P5-Electromagnetic Fields and Waves
Prof. Andrea C. Ferrari KWWS ZZZKWWS ZZZJHQJFDPDFXNQPVOHFWXUHQRWHVKWPOJHQJFDPDFXNQPVOHFWXUHQRWHVKWPO
CAMBRIDGE UNIVERSITY ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 1 1
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 ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 2 2 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 ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP 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 the characteristic impedance of a transmission cable
•Understand basic principles of EM wave propagation in free space, acrossCAMBRIDGE interfaces UNIVERSITY and the useELECTRONICNANOMATERIALS of antenn DEVICES AND ae DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 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 JohnCAMBRIDGE Wiley UNIVERSITY & Sons, 1992, 1997 ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENTShelf Mark: OF ENGINEERING NV 135 ANDSPECTROSCOPY MATERIALS GROUP 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 http://www-g.eng.cam.ac.uk/nms/lecturenotes.htmlCAMBRIDGE UNIVERSITY ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 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 ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP 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 ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 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 in some way).CAMBRIDGE UNIVERSITY ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 9 9
Examples
Power Plant
Consumer Home
CAMBRIDGE UNIVERSITY ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 10 10 Antenna CAMBRIDGE UNIVERSITY OpticalELECTRONICNANOMATERIALS Fibre DEVICES ANDLink DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 11 11
12
Pair of wires Co -ax cable
CAMBRIDGE UNIVERSITY ELECTRONICNANOMATERIALS DEVICES AND PCB tracks DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALSIC interconnects GROUP GROUP 12 Waveguides
CAMBRIDGE UNIVERSITY ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP 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 microwave frequency CAMBRIDGE UNIVERSITY ELECTRONICNANOMATERIALS DEVICES AND signals DEPARTMENT OF ENGINEERING AND MATERIALS GROUP SPECTROSCOPY GROUP 14 14 CAMBRIDGE UNIVERSITY ELECTRONICNANOMATERIALS DEVICES AND Microwave DEPARTMENTOven OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 15 15
Optical Fibres
CAMBRIDGE UNIVERSITY ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 16 16 Phasor Notation
A means A is complex Im
A=» eA{} +I mAj {} = Ae j∠ A Im{ A } A ∠ A » »e{ A } e A= A
CAMBRIDGE UNIVERSITY ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 17 17
jβ x Ae is short-hand for » e{Ae j(β x+ ω t ) }
which equals: Acos (β xt+ ω + ∠ A ) Proof ej(±θ ) =cos(θ ) ± j sin( θ ) then
Aejxt()βω+= Aee jAjxt ∠ ()( βω + = Ae jxtA βω ++∠ )
j(β x+ ω t +∠ A ) Ae=CAMBRIDGE Acos(βω UNIVERSITY xtAjA ++∠+ )ELECTRONICNANOMATERIALS sin( βω DEVICES xtA ++∠ AND ) DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 18 18 I.0.1 The Wave Equation
The generalised form of the wave equation is: ∂ 2 A =v2 ∇ 2 A ∂t 2
Where the Laplacian of a scalar A is:
∂2A ∂ 2 A ∂ 2 A ∇=2 A + + ∂x2 ∂ y 2 ∂ z 2 CAMBRIDGE UNIVERSITY ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 19 19
We will be looking at plane waves for which the wave equation is one-dimensional and appears as follows:
∂2A ∂ 2 A ∂2A1 ∂ 2 A = v 2 or = ∂t2 ∂ x 2 ∂x2 v 2 ∂ t 2
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) as in electromagneticCAMBRIDGE UNIVERSITY waves in free spaceELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 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 ∂2A ∂ 2 A ∂ 2 A =v2 + ∂t2 ∂ x 2 ∂ y 2
CAMBRIDGE UNIVERSITY ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP 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( x + vt ) CAMBRIDGE UNIVERSITYBackward movingELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 22 22 Direction of travel f( x− vt ) F(t+x/vf( x+) vt ) ∆x ∆x P 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 ELECTRONICNANOMATERIALS DEVICES AND Similarly, for DEPARTMENTx+vt ⇒ wave OF ENGINEERING is moving toANDSPECTROSCOPY left MATERIALS (Backward) GROUP GROUP 23 23
Verify that A = fx ( ± vt ) is general solution
∂A ∂2 A = ±vf'( x ± vt ) =v2 f''( x ± vt ) ∂t ∂t 2
2 ∂A ∂ A =f'( x ± vt ) =f''( x ± vt ) ∂x ∂x2
∂2A ∂ 2 A = v 2 ∂t2 ∂ x 2 CAMBRIDGE UNIVERSITY ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 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 ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 25 25
I.1.1 Telegrapher’s Equations Let us consider a short length, δδδx, of a wire pair
δ x This could, for example, represent a coaxial cable For a small δx, any function A(x) can be written as ∂A( x ) AxxAx(+δ ) ≈ () + δ x CAMBRIDGE∂x UNIVERSITY ELECTRONICNANOMATERIALS DEVICES AND In our caseDEPARTMENT A can OF ENGINEERINGbe Voltage (V)AND SPECTROSCOPYor MATERIALSCurrent GROUP GROUP (I) 26 26 Let us define L series/loop inductance per unit length [H/m] Lδ x I
V L δ x ∂I NANOMATERIALS AND CAMBRIDGEVL UNIVERSITY= Lδ x ELECTRONIC DEVICES DEPARTMENT OF ENGINEERING∂t ANDSPECTROSCOPY MATERIALS GROUP GROUP 27 27
C parallel/shunt capacitance per unit length [F/m]
I V C C Cδ x
δx
∂V I= Cδ x C C ∂t
CAMBRIDGE UNIVERSITY ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 28 28 ∂I VL = Lδ x ∂t ∂I I= I + δ x I B F ∂x
∂I I= − δ x C ∂x V L ∂V VC V= V + δ x C ∂x V= VC + V L
δ x
∂VC ∂∂V ∂∂ VV2 ∂ V ICx==δδ CxV( +=+ δδ xCx )NANOMATERIALS C () δδ xCx AND ≈ C ∂tCAMBRIDGE ∂∂ tx UNIVERSITY ∂∂∂ txtELECTRONIC DEVICES ∂ t DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 29 29
∂V ∂ I V= V − V V+δ xVL = − δ x C L ∂x ∂ t
∂I ∂ V I= I − I I+δ xI = − Cx δ F C ∂x ∂ t ∂V ∂ I = − L (1.1) ∂x ∂ t ∂I ∂ V = − C (1.2) ∂x ∂ t 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 early developmentCAMBRIDGE of long distance UNIVERSITY telegraphy (hencELECTRONICNANOMATERIALSe the DEVICES name) AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 30 30 I.1.2 Travelling Wave Equations Let us differentiate both (1.1) and (1.2) with respect to x ∂2V ∂ ∂I ∂2V = − L = LC (1.1a) ∂x2 ∂ t ∂x ∂t 2
2 ∂2I ∂ ∂V ∂ I = − C = LC 2 (1.2a) ∂x2 ∂ t ∂x ∂t ∂I Then in (1.1a) substitute using (1.2) ∂x ∂V Then in (1.2a)CAMBRIDGE substitute UNIVERSITY usingELECTRONICNANOMATERIALS (1.1) DEVICES AND DEPARTMENT OF ENGINEERING∂x ANDSPECTROSCOPY MATERIALS GROUP GROUP 31 31
∂2V ∂ 2 V = LC (1.1a) ∂x2 ∂ t 2 ∂2I ∂ 2 I = LC ∂x2 ∂ t 2 (1.2a) Same functional form as wave equation: ∂2A1 ∂ 2 A 1 = v2 = ∂x2 v 2 ∂ t 2 LC
We try a solution for V in (1.1a) of the form
CAMBRIDGE UNIVERSITY jβ x jt ω ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERINGV= Ae e ANDSPECTROSCOPY MATERIALS GROUP GROUP 32 32 −β2Aejxjtβω e = − ω 2 LC Ae jxjt βω e
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 VR eVe− jxβ Ve jx β e jt ω (1.4) ={( F + B ) }
R − jxβ jx β jt ω (1.5) I= eIe + IeNANOMATERIALS e AND {CAMBRIDGE( F UNIVERSITY B ELECTRONIC) } DEVICES DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP 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 [ ΩΩΩ/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 ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 34 34 Rδ x Lδ x ∂I I I+ δ x B ∂x ∂V V V V= V + δ x V R L C ∂x Gδ x Cδ x
δ x
V− VR − V L − V C = 0
CAMBRIDGE UNIVERSITY ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 35 35
∂I ∂ V VRxILx−δ − δ −+( V δ x ) = 0 ∂t ∂ x ∂V ∂ I = −RI + L ∂x ∂ t For simplicity we assume ∂I I= fxe( ) jω t =jfxeω( ) jω t = jI ω ∂t Then ∂V = −(R + jω LI ) ∂x ∂V ∂ I CAMBRIDGE UNIVERSITY= − L =ELECTRONICNANOMATERIALS − jω LI DEVICES AND Compare withDEPARTMENT (1.1) OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP ∂x ∂ t 36 36 Similarly, using Kirchoff’s current law to sum currents: ∂I IGxV−δ − jCxV ωδ −+( I δ x ) = 0 ∂x
∂I = −(G + jω CV ) ∂x
∂I ∂ V Compare with (1.2) =−C =− jCVω ∂x ∂ t
CAMBRIDGE UNIVERSITY ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP 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: (R+ jω L ) L' = in a lossy line jω C in a lossless line with: (G+ jω C ) C ' = in a lossy line jω Then β= ω LC In a lossless line corresponds to: 1 β'= (R + jLG ω )( + jC ω ) j CAMBRIDGE UNIVERSITY ELECTRONICNANOMATERIALSin DEVICESa lossy AND line DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 38 38 Substituting β ⇒ β’ into (1.4) and (1.5) and defining
γ=+(RjLGjC ω )( + ωαβ ) =+ j We get V=R eVe−()α +jxβ + Ve()α + jxβ e jω t (1.6) {( F B ) } I=R eIe−()α +jxβ + Ie()α + jxβ e jω t (1.7) {( F B ) }
γ is called propagation constant β is the phase constant
The real term α corresponds to the attenuation along the line and is known asCAMBRIDGE the attenuation UNIVERSITY constantELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 39 39
For a forward travelling wave:
x jωt -γx -αx j( ωt-βx) −α V = V F e e = V F e e V= VF e amplitude factor phase factor time variation Voltage
VF
x
CAMBRIDGE UNIVERSITY ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 40 40 Note:
At high frequencies: ωL>> R and ωC>> G :
γ=+(RjLGjC ω )( + ω ) ≈− ωω2 LC = j LC
Thus α ≈ 0 β≈ ω LC
The expressions approximate back to those for lossless lines
CAMBRIDGE UNIVERSITY ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 41 41
I.1.4 Wave velocity: v Our expressions for voltage and current contain 2 exponentials ± jβ x The one in terms of x: e gives the spatial dependence of the wave, hence the wavelength: 2π λ = β jω t The other: e gives the temporal dependence of the wave, hence its frequency: ω f = CAMBRIDGE UNIVERSITY2π ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 42 42 For a wave velocity v, wavelength λ and frequency f: v= f λ then ω2 π v = 2π β
since β= ω LC
1 v = (1.8)
CAMBRIDGE UNIVERSITYLC ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP 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 ? 2π From λ = and β= ω LC β 2π λ = ω LC 2π Then λ = = 23 metres 2π ⋅⋅ 10106 0.2210 ⋅⋅⋅− 6 8610 − 12
CAMBRIDGE UNIVERSITY ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 44 44 I.1.6 When must distances be accounted for in AC circuits?
λ λ 2 4
CAMBRIDGE UNIVERSITY λ ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP 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 ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 46 46 λ L ≥ Wave Relevant 16
λ L ≤ Wave Not Relevant 16
CAMBRIDGE UNIVERSITY ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP 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=0.5 µHm -1 C=60pFm -1
2π 2π From λ = And β= ω LC λ = β ω LC 2π Then λ = = 36.5 m 2π ⋅⋅ 5106 0.510 ⋅− 6 ⋅⋅ 6010 − 12
36.5 m is much greater than 200 mm (the size of the circuit board), so that wave theory is irrelevant.
Note: The problemCAMBRIDGE is even UNIVERSITY less relevant for ELECTRONICNANOMATERIALSmains frequencies DEVICES AND i.e. 50 Hz. This givesDEPARTMENT λ~3650 OF kmENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 48 48 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 ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 49 49
I.2.1 Lossless Lines Recalling the solutions for I & V (equations 1.4&1.5): VR eVe− jxβ Ve jx β e jt ω (1.4) ={( F + B ) } IR eIe− jxβ Ie jx β e jt ω (1.5) ={( F + B ) } Differentiating (1.5) with respect to x (1.4) with respect to t and multiplying by -C ∂I = Re− jβ I e − jβ x + jβ I e jβ x e jωt (2.1) ∂x {( F B ) }
∂V − jβ x jβ x jωt −CCAMBRIDGE= R e UNIVERSITY−Cjω V e − CjELECTRONICNANOMATERIALSω V e DEVICESe AND (2.2) DEPARTMENT OF( ENGINEERINGF AND MATERIALSB ) GROUP ∂t { SPECTROSCOPY GROUP} 50 50 According to the second Telegrapher’s equation: ∂I ∂ V = − C (1.2) ∂x ∂ t We can then equate (2.1) and (2.2):
Rej I e− jβ x j I e jβ x ejω t R e C j V e − jβ x Cj V e jβ x e jω t {(− β F + β B ) } = {(− ω F − ω B ) }
j(ω t+ β x ) Since e j(ω t− β x ) and e represent waves travelling in opposite directions, they can be treated separately.
This leads toCAMBRIDGE two independent UNIVERSITY expressionsELECTRONICNANOMATERIALS inDEVICES V AND and I DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 51 51
−jxβ − jx β −jIeβF = − CjVe ω F
V β F = I F Cω jxβ jx β IjeBβ= − CjVe ω B V β B = − I B Cω
Note: If we consider VF and VB to have the same sign then, due to the differentiation with respect to x, CAMBRIDGE UNIVERSITY ELECTRONICNANOMATERIALS DEVICES AND and DEPARTMENT have opposite OF ENGINEERING signs ANDSPECTROSCOPY MATERIALS GROUP GROUP I B IF 52 52 The characteristic impedance, Z 0 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 = Z0 is always positive IF β VF β Z = Since = 0 Cω IF Cω
From (1.3) β= ω LC L NANOMATERIALS(2.3) AND CAMBRIDGE UNIVERSITYZ0 = ELECTRONIC DEVICES DEPARTMENT OF ENGINEERINGC ANDSPECTROSCOPY MATERIALS GROUP GROUP 53 53
Units [β ] = m−1 [ω ] = s−1
F1 As⋅ s H1 Vs⋅ Ω ⋅ s [C ] = = = [L ] = = = mmVΩ ⋅ m m mA 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. Z 0 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 Z 0 has units of Ω. CAMBRIDGE UNIVERSITY ELECTRONICNANOMATERIALS DEVICES AND The characteristicDEPARTMENT impedance OF ENGINEERING does not dissipateANDSPECTROSCOPY power. MATERIALS It storesGROUP GROUP it 54 54 I.2.2 Lossy Lines V=R eVe−()α +jxβ + Ve()α + jx β e jω t (1.6) {( F B ) }
I=R eIe−()α +jxβ + Ie()α + jxβ e jω t (1.7) {( F B ) } γ=α + jβ = (R + j ω L )( G + j ω 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: (R+ jω L ) L' = in a lossy line jω C in a lossless line with: (G+ jω C ) CAMBRIDGEC UNIVERSITY' = ELECTRONICNANOMATERIALSin a lossy DEVICES lineAND DEPARTMENT OF ENGINEERINGjω ANDSPECTROSCOPY MATERIALS GROUP GROUP 55 55
L Thus Z = in a lossless line 0 C L' corresponds to Z = in a lossy line 0 C ' R+ jω L Z = 0 G+ jω C
Note: at high frequencies ωL>> R and ωC>> G , we recover the expression for lossless lines
L Z ≈ 0 C CAMBRIDGE UNIVERSITY ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 56 56 I.2.3 Summary 1) For a unidirectional wave:
V= Z0 I at all points
2) For any wave: V= − Z I VF= Z0 I F and B0 B
Hence VF and IF are in phase I are in antiphase VB and B
3) For a lossless line Z0 is real with units of ohms.
4) For a lossy line is complexNANOMATERIALS AND CAMBRIDGE UNIVERSITYZ 0 ELECTRONIC DEVICES DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 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 Z 0=V/I at all points, hence the probe behaves like a load of 50 Ω CAMBRIDGE UNIVERSITY ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 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 Z 0 = 75 Ω, what are the voltage and current 3 metres from the end of the cable?
− jxβ jx β A: From Equation 1.4: V= VeF + Ve B 2π 2 π β= = = π [m−1 ] λ 2m x = - 3m therefore: V=5 e+j3π + 1 e − j 3 π [volts] V V Since F B I F = and I B = − Z0 Z0
5 1 NANOMATERIALS AND Then CAMBRIDGEI= e +UNIVERSITYj3π − e − j 3 π [amps]ELECTRONIC DEVICES DEPARTMENT75 OF ENGINEERING 75 ANDSPECTROSCOPY MATERIALS GROUP GROUP 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 ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 60 60 I.3.1 Voltage reflection coefficient Consider a load added to the end of a transmission line
− jxβ jx β From Equation 1.4: V= VeF + Ve B
− jxβ jx β From Equation 1.5: I= IeF + Ie B At the load x=0, thus
V= VF + V B = V L
CAMBRIDGE UNIVERSITY ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENTII OF= ENGINEERING + I = I AND MATERIALS GROUP F B L SPECTROSCOPY GROUP 61 61
But: VV= + V = V = ZI F B L LL =ZL( I F + I B ) From our derivation of characteristic impedance:
VF VB I F = I B = − Z0 Z 0
IF and I B have opposite signs relative to V F and V B V− V Hence: F B VF + VB = ZL () IF + I B = Z L Z 0
V Z− Z B= L 0 V Z+ Z CAMBRIDGE UNIVERSITYF L 0 ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 62 62 The Voltage Reflection Coefficient, ρL , is defined as the complex amplitude of the reverse voltage wave divided by the complex amplitude of the forward voltage wave at the load:
VB ρL = (3.1a) VF
ZL − Z 0 ρ L = (3.1b) ZL + Z 0
CAMBRIDGE UNIVERSITY ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 63 63
I.3.2 Power Reflection At the load Vt()Re={ Vejω t } = V cos(ω tV +∠ ) It()Re={ Iejω t } = I cos(ω t +∠ I ) Instantaneous power dissipated at the load: PtVtItVI()= ()() = cos(ω t +∠ V )cos( ω t +∠ I ) Remembering the identity: 1 cos(AB )cos( )=[] cos( AB ++ ) cos( AB − ) 2 we get: 1 Pt()= VICAMBRIDGE cos(2ω tVI UNIVERSITY +∠+∠+ )cos( ∠−∠ELECTRONICNANOMATERIALS VI ) DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 2 64 64 Mean power dissipated at the load:
1T 1 1 * P= Ptdt() = VI cos( ∠−∠ V I ) = Re V I Av T ∫ 2 { } 0 2 * Where I is the complex conjugate of I * Thus I=» eI{} +I mIj {} = Ie j∠ I I=» eI{} −I mIj {} = Ie −j ∠ I
At the load:
V= VF + V B But, from (3.1a):
VB= ρ L V F
CAMBRIDGE UNIVERSITY ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENTV= OF V ENGINEERINGF (1 + ρ L ) ANDSPECTROSCOPY MATERIALS GROUP GROUP 65 65
Similarly: At the load: 1 VF V B III=+=F B ( VVF −= B ) 1 − Z0 Z 0 V F
VF I =(1 − ρ L ) Z 0
Hence: 2 1* 1 * VF V I =1 +ρ 1 − ρ ()L( L ) 2 2 Z 0
2 VF * 2 =1 +−−ρL ρ L ρ L ()NANOMATERIALS AND CAMBRIDGE2Z UNIVERSITY0 ELECTRONIC DEVICES DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 66 66 * ρ But ρ L is the complex conjugate of L
* is imaginary ρL− ρ L 2 1 * VF 2 so: ReV I = 1 − ρ power dissipated in the load {} ()L 2 2 Z 0
Therefore: 2 2 VF 2 VF Incident power= Reflected power= ρ L 2Z0 2Z0 The fraction of power reflected from the load is: 2 CAMBRIDGE UNIVERSITY ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERINGρ L ANDSPECTROSCOPY MATERIALS GROUP GROUP 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 VF+ V B VSWR = =
Minimum voltage VF− V B
The VSWR can be stated in terms of the reflection coefficient ρ L V B V 1+ 1+ B VF V 1+ ρ VSWR = =F = L
VB VB 1− ρ L CAMBRIDGE1− UNIVERSITY1− ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING VF ANDSPECTROSCOPY MATERIALS GROUP GROUP VF 68 68 Or alternatively (and more usefully) the reflection coefficient ρ L can be stated in terms of the VSWR (which can be measured) VSWR −1 ρ = (3.2) L VSWR +1
If there is total reflection then ρ L = 1 and VSWR is infinite.
Zero reflection leads to VSWR=1
CAMBRIDGE UNIVERSITY ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 69 69
I.3.4 Summary
•Full power transfer requires ρ L = 0
•When ρL = 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
ZL − Z 0 Since, from (3.1b):CAMBRIDGEρ UNIVERSITYL = ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERINGZ+ Z AND MATERIALS GROUP L 0 SPECTROSCOPY GROUP 70 70 2 •Fraction of power reflected = ρ L
•Reflections will set up standing waves.
The Voltage Standing Wave Ratio (VSWR) is given by:
1+ ρ VSWR = L
1− ρ L
CAMBRIDGE UNIVERSITY ELECTRONICNANOMATERIALS DEVICES AND DEPARTMENT OF ENGINEERING ANDSPECTROSCOPY MATERIALS GROUP GROUP 71 71