P5-Electromagnetic Fields and Waves
Prof. Andrea C. Ferrari
KWWSZZZ
KWWSZZZJHQJFDPDFXNQPVOHFWXUHQRWHVKWPO
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALS
GROUP
SPECTROSCOPY GROUP
1
1
2
2
6 Lectures
3 Main Sections
~2 lectures per subject
I Transmission Lines
I.0 The wave equation
I.1 Telegrapher’s Equations
I.2 Characteristic Impedance
I.3 Reflection
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALSGROUP
GROUP
SPECTROSCOPY
II Electromagnetic Waves in Free Space
II.1 Electromagnetic Fields
II.2 Electromagnetic Waves
II.3 Reflection and Refraction of Waves
III Antennae and Radio Transmission
III.1 Antennae
III.2 Radio
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALS
GROUP
SPECTROSCOPY GROUP
3
3
OBJECTIVES
As the frequency of electronic circuits rises, one can no
longer assume that voltages and currents are instantly
transmitted by a wire.
The objectives of this course are:
•Appreciate when a wave theory is needed
•Derive and solve simple transmission line problems
•Understand the importance of matching to
characteristic impedance of a transmission cable
the
•Understand basic principles of EM wave propagation in
free space, across
interfaces
of antennae
NANOMATERIALS
AND
CAMBRIDGE
UNIVERSITY and the use
ELECTRONIC
DEVICES
DEPARTMENT OF ENGINEERING
AND
MATERIALSGROUP
GROUP
SPECTROSCOPY
4
4
This course deals with transmission of electromagnetic waves
1) along a cable (i.e. a transmission line)
2) through free space (the ‘ether’).
In the first half of these lectures, we will derive the differential equations
which describe the propagation of a wave along a transmission line.
Then we will use these equations to demonstrate that these waves exhibit
reflection, have impedance, and transmit power.
In the second half of these lectures we will look at the behaviour of waves
in free space.
We will also consider different types of antennae for transmission and
reception of electromagnetic waves.
Reference: OLVER A.D.
Microwave and Optical Transmission
NANOMATERIALS
AND
CAMBRIDGE
DEVICES
John
WileyUNIVERSITY
& Sons, 1992, 1997ELECTRONIC
DEPARTMENT OF ENGINEERING
AND
MATERIALS
GROUP
Shelf Mark:
NV 135 SPECTROSCOPY GROUP
5
5
Handouts
The handouts have some gaps for you to fill
NOTE:
1) DO NOT PANIC IF YOU DO NOT
MANAGE TO WRITE DOWN IN “REAL TIME”
2) Prefer to just sit back and relax?
You will be able to Download a PDF of the complete
slides from
NANOMATERIALS
AND
CAMBRIDGE UNIVERSITY
ELECTRONIC
DEVICES
http://www-g.eng.cam.ac.uk/nms/lecturenotes.html
DEPARTMENT OF ENGINEERING
AND
MATERIALSGROUP
GROUP
SPECTROSCOPY
6
6
I.0 The Wave Equation
Aims
To recall basic phasors concepts
To introduce the generalised form of the wave equation
Objectives
At the end of this section you should be able to recognise
the generalized form of the wave equation, its general solution,
the propagation direction and velocity
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALS
GROUP
SPECTROSCOPY GROUP
7
7
I.0.0 Introduction
An ideal transmission line is defined as:
“a link between two points in which the signal at any point
equals the initiating signal”
i.e. transmission takes place instantaneously and there is
no attenuation
Real world transmission lines are not ideal, there is
attenuation and there are delays in transmission
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALSGROUP
GROUP
SPECTROSCOPY
8
8
A transmission line can be seen as a device for
propagating energy from one point to another
The propagation of energy is for one of two general
reasons:
1. Power transfer (e.g. for lighting, heating, performing
work) - examples are mains electricity, microwave
guides in a microwave oven, a fibre-optic illuminator.
2. Information transfer. examples are telephone,
radio, and fibre-optic links (in each case the energy
propagating down the transmission line is modulated
NANOMATERIALS
AND
CAMBRIDGE UNIVERSITY
ELECTRONIC
DEVICES
in some way).
DEPARTMENT OF ENGINEERING
AND
MATERIALS
GROUP
SPECTROSCOPY GROUP
9
9
Examples
Power
Plant
Consumer
Home
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALSGROUP
GROUP
SPECTROSCOPY
10
10
Antenna CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
NANOMATERIALS
AND
ELECTRONIC
DEVICES
Optical
Fibre
Link
AND
MATERIALSGROUP
GROUP
SPECTROSCOPY
11
11
12
Pair of wires
PCB tracks
Co-ax cable
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALSGROUP
GROUP
SPECTROSCOPY
12
IC interconnects
Waveguides
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALS
GROUP
SPECTROSCOPY GROUP
13
13
Mircostrip
Dielectric of thickness T, with a conductor deposited on the
bottom surface, and a strip of conductor of width W on the top
surface
Can be fabricated using Printed Circuit Board (PCB)
technology, and is used to convey NANOMATERIALS
microwave AND
frequency
CAMBRIDGE UNIVERSITY
ELECTRONIC DEVICES
signals
DEPARTMENT OF ENGINEERING
AND
MATERIALSGROUP
GROUP
SPECTROSCOPY
14
14
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
Microwave Oven
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALS
GROUP
SPECTROSCOPY GROUP
15
15
16
16
Optical Fibres
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALSGROUP
GROUP
SPECTROSCOPY
Phasor Notation
A
means A is complex
Im
I m{ A}
A = »e{ A} + Im{ A} j = A e j∠ A
A ∠A
»e
» e{ A}
A=A
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
Ae j β x
is short-hand for
which equals:
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALS
GROUP
SPECTROSCOPY GROUP
{
» e Ae j ( β x + ω t )
(
A cos β x + ω t + ∠ A
17
17
}
)
Proof
e
j ( ±θ )
= cos(θ ) ± j sin(θ )
then
Ae j ( β x + ω t ) = Ae j ∠ A e j ( β x + ω t ) = Ae j ( β x + ω t + ∠ A )
Ae
(
j β x +ω t + ∠ A
)
NANOMATERIALS
= CAMBRIDGE
A cos( βUNIVERSITY
x + ω t + ∠ A ) + ELECTRONIC
jA sin( βDEVICES
x +AND
ω t + ∠ A)
DEPARTMENT OF ENGINEERING
AND
MATERIALSGROUP
GROUP
SPECTROSCOPY
18
18
I.0.1 The Wave Equation
The generalised form of the wave equation is:
2
∂ A
2 2
=v ∇ A
2
∂t
Where the Laplacian of a scalar A is:
∂2 A ∂2 A ∂2 A
∇ A=
+ 2 + 2
2
∂x
∂y
∂z
2
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALS
GROUP
SPECTROSCOPY GROUP
19
19
We will be looking at plane waves for which the wave
equation is one-dimensional and appears as follows:
2
∂2 A
∂
A
2
=v
2
∂t
∂x 2
or
∂2 A 1 ∂2 A
= 2 2
2
∂x
v ∂t
Where A could be:
Either the Voltage (V) or the Current (I)
as in waves in a transmission line
Or the Electric Field (E) or Magnetic Field (H)
NANOMATERIALS
AND
CAMBRIDGE UNIVERSITY
ELECTRONIC
DEVICES
as in electromagnetic
waves
in
free
space
DEPARTMENT OF ENGINEERING
AND
MATERIALSGROUP
GROUP
SPECTROSCOPY
20
20
There are many other cases where the wave equation is used
For example
1) Waves on a string. These are planar waves where A
represents the amplitude of the wave
2) Waves in a membrane, where there is variation in both x and y,
and the equation is of the form
2
∂2 A
∂2 A 
2∂ A
=v  2 + 2 
∂t 2
∂y 
 ∂x
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALS
GROUP
SPECTROSCOPY GROUP
21
21
The constant v is called the wave speed.
This comes from the fact that the general solution to the
wave equation (D’Alembert solution,~1747) is
A = f ( x ± vt )
Note
A = f ( x − vt )
Forward moving
A = f ( xCAMBRIDGE
+ vt ) UNIVERSITY
Backward moving
NANOMATERIALS
AND
ELECTRONIC
DEVICES
DEPARTMENT OF ENGINEERING
AND
MATERIALSGROUP
GROUP
SPECTROSCOPY
22
22
Direction of travel
f ( x − vt )
F(t+x/v
f ( x +) vt )
P
∆x
∆x
t
t+∆t
t+∆t
t
x
x
Consider a fixed point, P, on the moving waveform, i.e. a
point with constant f
f(x-vt) will be constant if x-vt is constant
If t increases (t→t+∆t), x must also increase if x-vt is to be
constant
An x increase implies that the wave is moving to the
right (Forward)CAMBRIDGE UNIVERSITY
NANOMATERIALS
AND
ELECTRONIC
DEVICES
DEPARTMENT OF ENGINEERING
AND
MATERIALS
GROUP
GROUP
23
Similarly, for x+vt ⇒ wave is moving toSPECTROSCOPY
left (Backward)
Verify that A
23
= f ( x ± vt ) is general solution
∂A
= ±vf '( x ± vt )
∂t
∂2 A
= v 2 f ''( x ± vt )
2
∂t
∂2 A
= f ''( x ± vt )
2
∂x
∂A
= f '( x ± vt )
∂x
2
∂2 A
∂
A
2
=v
2
∂t
∂x 2
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALSGROUP
GROUP
SPECTROSCOPY
24
24
I.1 Electrical Waves
Aims
To derive the telegrapher’s equations
To account for losses in transmission lines
Objectives
At the end of this section you should be able to recognise
when the wave theory is relevant; to master the concepts of
wavelenght, wave velocity, period and phase; to describe the
propagation of waves in loss-less and lossy transmission lines
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALS
GROUP
SPECTROSCOPY GROUP
25
25
I.1.1 Telegrapher’s Equations
Let us consider a short length, δx, of a wire pair
δx
This could, for example, represent a coaxial cable
For a small δx, any function A(x) can be written as
∂A( x)
A( x + δ x) ≈ A(CAMBRIDGE
x) +
δx
∂x UNIVERSITY
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALSGROUP
GROUP
SPECTROSCOPY
DEPARTMENT OF ENGINEERING
In our case
A can be Voltage (V) or Current (I)
26
26
Let us define
L
series/loop inductance per unit length [H/m]
Lδ x
I
VL
δx
V = Lδ x
∂I
∂t
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALS
GROUP
SPECTROSCOPY GROUP
CAMBRIDGE UNIVERSITY
L
DEPARTMENT OF ENGINEERING
27
27
C parallel/shunt capacitance per unit length [F/m]
VC
IC
Cδ x
δx
I C = Cδ x
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
∂VC
∂t
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALSGROUP
GROUP
SPECTROSCOPY
28
28
VL = Lδ x
∂I
∂t
I
B
IF = I +
IC = −
VL
∂I
δx
∂x
∂I
δx
∂x
VC
VC = V +
V = VC + V L
∂V
δx
∂x
δx
I C = Cδ x
∂VC
∂
∂V
∂V
∂V
∂V
2
= CAMBRIDGE
C δ x (V UNIVERSITY
+
δ x ) = Cδ x
+
C
δ
x
≈
C
δ
x
(
)
NANOMATERIALS
AND
ELECTRONIC DEVICES
∂t DEPARTMENT
∂t OF ENGINEERING
∂x
∂tAND
∂x∂t GROUP
∂t
MATERIALS
GROUP
SPECTROSCOPY
29
VC = V − VL
I F = I − IC
V+
∂V
∂I
δ x =V − L δ x
∂x
∂t
I+
∂I
∂V
δx= I −
Cδ x
∂x
∂t
∂V
∂I
= −L
∂x
∂t
∂I
∂V
= −C
∂x
∂t
29
(1.1)
(1.2)
Eqs. (1.1),(1.2) are known as the “telegrapher’s equations”
They were derived in 1885 by Oliver Heaviside, and were crucial in the
NANOMATERIALS
AND
CAMBRIDGE UNIVERSITY
ELECTRONIC
DEVICES
early development
of
long
distance
telegraphy
(hence
the
name)
DEPARTMENT OF ENGINEERING
AND MATERIALS GROUP
SPECTROSCOPY GROUP
30
30
I.1.2 Travelling Wave Equations
Let us differentiate both (1.1) and (1.2) with respect to
x
∂ 2V
∂  ∂I 
= −L  
2
∂x
∂t  ∂x 
∂ 2V
= LC 2
∂t
(1.1a)
∂ I
∂  ∂V 
C
=
−


∂x 2
∂t  ∂x 
∂2I
= LC 2
∂t
(1.2a)
2
Then in (1.1a) substitute
∂I
∂x
using (1.2)
∂V
∂x
using (1.1)
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALS
GROUP
SPECTROSCOPY GROUP
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
Then in (1.2a) substitute
2
31
31
32
32
2
∂V
∂V
= LC 2
2
∂x
∂t
2
(1.1a)
2
∂ I
∂ I
= LC 2
2
∂x
∂t
(1.2a)
Same functional form as wave equation:
∂2 A 1 ∂2 A
= 2 2
2
∂x
v ∂t
1
v =
LC
2
We try a solution for V in (1.1a) of the form
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
V = Ae
jβ x
e
NANOMATERIALS AND
AND
MATERIALSGROUP
GROUP
SPECTROSCOPY
jω tELECTRONIC DEVICES
− β 2 Ae j β x e jω t = −ω 2 LC Ae j β x e jω t
Hence
β = ±ω LC
Phase Constant
(1.3)
Since β can be positive or negative, we obtain expressions for
voltage and current waves moving forward (subscript F) and
backward (subscript B) along the transmission line
{(
I = R e {( I
) }
V = R e V F e − j β x + V B e j β x e jω t
(1.4)
) }
− jβ x
jβ x
jω t
e
I
e
e
+
F
B
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
(1.5)
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALS
GROUP
SPECTROSCOPY GROUP
33
33
I.1.3 Lossy Transmission Lines
Thus far we considered a lossless transmission line. Therefore we
did not include any resistance along the line, nor any conductance
across the line.
If we now define
R= series resistance per unit length [Ω/m]
G= shunt conductance per unit length [S/m]
To derive the relevant expressions for a lossy transmission line our
equivalent circuit would become :
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALSGROUP
GROUP
SPECTROSCOPY
34
34
Rδ x
∂I
I+
δx
∂x
Lδ x
I
B
V
∂V
VC = V +
δx
∂x
VL
VR
Gδ x
Cδ x
δx
V − VR − VL − VC = 0
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALS
GROUP
SPECTROSCOPY GROUP
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
V − Rδ xI − Lδ x
35
35
36
36
∂I
∂V
− (V +
δ x) = 0
∂t
∂x
∂V
∂I 

= −  RI + L 
∂x
∂t 

For simplicity we assume
∂I
= jω f ( x)e jωt = jω I
∂t
I = f ( x)e jωt
Then
∂V
= −( R + jω L) I
∂x
∂V
∂I
= −L
∂x
∂t
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
Compare with (1.1)
NANOMATERIALS
AND
ELECTRONIC
DEVICES
=AND
−
j
ω
LI
MATERIALSGROUP
GROUP
SPECTROSCOPY
Similarly, using Kirchoff’s current law to sum currents:
I − Gδ xV − jω C δ xV − ( I +
∂I
δ x) = 0
∂x
∂I
= −(G + jωC )V
∂x
∂I
∂V
= −C
= − jω CV
∂x
∂t
Compare with (1.2)
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALS
GROUP
SPECTROSCOPY GROUP
37
37
Thus, we can write the expression for a lossy line starting
from that of a lossless line, if we substitute
L in a lossless line with:
L' =
C in a lossless line with:
( R + jω L)
jω
C'=
Then
β = ω LC
in a lossy line
(G + jωC )
jω
in a lossy line
In a lossless line corresponds to:
1
β'=
( R + jω L)(G + jωC )
j
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
in a lossy line
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALSGROUP
GROUP
SPECTROSCOPY
38
38
Substituting β ⇒ β’ into (1.4) and (1.5) and defining
γ = ( R + jω L)(G + jωC ) = α + j β
We get
{(
I = R e {( I
V = R e VF e
Fe
− (α + j β ) x
+ VB e (
− (α + j β ) x
+ I B e(
) }
)
)e }
α + jβ )x
e jω t
(1.6)
α + jβ x
jω t
(1.7)
γ is called propagation constant
β is the phase constant
The real term α corresponds to the attenuation along the line
NANOMATERIALS
AND
UNIVERSITY
ELECTRONIC
DEVICES
and is knownDEPARTMENT
asCAMBRIDGE
the attenuation
constant
OF ENGINEERING
AND
MATERIALS
GROUP
SPECTROSCOPY GROUP
39
39
For a forward travelling wave:
V = VF e −α x
V = VF e jωt e-γx = VF e-αx e j(ωt-βx)
amplitude factor
Voltage
VF
phase factor
time variation
x
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALSGROUP
GROUP
SPECTROSCOPY
40
40
Note:
ω L >> R and ωC >> G :
At high frequencies:
γ = ( R + jω L)(G + jωC ) ≈ −ω 2 LC = jω LC
Thus
α ≈0
β ≈ ω LC
The expressions approximate back to those for
lossless lines
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALS
GROUP
SPECTROSCOPY GROUP
41
41
I.1.4 Wave velocity: v
Our expressions for voltage and current contain 2 exponentials
The one in terms of x:
e± j β x
gives the spatial dependence of the wave, hence the wavelength:
λ=
2π
β
jω t
The other: e
gives the temporal dependence of the wave, hence its frequency:
ω
f =
CAMBRIDGE UNIVERSITY
2π
DEPARTMENT OF ENGINEERING
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALSGROUP
GROUP
SPECTROSCOPY
42
42
For a wave velocity v, wavelength λ and frequency f:
v= fλ
then
ω 2π
2π β
v=
β = ω LC
since
1
LC
v=
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
(1.8)
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALS
GROUP
SPECTROSCOPY GROUP
43
43
I.1.5 Example: Wavelength
An Ethernet cable has
L= 0.22 µHm-1 and C = 86 pFm-1.
What is the wavelength at 10 MHz ?
λ=
From
2π
and
β
λ=
Then
λ=
β = ω LC
2π
ω LC
2π
2π ⋅ 10 ⋅ 10 6 0.22 ⋅ 10 −6 ⋅ 86 ⋅ 10 −12
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
= 23metres
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALSGROUP
GROUP
SPECTROSCOPY
44
44
I.1.6 When must distances be accounted for in AC
circuits?
λ
λ
2
4
λ
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALS
GROUP
SPECTROSCOPY GROUP
45
45
Large ship is in serious trouble (as you can see) and we cannot ignore the
effect of the waves
A much smaller vessel caught in the same storm fares much better
If a circuit is one quarter of a wavelength across, then one end is at zero,
the other at a maximum
If a circuit is an eighth of a wavelength across, then the difference is
2 of the amplitude
In general, if the wavelength is long in comparison to our electrical circuit,
then we can use standard circuit analysis without considering
transmission line effects.
A good rule of thumb is for the wavelength to be a factor of 16 longer
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALSGROUP
GROUP
SPECTROSCOPY
46
46
L≥
L≤
λ
Wave Relevant
16
λ
Wave Not Relevant
16
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALS
GROUP
SPECTROSCOPY GROUP
47
47
I.1.7 Example: When is wave theory relevant?
A designer is creating a circuit which has a clock rate of 5MHz and has
200mm long tracks for which the inductance (L) and capacitance (C) per
unit length are:
L=0.5µHm-1
From
Then
λ=
C=60pFm-1
2π
β
λ=
And
β = ω LC
λ=
2π
2π ⋅ 5 ⋅10
6
−6
0.5 ⋅10 ⋅ 60 ⋅10
−12
2π
ω LC
= 36.5m
36.5 m is much greater than 200 mm (the size of the circuit board),
so that wave theory is irrelevant.
NANOMATERIALS
AND
CAMBRIDGE
UNIVERSITY
DEVICES
Note: The problem
is even
less relevant for ELECTRONIC
mains frequencies
i.e.
DEPARTMENT OF ENGINEERING
AND
MATERIALSGROUP
GROUP
SPECTROSCOPY
48
48
50 Hz. This gives λ~3650 km
I.2 Characteristic Impedance
Aims
To define and derive the characteristic impedance for
lossless and lossy lines
Objectives
At the end of this section you should be able to describe the
forward and backward waves in a transmission line and
calculate the characteristic impedance
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALS
GROUP
SPECTROSCOPY GROUP
49
49
I.2.1 Lossless Lines
Recalling the solutions for I & V (equations 1.4&1.5):
{(
I = R e {( I
) }
)e }
V = R e V F e − j β x + V B e j β x e jω t
− jβ x
jβ x
e
I
e
+
F
B
(1.4)
jω t
(1.5)
Differentiating (1.5) with respect to x
(1.4) with respect to t and multiplying by -C
∂I
= Re − j β I F e − jβ x + j β I B e j β x e jωt
∂x
) }
(2.1)
∂VCAMBRIDGE UNIVERSITY − j β x
j β x DEVICES
jω
t
NANOMATERIALS
AND
ELECTRONIC
=
R
e
−
Cj
ω
V
e
−
Cj
ω
V
e
e
F
B
DEPARTMENT OF ENGINEERING
AND
MATERIALS
GROUP
SPECTROSCOPY
GROUP
∂t
(2.2)
{(
−C
{(
) }
50
50
According to the second Telegrapher’s equation:
∂I
∂V
= −C
∂x
∂t
(1.2)
We can then equate (2.1) and (2.2):
Re
{( − j β I
Since
F
) }
e − j β x + j β I B e j β x e jω t = R e
e
j (ω t − β x )
and
e
{( −C jω V e
− jβ x
F
) }
− C jω V B e j β x e jω t
j (ω t + β x )
represent waves travelling in opposite directions, they can
be treated separately.
NANOMATERIALS
AND
UNIVERSITY
ELECTRONIC
DEVICES
This leadsDEPARTMENT
toCAMBRIDGE
two independent
expressions
in
V
and I
OF ENGINEERING
AND
MATERIALS
GROUP
SPECTROSCOPY GROUP
51
51
− j β I F e − jβ x = −Cjω VF e − j β x
VF
β
=
I F Cω
I B jβ e
jβ x
= − Cjω V B e
jβ x
β
VB
=−
Cω
IB
Note: If we consider VF and VB to have the same sign
then, due to the differentiation with respect to x,
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
I B and I F have opposite signs
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALSGROUP
GROUP
SPECTROSCOPY
52
52
The characteristic impedance, Z0 is defined as the ratio
between the voltage and the current of a unidirectional
forward wave on a transmission line at any point, with no
reflection:
VF
Z0 =
IF
Since
Z0 is always positive
β
VF
=
I F Cω
From (1.3)
Z0 =
β
Cω
β = ω LC
L
DEPARTMENT OF ENGINEERING
C
Z0 =
CAMBRIDGE UNIVERSITY
(2.3)
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALS
GROUP
SPECTROSCOPY GROUP
53
53
Units
[ β ] = m −1
[ω ] = s −1
F 1 A⋅ s
s
[C ] = =
=
m m V
Ω⋅m
[ L] =
H 1 V ⋅s Ω⋅s
=
=
m m A
m
V   L 
[ Z0 ] =   =   = Ω
I   C
Z0 is the total impedance of a line of any length if there are no reflections ⇒
I and V in phase everywhere. Z0 is real
If there are reflections, the current and voltage of the advancing wave are
again in phase, but not necessarily with the current and voltage of the
retreating wave
The lossless line has no resistors. Yet Z0 has units of Ω.
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALSGROUP
GROUP
SPECTROSCOPY
The characteristic impedance does not dissipate power. It stores it
54
54
I.2.2 Lossy Lines
{(
I = R e {( I
V = R e VF e
F
e
− (α + j β ) x
− (α + j β ) x
+ VB e (
+ I B e(
)e }
)
)e }
α + jβ )x
α + jβ x
jω t
(1.6)
jω t
(1.7)
γ = α + j β = ( R + jω L )(G + jω C )
Remembering that we can write the expressions for a lossy line starting
from those of a lossless line, if we substitute
L in a lossless line with:
L' =
( R + jω L)
jω
in a lossy line
C in a lossless line with:
C'=
(G + jωC )
jω
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
Thus
Z0 =
L
C
corresponds to
NANOMATERIALS
AND
ELECTRONIC
DEVICES
in a lossy
line
AND
MATERIALS
GROUP
SPECTROSCOPY GROUP
55
55
in a lossless line
Z0 =
Z0 =
Note: at high frequencies
L'
C'
in a lossy line
R + jω L
G + jωC
and
ω L >> R
ω C >> G ,
we recover the expression for lossless lines
L
Z0 ≈
C
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALSGROUP
GROUP
SPECTROSCOPY
56
56
I.2.3 Summary
1) For a unidirectional wave:
V = Z 0 I at all points
2) For any wave:
VF = Z 0 I F
Hence
and
VB = − Z 0 I B
VF and
IF
are in phase
VB and
IB
are in antiphase
3) For a lossless line Z0 is real with units of ohms.
4) For a lossy
line UNIVERSITY
NANOMATERIALS
AND
CAMBRIDGE
DEVICES
Z is complexELECTRONIC
0
DEPARTMENT OF ENGINEERING
AND
MATERIALSGROUP
GROUP
SPECTROSCOPY
57
57
I.2.4 Characteristic Impedance – Example 1
Q: We wish to examine a circuit using an oscilloscope.
The oscilloscope probe is on an infinitely long cable and
has a characteristic impedance of 50 Ω.
What load does the probe add to the circuit?
A:
1) Since the cable is infinitely long there are no reflections
2)For a unidirectional wave with no reflections
Z0=V/I
at all points, hence the probe behaves like a load of
NANOMATERIALS
AND
CAMBRIDGE UNIVERSITY
ELECTRONIC
DEVICES
50Ω
DEPARTMENT OF ENGINEERING
AND
MATERIALSGROUP
GROUP
SPECTROSCOPY
58
58
I.2.5 Characteristic Impedance – Example 2
Q: A wave of VF = 5 volts with a wavelength λ=2 metres
has a reflected wave of VB = 1 volt
If Z0 = 75Ω, what are the voltage and current 3 metres from the
end of the cable?
V = VF e − j β x + VB e j β x
A: From Equation 1.4:
β=
2π
λ
=
2π
= π [ m −1 ]
2m
x = - 3m therefore:
Since
Then
IF =
VF
Z0
V = 5e + j 3π + 1e − j 3π [volts]
VB
IB = −
and
Z0
5 j 3π 1 − j 3π
CAMBRIDGE
I = eOF+UNIVERSITY
− e
DEPARTMENT
ENGINEERING
75
75
NANOMATERIALS AND
ELECTRONIC DEVICES
[amps]
AND
MATERIALSGROUP
GROUP
SPECTROSCOPY
59
59
I.3 Reflection
Aims
To introduce the concept of voltage reflection coefficient and
its relation to the reflected power at the load
Objectives
At the end of this section you should be able to calculate the
voltage reflection coefficient, the incident and reflected power
on the load, the conditions for ringing and quarter wave
matching
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALSGROUP
GROUP
SPECTROSCOPY
60
60
I.3.1 Voltage reflection coefficient
Consider a load added to the end of a transmission line
From Equation 1.4:
V = VF e − jβ x + VB e jβ x
From Equation 1.5:
I = I F e− jβ x + I Be jβ x
At the load x=0, thus
V = VF + VB = VL
NANOMATERIALS
AND
ELECTRONIC
DEVICES
I = I + I = I L AND
MATERIALS
GROUP
SPECTROSCOPY GROUP
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
F
B
But:
V = VF + VB = VL = Z L I L
(
= ZL IF + IB
61
61
62
62
)
From our derivation of characteristic impedance:
IF =
VF
Z0
IB = −
VB
Z0
IF and IB have opposite signs relative to VF and VB
Hence:
(
)
VF + VB = Z L I F + I B = Z L
VB Z L − Z 0
=
VF Z L + Z 0
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
VF − VB
Z0
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALSGROUP
GROUP
SPECTROSCOPY
ρ L , is defined as
The Voltage Reflection Coefficient,
the complex amplitude of the reverse voltage wave
divided by the complex amplitude of the forward voltage
wave at the load:
ρL =
VB
VF
(3.1a)
Z L − Z0
ρL =
Z L + Z0
(3.1b)
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALS
GROUP
SPECTROSCOPY GROUP
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
63
63
64
64
I.3.2 Power Reflection
At the load
{
}
{
}
V (t ) = Re V e jω t = V cos(ω t + ∠V )
I (t ) = Re Ie jω t = I cos(ωt + ∠ I )
Instantaneous power dissipated at the load:
P (t ) = V (t ) I (t ) = VI cos(ω t + ∠V ) cos(ω t + ∠ I )
Remembering the identity:
cos( A)cos( B) =
1
[cos( A + B) + cos( A − B)]
2
we get:
1 CAMBRIDGE UNIVERSITY
NANOMATERIALS
AND
ELECTRONIC
DEVICES
cos(2ωOF

P (t ) = VIDEPARTMENT
t
+
∠
V
+
∠
I
)
+
cos(
∠
V
−
∠
I
)
ENGINEERING
AND
MATERIALS
GROUP
SPECTROSCOPY
GROUP
2 
Mean power dissipated at the load:
{ }
*
1
= Re V I
2
T
1
1
PAv = ∫ P (t ) dt = VI cos( ∠V − ∠ I )
T 0
2
Where I
Thus
*
is the complex conjugate of
I
*
I = »e{I } − Im{I } j = I e − j∠ I
I = »e{I } + Im{I } j = I e j∠ I
At the load:
V = VF + VB
But, from (3.1a):
VB = ρ L VF
V = VF (1 + ρ L )
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALS
GROUP
SPECTROSCOPY GROUP
65
65
66
66
Similarly:
At the load:
1
VF
(V F − V B ) =
I = IF + IB =
Z0
Z0
I=
Hence:
VF
(1 − ρ L )
Z0
VF
*
*
1
1
V I = 1+ ρ L 1− ρ L
Z0
2
2
VF
2Z
2
)
)(
(
=
 VB 
1 −

V
F 

2
(1 + ρ
*
L
− ρL − ρL
CAMBRIDGE UNIVERSITY
0
DEPARTMENT OF ENGINEERING
2
)
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALSGROUP
GROUP
SPECTROSCOPY
*
But
ρL
ρL
is the complex conjugate of
*
is imaginary
ρL − ρL
2
so:
VF
*
1
Re V I =
1− ρ L
2
2Z0
(
{ }
Therefore:
Incident power=
VF
2
) power dissipated in the load
2
Reflected power=
2Z 0
ρL
2
VF
2
2Z 0
The fraction of power reflected from the load is:
2
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALS
GROUP
SPECTROSCOPY GROUP
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
L
ρ
67
67
I.3.3 Standing Waves
Reflections result in standing waves being set up in the
transmission line. The Voltage Standing Wave Ratio (VSWR) is a
measurement of the ratio of the maximum voltage to the minimum
voltage.
Maximum voltage V + V
=
VSWR =
Minimum voltage V − V
F
B
F
B
ρL
The VSWR can be stated in terms of the reflection coefficient
V
V
V
1+ ρ
V
VSWR =
=
=
1− ρ
V
V
1
−
1 − UNIVERSITY
NANOMATERIALS
AND
CAMBRIDGE
ELECTRONIC
DEVICES
V
DEPARTMENT OFV
ENGINEERING
AND
MATERIALSGROUP
GROUP
SPECTROSCOPY
1+
B
1+
B
F
F
L
B
B
L
F
F
68
68
ρL
Or alternatively (and more usefully) the reflection coefficient
can be stated in terms of the VSWR (which can be measured)
VSWR − 1
ρ =
VSWR + 1
L
If there is total reflection then
(3.2)
ρ L = 1 and VSWR is infinite.
Zero reflection leads to VSWR=1
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALS
GROUP
SPECTROSCOPY GROUP
69
69
70
70
I.3.4 Summary
ρL = 0
•Full power transfer requires
•When ρ L = 0 a load is said to be “matched”
•The advantages of matching are that:
1) We get all the power to the load
2) There are no echoes
•The simplest way to match a line to a load is to set
Z0 = Z L
i.e. so that the load equals the characteristic impedance
Z L − Z0
Z L + Z0
DEPARTMENT OF ENGINEERING
ρL =
Since, from (3.1b):
CAMBRIDGE UNIVERSITY
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALSGROUP
GROUP
SPECTROSCOPY
•Fraction of power reflected =
ρL
2
•Reflections will set up standing waves.
The Voltage Standing Wave Ratio (VSWR) is given by:
1+ ρ
VSWR =
1− ρ
CAMBRIDGE UNIVERSITY
DEPARTMENT OF ENGINEERING
L
L
NANOMATERIALS
AND
ELECTRONIC
DEVICES
AND
MATERIALS
GROUP
SPECTROSCOPY GROUP
71
71