9/28/2017
Course Instructor
Dr. Raymond C. Rumpf
Office: A‐337
Phone: (915) 747‐6958
E‐Mail: rcrumpf@utep.edu
EE 4347
Applied Electromagnetics
Topic 4b
Transmission Line
Behavior
 TheseLine
notes
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Transmission
Behavior
Slide 1
Lecture Outline
•
•
•
•
•
•
Scattering at an Impedance Discontinuity
Power on a Transmission Line
Voltage Standing Wave Ratio (VSWR)
Input Impedance, Zin
Parameter Relations
Special Cases of Terminated Transmission Lines
– Shorted line (ZL = 0)
– Open‐circuit line (ZL = )
– Matched line (ZL = Z0)
Transmission Line Behavior
Slide 2
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Scattering at an
Impedance
Discontinuity
Transmission Line Behavior
Slide 3
Problem Setup
Transmission Line 2
Transmission Line 1
 1 , Z1
 2 , Z2
z0
z
?
We will get a reflection
Transmission Line Behavior
Slide 4
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Incorporate Reflected Wave
Transmission Line 2
Transmission Line 1
 1 , Z1
 2 , Z2
z
z0
V1  z   V1 e1z  V1 e1z
I1  z  
V2  z   V2 e 2 z
V1 1z V1 1z
e 
e
Z1
Z1
I2  z  
V2  2 z
e
Z2
Transmission Line Behavior
Slide 5
Enforce Boundary Conditions (1 of 2)
Transmission Line 2
Transmission Line 1
 1 , Z1
 2 , Z2
z0
z
V1  z   V2  z 
  1z
1
V e
 V1 e1z  V2 e 2 z
I1  z   I 2  z 
Boundary conditions require the
voltage and current on either side
of the interface to be equal.

1
V 1z V1 1z V2  2 z
e 
e 
e
Z1
Z1
Z2
Transmission Line Behavior
Slide 6
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Enforce Boundary Conditions (2 of 2)
Transmission Line 2
Transmission Line 1
 1 , Z1
 2 , Z2
z
z0
V1  0   V2  0 

1
V  V1  V2
The interface occurs at z = 0.
I1  0   I 2  0 

1
V
V V
 1  2
Z1 Z1 Z 2
Transmission Line Behavior
Slide 7
Reflection Coefficient, 
Enforcing the boundary conditions at z = 0 gave us
V1  V1  V2
Eq. 1
V1 V1 V2


Z1 Z1 Z 2
Eq.  2 
Substitute Eq. (1) into Eq. (2) to eliminate V2.
V1 V1 V1  V1


Z1 Z1
Z2
Solve this new expression for V1 V1.
1  1  1  1 
V1  V1  V1  V1
Z1
Z1
Z2
Z2
1  1  1  1 
V1  V1  V1  V1
Z1
Z2
Z1
Z2
 1
1    1
1  
   V1     V1
 Z1 Z 2 
 Z1 Z 2 
 Z 2  Z1 V1   Z 2  Z1 V1
V1 Z 2  Z1

V1 Z 2  Z1
Transmission Line Behavior
V1 Z 2  Z1
  
V1
Z 2  Z1
Slide 8
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Revised Equations for V(z) and I(z)
The total voltage and current in any section of line was written as
V  z   V0 e z  V0 e z
I  z 
V0  z V0  z
e 
e
Z0
Z0
Using the concept of the reflection coefficient , these equations can
now be written as
V  z   V0 e  z  V0 e z  V0  e  z  e z 
I  z 
V0  z V0  z V0  z
e 
e 
 e  e z 
Z0
Z0
Z0
Reflection coefficient
at the load

V0 Z 2  Z1

V0 Z 2  Z1
Transmission Line Behavior
Slide 9
Power on a
Transmission Line
Transmission Line Behavior
Slide 10
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Power Flowing Along Length of Line
The RMS power flowing at a distance z from the load is
Pavg  z  
1
Re V  z  I *  z  
2 
* is complex conjugate
This equation is valid for any
line, even those with loss.
For lossless lines (not lossless loads), we have
V  z   V0  e  j  z   L e j  z 
I  z 
V0  j  z
 e   L e j z 
Z0
Substituting these equations into our expression for Pavg(z) gives
*


V0   j  z
*

1    j z
j z
  Le  
  L e j z  
Pavg  z   Re V0  e
e


2 
Z0


Pavg 
V0
2Z 0
1   
2
L
Notice that the z dependence vanished. This is because
power flows uniformly without decay in lossless lines.
Transmission Line Behavior
Slide 11
Voltage Standing
Wave Ratio (VSWR)
Transmission Line Behavior
Slide 12
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Voltage Standing Wave Ratio (VSWR)
The VSWR is essentially the same concept as the standing wave ratio
(SWR) discussed along with waves. The only difference is that it
describes voltage and current instead of electromagnetic fields.
VSWR 
max V  z 
min V  z 

max I  z 
min I  z 
Transmission Line Behavior
Slide 13
Derivation of VSWR (1 of 2)
We start with our expression for waves travelling in opposite
directions on a transmission line. We will assume a lossless line.
V  z   V0  e  j  z   L e j  z 
I  z 
V0  j  z
  L e j z 
e

Z0
The magnitude of the voltage signal V(z) is
V  z   V0 e  j  z   L e j  z  V0 1   L e j 2  z
By inspection of this equation, we determine the maximum and
minimum values of this function.
Vmax  max V  z   V0 1   L
Vmin  min V  z   V0 1   L
Transmission Line Behavior


Slide 14
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Derivation of VSWR (2 of 2)
The VSWR is therefore
VSWR 
max V  z 
min V  z 

V0 1   L

0
V
1  
L



VSWR 
1  L
1  L
The VSWR is an easily measured quantity and we can calculate the
magnitude of the reflection coefficient || from the VSWR.
L 
VSWR  1
VSWR  1
Transmission Line Behavior
Slide 15
Animation of VSWR (1 of 6)
Case 1: 50  transmission line terminated with a short‐circuit load.
 L  1
Transmission Line Behavior
Slide 16
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Animation of VSWR (2 of 6)
Case 2: 50  transmission line terminated with an open‐circuit load.
 L  1
Transmission Line Behavior
Slide 17
Animation of VSWR (3 of 6)
Case 3: 50  transmission line terminated with a 16.5  load.
Z 0  Z L   L  0.5
Transmission Line Behavior
Slide 18
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Animation of VSWR (4 of 6)
Case 4: 50  transmission line terminated with a 150  load.
Z 0  Z L   L  0.5
Transmission Line Behavior
Slide 19
Animation of VSWR (5 of 6)
Case 5: 50  transmission line terminated with an RL load.
Transmission Line Behavior
Slide 20
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Animation of VSWR (6 of 6)
Case 6: 50  transmission line terminated with an RC load.
Transmission Line Behavior
Slide 21
Input Impedance, Zin
Transmission Line Behavior
Slide 22
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Problem Setup
Generator
Transmission Line
Load
Zg
Vg
 , Z0
Z in
ZL
z0
z  
Generator
Input Impedance
The input impedance Zin is the
impedance observed by the generator.
Zg
Vg
z
Z in
z  
Z in
z
The input impedance Zin is NOT
necessarily the line’s characteristic
impedance Z0 or the load impedance ZL.
Transmission Line Behavior
Slide 23
Animation of Impedance Transformation
  Z2

Z in   m   0
2  ZL
4
Input impedance inverts
Z in  125  j125 
Transmission Line Behavior
 
Z in  m   Z L
 2
Input impedance repeats
Z in  10  j10 
Slide 24
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Derivation of Input Impedance, Zin (1 of 2)
The reflection coefficient at any point z from the load is
Backward Wave
 z 
V0 e z V0 2 z
e

V0 e  z V0
Forward Wave
This means that from the perspective of the generator, the reflection
going into the transmission line will change depending on the length
of the transmission line. This can only happen of the input
impedance to the transmission line is changing.
Transmission Line Behavior
Slide 25
Derivation of Input Impedance, Zin (2 of 2)
We define the impedance of the line at position z to be
Z  z 
V  z
I  z
We previously wrote V(z) and I(z) as
V  z   V0  e  z   L e z 
I  z 

0
V
 e   z   L e z 
Z0
L 
V0 Z L  Z 0

V0 Z L  Z 0
Substituting in our expressions for V(z) and I(z) gives
Z  z 
V0  e  z   L e z 
V0  z
 e   L e z 
Z0
 Z0
e   z   L e z
e   z   L e z
It makes sense that the impedance is not
a function of voltage in a linear system.
Transmission Line Behavior
Slide 26
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Sanity Check: Input Impedance at Load
The input impedance at the load can be determined by setting z = 0 in
our previous equation.
Z in  0   Z 0
 Z0
e0   L e0
e 0   L e0
1 L
1  L
ZL  Z0
ZL  Z0
 Z0
Z  Z0
1 L
ZL  Z0
1
 Z0
ZL  Z0  ZL  Z0
ZL  Z0  ZL  Z0
 Z0
2Z L
2Z 0
 ZL
Transmission Line Behavior
Slide 27
Input Impedance at z  
The input impedance at location z   is
Z in     Z 0
e     Le  
e    L e   

Z
0 
 
  
e   L e 
e     Le  
  
 
A Note About Sign: Backing away from the load, z becomes negative.
However, we defined z   so  stays positive in this equation and
for equations that follow.
Transmission Line Behavior
Slide 28
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Impedance Transformation Formula (1 of 2)
Recall that
Z     Z 0
e    L e   
e    L e   
L 
Z L  Z0
Z L  Z0
We can eliminate L from the input impedance equation by
substituting in our expression for L.
Z L  Z 0  
e
Z L  e   e      Z 0  e    e    
Z L  Z0
Z in     Z 0
 Z0
Z  Z 0  
Z L  e   e      Z 0  e    e    
e   L
e
Z L  Z0
e  
Transmission Line Behavior
Slide 29
Impedance Transformation Formula (2 of 2)
Now recall the definitions of hyperbolic sine and cosine functions.
sinh  z  
e z  e z
2
cosh  z  
e z  e z
2
This lets us write the input impedance expression as
sinh    
Z  2 cosh      Z 0  2sinh    
cosh    
Z in     Z 0 L
 Z0
sinh
Z L  2sinh      Z 0  2 cosh    
    Z
ZL
0
cosh    
Z L  Z0
Recognizing that tanh(z) = sinh(z)/cosh(z), our expression reduces to
Z in     Z 0
Transmission Line Behavior
Z L  Z 0 tanh    
Z 0  Z L tanh    
Slide 30
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Input Transformation for Lossless Line
The lossless line has
 0
  j
Putting these values into our impedance transformation formula gives
Z in     Z 0
Z L  Z 0 tanh  j   
Z 0  Z L tanh  j   
Recognizing that tanh(jz) = jtan(z), our expression for lossless lines
becomes
Z in     Z 0
Z L  jZ 0 tan    
Z 0  jZ L tan    
Transmission Line Behavior
Slide 31
Input Impedance Repeats for Lossless
Lines
For lossless lines, the tan     function in the impedance
transformation equation tells us that the function is periodic and
repeats.
The function repeats every integer multiple of .
   m
m  , , 3, 2, 1, 0,1, 2,3, , 
Recognizing that  = 2/, the above expression leads to
m

2
Note:  is the wavelength in the transmission line, not the free space
wavelength 0.
This means the input impedance repeats for every half‐wavelength
long the transmission line is.
We will revisit this when we cover Smith charts, which will give you a way to
visualize the impedance transformation phenomenon.
Transmission Line Behavior
Slide 32
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Example: Impedance Transformation (1 of 3)
A transmission line with 50  characteristic impedance is connected to
a 10 nF capacitor as the load. If the phase constant of the transmission
line is  = 60 m-1, what is the input impedance Zin of a 1 inch section of
line operating at 4.0 GHz? What equivalent circuit would the source
see?
Transmission Line
Load
50 
Z in
10 nF
z0
z  
z
1 inch
Transmission Line Behavior
Slide 33
Example: Impedance Transformation (2 of 3)
Loss was not specified so we assume a lossless transmission line. Our
impedance transformation equation is therefore
Zin     Z 0
Z L  jZ 0 tan    
Z 0  jZ L tan    
The variables in this equation are
Z 0  50 
 2.54 cm   1 m 

  1.524
 1 inch   100 cm 
1
1
1
ZL 


  j 0.004 
9 1
jC j 2 fC j 2  4.0  10 s 10 109 F 
    60 m 1  1 inch  
Transmission Line Behavior
Slide 34
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Example: Impedance Transformation (3 of 3)
Substituting in the values of our variables gives
   j 0.004    j  50   tan  60  0.0254  
3
Z in   50   
  j1.07  10 
50


j

j
0.004

tan
60

0.0254








The input impedance is purely imaginary and positive. Thus, the input
impedance looks like an inductor to the generator.
Z in  j Leq
Leq 
Z in
Z in
j1.07  103 


 4.24  108 H  42.4 nH
9 1
j j 2 f
j 2  4.0 10 s 
Transmission Line Behavior
Slide 35
Parameter Relations
Transmission Line Behavior
Slide 36
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Vmax, Vmin, Imax & Imin in Terms of VSWR
Vmax and Vmin
2 VSWR
VSWR  1
2
 V0 1   L   V0
VSWR  1
Vmax  V0 1   L   V0
Vmin
Imax and Imin
I max
V0
V0 2 VSWR

1   L   Z VSWR  1
Z0
0
I min 
V0
Z0
1    
L
V0
2
Z 0 VSWR  1
Transmission Line Behavior
Slide 37
Z0 in Terms of VSWR
The characteristic impedance Z0 can be calculated from Vmax and Imax
or Vmin and Imin.
Z0 
Vmax Vmin

I max I min
The input impedance Zin repeats as you back away from the load. We
can calculate the maximum and minimum impedance as
max  Z in  
Vmax
 Z 0  VSWR
I min
Z0
V
min  Z in   min 
I max VSWR
Transmission Line Behavior
min  Z in   Z in  max  Z in 
Slide 38
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Example (1 of 3)
A 50  impedance transmission line is connected to an antenna with
a 72  input impedance. A source provides an input signal of 24 V
peak‐to‐peak.
What is the reflection coefficient at the antenna?
Z  Z 0  72     50  
In this case, the antenna is the load.
L  L

 0.1803
Z L  Z 0  72     50  
What fraction of the input power is delivered to the antenna?
2
2
R   L  0.1803  0.0325
T  1  R  1  0.0325  0.9675  96.7%
Despite the mismatch, almost all power is
still delivered to the antenna. This still
does not mean the antenna will radiate!
What is the VSWR on the line feeding the antenna?
VSWR 
1   L 1  0.1803

 1.44
1   L 1  0.1803
VSWR dB  20 log10  VSWR   20 log10 1.44   3.17 dB
Transmission Line Behavior
Slide 39
Example (2 of 3)
What is the minimum and maximum voltage on the line?
First, we need to convert voltage peak‐to‐peak Vp-p to voltage magnitude V0.
V0 
Vp-p
2

24 V
 12 V
2
Now we are in a position to calculate Vmin and Vmax.

0
Vmin  V

0
Vmax  V
1     12 V 1  0.1803   9.84 V
1     12 V 1  0.1803   14.16 V
L
L
When we are utilizing
high voltages, we want to
be sure Vmax will not
cause arcing or any other
breakdown problems.
What is the minimum and maximum current on the line?
V
9.84 V
 0.1967 A
I min  min 
Z0
50 
At high power, we want
to be sure Imax will not
Vmax 14.16 V

 0.2833 A
I max 
cause heating problems.
Z0
50 
Transmission Line Behavior
Slide 40
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9/28/2017
Example (3 of 3)
What is the total range of input impedances a source could see?
min  Zin  
Vmin
9.84 V

 34.72 
I max 0.2833 A
max  Z in  
Vmax 14.16 V

 72 
I min 0.1967 A
min  Z in   Z in  max  Z in 
34.72   Z in  72 
Transmission Line Behavior
Slide 41
Special Cases of
Terminated
Transmission Lines
Transmission Line Behavior
Slide 42
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Shorted Line, ZL = 0
Reflection from Load
Voltage Standing Wave Ratio
 L  1
VSWR  
Vmin and Vmax
There exists V(z) = 0.
Imin and Imax
Vmin  0
I min  0

0
Vmax  2 V
I max 
Z0
min[Zin] and max[Zin]
Input Impedance
 Z tanh  
Z in      0
 jZ 0 tan  
2 V0
lossy
lossless
min  Z in   0
short circuit
max  Z in   
open circuit
Note 1: Zin for the lossless line is purely
imaginary. This means it is purely
reactive and no dissipation occurs in the
line. The input impedance alternates
between being capacitive and inductive
as you back away from the load.
Note 2: The shorted line behaves much
the same way as the open‐circuit line.
We also observe that
Z in,short Z in,open  Z 02
Transmission Line Behavior
Slide 43
Open‐Circuit Line, ZL = 
Reflection from Load
Voltage Standing Wave Ratio
 L  1
VSWR  
Vmin and Vmax
Imin and Imax
Vmin  0
I min  0
Vmax  2 V0
I max 
lossy
lossless
Note 1: Zin for the lossless line is purely
imaginary. This means it is purely
reactive and no dissipation occurs in the
line. The input impedance alternates
between being capacitive and inductive
as you back away from the load.
Transmission Line Behavior
2 V0
Z0
min[Zin] and max[Zin]
Input Impedance
 Z coth  
Z in      0
 jZ 0 cot  
There exists V(z) = 0.
min  Z in   0
short circuit
max  Z in   
open circuit
Note 2: The open‐circuit line behaves
much the same way as the shorted line.
We also observe that
Z in,short Z in,open  Z 02
Slide 44
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Matched Line, ZL = Z0
Reflection from Load
L  0
Vmin and Vmax
Vmin  Vmax  V0
Input Impedance
Z in     Z 0
Voltage Standing Wave Ratio
VSWR  1
because Vmax = Vmin
Imin and Imax
I min  I max  V0 Z 0
min[Zin] and max[Zin]
min  Z in   max  Z in   Z 0
Note: F the matched line, there are no
reflections and all of the power is
delivered to the load.
Transmission Line Behavior
Slide 45
23