Download Reflection Coefficient

Derivation and
Use of Reflection
Coefficient
Transmission Line Model
IF
ZS
+
V_S
Z0
IT
+
+
VF
VT
_
_
ZT
x
1. Forward Power (positive x direction): VF = IF Z0
2. Matched source of power. Assume ZS = Z0 = R + j0
3. Terminating impedance: ZT = RT + jXT ; VT = IT ZT
4.What’s wrong with this picture?
Terminating Condition ( x = 0 )
IF
ZS
+
V_S
Z0
IT
+
+
VF
VT
_
_
ZT
x
At x = 0, IF(0) = IT and VF(0) = VT if and only if ZT = Z0 !!
But . . . ZT can be ANYTHING!!!
Therefore, there must exist additional voltages and currents in the transmission line such that
the Total Line Current and Total Line Voltage at x = 0 satisfy continuity:
IL(0) = IT = IF(0) + [?]
and
VL(0) = VT = VF(0) + [?]
The mystery terms in the brackets must also correspond to voltages and
currents associated with the transmission of power in the line.
We have already accounted for power traveling in the positive x direction
(forward power).
The mystery terms in the brackets must therefore correspond to voltages
and currents associated with the transmission of power in the opposite
(negative x) direction.
This represents the portion of forward
power not absorbed by the terminating
load, and “reflected” back toward the
source.
IR
+
Z0
VR
The voltage and current associated
with the reflected power must still
obey:
_
x
VR = IRZ0
Now we do the Math!
IL = (IF – IR)
+
ZS
+
V_S
Z0
VL = (VF + VR)
_
IT
+
ZT
VT
_
x
The total voltage and current in the transmission line is the superposition (sum)
of the voltages and currents associated with the forward and reflected power.
The continuity condition at the termination (x = 0) is:
VF 0 V R 0
VT VL 0 VF 0 V R 0
ZT 


 Z0 





I T I L 0 I F 0  I R 0
V
0

V
0
R
 F

(Continuing)
ZT 
V 0 V R 0
VT VL 0 VF 0 V R 0


 Z0  F





I T I L 0 I F 0  I R 0
V
0

V
0
R
 F

…by inspection:
Define
Normalized
Impedance:
Define
Reflection
Coefficient:
Z
ˆ
Z
Z0
VR  x 
 x  
VF  x 
QED:
1  T
ˆ
ZT 
1  T
ZT 1  V R 0 VF 0

Z0 1  V R 0 VF 0
Rewrite the above,
and define T = (0)
1  0 1  T
ZˆT  


1  0 1  T
ZˆT  1 ZT  Z 0
T 

ZˆT  1 ZT  Z 0
Recap . . .
• Based on fundamental principles, we have determined
that the voltage and current that we would measure on
a transmission line is the superposition of two sets of
voltages and currents, corresponding to power
flowing in both directions.
• We have defined the concept of normalized
impedance.
• We have defined the concept of reflection coefficient,
and determined how to compute it’s value at the
termination point of a transmission line, given the
terminating impedance.
Observations
• The forgoing results are independent of the
mathematical form of the voltage function. It can be
any complex function of time and position. However,
having invoked the concept of Impedance, a
sinusoidal (complex exponential) function of time is
implied.
• Our definition of reflection coefficient constrains its
magnitude to unity or less, thus it may be represented
by a vector (point) on the complex plane lying within
or on the “Unit Circle”.
The quintessential
computation when working
with transmission lines is the
determination of the
Impedance (or Admittance)
presented by a line of length
l terminated by ZT.
By definition, Line
Impedance is the ratio of
total Line Voltage to total
Line Current:
ZL: Line Impedance
IL(x = -l)
IT
+
ZL(x = -l)
VL(x = -l)
_
+
Z0
ZT
VT
_
x
(x = -l)
(x = 0)
1l VL l l
1  V R  l  VF  l 
VF  l  V R Zl   l  VZF 
R
L  Z0
0
Z L  l  
   Z0 




1



l
L










I F  l   I R  l 
V

l

V

l
1

V

l
V

l
F
R
R
 F



Line Reflection Coefficient
I order to determine Reflection coefficient at any point, x,
along the line, we will incorporate the forms for the Forward
and Reflected Voltage waveforms determined previously:
VR t , x  V  e j t  x  V  2 jx
 x  
  j t x    e
VF t , x  V e
V
At the termination point (x = 0 ),
Substituting,
x   T e
V
0  T  
V
2 jx
We see from the above that the reflection coefficient, when considered as a
Polar (not Cartesian) complex vector, is “well behaved”, in the sense that it
is a vector of fixed magnitude < 1, with only its complex phase angle
affected by the distance from the line termination to the point of interest.
Conceptual Solution to the
Fundamental Problem
Find the impedance presented by a transmission line having characteristic
impedance Z0 and length l, when terminated with impedance ZT.
Step 1
Determine Reflection Coefficient at
the Termination.
Step 2
Rotate the Reflection Coefficient
Vector by an angle corresponding to
the “electrical length” of the line.
Step 3
Determine Line Impedance from the
rotated Reflection Coefficient vector.
ZT  Z 0
T 
ZT  Z 0
l  T  2l 
1  l 
Zl  Z0 

1  l 
The Smith Chart
The Smith Chart is a
graphical tool which
performs the
computations of steps 1
and 3 graphically.
Early Smith Charts were
scaled to a specific
transmission line
characteristic
impedance. Now, all
Smith charts are
normalized to a
characteristic impedance
of 1 ohm, so the
practical procedure
using a smith chart has
five steps:
1.
Normalize Terminating Impedance
2.
Draw Reflection Coefficient
vector corresponding to
Normalized Terminating
Impedance. (Old step 1)
3.
Rotate the Reflection Coefficient
Vector by an angle corresponding
to the “electrical length” of the
line. (Old step 2)
4.
Read the Normalized Line
Impedance from the tip of the
rotated Reflection Coefficient
vector. (Old step 3)
5.
De-normalize the Line Impedance.
ZT
ˆ
ZT 
Z0
Z L  Z0 Zˆ L
How does it work?
Consider the fundamental equation relating
Impedance to Reflection coefficient:
Z = R +jX
Z  Z0

 
Z  Z0
  
If we take every possible value of Z from the complex Z Plane (RHS), and
plot the Reflection Coefficient according to the equation above, you get
every possible point inside (or on) the unit circle.
Reflection Coefficient:
Special Cases
Short Circuit
ZT  0
T  1 R
Since  = 1, the magnitude of the
forward and reflected voltage waves
are equal. Since their phase difference
is , they will sum to zero volts at the
termination point.
Open Circuit
ZT  
T  10 R
Since  = 1, the magnitude of the
forward and reflected voltage waves
are equal. Since their phase difference
is 0, they will sum to a voltage
maximum at the termination point.
Z  Z0

 
Z  Z0
Shorted Quarter Wave Line
Consider a line, terminated with a short circuit, having length
l such that: -2l = -, or l = (/2)/.
Recall that  = /vp = 2f/vp = 2/l.
Thus l = l/4 (quarter wave).
Zl  
Z0
ZT  0
T
l
x
(x = -l = -l/4)
(x = 0)
We see that a distance on the transmission line corresponding to one quarter wavelength represents a
rotation of  radians of the reflection coefficient vector. A short circuit is “transformed” to an open
circuit, and vice versa.
Standing Wave Example
Consider a line terminated with impedance ZT = 3Z0 . We have
ZS
+
V_S
Z0
+
+
VL
VT
_
_
T 
ZT = 3 Z0
  0.5 
x
At x = -l/2 (an additional quarter wave away from the load),
the phase difference will be 2 (or zero), and there will be
another voltage maximum. This variation in amplitude of VL
as we move along the line is called a standing wave, and is
characterized by the “Standing Wave Ratio” (SWR), which is
the ratio of Voltage maximum to voltage minimum:
SWR 
VL (max) 1   VF 1   


1   VF 1   
VL (min)
ZT  Z 0
 0.50
ZT  Z 0
VR
VF
This means that the amplitude of the
reflected wave is half that of the
forward wave. Since the phase
difference is zero at x = 0, the amplitudes
add, yielding a voltage maximum:
|VL(max)| = (1+)| VF | = 1.5 |VF|
One quarter wave distant from the
termination, x = -l/4, the reflection
coefficient vector will have rotated by 
radians, the reflected wave will be exactly
out of phase with the forward wave and
the amplitudes will subtract, yielding a
voltage minimum:
|VL(min)| = (1-)| VF | = 0.5 |VF|
 and SWR
V (max) 1   VF 1   
SWR  L


1   VF 1   
VL (min)
If the phase angle of  is 0
(positive real axis ), then

SWR  1
SWR  1
SWR  1
 

SWR  1
Zˆ  1
Zˆ  1
Thus the constant  circle crosses the positive real axis where
Zˆ  Rˆ  j 0  SWR
Standing Wave Measurements
Consider a 50 ohm line (eR = 1.69) terminated with an unknown impedance ZT . We
measure a voltage maximum of 6.5 VAC at a point 0.3 m from the termination, and a
voltage minimum of 2.7 VAC at a point 1.1 meters from the termination.
ZS
+
V_S
+
+
VL(min)
Z0
_
ZT = ?
VL(max)
The “electrical” distance (l) from the
termination to the voltage maximum is
0.3/3.2 = 0.09375 l  33.75o  0.59R
_
x
-1.1
SWR 
-0.3
1   
VL (max) 6.5

 2.41 
VL (min) 2.7
1   
SWR  1 1.41

 0.414
SWR  1 3.41
ZˆT  1
0
T  0.4142  33.75  
Zˆ  1
0
 = 0.414
If eR = 1.69, vp = ceR(-1/2) = c/1.3
f = vp / l = c/(1.3)(3.2 m)= 72.116 Mhz
67.5O
T

T
Since voltage max/min are
always separated by l/4,
l = 4(1.1 - 0.3) = 3.2 m.
@Vmin
@Vmax
1  T   0.97  0.89 j
ZˆT 
1  T 
ZT  ZˆT Z0  48.53  44.66j