Lecture 2: Transmission Line Theory 0 0 = - =

Lecture 2: Transmission Line Theory
Applying Kirchoff’s voltage and current laws to the
lumped-element circuit, we get
Lumped-element model of transmission lines
i(z,t)
v ( z , t )  Z  z  i ( z , t )  v ( z  z , t )
i ( z , t )  Y  z  v ( z  z , t )  i ( z  z , t )
Z and Y are the impedance and
admittance per unit length in zdirection.
+ v(z,t)
-
Z = R + jL and Y = G + jC,
where
I
Dividing both equations by z and taking z->0, we get
dV/dz= -ZI and dI/dz= -YV, simultaneous solution of which yields
Z
Z = R + j L
V
Y
Y = G + j C
Expression of the transmission line for
a small section between z and z+z
R is the series resistance per unit
length z, /m
L is the series inductance per unit
length z, H/m
G is the shunt conductance per unit
length z, S/m
C is the shunt capacitance per unit
length z, F/m
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transmission line, ZY=(R+jL)(G+jC)=2, therefore, we have
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Wave Propagation along a Transmission Line
The solution of these equations is in the form of waves in the +z and -z
direction, which for sinusoidal excitation take the form
V(z) = V+e jt-z+ V- e jt+z and I(z) = I+ e jt-z+ I- e jt+z
d 2V
dz 2
  2V  0
d 2I
dz 2
 I  0
2
Wave
Equations!
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For a single wave solution in one direction, the ratio V(z)/I(z) is the
same everywhere on the line given by
V  ( z )  V  ( z ) Z ( R  j L )
R  j L
 
 

 Z0

I ( z)
I ( z)
G  j C


Zo is defined as the characteristic impedance.
The propagation constant  is given by
 =  + j = ZY .
which for a lossless line is a real number
For L>>R and C>>G (low or zero loss case),
LC
 = so
 = 2 ,
V
* To distinguish it from the free-space wavelength nomenclature  or o, the
wavelength on a waveguide or coaxial transmission line is often referred to as the
guide wavelength g.
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d2V
d2I
= ZYI; z here represents distance along the
2 = ZYV and
dz
dz 2
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Z
L
Zo = I  = Y = C , where L and C are the inductance and
capacitance per unit length.
Thus we can rewrite the current equation as
I(z) = I+e j(t-z) + I-e j(t+z) =
V
Zo
e j(t-z) -
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V
j(t+z)
Zo e
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If we solve for Zo of coaxial and microstrip line, we have
Transmission Line Parameters
Determine L, C, R, G for a transmission line by Solving Maxwell’s equations
(see pp. 52-57 in Pozar’s book for details)
Zo =
377
2 
ln(b/a) for coaxial line (note use of ln and log10 in
r
different references), and Zo 
377
d/W for microstrip line,
r
ignoring fringing fields.
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Transmission Line Discontinuities and Load Impedances
Termination of transmission lines
Infinite lines: the wave in the +z direction will continue
indefinitely and never return in the -z direction.
+z
-z
Z o
Z L Z o
M is m a tc h e d L o a d C r e a te s R e f le c te d W a v e
Zo
+z
At the load junction (z=0) we have,
Matched termination: all power of +z wave delivered to the
load, no reflected wave
Zo
+z
1
(V   V  )
Z0
For a given load impedance ZL, the load boundary condition is
V ( 0)  Z L I ( 0)
Therefore, we have
Z = Zo
Z
V   V   L (V   V  )
Z0
Matched Termination Same as Infinite Line
Boundary conditions at a matched load are the same as for the
infinite transmission line.
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I ( 0) 
V ( 0)  V   V 
Infinite Transmission Line
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
V  Z L  Z0

V  Z L  Z0
Load reflection coefficient
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What happens when waves exist in both +z and -z
directions along a transmission line?
The Complex Reflection Coefficient 
Full Wave Equations:
V(z) = V+ ej(t-z) + V- ej(t+z), and
• Both waves are coherent and interference patterns exit
• The interference pattern will be stationary with respect to
the point of reflection, and will thus be a standing
wave such as may be found on the strings of musical
instrument. The standing wave interference pattern is
present both in the resulting V(z) and I(z).
V+
VI(z) = Z ej(t-z) - Z ej(t+z).
o
o
consider a complex load impedance ZL terminating a transmission
line Zo, the magnitude of the -z wave is related to that of the +z
wave at the termination by a complex quantity defined as the
reflection coefficient , defined such that
V- = V+, where
=
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At the reflection point (z=0)
V- = LV+ and
VL = V++V- = V+(1+L )
(resulting wave)
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V L = (1+ )V +
V - = V +
V- -Ij
j
V+ = I+ = ||e =  e = /
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If we move in the -z direction (away from the load) by a
distance l (i.e. z = -l ), we have
Vo e j  z Vo e  j  l
( z )    j  z   j  l   L e 2 j  l
Vo e
Vo e
V(z) = 1+(z) = 1+(-l) = 1+L e-2jl = 1+|| ej(-2l)
1+ 
V+
1+ 
-2
z
θ-2βl
1+ 
V(z) varies from a maximum
of V+(1+||) to a minimum of
V+(1-||) and the distance
between two successive
maxima (or minima) is
2l=2l=/2

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Vector Form
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
1

Normalized expression (V+=1)
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11

-2 zl
1
v min
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v ma x
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Standing Wave Ratio (SWR) (or VSWR)
vmax 1+|L| 1+
SWR = v
= 1-| | =
. Note that this can be solved for , yielding
L
1-
min
SWR-1
 = SWR+1 , so if we know SWR we know .
SWR characterize the
degree of impedance
mismatch!
For a matched load =0, SWR =1 and the voltage on the line is just
V(z) = V+ for all z; under such a condition the line is termed flat
Return Loss
P|V-|2
2
P+ = |V+|2 =  , the ratio of the power in the reflected wave to that in the incident wave
or, expressed as a loss (a positive number) in dB
RL = -10 log10 2 = -20 log10 dB) = -20 log10  (dB).
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