.)
r a
bz
0 0
1 1ln
2r
bZ
a
00
0
0 376.7303
-120
-70
8.8541878 10 [F/m]
= 4 10 [H/m] ( )µ
exact
(intrinsic impedance of free space)
Twin Line
10 0
1 1cosh
2r
dZ
a
d
a = radius of wires
r
10 0
1
F/m cosh H/m2
cosh2
r dC L
d aa
0 0
1 1ln
r
dZ
a
a d
1 2cosh ln 1 ln 2xx x x x
Twin Line (cont.)
coaxial cable
0 300Z
twin line
0 75Z
These are the common values used for TV.
75-300 [] transformer
300 [] twin line 75 [] coax
w
hr
Microstrip Line
0
0
,
,
r
wC w h
hh
L w hw
0 0
1,
r
hZ
w
w h
parallel-plate formulas:
Microstrip Line (cont.)
More accurate CAD formulas:
w
hr
w
hr
0
120
/ 1.393 0.667 ln / 1.444effr
Zw h w h
1 1 11 /
2 2 4.6 /1 12 /
eff r r rr
t h
w hh w
( / 1)w h
( / 1)w h
Note: the effective relative permittivity accounts for the fact that some of the field exists outside the substrate, in the air region. The effective width w' accounts for the strip thickness.
21 ln
t hw w
t
t = strip thickness
Some Comments
Transmission-line theory is valid at any frequency, and for any type of waveform (assuming an ideal transmission line).
Transmission-line theory is perfectly consistent with Maxwell's equations (although we work with voltage and current, rather than electric and magnetic fields).
Circuit theory does not view two wires as a "transmission line": it cannot predict effects such as signal propagation, distortion, etc.