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*Note: The information below can be referenced to: Carr, J., Practical Antenna
Handbook, Tab Books, Blue Ridge Summit, PA, 1989, ISBN: 0-8306-9270-3.
Goeckner, M., (Class Notes).
CHAPTER 7 TRANSMISSION LINES
Introduction
A transmission line is defined as two or more conductors embedded in a dielectric system
that transports electromagnetic energy from one point to another. As an example,
transmission lines can be used between an exciter output and a transmitter input, and
between the transmitter and an antenna. Transmission lines are complex networks
containing the equivalent of all three basic electrical components: resistance, capacitance
and inductance. So in light of this fact, transmission lines must be analyzed in terms of
an RLC network.
There are several basic types of transmission lines and some are shown in Figure 7.1.The
simplest form is the parallel line shown in Figure 7.1a. The two conductors, of diameter
d, are separated by a dielectric by a spacing S. Parallel lines have been used at VLF,
MW, and HF frequencies. Antennas into the low-VHF have also found use of parallel
lines. Another form of the transmission line, that finds considerable application at
microwave frequencies, is the coaxial cable shown in Figure 7.1b. This form of line
consists of two cylindrical conductors sharing the same axis (hence “co-axial”) and
separated by a dielectric. In flexible cables at low frequencies the dielectric may be a
polyethylene or polyethylene foam and at higher frequencies Teflon and other materials
are used. Also used in some applications are dry air and dry nitrogen. Stripline, also
called microstripline, is another form of transmission line used at high UHF and
microwave frequencies. This type is shown in Figure 7.1c.The stripline consists of a
critically sized conductor over a ground plane conductor, and separated from it by a
dielectric. Some are sandwiched between two ground planes, and are separated from
each by a dielectric.
Outer
Conductor
Strip Line
W
S
d
d
(a) Parallel line
Dielectric
(b) Coaxial
Cable
Notes by: Debbie Prestridge
Inner
Conductor
Ground Plane
(c) Microstrip
Line
1
Characteristics of Wave Propagation in Transmission Lines
Consider the simplest case using a pair of infinitely long plane conductors separated by a
dielectric as shown in Figure 7.2.
W
X
Conductor
d
Z out of the
page
Y
Conductor 1
Dielectric
Conductor 2
Cut thru here
By applying a bias between the conductors will result in an electric field:
Ignore fringe
E≈
E=
→ E = E (z, t) only!
Figure 7.2
We can use Maxell’s Equations to further examine the propagation characteristics.

 
 
t


t
(A1)  z  x     t  y
Notes by: Debbie Prestridge
2
(A2)  z  y    t  x
     t D    t
(B1)  z  y     t  x
(B2)  z  x    t  y
Notice:
(A1) & (B1) contain  x &  y
(A2) & (B2) contain  y &  x
Why? They are independent of each other. The solutions to these independent equations
can be found by considering the boundary conditions.
n   1   2  0
n   D1  D2  ρ
Let E1 be the field in the dielectric and E2 be the field in the conductor.
Assume a Perfect Conductor
E2, dielectric Tangential = 0 = E1, conductor Tangential
Since the tangential part of E is Ey
 y
= 0 (everywhere in the dielectric & this requires ignoring fringing
 y  0 
x
fields-the extension of electric lines of force between the outer edges of capacitor plates,
also known as edging effect.)
  y  0 From equations (A2) & (B2)
Finally,
n   D1  D2  ρ  ρs = ε Ex *on the surfaces & D1 = 0 (in the conductor)
Now, do the same thing for the magnetic field:
n   1  2  0
n   1   2  
J
s
This will give
Js = Hy on the lower conductor ( n  x )

Js = -Hy on the upper conductor ( n   x )
Now combine the two remaining equations:
Notes by: Debbie Prestridge
3
(A1)  z  x     t  y
(B1)  z  y     t  x
 z  z  x  
2
z
 x    t  x  y
     t2  x
*This is the wave equation!
 Ex = Ex (t – z/vph) + Ex (t + z/vph)
vph = (με) -1/2
Likewise, you can also show that:
 2z  y      t2  y
 Hy = Hy (t – z/vph) + Hy (t + z/vph)


Note:     Propagation! Thus two important things have been found:
1. Electromagnetic waves propagate in a 2-conductor (loss-less) transmission line
with a phase velocity determined by the dielectric between the conductors. This
is also true for coaxial.


2.     Propagation- this is known as a transverse electromagnetic wave (TEM).
(Ez = 0 = Hz)
Voltage can be defined as:
d
V (z, t) =     dl    x d
0
And the total current can also be determined through the upper/ (lower) plate:
 J  ds  t    ds     dl
Where ds is just one thin metal line so that,
∫J·ds = I = ωJ (z, t)| plate
 t  E·ds = 0 (E = 0, in metal)
   dl   
y
(Hy = 0, above the plate)
 I = ω J (z, t) = -ω Hy (z, t)
 x  V d
y   
Now place these in equations (A1) & (B1):
Notes by: Debbie Prestridge
4
 V 
I
A1)  z ( x )   z 
     t ( y )    t  
 d 

B1)
 V 
I

 z ( y )   z      t ( x )    t 

 d 
Calculate-using some algebrad
 t    L  t 
A1)  zV 

 t
 tV   C  t V
d
  RQ  C 1 Q ) Notice that L΄ is an effective Inductance per unit
(Note: V = LQ
length and C΄ is an effective capacitance per unit length. As with E & H fields they
can be combined in the equations above to give wave equations:
B2)
z  

2
zV

2
z
 L C  2t V
And
I  L  C  2t I
Phase velocity is νph = (L΄C΄) -1/2 = (εμ) -1/2
* V (z, t) = V+ (t – z/vph) + V_ (t + z/vph)
** I (z, t) = I+ (t – z/vph) + I_ (t + z/vph)
↑
↑
(Forward motion +z) (Backward motion –z)
V & I are lumped by (A1) & (B1)
1
1
V' 
V'
(A1)   z V  
v ph
v ph

  L   '   '
integrating ωrt (time)

 
  

Noting that vph = (L΄C΄)-1/2 and
C
1
V  ( const )  V   const
L
z
C
1
V  ( const )   V   const
L
z
Notes by: Debbie Prestridge
5
This looks like standard circuit theory where
C
is the characteristic impedance
L
d
 d
= 
Zo =

 
Now it is possible to solve for V+ & V_

V+ (t –z/vph) = 1/2[V (z, t) + Z I (z, t)]
V- (t +z/vph) =1/2 [V (z, t) - Z I (z, t)]
The traveling currents are:
I+ = 1/zo V+
I_ = 1/zo V_
Coaxial Cables
This model can apply to other geometries-the most important is probably co-axial cables.
E
2a
b

o
So that,
  r r
b
b
V     dl    ln
a
a
(Eo = q/2πεL, if a charge is in the center)
From other Maxell’s Equations:
     t    t  

r
r
z
1
       

 r  z 
z r
t 
r
r
0
0
Notes by: Debbie Prestridge
6

r
r
1
  H   t    r
r
0


r 
z

z    z   r    t  r r
0

 ; By   dl  J  ds   t D  ds
   



(D = 0, in the conductor)
= HΦ 2πr = I
Plug the following into the above equations and solve:
V (z, t) = V+ + V_

 b
ln 
 a 
V
V
I (z, t) =    where Z 


2

For RG-58/u
εr ≈ 2.26
μr ≈ 1
a ≈ 0.406 mm
b ≈ 1.48 mm
And Z ≈ 51.6 Ω
This sort of impedance level is typical for most coaxial cables.
Reflections & Transmission of Pulses
Reflections occur when a wave hits a barrier. We see this when water waves hit the shore
or light hits a mirror. In both cases, part of the wave is reflected and part is lost (typically
giving up energy to the reflecting surface). The same thing occurs in a transmission line.
Consider the following in Figure 7.3:
I_
I+
Ii
RL
VL = RLIL
Figure 7.3
In Figure 7.3 the transmission line is terminated by a load (Resistance, RL or Impedance
ZL). What is the voltage across the load? It has been determined that –
Notes by: Debbie Prestridge
7
V (z, t) = V+ (z, t) + V_ (z, t)
So that at the load (Z = ZL)
V (ZL, t) = VL (t)
= V+ (ZL, t) + V_ (ZL, t)
= V+ + V_
And likewise –
I (z, t) = I+ (Z, t) + I_ (Z, t) so, that
IL (z, t) = I+ (ZL, t) + I_ (ZL, t)
Now VL = RLIL (t) and from before   
V
V
V
V 
,  
so that, V  V  RL     
Zo
Zo
 Zo Zo 

R

V Z o  R L
R 

Rearrange: V+ + V_ = 1  L   V  L  1 
  
V R L  Z o
Zo 

 Zo

The term on the right hand side of the equation is independent of time so
V
is also time
V
independent.
V
The ratio  = ρ tells how much of the forward propagating wave is reflected.
V
If we were to use ZL instead of RL, we would get –

Z L  Zo
Z L  Zo
What happens to the rest of the energy? It gets ‘passed’ through the load and the
VL
transmitted portion of V+ that goes through the load is VL.  We want
V
VL = V+ + V_

VL
V
2 RL
 1
 1    
And τ is defined as the transmission coefficient.
V
V
RL  Zo


Notes by: Debbie Prestridge
2R L
(Transmission coefficient)
RL  Zo
Z L  Zo
(Reflection coefficient)
Z L  Zo
8
There are several regions that ρ and τ fall into depending on RL & Zo.
*Note: RL can be ‘ZL’
Relationship
ρ
τ
Comment
between RL & ZL
RL = 0
-1
0
The forward V+ is
(shorted circuit)
all reflected as –V+
 V =V+ + V_ = 0
RL = ∞
1
2
The forward V+ is
reflected as + V+
 V = 2 V+
RL > Zo
1> ρ >0
()
Reflected Pulse
adds to incident
pulse (constructive
interference)
RL<0
ρ<0
()
Reflected pulse
subtracts from V+
(destructive
interference) Can
also be thought of as
reflecting an
inverted wave.
RL = Zo
ρ=0
1
No reflected V+
(matched Load)
Form of WavesTake a periodic function and model these waves as sines & cosines (Fourier series).
These are two other temporal variations that are very important to digital signals. The
first important is turning off an applied (steady state) voltage – ‘precharged’; the second
is sending a signal pulse down line.

Turning off a “precharged” line
Notes by: Debbie Prestridge
9
Switch
+++++++
---------
l
1
U1
Vo
2
RL
V (z, t < 0)

----------
- l/2
z=ρ
l/2
Z
- l/2
*Note: RL (ZL ← will be shown later)
l/2
Figure 7.4
Prior to t = 0 (when we close the switch)
V V

I (z, t < 0) = 0 = I+ + I_ =
Zo Zo
V (z, t < 0) = Vo = V+ + V_
1
 V  z , t  0  Vo
2
1
V  z , t  0  Vo
2
On the side that the load is, the reflection coefficients is
Z  Zo
L  L
Z L  Zo
On the source side the reflection coefficient is
Z s  Zo
Z s  Zo
*Note: (Having ρL ≠ ρs for digital systems can cause problems)
For now, assume Zs = ∞  all energy (voltages) traveling toward the source is reflected.
Assume now that the switch on  L is closed to t = 0. What is the voltage across  L as
a function of time?
s 
 L = Zo  ρL = 0 (no reflected wave)
1
At t = 0 V L  Vo ; only V is found at the load as V_ is moving toward the source. This
2
voltage level remains constant until the tail end of V_ traveling from  L to Zs, where it is
reflected, and back to  L where it is transmitted through  L  the time it takes is
2t loop 
2
; t = time to make 1 way transit.
v loop
Notes by: Debbie Prestridge
10
 L = Zo
Figure 7.5
V
Vo
Vo/z
2  /v
t
 L > Zo  0 <  < 1
Now only part of the signal is reflected

1
1
Vo  Vo  Vo 1   
2
2
2
1
V 1   
Vo
4 o
4tloop
2tloop
1
V 1    2
8 o
6tloop
Figure 7.6
In general:
V
V L o  n  11    for 2n  1t loop  t  2nt loop & t  0
2
Finally,
 L < Zo   < 0  reflects inverted wave
VL
Vo

V
1
Vo  Vo  o 1   
2
2
2
Ringing
6tloop
2tloop
(Figure 7.7)
Notes by: Debbie Prestridge
4tloop
Vo
 1   
2 
11
With the same general equation (which in fact holds for all three cases).
V
V L  o 1   n 1t  0 & 2 n  1 t loop  t  2nt loop
2
Now the opposite situation from what we have just looked at is charging the line.
Turning on an uncharged line
Now consider Figure 7.8 –
1
U3
2
RL (or ZL)
Zs
Figure 7.8
If we assume ρs = -1 (Zs = 0) and R L  0 < ρs< 1
Then we see a series of pictures:
0 < t < tloop (Figure 7.9)
V
Vo
Z

Figure 7.9
tloop < t < 2 tloop (Figure 7.10)
V
Vo
Z
Figure 7.10
Notes by: Debbie Prestridge

12
2tloop < t < 3 (Figure 7.11)
V
  Vo
Vo
Figure 7.11

These examples can be generalized (Figure 7.12)
Time
V+



V  V ; t   t loop ; 
v




t=0
ρs
Z=0
Z=
L΄
Z =L
Figure 7.12
At any point Z  V (z, t) = V+ (z, t) + V (z, t)
Note that the initial wave form becomes distorted by the direct and reflected waves and
2  z 

 (=V_)
V (z, t) = V+ (0, t = z/v) +  V+  0,t 

v 
2  z 

  sec ond reflected
+ s  V+  0, t 

v 
3  z 

  third reflected
+ s 2 V+  0,t 

v 
+ ……
This is independent of the form V+!
Notes by: Debbie Prestridge
13
Two more quick examples;
Connection of two transmission Lines (Figure 7.13)
Figure 7.13
+
++
Zo1 v1
-
Zo2 v2
-For now let V_ _ =0
(not always the case)
V+ + V_ = V++
I+ + I_ = I++
Then
V
 V
V
; 
;   
Z o1
Z o1
Zo2
Combining the above equations and doing a little algebra for a result of:
From before 
V Z o 2  Z o1


V  Z o 2  Z o1
*Note: Looks like a load impedance
“Damaged” coupled transmission lines
Notice where we have a loss in Figure 7.13:
Line #1
Line #2
+
Zo1 v1
+ +
-
RL
L
Zo2 v2
(Figure 7.13)
V+ + V_ = V++ = VL
I+ + I_ = I++ = IL
I+=
V
Z o1
I_=
 V
Z o1
Notes by: Debbie Prestridge
I++=
V 
Z o2
IL=
V
RL
14
Again some fraction of the voltage wave is repeated. Running through the algebra we
find.
1  1
1 



Z o1  Zo 2 R L 
V


V
1  1
1 



Z o1  Z o 2 R L 

1
1
1


Zeff Z o 2 R L
*Zeff is the effective impedance seen by the line #1.
Definition: Effective Impedance is the sum of the actual impedance of the circuit
and the reflected impedance of the load.
Power in Transmission Lines
The power flux is given by the Poynting vector –
       (   x  y z)
Ex 
V

,y 
d

The total power is   ower       ds
=
V 
d
= VI

ds  d x d y z
dx d y
where V (z, t), I (v, t)
All this is from circuit theory…
What do our power (total) equations ‘say’ about reflected/transmitted power?
Notes by: Debbie Prestridge
15
Name
Voltage Reflection
Voltage Transmission
Current Reflection
Current Transmission
Incident Power
Reflected Power
Delivered Power
Notes by: Debbie Prestridge
Equation
V _ RL  Zo

V R L  Z o
V
2 RL
V  V 
  
 1  
V R L  Z o
V
_


  _

1   L 



V 2
   V   
Zo
   V     V (    )   2
 L  V L  L   V (1   )   (1   )V (1   ) 
 (1   ) 
      L
16