Transmission Lines

Chapter 2 Transmission Lines

Contents 2.1 Introduction...... 2-3 2.1.1 The Role of Wavelength...... 2-4 2.1.2 Propagation Modes...... 2-6 2.2 The Lumped Element Model...... 2-7 2.2.1 Coax Detail...... 2-9 2.2.2 TEM Line Facts...... 2-10 2.3 The Transmission Line Equations...... 2-13 2.4 Wave Propagation on a Transmission Line...... 2-15 2.4.1 Getting to the 1D Wave Equation...... 2-15 2.4.2 Solving the Wave Equation (Phasor Form).... 2-17 2.4.3 Returning to the Time Domain...... 2-20 2.5 The Lossless Microstrip Line...... 2-22 2.5.1 Analysis and Synthesis of Microstrip...... 2-23 2.5.2 Common Mstrip Materials...... 2-27 2.6 Lossless Tline General Case...... 2-33 2.6.1 Voltage Reﬂection Coefﬁcient...... 2-36 2.6.2 Standing Waves...... 2-42

2-1 CHAPTER 2. TRANSMISSION LINES

2.6.3 VSWR...... 2-47 2.7 Wave Impedance of a Lossless Line...... 2-52 2.8 Special Cases of the Lossless Line...... 2-56 2.8.1 Short-Circuited Line...... 2-56 2.8.2 Open-Circuited Line...... 2-61 2.8.3 Lines of Length a Multiple of =2 ...... 2-64 2.8.4 Quarter-Wavelength Transformer...... 2-65

2.8.5 Matched Line: ZL Z0 ...... 2-67 D 2.9 Power Flow in a Lossless Line...... 2-69 2.9.1 Instantaneous Power...... 2-69 2.9.2 Time-Averaged Power...... 2-70 2.10 The Smith Chart...... 2-72 2.10.1 Parametric Equations and the -Plane...... 2-73 2.10.2 Wave Impedance...... 2-77 2.10.3 Impedance/Admittance Transformation..... 2-80 2.11 Impedance Matching...... 2-81 2.11.1 Quarter-wave Transformer Matching...... 2-83 2.11.2 Lumped Element Matching...... 2-94 2.11.3 Single-Stub Matching...... 2-103 2.12 Transients on Transmission Lines...... 2-110 2.12.1 Response to a Step Function...... 2-111 2.12.2 Dr. Wickert’s Archive Notes...... 2-113 2.12.3 The Time-Domain Reﬂectometer...... 2-116

2-2 2.1. INTRODUCTION

2.1 Introduction

In this second chapter your knowledge of circuit theory is connected into the study transmission lines having voltage and current along the line in terms of 1D traveling waves. The transmission line is a two-port circuit used to connect a generator or transmitter signal to a receiving load over a distance. In simple terms power transfer takes place.

Z g A B + ~ Sending-end Receiving-end V Transmission line Z g port port L − A' B' Generator circuit Load circuit

Figure 2-1 Atransmissionlineisatwo-portnetworkconnectingagenerator circuit at the sending end to a load at the receiving end. Figure 2.1: Two-port model of a transmission line.

In the beginning the transmission line is developed as a lumped element circuit, but then a limit is taken to convert the circuit model into a distributed element circuit

– Distributed element means that element values such as R, L, and C become R, L, and C per unit length of the line, e.g, /m, H/m, and F/m respectively

In the lumped element model we have a pair of differential equations to describe the voltage and current along the line

Under the limiting argument which converts the circuit to distributed element form, we now have a pair of partial differential equations, which when solved yield a solution that is a 1D traveling wave

2-3 CHAPTER 2. TRANSMISSION LINES

Key concepts developed include: wave propagation, standing waves, and power transfer

Returning to Figure 2.1, we note that sinusoidal steady-state is implied as the source voltage is the phasor V , the source Qg impedance Zg is a Thévenin equivalent and ZL is the load impedance

– The source may be an RF transmitter connecting to an antenna – The source/load may be a pair at each end of the line when connecting to an Ethernet hub in wired a computer network – The source may be power collected by an antenna and via cable to the receiver electronics, e.g., satellite TV (Dish Network)

2.1.1 The Role of Wavelength When circuits are interconnected with wires (think protoboard or a printed circuit board (PCB)), is a transmission line present?

The answer is yes, for better or worse As long the circuit interconnect lengths are small compared with the wavelength of the signals present in the circuit, lumped element circuit characteristics prevail

– Ulaby suggests transmission line effects may be ignored when l= . 0:01 ( I have been content with a =10 limit)

2-4 2.1. INTRODUCTION

– Lumped element capacitance and inductance (parasitics) due to interconnects may still alter circuit performance

Power Loss and Dispersion

Transmission lines may also be dispersive, which means the propagation velocity on the line is not constant with frequency

For example the frequency components of square wave (recall odd harmonics only) each propagate at a different velocity, meaning the waveform becomes smeared

Dispersion is very important to high speed digital transmission (ﬁber optic and wired networks alike)

The longer the line, the greater the impact

Dispersionless line

Short dispersive line

Long dispersive line

FigureFigure 2.2: 2-3 The impactAdispersionlesslinedoesnotdistort of transmission line dispersion. signals passing through it regardless of its length, 2-5 whereas a dispersive line distorts the shape of the input pulses because the different frequency components propagate at different velocities. The degree of distortion is proportional to the length of the dispersive line. CHAPTER 2. TRANSMISSION LINES

2.1.2 Propagation Modes When a time-varying signal such as sinusoid connected to (or launched on) a transmission line, a propagation mode is estab- lished

Recall that both electric and magnetic ﬁelds will be present (why?)

Two mode types as: (1) transvere electromagnetic (TEM) and (2) non-TEM or higher-order

Metal Metal w 2b 2a d h D Dielectric spacing Dielectric spacing Dielectric spacing (a) Coaxial line (b) Two-wire line (c) Parallel-plate line

Metal strip conductor Metal Metal w

h Metal ground plane Dielectric spacing Metal ground plane Dielectric spacing Dielectric spacing (d) Strip line (e) Microstrip line (f) Coplanar waveguide TEM Transmission Lines

Metal

Concentric dielectric layers

(g) Rectangular waveguide (h) Optical fiber Higher-Order Transmission Lines

FigureFigure 2-4 2.3:Afewexamplesoftransverseelectromagnetic(TEM)andhigh A collection classical TEM transmissioner-order transmission lines (a)–(c), lines. mircowave circuit TEM lines (d)–(f), and non-TEM lines (g) & (h).

2-6 2.2. THE LUMPED ELEMENT MODEL

TEM, which means electric and magnetic ﬁelds are transverse to the direction of propagation, is the exclusive study of Chap- ter 2

Magnetic field lines Electric field lines

Rg + RL Vg Coaxial line − Generator Load Cross section

Figure 2-5 In a coaxial line, the electric field is in the radial directioFigurenbetweentheinnerandouterconductors,andthe 2.4: The transverse fields of the coax. magnetic field forms circles around the inner conductor. The coaxial line is a transverse electromagnetic (TEM) transmission line because both the electric and magnetic fields are orthogonalnon-TEM to the direction means of propagation one or more between field the components generator and lies in the di- the load. rection of propagation, e.g. metal or optical fiber waveguides

We start with a lumped element model of a TEM line and de- rive the telegrapher’s equations

2.2 The Lumped Element Model

TEM transmission lines all exhibit axial symmetry From a circuit theory perspective the line is completely described by four distributed parameter quantities:

– R0: The effective series resistance per unit length in /m (all conductors that make up the line cross-section are included)

2-7 CHAPTER 2. TRANSMISSION LINES

– L0: The effective series inductance per unit length in H/m (again all conductors that make up the line cross-section are included)

– G0: The effective shunt conductance of the line insulation (air and/or dielectric) per unit length in S/m (mhos/m?)

– C 0: The effective shunt capacitance per unit length between the line conductors in F/m

We will see that these four parameters are calculated from (1) the line cross-sectional geometry (see 2.4) and (2) the EM constitutive parameters (c and c for conductors and , , and insulation/dielectric)

For the three classical line types the four parameters are calculated using the table below (also see the text Java modules1)

TableTable 2-1 2.1:Transmission-lineR0, L0, G0, and parametersC 0 forR′, theL′, G four′,and classicalC ′ for three linetypes oftypes. lines.

Parameter Coaxial Two-Wire Parallel-Plate Unit Rs 1 1 2Rs 2Rs R′ + Ω/m 2π a b πd w % &

µ µ 2 µh L′ ln(b/a) ln (D/d)+ (D/d) 1 H/m 2π π − w ' ( ) 2πσ πσ σw G′ S/m ln(b/a) ln (D/d)+ (D/d)2 1 h − * + , 2πε πε εw C ′ F/m ln(b/a) ln (D/d)+ (D/d)2 1 h − * , Notes: (1) Refer to Fig. 2-4 for deﬁnitions of+ dimensions. (2) µ,ε,andσ pertain to the insulating material between the conductors. (3) Rs = π f µc/σc.(4)µc and σc pertain to the conductors. (5) If (D/d)2 1, then ln (D/d)+ (D/d)2 1 ln(2D/d). ≫ + − ≈ - + . 1 http://em7e.eecs.umich.edu/ulaby_modules.html

2-8 2.2. THE LUMPED ELEMENT MODEL

2.2.1 Coax Detail

The formula for R0 (see text Chapter 7) is Â Ã Rs 1 1 R0 (/m) D 2 a C b

where Rs is the effective surface resistance of the line

In text Chapter 7 it is shown that s fc Rs ./ D c – Note: The series resistance of the line increases as the pf

– Also Note: If c fc R0 0 '

The formula for L0 (see text Chapter 5) is ÂbÃ L0 ln (H/m) D 2 a where is the joint inductance of both conductors in the line cross-section

The formula for G0 (see text Chapter 4) is given by 2 G0 (S/m) D ln.b=a/

Note: The non-zero conductivity of the line allows current to ﬂow between the conductors (lossless dielectric 0 and ! D G0 0) D 2-9 CHAPTER 2. TRANSMISSION LINES

The formula for C 0 (see Chapter 4), which is the shunt capacitance between the conductors, is 2 C 0 F/m D ln.b=a/

The voltage difference between the conductors sets up equal and opposite charges and it is the ratio of the charge to voltage difference that deﬁnes the capacitance

2.2.2 TEM Line Facts Velocity of propagation relationship: 1 1 L0C 0 up D ) D p D pL0C 0

– Recall from Chapter 1 c 1=p00 D Secondly, G0

C 0 D

Example 2.1: Coax Parameter Calculation Using Module 2.2

A coaxial air line is operating at 1 MHz The inner radius is a 6 mm and the outer radius is b 12 D D mm

Assume copper for the conductors ( 1) 2-10 2.2. THE LUMPED ELEMENT MODEL

Use the text Module 2.2 to obtain the transmission line param-

eters



 4  parms.

Figure 2.5: Transmission line parameters obtained from text Module 2.2

As an alternative, it is not much work to do the same calculation in the Jupyter notebook

With hand coding many options exist to do more than just obtain R0, L0, G0, and C 0

Soon we will see other tline characteristics derived from the four distributed element parameters

2-11 CHAPTER 2. TRANSMISSION LINES

Figure 2.6: Transmission line parameters from Python function coax_parameters()

2-12 2.3. THE TRANSMISSION LINE EQUATIONS

2.3 The Transmission Line Equations

The moment you have been waiting for: solving the tline equations starting from a differential length of line (classical!)

Node Node N i(z, t) N + 1 i(z + ∆z, t) + + R' ∆z L' ∆z

υ(z, t) G' ∆z C' ∆z υ(z + ∆z, t)

− − ∆z Figure 2.7: A differential section of TEM transmission line, z Figure 2-8 Equivalent circuit of a two-conductor readytransmission for Kirchoff’s line laws. of differential length ∆z. The goal of this section is to obtain the voltage across the line v.z; t/ and the current through the line i.z; t/ for any l Ä z < 0 and t value

– Note: It is customary to let the line length be l and have z 0 at the load end and z l at the source end (more D D later)

From Kirchoff’s voltage law we sum voltage drops around the loop to zero: @i.z; t/ v.z; t/ R0z i.z; t/ L0z v.z z; t/ 0 @t C D

2-13 CHAPTER 2. TRANSMISSION LINES

Getting set for form a derivative in the limit, we divide by z everywhere are rearrange Äv.z z; t/ v.z; t/ @i.z; t/ C D R0i.z; t/ L0 z D C @t Taking the limit as z 0 gives ! @v.z; t/ @i.z; t/ R0i.z; t/ L0 @z D C @t From Kirchoff’s current law we sum current entering minus current leaving node N 1 to zero: C @v.z ız; t/ i.z; t/ G0z v.z z; t/ C 0z C C @t i.z z; t/ 0 C D Rearranging in similar fashion to the voltage equation, and taking the limit as z 0, yields ! @i.z; t/ @v.z; t/ G0v.z; t/ C 0 @z D C @t The above blue-boxed equations are the telegrapher’s equations in the time domain

– A full time-domain solution is available, but will be de- ferred to later in the chapter (I like personally like starting here) A sinusoidal steady-state solution (frequency domain) is possible to if you let v.z; t/ ReV .z/ej!t D Q i.z; t/ ReI .z/ej!t ; D Q

2-14 2.4. WAVE PROPAGATION ON A TRANSMISSION LINE

with V .z/ and I .z/ being the phasor components correspond- Q Q ing to v.z; t/ and i.z; t/ respectively

The phasor form of the telegrapher’s equations is simply

dV .z/ Q R0 j!L0 I .z/ dz D C Q dI .z/ Q G0 j!C 0 V .z/ dz D C Q

Note: @ d since t is suppressed in the phasor form ! Solving the 1D wave equation is next

2.4 Wave Propagation on a Transmis- sion Line

The phasor telegrapher equations are coupled, that is one equation can be inserted into the other to eliminate either V .z/ or Q I .z/ and in return have a second-order differential equation in Q z

2.4.1 Getting to the 1D Wave Equation

To eliminate I .z/ differentiate the ﬁrst equation with respect Q to z so that dI .z/=dz is found in both equations Q 2 d V .z/ dI .z/ Q R0 j!L0 Q ; dz2 D C dz

2-15 CHAPTER 2. TRANSMISSION LINES

then replace dI .z/=dz with the right side of the second equa- Q tion 2 d V .z/ Q R0 j!L0 G0 j!C 0 V .z/ 0 dz2 C C Q D or in the form of the phasor-based 1D voltage wave equation

d 2V .z/ Q 2V .z/ 0; dz2 Q D where is the complex propagation constant q R j!L G j!C D 0 C 0 0 C 0

Working with the telegrapher’s equations to solve for I .z/ re- Q sults in the current wave equation

d 2I .z/ Q 2I .z/ 0 dz2 Q D Note: When is expanded out it becomes ˛ jˇ D C where as seen in Chapter 1, ˛ is the wave attenuation constant in Np/m and ˇ is the phase constant, which (recall) equals 2=

In full detail Âq Ã ˛ Re R j!L G j!C (Np/m) D 0 C 0 0 C 0 Âq Ã ˇ Re R j!L G j!C (rad/m) D 0 C 0 0 C 0

2-16 2.4. WAVE PROPAGATION ON A TRANSMISSION LINE

– For low loss lines the equations for ˛ and ˇ simplify (more later) – For now just know that for a passive tline ˛ 0 2.4.2 Solving the Wave Equation (Phasor Form) The general solutions of the two wave equations each involve a pair of exponentials (think back to 2nd-order differential equations)

V .z/ V e z V e z (V) Q 0C 0 D z C z I .z/ I Ce I e (A) Q D 0 C 0 z where it becomes clear later that e is a wave propagating in the z direction and e z is a wave propagating in the z C direction

Example 2.2: Verify the General Solution Satisﬁes Wave Equa- tion

z z Is it really true that V .z/ V Ce V e satisﬁes Q D 0 C 0 d 2V .z/=dz2 2V .z/ 0? Q Q D Form d 2V .z/=z2 and see: Q d z z V .z/ V Ce V e dz Q D 0 C 0 2 d 2 z 2 z also 2 V .z/ V Ce V e V .z/ X dz2 Q D 0 C 0 D Q

2-17 CHAPTER 2. TRANSMISSION LINES

The variables .V C;I C/ and .V ;I / are unknowns that will 0 0 0 0 be found using boundary conditions

Zg + + −γz + (V0 , I0 )e Incident wave Vg ZL − − γz − (V0 , I0 )e Reflected wave z

FigureFigure 2.8: 2-9 IncidentIn general, and reflected a transmission traveling line waves can support on a tline. two traveling waves, an incident wave (with voltage How do you know the solution+ works?+ and current amplitudes (V0 , I0 )) traveling along the +z direction (towards the load) and a reflected wave (with Plug it into the voltage and current wave equations and see (V0−, I0−)) traveling along the z direction (towards the source). − You can eliminate two (I C and I ) of the four unknowns by 0 0 inserting the general solution for V .z/ into the first of the tele- Q grapher’s equations, i.e.,

V .z/ d z z z z Q V Ce V e V Ce V e dz D dz 0 C 0 D 0 0 also R0 j!L0 I .z/; D C Q so solving for I .z/ yields Q z z I .z/ V Ce V e Q D R j!L 0 0 0 C 0 also z z I Ce I e D 0 C 0 where the last line follows from the second general wave equation solution

What just happened? Confused? 2-18 2.4. WAVE PROPAGATION ON A TRANSMISSION LINE

It appears that V C R0 j!L0 V 0 must C must 0 I0C D D I0

We deﬁne s R0 j!L0 R0 j!L0 Z0 C C ./ D D G j!C 0 C 0 Note: The units look good as V=I is R from Ohm’s law

Furthermore, Z0 involves the ratio of incident or reﬂected voltage over current waves, not the total voltage and current, V .z/ Q and I .z/ Q Summarizing: The general voltage and current solutions are now down to just two unknowns

z z V .z/ V Ce V e (V) Q D 0 C 0 V0C z V0 z I .z/ e e (A) Q D Z0 Z0

Secondly, Z0 although clearly a function of !, it is the line cross-sectional geometry and constitutive parameters that really control its value

Finally, when source Zg and load ZL boundary conditions are applied, you will be able to solve for the complex quantities

V0C and V0

2-19 CHAPTER 2. TRANSMISSION LINES

2.4.3 Returning to the Time Domain For the sinusoidal steady-state solution, a return to the time jC domain is possible by writing V0C V0C e and V0 j D j j D V e C, then plug into the cosine form: j 0 j v.z; t/ ReV .z/ej!t Q D ˛z V0C e cos !t ˇz C D j j ˛z C V e cos !t ˇz C j 0 j C C When you consider LTspice simulations using .tran a little bit later, this is what you will be observing

Also, the ﬁrst term travels in the z direction while the second C term travels in the z direction The velocity of propagation for both waves is of course ! up f D D ˇ

Observe that just as in HMWK problem 1.8, a standing wave pattern results when the two propagating waves superimpose

Solving for V C and V is still an open problem 0 0

Example 2.3: No Speciﬁc Geometry Air Line

In this example we assume that the two conductors of a transmission line are in air

Furthermore, the air assumption is taken to mean 0 D 2-20 2.4. WAVE PROPAGATION ON A TRANSMISSION LINE

The conductors are perfect, that is Rs 0 D

The requirements are to ﬁnd L0 and C 0, given that Z0 50 !, D ˇ 20 rad/m, and the operating frequency is f 700 MHz D D

The ﬁrst set of assumptions tell us that R0 G0 0 D D

The expressions for ˇ and Z0 in terms of the tline distributed element parameters, simplify greatly under the lossless conditions hp i ˇ Im .0 j!L /.0 j!C / !pL C D C 0 C 0 D 0 0 s r 0 j!L0 L0 Z0 C ; D 0 j!C D C C 0 0 so with two equations involving the unknowns L0 and C 0 we can solve for them using the Z0, ˇ, and f

Start by forming ˇ=Z0 ˇ !C 0 2 f C 0; Z0 D D which implies that ˇ 20 C 9:09 10 11 (F/m) 0 8 D 2f Z0 D 2 7 10 50 D „ ƒ‚ … 90:9 (pF/m)

With C 0 found, the Z0 expression can be used to ﬁnd L0 2 2 11 7 L0 Z0C 0 50 9:09 10 2:27 10 (H/m) D D D „ ƒ‚ … 227 (nH/m)

2-21 CHAPTER 2. TRANSMISSION LINES

2.5 The Lossless Microstrip Line

Conducting μ , σ strip ( c c) Dielectric insulator (ε, μ, σ) w

h t Metal thickness w assumed zero s = ---- h μ , σ Conducting ground plane ( c c) Table 2.2: Microstrip geometry.

In RF/microwave circuit design the microstrip structure of Fig- ure 2.3e and also shown immediately above, is very common

The tline geometry ﬁts well with surface-mount PCB design of today

– A narrow strip of width w sits on top of a dielectric substrate of height h, which in turn sits over a ground plane – Surface-mount components can be mounted on the top of the substrate

A downside of microstrip is that the air-dielectric interface gives rise to a small axial ﬁeld component, making the propagation mode quasi-TEM

The nature of quasi-TEM is that the line introduces some dispersion (recall this means constant up versus frequency)

2-22 Conducting

strip (μc , σc) Dielectric insulator (ε, μ, σ) w

Conducting ground plane (2.5.μc , σ THEc) LOSSLESS MICROSTRIP LINE (a) Longitudinal view

(b) Cross-sectional view with E and B field lines Figure 2.9: Electric and Magnetic ﬁelds of mircostrip

In the discussion and analysis of the text, the dielectric ﬁlling material is assumed to be lossless ( 0) D The conductors, strip on top and ground plane on the bottom, are assumed perfect ( ) 1 (c) Microwave circuit Finally, the permeability is 0, that is the material is Figure 2-10 Microstrip line: (a)D longitudinal view, (b) nonmagneticcross-sectional view, and (c) circuit example. (Courtesy of Prof. Gabriel Rebeiz, U. California at San Diego.) – Note there are applications involving microstrip where magnetic materials are used, just not discussed here

2.5.1 Analysis and Synthesis of Microstrip

For tlines with a homogeneous dielectric ﬁlling (also nonmag- netic), e.g., coax and stripline of Figure 2.3, the velocity of wave propagation or phase velocity, is always

also 1 c up vp D D p00r D pr

Considering Figure 2.9, the electric ﬁeld lines are mostly in the dielectric, but some are topside in air

2-23 CHAPTER 2. TRANSMISSION LINES

In the analysis of microstrip the mixture of air and dielectric is managed by deﬁning eff, the effective relative permittivity, hence the phase velocity is written as

c up vp D D peff

The detailed analysis of microstrip can be found in text books such as2

There are three basic equations you need to be familiar with when working with microstrip:

1. An equation that ﬁnds eff given the strip aspect ratio s D w=h and r

2. An equation that takes s together with eff to ﬁnd the line characteristic impedance Z0

3. An equation that takes Z0 together with r to ﬁnd s D w=h; generally eff is also found along the way

In microwave circuit design various Z0 values will needed along with the corresponding eff values (3) is very valuable! )

The equations are empirical function ﬁts to detailed ﬁeld analysis results for tline inductance and capacitance per unit length, as aspect ratio, s, changes

2D. H. Schrader, Microwave Circuit Analysis, Prentice Hall, 1995

2-24 2.5. THE LOSSLESS MICROSTRIP LINE

Obtain eff From w=h and r The empirical formula for zero thickness strips (t=h 0) is D given in terms of s w=h by D Â ÃÂ Ã xy er 1 r 1 10 eff C 1 ; D 2 C 2 C s where Ä r 0:9 x 0:56 D r 3 CÂ 4 4 2 Ã s 3:7 10 s y 1 0:02 ln C D C s4 0:43 C 4 3 0:05 ln.1 1:7 10 s / C C Chapter 2 notebook Python function implementation:

Figure 2.10: A Python function that ﬁnds eff given s w=h and r . D

Obtain Z0 From w=h and r Design for zero thickness strips is given by " # 60 6 .2 6/e t r 4 Z ln 1 ; 0 C 2 D peff s C C s

2-25 CHAPTER 2. TRANSMISSION LINES

where Â30:67Ã0:75 t D s

Chapter 2 notebook Python function implementation:

Figure 2.11: A Python function that ﬁnds Z0 given s w=h and r . D

Obtain w=h From Z0 and r

Design for zero thickness strips is given by the piecewise solution 8 2 h ˆ .q 1/ ln.2q 1/ ˆ ˆ r 1 w <ˆ ln.q 1/ C 2r h D 0:52Ái ˆ 0:29 ;Z0 .44 2r / ˆ C r Ä ˆ 8ep :e2p 2;Z0 > .44 2r /; 2-26 2.5. THE LOSSLESS MICROSTRIP LINE

where

602 q D Z0pr r Â ÃÂ Ã r 1 Z0 r 1 0:12 p C 0:23 D 2 60 C r 1 C r C Chapter 2 notebook Python function implementation:

Figure 2.12: A Python function that finds s w=h given Z0 and r . D 2.5.2 Common Mstrip Materials FR-4 made of epoxy fiberglass with a fire retardant property; r 4:6; multilayer PCB designs are no problem; material loss is significant with tan ı 0:0180 (loss tangent) D 2-27 CHAPTER 2. TRANSMISSION LINES

Microfiber PTFE: Is Polytetrafluoroethylene or Teflonr, with microfibers for reinforcing; Rodgers3 RT_Duroid 5880 material has r 2:20 (for design) and is very low loss with D tan ı 0:0004 D

Ceramic-ﬁlled PTFE: Uses ceramic material to increase r up to a range of 3.0 to 10.2; Rogers RO30104 material has r 11:2 (for design) and tan ı 0:0022 D D 5 Alumina : Aluminum oxide (Al2O4) in 99.5% concentration is used for thin-ﬁlm microwave circuits; r 9:9 and very low D loss tan ı 0:0001 D Silicon: Used for monolithic microwave integrated circuits (MMICs); r 11:9 and very lossy D Gallium Arsenide6: Used for monolithic microwave integrated circuits (MMICs); r 12:88 and very low loss tan ı D D 0:0004

Sapphire: Crystalline alumina; r 9:0 (text) and tan ı < D 0:0015

3http://www.rogerscorp.com/documents/606/acs/RT-duroid-5870-5880-Data- Sheet.pdf 4https://www.rogerscorp.com/documents/722/acs/RO3000-Laminate-Data-Sheet- RO3003-RO3006-RO3010-RO3035.pdf 5http://www.microwaves101.com/encyclopedias/alumina-99-5 6http://www.microwaves101.com/encyclopedias/gallium-arsenide

2-28 2.5. THE LOSSLESS MICROSTRIP LINE

Example 2.4: 50 Mstrip on FR-4

Design a 50 microstrip on FR-4 material having thickness of 1/16 inch. Assume r 4:6. Find the strip width in mils, D eff, and the length a =4 line operating at 2.4 GHz

The equations in the book can be used straight away, but here we will use the Python functions that are in Chapter 2 IPYNB

To make things easy ﬁrst write a function that calculates addi- tional tline parameters

Figure 2.13: The mstrip_extra function.

2-29 CHAPTER 2. TRANSMISSION LINES

A mil is a thousandth of an inch and most PCB shops use mils Here h 1=16 1000 62:5 mils (also 2.54 cm/in) D D The answer is worked out below in the IPYNB

Figure 2.14: The desired calculations plus some extras.

In summary, the line width W 115:4 mils (2.931 mm), D eff 3:460, and g=4 661 mils (1.680 cm) D D

Example 2.5: Design/Analysis Charts for PCB Mstrip

Produce some useful design/analysis charts for use on quizzes and exams

The relative permittivity values are taken from plastic sub- strates used in PCBs

2-30 2.5. THE LOSSLESS MICROSTRIP LINE

We obtain plots of Z0 versus s w=h and eff versus w=h D using the Python functions described earlier.

Figure 2.15: Python commands followed by Z0 versus w=h plot.

2-31 CHAPTER 2. TRANSMISSION LINES

Figure 2.16: Python commands followed by eff versus w=h plot. Larger versions of the plots themselves are available for down- load (soon)

2-32 2.6. LOSSLESS TLINE GENERAL CASE

2.6 The Lossless Transmission Line: General Considerations

Up to this point we have seen that Z0 and play a central role in the characterization of a transmission line

Modeling with a low-loss dielectric and high conductivity conductors is a good place to start a design

Under the assumption that R0 G0 0 we have D D ˛ jˇ j!pL C D C D 0 0 so ˛ 0 (Lossless) D ˇ !pL C (Lossless) D 0 0 The characteristic impedance becomes r L0 Z0 (Lossless) D C 0 Summarizing some other useful results into one location ˇ !p (rad/m) D 1 c up (m/s) D p D pr also up c 1 0 g (m) D D f D f pr D pr

Note: The notation g refers to wavelength in a guided medium, most often a waveguide or a TEM transmission line; 0 c=f D is of course the free space wavelength

2-33 CHAPTER 2. TRANSMISSION LINES

The table below summarizes lossless line parameters, , up D vp, and Z0, for the four classical tline structures originally shown in Figure 2.3(a)–(d).

Table 2.3: Summary of lossless line parameters for the four classical line types. Table 2-2 Characteristic parameters of transmission lines.

Propagation Phase Characteristic Constant Velocity Impedance γ = α + jβ up Z0

(R′ + jωL′) General case γ = (R′ + jωL′)(G′ + jωC ′) up = ω/β Z0 = /(G′ + jωC ′) + Lossless α = 0, β = ω√εr/cup = c/√εr Z0 = L′/C ′ (R = G = 0) ′ ′ +

Lossless coaxial α = 0, β = ω√εr/cup = c/√εr Z0 =(60/√εr)ln(b/a)

Lossless α = 0, β = ω√εr/cup = c/√εr Z0 =(120/√εr) two-wire ln[(D/d)+ (D/d)2 1] · − Z0 (120/√εr)ln+(2D/d), ≈ if D d ≫

Lossless α = 0, β = ω√εr/cup = c/√εr Z0 =(120π/√εr)(h/w) parallel-plate

Notes: (1) µ = µ0, ε = εrε0, c = 1/√µ0ε0,and µ0/ε0 (120π) Ω,whereεr is the relative permittivity of insulating material. (2) For coaxial line, a and b are radii≈ of inner and outer conductors. (3) For two-wire line, d = wire diameter and D = separation between+ wire centers. (4) For parallel-plate line, w = width of plate and h = separation between the plates.

Example 2.6: Solve for r Given f and g

The wavelength in a lossless transmission line is known to be 5.828 cm at a frequency of 2.4 GHz

Find r of the tline insulating material

We know that g 0=pr and 0 c=f , so putting these D D 2-34 2.6. LOSSLESS TLINE GENERAL CASE

two equations together we have

Â Ã2 Â Ã2 0 c=f r D g D g Â3 108=2:4 109 Ã2 4:60 D 0:05828 D

Example 2.7: Line Parameters from r and L0

Given a lossless line has r 6 and the line inductance is D L0 0:8 H/m D

Find up, C 0 and Z0

To ﬁnd up we use the fact that 8 c 3 10 8 up 1:225 10 (m/s) D pr D p6 D

also To ﬁnd C 0 we use the fact that ˇ !pL C !p, so D 0 0 D 7 12 0r 4 10 8:85 10 6 C 0 83:42 (pF/m) D L D 0:8 10 6 D 0

Finally, ﬁnding Z0 makes use of 6 L0 0:8 10 Z0 97:94 ./ D C D 83:42 10 12 D 0

2-35 CHAPTER 2. TRANSMISSION LINES

2.6.1 Voltage Reﬂection Coefﬁcient In this subsection we impose the boundary conditions of the following circuit

~ I i Transmission line Zg ~ + + IL + ~ ~ ~ Vg Vi Z0 VL ZL − −− Generator Load z z = −lz = 0 d d = l d = 0 Figure 2.17: Boundary conditions imposed by placing a lossless tline Figure 2-12 Transmission line of length l connected betweenon a onegenerator end to and a generator load. circuit and on the other end to a load Z .Theloadislocatedatz = 0andthe The total voltageL and current, under the lossless assumption, generator terminals are at z = l.Coordinated is deﬁned makesas d = z. − − ˇz ˇz V .z/ V Ce V e Q D 0 C 0 V0C ˇz V0 ˇz I .z/ e e Q D Z0 Z0 and in particular at the load end of the line, where z 0, we D have also VL V .z 0/ V C V Q D Q D D 0 C 0 also V0C V0 IL I .z 0/ Q D Q D D Z0 Z0

2-36 2.6. LOSSLESS TLINE GENERAL CASE

From Ohms law for impedances it is also true that

must VL ZL Q ; Q D I QL

so rearranging to solve for V0 in terms of V0C results in Â Ã ZL Z0 V0 V0C D ZL Z0 C which establishes a relationship between the incident voltage

wave amplitude V0C and the reﬂected voltage wave amplitude V0

Definition: The voltage reflection coefficient , is Â Ã V ZL Z0 zL 1 0 ; D V C Á ZL Z0 D zL 1 0 C C

where zL ZL=Z0 is the normalized load impedance D

A similar relationship holds for I =I C, except due to V D=I C 0 0 0 0 D V =I C, we have 0 0 I V 0 0 I0C D V0C D

It is also worth noting that since ZL is in general complex, is complex with polar form ejr j j

– For a passive load ZL it turns out that 1 j j Ä 2-37 CHAPTER 2. TRANSMISSION LINES

Matched Load When V 0 L 0 ZL Z0 0 D ) D ) D making the load matched to the tline characteristic impedance.

Reﬂection Coefﬁcient Special Cases

Table 2.4: Reflection coefficient under special cases (r R=Z0 and Table 2-3 Magnitude and phase of the reflection coefficient for various types of loads. TheD normalized load impedance zL = ZL/Z0 =(R +jxjX)/ZjX=Z0 = r + jx0).,wherer = R/Z0 and x = X/Z0 are the real and imaginary parts of zL,respectively. D Reflection Coefficient Γ = Γ e jθr | | Load Γ θr | | 2 2 1/2 (r 1) + x 1 x 1 x Z0 Z =(r + jx)Z − tan− tan− L 0 (r + 1)2 + x2 r 1 − r + 1 ' ) % − & % &

Z0 Z0 0(noreﬂection) irrelevant

Z0 (short) 1 180 (phase opposition) ± ◦

Z0 (open) 1 0 (in-phase)

1 Z0 jX = jωL 1 180 2tan x ± ◦ − −

j 1 Z0 jX = − 1 180◦ + 2tan− x ωC ±

2-38 2.6. LOSSLESS TLINE GENERAL CASE

Example 2.8: Series RC Load

Transmission line A

RL 50 Ω Z0 = 100 Ω CL 10 pF

Figure 2.18: RC load on a Z0 100 lossless tline. Figure 2-13 RC loadD (Example 2-3).

Consider a Z0 100 lossless tline driving a load ZL com- D posed of resistor RL 50 in series with capacitor CL 10 D D pF

Find L and check the result using LTspice GammaL Here we use Python, MATLAB or Excel would also work

Figure 2.19: Python calcuation of L.

LTspice has both loss and lossy transmission line models To make LTspice work in this problem we use an ideal tlie having Z0 100 , arbitary line length speciﬁed by a time delay D Td l=up, and source impedance Zg Z0 LTspice can only D D 2-39 CHAPTER 2. TRANSMISSION LINES

measure voltage and currents at circuit nodes and branches, respectively

We desire V C, V , and L o 0

From the original boundary conditions at the load,

VL V C V Q D 0 C 0 V0C V0 IL Q D Z0 Z0 or ILZ0 V C V Q D 0 0

By adding and subtracting the ﬁrst and last equations above, we obtain

VL ILZ0 V C Q C Q 0 D 2 VL ILZ0 V Q Q 0 D 2 and also the relationship

V0 VL ILZ0 L Q Q D V C D VL ILZ0 0 Q C Q

In this example we use the L equation above to post-process the spice voltage and current at the load as shown in the ﬁgures below

2-40 2.6. LOSSLESS TLINE GENERAL CASE

Source Termination is Matched Simulation Focus

I = I R2 L

Lossline with length given as delay time, i.e., Td = l/up

– ˜ ˜ V0 VL – ILZ0  ==------L + V˜ + I˜ Z V0 L L 0 Figure 2.20: LTspice model emplying a lossless transmission driven by a source having matching impedance of Zg Z0 100 . D D

Phase L Mag plot formula

 Magnitude and L L phase values overlapping here

Note: Right-click plot, choose manual scale, then choose linear for vertical LTspice Agrees! Figure 2.21: LTspice AC steady-state results with total voltage and current converted to the voltage reﬂection coefﬁcient (use plot linear mode).

The results from both calculation means agree!

2-41 CHAPTER 2. TRANSMISSION LINES

Example 2.9: Series RC Load

Consider the case of a purely reactive load ZL jXL D Calculating the reﬂection coefﬁcient we have jXL Z0 .Z0 jXL/ L D jXL Z0 D Z0 jXL Cj 2 C 1e ; D

where .Z0 jXL/ D† C

The signiﬁcant result is that L 1 independent of the value j j D of XL

The angle of the angle does change with changes in Xl relative to Z0

2.6.2 Standing Waves

Knowing now that V V D means that we can write 0 D 0 jˇz jˇz V .z/ V C e e Q D 0 C V0C jˇz jˇz V .z/ e e Q D Z0

As we move along the line for l z 0, a curiosity is what Ä Ä is the nature of V .z/ and I .z/ j Q j Q 2-42 2.6. LOSSLESS TLINE GENERAL CASE

To answer that we do some math analysis, such as q V .z/ V .z/ V .z/ j Q j D Q Q jˇz jˇz jˇz jˇz1=2 V C e e e e D j 0 j C C 2 1=2 V C 1 2 cos.2ˇx r / ; D j 0 j C j j C j j where in polar form ejr D j j

A similar analysis for I .z/ can be performed with V C Q j 0 j ! V C =Z0 and ﬂipping the sign of the cosine term j 0 j

It is convenient to view the voltage and current magnitude in terms of a positive distance d back from the load, so we let d z D

Finally, we have

2 1=2 V .d/ V C 1 2 cos 2ˇd r j Q j D j 0 j C j j C j j V0C 2 1=2 I .d/ j j 1 2 cos 2ˇd r j Q j D Z0 C j j j j

These two expressions generate what is known as the standing wave pattern on the line

Using a slotted-line, you can physically measure the voltage amplitude (magnitude) along the line by sliding a carriage along a track that has a probe inserted into a slit cut into an air-line

2-43 CHAPTER 2. TRANSMISSION LINES

Sliding probe To detector Probe tip Slit

Z ~ + g Vg ZL − 40 cm 30 cm 20 cm 10 cm

Figure 2.22: Slotted coaxial line with E-ﬁeld probe to measure the Figure 2-16 Slotted coaxial line (Example 2-6). standing wave magnitude.

The expressions just derived are implemented in the Chapter 2 IPYNB

Figure 2.23: Python code for plotting standing wave patterns.

2-44 2.6. LOSSLESS TLINE GENERAL CASE

+ max V = 10 0  Z = 50 --- 0 2 Load min End

max min/max correspond min Load End

Figure 2.24: Plots of V .d/ and I .z/ for 0:8 45ı. j Q j Q j D †

The 2ˇ term is responsible for the period of the pattern being =2

Clearly the voltage maximum (constructive interference) value corresponds to V C.1 / 0 C j j Similarly, voltage minimum (destructive interference) value corresponds to V C.1 / 0 j j Special Cases:

– Matched 0, is a ﬂat line at V C ) D j 0 j – Short 1, begins with V .0/ 0 an d the ) D j Q j D maximum values going to 2 V C and minimums going to j 0 j zero (looks like a half-wave rectiﬁed sine wave)

2-45 CHAPTER 2. TRANSMISSION LINES

– Open 1, begins with V .0/ 2 V C and min- ) D j Q j D j 0 j imum values going to zero (phased 90ı relative the short case)

~ |V(d)|

Matched line + |V0 | d λ 3λ λ λ 0 4 2 4 (a) ZL = Z0 Short-circuited line ~ |V(d)| λ/2 + 2|V0 |

d λ 3λ λ λ 0 4 2 4 (b) ZL = 0 (short circuit)

Open-circuited line ~ |V(d)| λ/2 + 2|V0 |

d λ 3λ λ λ 0 4 2 4

Figure 2.25:Figure Standing-waves 2-15 Voltage standing-wave for matched, patterns short, for and (a) open a cases. matched load, (b) a short-circuited line, and (c) an open- circuited line. Maximum and Minimum Values

The maximum value of the voltage magnitude, V max V C Œ1 j Q j D j 0 j C occurs when the argument of the cos./ term is an even j j 2-46 2.6. LOSSLESS TLINE GENERAL CASE

multiple of i.e.,

2ˇdmax r 0; 2; : : : D

Similarly the voltage magnitude minimum, V min V C Œ1 j Q j D j 0 j , occurs when the argument of the cos./ term is an odd j j multiple of i.e.,

2ˇdmin r ; 3; : : : D

The maximum value of the current magnitude, I max I C Œ1 j Qj D j 0 j C occurs when the argument of the cos./ term is an odd mul- j j tiple of i.e.,

2ˇdmax r ; 3; : : : D

The maximum value of the current magnitude, I min I C Œ1 j Qj D j 0 j occurs when the argument of the cos./ term is an even j j multiple of i.e.,

2ˇdmin r 0; 2; : : : D In all cases minimum and maximum values are separated by =4 and voltage maxim correspond to current minimum, etc.

2.6.3 VSWR

The ratio of V max to V max is given the name voltage standing- j Q j j Q j wave ratio (VSWR) or (SWR) and in the text is denoted by S

V max 1 S j Q j C j j D V min D 1 j Q j j j also S 1 j j D S 1 C 2-47 CHAPTER 2. TRANSMISSION LINES

VSWR is an indicator of the mismatch between Z0 and ZL Note: 1 S 1 and 1 S D ) D j j D ) D 1

Example 2.10: from VSWR j j Given VSWR S 5 ﬁnd D D j j 5 1 0:667 j j D 5 1 D C

Example 2.11: Find L in Parallel RL Giving S 5 D

Z0 = 100

Figure 2.26: An RL Load Terminating a Z0 100 tline. D Consider the circuit shown above with L unknown, S VSWR D D 5, and f 100 MHz D To start with we know that 1 S 1 S C j j ; Á 1 ) j j D S 1 j j C so 5 1 2 j j D 5 1 D 3 C 2-48 2.6. LOSSLESS TLINE GENERAL CASE

Now to solve for in terms of L we ﬁrst ﬁnd ZL to be RL j!L ZL ;! 2f D RL j!L D C next plugging into the expression for

RL j!L R j!L Z0 LC D RL j!L R j!L Z0 LC C j!RLL .Z0RL/ j!Z0L D j!RLL .Z0RL/ j!Z0L C Z0RL j!.Z0 RL/L D Z0RL j!.Z0 RL/L C To isolate L ﬁnd 2 and then rearrange j j .Z R /2 !2.Z R /2L2 2 0 L 0 L 2 C 2 2 2 j j D .Z0RL/ ! .Z0 RL/ L C C so working the algebra

s 2 2 .Z0RL/ .1 / L j j 2 2 2 2 D ! .Z0 RL/ .Z0 RL/ C j j Using Python for the calculations (also in Chapter IPYNB) we arrive at L 68:489 (nH) D

Figure 2.27: Calculation of L.

2-49 CHAPTER 2. TRANSMISSION LINES

A less elegant approach is to numerically ﬁnd L by plotting j j versus L and then numerically ﬁnd the L value which makes 2=3 j j D

Figure 2.28: Plotting of versus L. j j

A precise numerical solution can be found by using a solver such as fsolve, which is available in the scipy.optimize package

We start by forming an objective function that will instead cross through zero at the desired L value, that is search for L such that Fobj.L/ .L/ 2=3 0 D j j D 2-50 2.6. LOSSLESS TLINE GENERAL CASE

Figure 2.29: Using fslove to numerically search for L.

The results is the same in both cases Just pulling the L value from the plot is easier still, and quite accurate when you zoom in

2-51 CHAPTER 2. TRANSMISSION LINES

2.7 Wave Impedance of a Lossless Line

Anywhere along the line we can form the ratio V .d/=I .d/, Q Q hence this voltage to current ratio is given the name wave impedance

Drilling into this definition reveals jˇd jˇd V .d/ V C e e Z.d/ Q 0 C Z jˇd jˇd 0 D I .d/ D V C e e Q 0 Ä j 2ˇd 1 e Z0 C ./ D 1 e j 2ˇd If we define the reflection coefficient at location d z as D j 2ˇd .d/ e Á then we can also write Ä 1 d Z.d/ Z0 C ./ D 1 d To be clear: – Z.d/ is the ratio of the total voltage to the total current (both incident and reflected)

– Z0 is the ratio of just the incident voltage over the incident current (minus sign on reﬂected voltage over re- ﬂected current)

A great use of the wave impedance concept is in equivalent circuits as shown below

2-52 2.7. WAVE IMPEDANCE OF A LOSSLESS LINE

ABC Z + g ~ Z Z Vg Z(d) 0 L − A′ B′ C′

d = l d 0 (a) Actual circuit

AB Z + g ~ Vg Z(d ) − A′ B′ (b) Equivalent circuit Figure 2.30: Signiﬁcance of the wave impedance in tline circuit Figure 2-17 The segment to the right of terminals BB ′ modeling.can be replaced with a discrete impedance equal to the wave impedance Z(d).

Line Input Impedance

As a special case consider the line input impedance Ä 1 l Zin Z.l d/ Z0 C D D D 1 l

Noting that j 2ˇl l e D 2-53 CHAPTER 2. TRANSMISSION LINES

allows us to write

Â Ã ZL cos.ˇl/ jZ0 sin.ˇl/ Zin Z0 C D Z0 cos.ˇl/ jZL sin.ˇl/ Â C Ã ZL jZ0 tan.ˇl/ Z0 C D Z0 jZL tan.ˇl/ C also making use of Euler’s identity and then dividing top and bottom by cos./

Application: Find the voltage at the source end of the line Vi V .z l/ (remember here the notation assumes z) Q D Q D using Zin, then go on to ﬁnd the incident voltage V0C

– From simple circuit analysis (voltage divider)

VgZin Vi Ii Zin Q Q D Q D Zg Zin C

– It is also true that

jˇl jˇl Vi V .z l/ V C e e Q D Q D D 0 C

– Solving for V0C yields

! V Z Â 1 Ã V Qg in 0C jˇl jˇl D Zg Zin e e C C

2-54 2.7. WAVE IMPEDANCE OF A LOSSLESS LINE

~ I i A Transmission line Zg + ~ + IL + ~ ~ ~ Vg Vi Zin Z0 VL ZL − −A′ − Generator Load z = −l z = 0 d = l d = 0

~ Ii A Z g + + ~ ~ Vg Vi Zin − − A′

Figure 2.31: How to obtain V using Z to obtain V and I , and Figure 2-18 At the0 generatorC end,in the terminatedQi Qi then V using V . 0CtransmissionQi line can be replaced with the input impedance of the line Zin. Finally obtaining V C completes the solution of the 1D trans- 0 mission line wave equation; everything is known!

2-55 CHAPTER 2. TRANSMISSION LINES

2.8 Special Cases of the Lossless Line

In this section we consider the input impedance of a lossless transmission line under the special cases of: (i) a short at the load, (ii) an open at the load, and (iii) a pure resistive load RL Z0 at the load ¤ This study opens the door to ﬁlter design and impedance matching circuits, and the use of the Smith Chart

2.8.1 Short-Circuited Line

For a short circuited line ZL 0 so at distance d away from D the load the voltage and current are given by

1 jˇd ‚…„ƒ jˇd Vsc.d/ V C e e 2jV C sin ˇd Q D 0 C D 0 V0C jˇd jˇd 2V0C Isc.d/ e e cos ˇd Q D Z0 „ƒ‚… D Z0 1 The input impedance of a length l line is sc ˇ Zin .l/ Zin.l/ˇZ 0 jZ0 tan.ˇl/ D LD D The impedance is purely reactive, but changing with d as we move back from the short circuit load (d increasing)

The ﬁgure below shows that the reactance is positive, as in an inductor, part of the time and negative, as in a capacitor, at other times

2-56 2.8. SPECIAL CASES OF THE LOSSLESS LINE

sc Z short Zin 0 circuit

d l (a) 0 ~ Vsc(d) + 2jV0 Voltage 1

d 0 λ 3λ λ λ 4 2 4 −1 (b) ~ Isc(d) Z0 + 2V0 1 Current

d 0 λ 3λ λ λ 4 2 4 −1 (c) sc Zin jZ0 Impedance

l 0 λ 3λ λ λ 4 2 4

(d)

Figure 2.32:Figure Short-circuited 2-19 Transmission line line showing terminated voltage, in a short current and sc Z =jZ0 versuscircuit:l. (a) schematic representation, (b) normalized in voltage on the line, (c) normalized current, and (d) normalized input impedance. 2-57 CHAPTER 2. TRANSMISSION LINES

Inductor Like: Set

j!Leq jZ0 tan.ˇl/ D or Z0 tan.ˇl/ Leq (H) D ! provided tan.ˇl/ > 0

Capacitor Like: Set 1 jZ0 tan.ˇl/ j!Ceq D or 1 Ceq (F) D Z0! tan.ˇl/ provided tan.ˇl/ < 0

sc Z short Zin 0 circuit

sc 1 Zin Zc = jωCeq

Figure 2.33:Figure Short-circuited 2-20 Shorted line line behaves as equivalent as a capacitor capacitor depending upon the(Example line length 2-8).l.

2-58 2.8. SPECIAL CASES OF THE LOSSLESS LINE

Impedance versus Frequency In microwave circuit design considering the Zsc relationship in versus frequency of interest, as circuits typically operate over a band of frequencies

To take this view we write elect. length f ˇl D D 2 f0

where is line electrical length in radians and f0 is the corresponding quarter-wave frequency of the line

The quarter-wave frequency f0 follows from the fact that 0 D up=f0 or f0 up=0 D Now, 2 2 u pi f ˇl 0 p D 4 D up=f 4f0 D 2 f0 Expressing Zsc as function frequency, we have in Â Ã sc f Zin .f / jZ0 tan./ jZ0 tan D D 2 f0

– Note: At f f 0 we have =2 rad or 90ı, which D C D both correspond to 0=4

sc Example 2.12: Z versus f for Z0 100 in D Here we put together a simple model in LTspice to see how the impedance (pure reactance here) varies with frequency relative to the quarter-wave frequency

2-59 CHAPTER 2. TRANSMISSION LINES

Here we choose f0 100 MHz which has period of T D D 10ns, so the corresponding tline time delay is T=4 2:5 ns D

Figure 2.34: LTspice schematic used to study the input impedance versus frequency of a quarter-wave short circuit line at 100 MHz.

Using a linear plotting scale in LTspice we plot the line input impedance

Figure 2.35: Input impedance plot versus frequency, which clearly shows the response repeats at odd multiples of f0, the quarter-wave frequency.

2-60 2.8. SPECIAL CASES OF THE LOSSLESS LINE

Note: At the quarter-wave frequency the short circuit appears an open circuit as the impedance magnitude goes to inﬁnity

2.8.2 Open-Circuited Line

For the case of ZL the analysis is similar to that of the D 1 short circuit

At distance d away from the load the voltage and current are given by

1 C jˇd ‚…„ƒ jˇd Vsc.d/ V C e e 2jV C cos ˇd Q D 0 C D 0 V0C jˇd jˇd 2V0C Isc.d/ e e sin ˇd Q D Z0 „ƒ‚… D Z0 1 C

The input impedance of a length l line is oc ˇ Zin .l/ Zin.l/ˇZ jZ0 cot.ˇl/ D LD1 D

The impedance is purely reactive, but changing with d as we move back from the short circuit load (d increasing)

The ﬁgure below shows that the reactance is negative, as in a capacitor, part of the time and positive, as in a capacitor, at other times

2-61 CHAPTER 2. TRANSMISSION LINES

oc Zin Z0

d (a) l 0 ~ Voc(d) + 2V0 Voltage 1

d 0 (b) λ 3λ λ λ 4 2 4 −1

~ Ioc(d) Z0 + Current 2jV0 1

d 0 (c) λ 3λ λ λ 4 2 4 −1

oc Zin Impedance jZ0

l 0 (d) λ 3λ λ λ 4 2 4

Figure 2.36:Figure Open-circuited 2-21 Transmission line line showing terminated voltage, in an open current and oc Z =jZ0 versuscircuit:l. (a) schematic representation, (b) normalized in voltage on the line, (c) normalized current, and 2-62 (d) normalized input impedance. 2.8. SPECIAL CASES OF THE LOSSLESS LINE

Impedance versus Frequency

We can again view the open circuit impedance as a function of frequency with the line length ﬁxed at some quarter-wave frequency f0 0=4 $ Expressing Zoc as function frequency, we have in Â Ã oc f Zin .f / jZ0 cot./ jZ0 cot D D 2 f0

oc Example 2.13: Z versus f for Z0 100 in D We again put together an LTspice model for the open-circuit case, to see how the impedance (pure reactance here) varies with frequency relative to the quarter-wave frequency

Here we choose f0 100 MHz which has period of T D D 10ns, so the corresponding tline time delay is T=4 2:5 ns D

Figure 2.37: LTspice schematic used to study the input impedance versus frequency of a quarter-wave open circuit line at 100 MHz.

2-63 CHAPTER 2. TRANSMISSION LINES

Using a linear plotting scale in LTspice we plot the line input impedance

Figure 2.38: Input impedance plot versus frequency, which clearly shows the response repeats at odd multiples of f0, the quarter-wave frequency.

Note: At the quarter-wave frequency the open circuit appears a short circuit as the impedance magnitude goes to zero

2.8.3 Lines of Length a Multiple of =2 =2 periodicity has been seen since the ﬁrst plot of standing waves back in the homework problem of Chapter 1

With regard to line input impedance this is formalized by considering l n=2 in tan ˇl; n 1; 2; : : : D D tan.ˇ n=2/ tan .2=/ n=2 tan.n/ 0 D D D 2-64 2.8. SPECIAL CASES OF THE LOSSLESS LINE

Thus Ä ZL jZ0 0 Zin Z0 C ZL D Z0 jZL 0 D C

A =2 (and multiples) length line transfers the load impedance to the source end

2.8.4 Quarter-Wavelength Transformer

Suppose l =4 n=2; n 0; 1; : : : D C D 2 ˇl ; D 2 D 2

so as tan.=2/ ! 1

2 Z0 Zin ; l n ; n 0; 1; 2; : : : D ZL D 4 C 2 D

Example 2.14: Matching a Real Impedance Load (or tline)

Suppose a 50 feedline needs to be matched to a 100 load

To eliminate reﬂections (at a single frequency) we insert a =4 section of transmission line to act as an impedance transformer

p Z02 Zin ZL p50 100 70:7 D D D 2-65 CHAPTER 2. TRANSMISSION LINES

Feedline A λ/4 transformer

Z01 = 50 Ω Zin Z02 ZL = 100 Ω

A' λ/4

Figure 2.39: Quarter-wave matching of 100 to 50 via Z02 D 70:7 . Figure 2-22 Conﬁguration for Example 2-10.

Perfect matching here means that 0 looking into the in- D terface at .A; A0/ is zero

As the quarter-wave operating frequency changes the matching property deviates from the perfect match ( 0) ¤ To see this consider an LTspice simulation:

Figure 2.40: LTspice schematic used to study versus frequency of a quarter-wave impedance transformer circuit.

2-66 2.8. SPECIAL CASES OF THE LOSSLESS LINE

Figure 2.41: Plot of versus frequency of a quarter-wave impedance transformer circuit having f0 100 MHz. D

Note: At multiples of the =2 frequency the line transfers ZL to the source end and we simply see the full mismatch of D .100 50/=.100 50/ 1=3; surprised? C D

2.8.5 Matched Line: ZL Z0 D

When a line is perfectly matched to the load, ZL Z0, there D is no standing wave pattern and all of the power ﬂowing from the source is delivered to the load; why? ( 0 for all values D of d)

2-67 CHAPTER 2. TRANSMISSION LINES

Table 2.5: Summary of standing wave and input impedance phenom- ina. Table 2-4 Properties of standing waves on a lossless transmission line.

+ Voltage maximum V max = V0 [1 + Γ ] | | | +| | | Voltage minimum V min = V0 [1 Γ ] |#| | | − | | θrλ nλ Positions of voltage maxima (also d#max = + , n = 0,1,2,... positions of current minima) 4π 2 θrλ , if 0 θr π 4 ≤ ≤ Position of first maximum (also dmax = ⎧ π position of first current minimum) ⎪ θrλ λ ⎨ + , if π θr 0 4π 2 − ≤ ≤ ⎪ ⎩θrλ (2n + 1)λ Positions of voltage minima (also dmin = + , n = 0,1,2,... positions of current maxima) 4π 4 λ θ Position of first minimum (also d = 1 + r min 4 π position of first current maximum) % &

zL + j tanβl 1 + Γl Input impedance Z = Z0 = Z0 in 1 + jz tanβl 1 Γ % L & % − l & Positions at which Zin is real at voltage maxima and minima 1 + Γ Z at voltage maxima Z = Z0 | | in in 1 Γ % − | |& 1 Γ Z at voltage minima Z = Z0 − | | in in 1 + Γ % | |& sc Zin of short-circuited line Zin = jZ0 tanβl Z of open-circuited line Zoc = jZ cotβl in in − 0 Zin of line of length l = nλ/2 Zin = ZL, n = 0,1,2,... 2 Zin of line of length l = λ/4 + nλ/2 Zin = Z0 /ZL, n = 0,1,2,...

Zin of matched line Zin = Z0

+ jθr j2β l V = amplitude of incident wave; Γ = Γ e with π < θr < π; θr in radians; Γ = Γe− . | 0 | | | − l

2-68 2.9. POWER FLOW IN A LOSSLESS LINE

2.9 Power Flow in a Lossless Line

Fundamentally the power ﬂow in a tline is related to the time domain voltage and current v.d; t/ and i.d; t/

The phasor quantity V .d/ is related to v.d; t/ via Q v.d; t/ ReV .d/ej!t D Q jC jˇd jr jˇd Re V0C e e e e D j Ä j C j j V C cos !t ˇd C D j 0 j C C cos.!t ˇd C r / C j j C C and similarly for i.d; t/ Ä V0C i.d; t/ j j cos !t ˇd C D Z0 C C cos.!t ˇd C r / j j C C

j j where V C V C e C and e r 0 D j 0 j D j j 2.9.1 Instantaneous Power Instantaneous power is the product P.d; t/ v.d; t/ i.d; t/ D which can be written as

P.d; t/ P i .d; t/ P r .d; t/ (book has + sign) D 2-69 CHAPTER 2. TRANSMISSION LINES

where 2 i V0C P .d; t/ j j 1 cos.2!t 2ˇd 2C/ D 2Z0 C C C 2 r 2 V0C P .d; t/ j j 1 cos.2!t 2ˇd 2C 2r / D j j 2Z0 C C C (book has minus on reﬂected power term)

In the average power calculation, up next, the 2! terms average to zero

2.9.2 Time-Averaged Power From circuits and systems, the time averaged power is the primary interest

For a periodic signal of period T 2=!, the average power D in W is 1 Z T Pav.d/ P.d; t/ dt D T 0 The double frequency term will average to zero (why?) yield- ing i r Pav Pav ‚…„ƒ2 ‚ …„ ƒ2 V0C 2 V0C Pav j j j j D 2Z0 j j 2Z0 2 V C j 0 j 1 2 (W) D 2Z0 j j Note: The power ﬂow is independent of d for both the incident and reﬂected terms

2-70 2.9. POWER FLOW IN A LOSSLESS LINE

Transmission line Zg i + Pav ~ ZL Vg r 2 i P = |Γ| P − av av

d = l d = 0 Figure 2.42: Summary of power ﬂow in a loss transmission line. Figure 2-23 The time-average power reﬂected by a load connected to a lossless transmission line is equal to the Phasorincident Domain power Power multiplied Calculation by Γ 2. | | Direct calculation of the power in the phasor domain is possible using 1 Pav Re V I D 2 Q Q

Example 2.15: Reﬂected Power

Given 100 (W) of power sent down a tline toward the load and 0:2, how much power is returned? j j D According to the average power expressions, P r 2P i 0:04 100 4 (W) av D j j av D D

2-71 CHAPTER 2. TRANSMISSION LINES

2.10 The Smith Chart

The Smith chart, developed in 1939 by P.H. Smith, is a widely used graphical tool for working and visualizing transmission line theory problems.

In mathematical terms the Smith chart is a transformation between an impedance Z and the reﬂection coefﬁcient , by virtue of the relation

Z Z0 D Z Z0 C

In complex variable theory this is known as a bilinear transformation

Assuming Z0 is real the transformation maps the Z-plane to the -plane as shown below:

Im Im Z − Z z −1 Γ = 0 = j < Z + Z0 z +1 Re(Z) 0 Re(Z) < 0 −1 Re 1+ Γ Re z = 1 1− Γ − j Re(Z) = 0 z-Plane where z = Z Z0

Figure 2.43: Smith chart as a mapping from Z to .

2-72 2.10. THE SMITH CHART

When working with the Smith chart impedances are normalized by Z0, i.e. z Z=Z0, so D z 1 1 and z C D z 1 D 1 C 2.10.1 Parametric Equations and the -Plane To better understand the transformation let z r jx D C .r 1/ jx Re./ j Im./ U jV C D C D C D .r 1/2 jx C C The real and imaginary parts are r2 1 x2 U C D .r 1/2 x2 C 2x C V D .r 1/2 x2 C C Now, eliminate x from the above expressions for U and V Â r Ã2 Â 1 Ã2 U V 2 r 1 C D r 1 C C Thus for a constant r we have a circle of radius 1=.r 1/ in C the U V plane (-plane) with center at

.U0;V0/ .r=.r 1/; 0/ D C Putting the two circle types together results the high-level Smith chart shown below

2-73 CHAPTER 2. TRANSMISSION LINES

Impedance − Plane Γ − Plane

Constant resistance lines V x +x

0 0.5 1 3 0 0.5 1 3 r U −x unit circle Z = R + jX and Γ =U + jV Z Z0 = z = r + jx Figure 2.44: Constant r circles in the -plane. By eliminating r from the U and V equations we ﬁnd 1 Â1Ã2 .U 1/2 .V / C x D x

Thus for constant x we have a circle of radius 1=x in the U V plane with center at

.U0;V0/ .1; 1=x/ D Impedance − Plane Γ − Plane

V 1 x Constant reactance lines 0.5 3 3

1 0.5 0 −0.5 r 1 −1 U

unit circle −3 −3 Z = R + jX and −0.5 −1 Γ = + Z Z0 = z = r + jx U jV Figure 2.45: Constant x circles in the -plane.

2-74 2.10. THE SMITH CHART

Next place the constant resistance and constant reactance circles on the same unit circle -plane

V = Γi

xL = 2

xL = 0.5 xL = 1

rL = 0, Γ = −1 Short circuit load

xL = 0 U = Γr zL =1, Γ = 0 Matched load rL = ∞, Γ = +1 rL = 0.5 Open circuit load x = −1 L rL = 2

r = 1 x = −0.5 L rL = 0 L

xL = −2

Figure 2.46: A few constant r and constant x circles on he unit circle -plane.

Example 2.16: Getting Acquainted

Suppose that ZL 50 j 50 and ﬁnd Zin and S for a line D C having electrical length of 45ı (recall ˇd ) and Z0 50 D D 2-75 CHAPTER 2. TRANSMISSION LINES

0.12 0.13 0.11 0.14 0.38 0.37 0.15 Three outer scales: 0.1 0.39 0.36 90 zL = 2 − j1 0.4 100 80 0.35 0.16 0.09 θ 45 r in degrees 50 70 0.34 0.41 110 40 1.0

55 0.9 0.17 1.2

0.08 0.8 λ towards generator 35 0.33 1.4 60

0.42 120 0.7 60 o) / Y 0.18 +jB 1.6

E ( 30 λ towards load 0.07 0.6 C 0.32 A N T 0.2 1.8 0.43 EP Γ 65 SC 50 130 SU

E 2.0 0.19 0.5 IV 0.06 IT 0.31 C 25 A 0.44 P A C 70 R O 0.4 40

), 0.2 140 0.4 o Z 0.05 / 0.3 jX 20 0.45 (+ T

75 N 3.0 E 0.6 N 0.21 O P

30 0.29 0.04 0.3 M 0.46 150 O C > E 0.8 15 — C 80 4.0 R N O A 0.22 T T 0.28 A C 1.0 R A E

E 0.47 5.0 R 1.0 N 0.2 20 E E

85 G 160 V I

T 0.8 10 D C R U 0.23 A 0.27

D A

N N A I 0.6 O 0.48 90 N T G

G S L

E L H

10 E T O

0.1 O

G F 170 0.4 F N T 0.24 0.26 R E R

A L E

E F N 0.49

L V

20 S

A M 0.2 C

W I T

S > 50 O I

— 0.25 0.25

C 0.1 0.2 0.3 1.6 1.8 2.0 0.9 0.4 0.5 0.6 0.8 1.0 1.2 1.4 0.7 3.0 4.0 5.0 10 50

20 0.25λ

C O

λ 0.0 0.0

0 180

F E

50 F

F I

RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) F

—

C < I

0.2 E I

D 20 E

0.24

T A

0.26

N L

Γ = ∠ − ° I

0.49 0.45 26.2

D E

R 0.1 0.4 G

-170 10 G

Move towards the generator R W

E R

E E

S 0.23 E

0.27

S ) 0.6

-90 o

0.48

B j

-10

N − ° (

E 26.2

0.8

-20

E L

E C

-85

-160

V 0.2 N 0.22

5.0

A 1.0 A 0.28

P 1.0

0.47

—

4.0

-80 -15

0.8 E

0.21 V

-30 I

T 0.29

0.04

-150 0.3 U

0.46 D

0.6 I

3.0

O -75

)

0.2

-20

/ 0.3

0.05

X -40 j

(

0.45 0.4 T

-140 0.4

-70 0.19 O

0.06

0.31

-25 C

0.44 C

-50

0.5 N

2.0

-130

0.18

0.2 -65 E

0.07 E

1.8

0.32 I

-30 C

0.43 0.6 P

1.6

-60

0.17

-60 -120

0.08 0.7

0.33

1.4 -35

0.42

0.8 0.16

-55

1.2 -70

0.09 -110 0.9

1.0 -40

0.34

-50

-45

0.15 0.41

-80 0.1

-100

-90

0.14 0.35

0.11

0.4

0.13 0.12

0.36

0.39

0.37 0.38

RADIALLY SCALED PARAMETERS RTN. LOSS [dB]SWR TOW A R D L OA D —> <— TOW A R D GEN ER A TOR ∞10040 20 10 4 53 2.5 2 1.8 1.6 1.4 1.2 1.1 1 15 10 7 4 35 2 1 ATTEN. [dB] RFL. COEFF, dBSP RFL. COEFF, E or I ∞40 30 20 15 10 8 1 1 12 3 4 1.1 5 1.2 6 1.3 1.4 1.6 1.8 2 3 4 5 10 20 ∞ S.W. LOSS COEFF

0 1 2 3 4 5 6 7 8 9 10 12 14 20 30∞ 0 0.1 0.2 0.4 0.6 0.8 1 1.5 2 3 4 5 6 10 15∞ RFL. LOSS [dB] 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.05 0.01 0 0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 2.5 3 4 5 10 ∞ S.W. PEAK (CONST. P)

1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 1 0.99 0.95 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 TRANSM. COEFF, P

CENTER 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 TRANSM. COEFF, E or I

ORIGIN Use this scale for Γ 0.447

Figure 2.47: Pointing out some of the scales found on the Smith chart.

2-76 2.10. THE SMITH CHART

2.10.2 Wave Impedance Transmission calculations using the Smith chart follow from the basic lossless line result that the reﬂection coefﬁcient looking into a length d line is

j 2ˇd j.r 2ˇd / d e e ; D D where as before ejr is the reﬂection coefﬁcient of the D j j load

– Think of as a vector in the -plane, with origin at the chart center .U; V / .0; 0/ D – To ﬁnd d you rotate the tip of the vector clockwise by ˇd degrees, as a negative angle in the complex plane D is indeed clockwise – In Smith chart terminology this is called moving towards the generator – Moving in the opposite direction, positive angle shift of the vector, is called moving towards the load

Another impact of the 2ˇd factor is that d =2 is one trip D around the chart (follows from =2 periodicity of tline input impedance)

Once d is found, it then follows that Z.d/ 1 d z.d/ C D Z0 D 1 d

2-77 CHAPTER 2. TRANSMISSION LINES

Additionally, since 1 S VSWR C j j D D 1 j traces out a constant VSWR circle in the -plane There are scales on the chart for directly reading S or the r > 1 axis scale can also be used (see next example)

Example 2.17: Find Zin

Suppose that ZL 50 j 50 and ﬁnd Zin and S for a line D C having electrical length of 45ı (recall ˇd ) and Z0 50 D D Normalize by Z0 50 we plot zL 1 j1 on the chart D D C Next we note the wavelengths toward generator value; here it is 0:162 Draw a line through the origin that passes through the wavelengths toward generator value of 0:162 0:125 0:287, C D this is angle location of d

Draw a circle centered on the origin that passes through zL and rotates clockwise to intersect the radial line passing through 0:287

The intersection point is z.d 0:125/ 2 j1 D D The value Zin Z0 z.0:125/ 100 j 50 D D The VSWR is read from the r > 1 axis to be 2.62; note the voltage minima can be read from the r < 1 axis

2-78 2.10. THE SMITH CHART

λ 8

0.12 0.13 0.11 0.14 0.162λ Γ 0.38 0.37 0.15 d Three outer scales: 0.1 0.39 0.36 90 zL = 1 + j1 0.4 100 80 0.35 0.16 0.09 θ 45 r in degrees 50 70 0.34 0.41 110 40 1.0

55 0.9 0.17 1.2

0.08 0.8 λ towards generator 35 0.33 1.4 60

0.42 120 0.7 60 o) / Y 0.18 +jB 1.6

E ( 30 λ 0.07 0.6 towards load N C 0.32 TA 0.43 EP 0.2 1.8 Γ 65 SC 50 130 SU

E 2.0 0.19 0.5 IV 0.06 IT 0.31 C 25 A 0.44 P A C 70 R O 0.4 40

), 0.2 140 0.4 o Z 0.05 / 0.3 jX 20 0.45 (+ T

75 N 3.0 E 0.6 N 0.21 O P

30 0.29 0.04 0.3 M 0.46 150 O C > E 0.8 15 — C 80 4.0 0.125λ R N O A 0.22 T T 0.28 A C 1.0 R A E

E 0.47 5.0 R 1.0 N 0.2 20 E Γ = ∠ ° E 0.447 63.4 85 G 160 V I

T 0.8 10 D C R U 0.23 A 0.27

D A

N N A I 0.6 O 0.48 90 N T G

G S L

E L H

10 E T O

0.1 O

G F 170 0.4 F N T 0.24 0.26 R E R

A L E

E F N 0.49

L V

20 S

A M 0.2 C

W I T

S > 50 O I

— 0.25 0.25

0.1 0.2 0.3 1.6 1.8 2.0 0.9 0.4 0.5 0.6 0.8 1.0 1.2 1.4 0.7 3.0 4.0 5.0 10 50 20

C O 0.0 0.0

180

F E

50 F

F I

RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) F

—

C < I

0.2 VSWR E I

D 20 E

0.24

T A

0.26

0.49

D E

R 0.1 0.4 G

-170 10

E R

E E

S 0.23 E

0.27

S ) 0.6

-90 o

0.48

j λ = −

-10 - N z(0.125 ) 2 j1

(

E 0.8

-20

E L

E C

-85

-160

V 0.2 N 0.22

5.0

A 1.0 A 0.28

P 1.0

0.47

—

4.0

-80 -15

0.8 E

0.21 V

-30 I

T 0.29

0.04

-150 0.3 U

0.46 D

0.6 I

3.0

O -75

)

0.2

-20

/ 0.3

0.05 λ + λ

X -40

j 0.162 0.125 -

(

0.45 0.4 T

-140 0.4

-70 0.19

O = 0.287λ P

0.06

0.31

-25 C

0.44 C

-50

0.5 N

2.0

-130

0.18

0.2 -65 E

0.07 E

1.8

0.32 I

-30 C

0.43 0.6 P

1.6

-60

0.17

-60 -120

0.08 0.7

0.33

1.4 -35

0.42

0.8 0.16

-55

1.2 -70

0.09 -110 0.9

1.0 -40

0.34

-50

-45

0.15 0.41

-80 0.1

-100

-90

0.14 0.35

0.11

0.4

0.13 0.12

0.36

0.39

0.37 0.38

1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 1 0.99 0.95 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 TRANSM. COEFF, P

CENTER 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 TRANSM. COEFF, E or I ORIGIN Use this scale for Γ

Figure 2.48: Finding z.d =8/ when zL 1 j1. D D C

2-79 CHAPTER 2. TRANSMISSION LINES

2.10.3 Impedance/Admittance Transformation

Converting an impedance to an admittance is simply a matter of rotating the normalized impedance value, z plotted on the chart by 180ı, then de-normalizing by multiplying the value found, y, by Y0 1=Z0 D

To prove this consider . . . (see book)

Load impedance zL

Load admittance yL

Figure 2-29FigurePoint A 2.49:represents Converting a normalizedz load zL0:6= 0.6 +j1:4j1.4.to Itsy corresponding0:25 normalizedj 0:6. admittance is y = 0.25 j0.6, and it is at point B. L L L − D C D 2-80 2.11. IMPEDANCE MATCHING

2.11 Impedance Matching

The Smith chart is often used to design impedance matching circuits

By impedance matching, we mean making the load ZL appear to be Z0, like in the design of the quarterwave transformer

Here the load impedance is in general complex

Feedline M A Zg + ~ V Z Z Matching Z g 0 in network L − M' A' Generator Load Figure 2.50: The general impedance matching problem. Figure 2-32 The function of a matching network is to transform the load impedance ZL such that the input impedance Zin looking into the network is equal to Z0 of the feedline. The circuit elements used n a matching network may be lumped L and C and/or lossless transmission line sections and/or open and short circuit stubs that attach to the through circuit

2-81 CHAPTER 2. TRANSMISSION LINES

Feedline M y(d2) λ/4 transformer A Feedline M A

Z0 Zin Z02 ZL A' Z Z Z Z M' 0 in L 0 L M' d2 A' (a) If ZL is real: in-series λ/4 transformer inserted at AA'

(d) In-parallel insertion of inductor at distance d2 Z(d) Feedline M λ/4 d A y (l ) B s 1 Feedline M d1 A Z01 Zin Z02 Z01 ZL B' M' A' Z0 Z0 ZL

(b) If ZL = complex: in-series λ/4 transformer inserted M' A' at d = dmax or d = dmin

Z0 y(d1) Feedline MA l1

Z0 Zin C Z0 ZL M' d1 A'

FigureFigure 2-332.51:Five A examples collection of in-series and of in-parallel matching matching circuits. networks.

Working through all the possible combinations is beyond the scope of this class

Here we will explore a few examples The goal in all cases is to drive the impedance seen looking into the matching circuit to be Z0 on the Smith chart, that is 0 j 0, also known as the match point D C The Smith always ﬁnd a graphical solution, but pure analytical solutions are also possible

For the network topologies chosen here there are two possible solutions per approach

2-82 Feedline M y(d2) λ/4 transformer A Feedline M A

Z0 Zin Z02 ZL A' Z Z L Z Z M' 2.11. IMPEDANCE MATCHING 0 in 0 L M' d2 A' (a) If Z is real: in-series λ/4 transformer inserted at AA' 2.11.1 Quarter-waveL Transformer Matching (d) In-parallel insertion of inductor at distance d2 Z(d) Feedline M λ/4 d A y (l ) B s 1 Feedline M d1 A Z01 Zin Z02 Z01 ZL B' M' A' Z0 Z0 ZL (b) If Z = complex: in-series λ/4 transformer inserted M' A' Figure 2.52:L A quarter-wave matching approach where d dmax or at d = d or d = d D dmin for complex-to-realmax conversion.min Z0 Analytical Solution y(d1) Feedline MA For the network topology shown above, originally in 2.50 cir- l 1 cuits (a), you start from the load, ZL and move toward the generatorZ0 Zuntilin either dmin orCdmax is reachedZ0 ZL

– Yes, there are twoM' possible solutionsd1 A'

(c) In-parallel– Recall: dinsertionmax corresponds of capacitor to the at ﬁrst distance voltage d1 maximum (e) In-parallel insertion of a short-circuited stub along the line and dmin corresponds to the ﬁrst voltage minimum along the line Figure 2-33 Five examples of in-series and in-parallel matching networks. From earlier analysis 8 Á rı < ; 0 ı < 180ı 720ı r dmax Á Ä D 1 rı : ; 180ı rı < 0 2 C 720ı Ä Â Ã 1 rı dmin D 4 C 720ı

2-83 CHAPTER 2. TRANSMISSION LINES

The next and ﬁnal step is ﬁnd the quarter-wave section characteristic impedance

Choose p p p Z02 Z01 Z.d/ Z01 z.d/Z01 Z01 z.d/ D D D The normalized impedance z.d/ corresponds to S = VSWR: for dmax it is S, while for dmin it is 1=S (this is very obvious on the Smith chart)

In summary: ( Z01pS; d dmax Z02 p D D Z01 1=S; d dmin D – Recall: S .1 /=.1 / D C j j j j

Example 2.18: A series RC Load

1/4 푇 = 푑 100 × 106 0.092 푇푑 = 6 푍02 100 × 10

Figure 2.53: LTspice schematic of the dmin solution.

2-84 2.11. IMPEDANCE MATCHING

The load consists of R 25 in series with C 60 pF D D capacitor

The match is to be made at f 100 MHz which has a period D of 10 ns

Solution 1: Choose the dmin case and use Python to obtain the numerical values

Figure 2.54: Solution 1 calculation in a Jupyter notebook. The design values found above are used in the original LTspice schematic shown earlier

A frequency response plot of 20 log in is obtained and com- 10 j j pared with 20 log , which is the no matching case 10 j j – Note: This quantity is related to the Return Loss of a one- port network

2-85 CHAPTER 2. TRANSMISSION LINES

– Formally,

Return Loss RL 20 log (dB) D Á 10 j j – Loss is always a positive quantity, but in LTspice it is more convenient to plot the negative of RL

Here the 20 dB RL bandwidth is found to be 10:34 MHz The 푑min Solution

20 log10 |Γ| = -Return Loss (dB)

20 dB return-loss Bandwidth ~10.34 MHz 20 log10 |Γin| = -Return Loss (dB)

Figure 2.55: Frequency response plot of solution 1 as the negative of the Return Loss

At 100 MHz the match is perfect as RT approaches (in the- 1 ory with no component error)

Without any matching circuit RT is about 5 dB, very poor

2-86 2.11. IMPEDANCE MATCHING

Moving on to solution 2, the next three ﬁgures show the design, the ﬁnal schematic, and the frequency response

Figure 2.56: Solution 2 calculation in a Jupyter notebook.

1/4 푇 = 푑 100 × 106 0.342 푇푑 = 6 푍02 100 × 10

Figure 2.57: Solution 2 LTspice schematic for quarter-wave matching.

2-87 CHAPTER 2. TRANSMISSION LINES

The 푑max Solution

20 dB return-loss Bandwidth ~5.21 MHz

20 log10 |Γin| 20 log10 |Γ| = -Return Loss (dB) = -Return Loss (dB)

Figure 2.58: Frequency response plot of solution 2 as the negative of the Return Loss

Here the RL bandwidth is only 5.21 MHz; likely due to dmax being a =2 longer line section

Smith Chart Solution

The tutorial of text Module 2.7 does a great job of explaining the steps

2-88 2.11. IMPEDANCE MATCHING

Figure 2.59: Quarter-wave match Smith chart: Applet set-up

Figure 2.60: Quarter-wave match Smith chart: Step 1

2-89 CHAPTER 2. TRANSMISSION LINES

Figure 2.61: Quarter-wave match Smith chart: Solution 1

Figure 2.62: Quarter-wave match Smith chart: Step 2

2-90 2.11. IMPEDANCE MATCHING

Figure 2.63: Quarter-wave match Smith chart: Step 3

Figure 2.64: Quarter-wave match Smith chart: Step 4

2-91 CHAPTER 2. TRANSMISSION LINES

Figure 2.65: Quarter-wave match Smith chart: Step 5

Figure 2.66: Quarter-wave match Smith chart: Step 6

2-92 2.11. IMPEDANCE MATCHING

Figure 2.67: Quarter-wave match Smith chart: Solution 2 set-up

Figure 2.68: Quarter-wave match Smith chart: Solution 2

2-93 Feedline M y(d2) λ/4 transformer A Feedline M A

Z0 Zin Z02 ZL A' Z Z Z Z M' 0 in L 0 L M' d2 A' (a) If ZL is real: in-series λ/4 transformer inserted at AA'

(d) In-parallel insertion of inductor at distance d2 Z(d) Feedline M λ/4 d A y (l ) B s 1 Feedline M d1 A Z01 Zin Z02 Z01 ZL B' M' A' Z0 Z0 ZL CHAPTER 2. TRANSMISSION LINES (b) If ZL = complex: in-series λ/4 transformer inserted M' A' at d = dmax or d = dmin 2.11.2 Lumped Element Matching Z0 y(d1) Feedline MA l1

Z0 Zin C Z0 ZL M' d1 A'

(c)Figure In-parallel 2.69: Lumpedinsertion element of capacitor matching at usingdistance a shunt d1 C (e) In-parallel insertion of a short-circuited stub Feedline M y(d2) λ/4 transformer A Feedline FigureM 2-33 Five examplesA of in-series and in-parallel matching networks.

Z0 Zin Z02 ZL A' Z Z Z Z M' 0 in L 0 L M' d2 A' (a) If ZL is real: in-series λ/4 transformer inserted at AA'

(d)Figure In-parallel 2.70: Lumpedinsertion element of inductor matching at distance using a shuntd2 L Z(d) In 2.50 circuits (c) and (d) consider a special case of lumped Feedline M λ/4 d A element matching,y (l which) employs a series transmission line B s 1 Feedlineconnected to the loadM with a shunt Ldor1 C across theA line Z01 Zin Z02 Z01 ZL As shunt elements are added across the through line, all of the B' analysis is done using admittances since admittances add in a M' A' Z0 Z0 ZL parallel circuit, i.e., referring to either 2.69 or refﬁg:num37c (b) If Z = complex: in-series λ/4 transformer inserted M' A' L 1 at d = d or d = d Zin max min D 1 1 Zd Zs C Z Yin Yd Ys0 y(d1) D C

Feedline MA2-94 l1

Z0 Zin C Z0 ZL M' d1 A'

Figure 2-33 Five examples of in-series and in-parallel matching networks. 2.11. IMPEDANCE MATCHING

where Yd is the admittance looking into the series line of length d1 or d2 and Ys is the admittance of either a shunt C, i.e., YC D j 2f C , or a shunt L, i.e., Ys 1=.j 2fL/ D

In practice Yd Gd jBd is complex (a conductance and a D C susceptance), we will force to be real shortly, and Ys is pure imaginary (inductor or capacitor), so Ys jBs D

The objective is to force Yin Y0 j 0 or in normalized form D C yin 1 j 0, meaning that D C

gd 1 real-part condition D bs bd imaginary-part condition D

where gd gD=Y0, bs Bs=Y0, and bd Bd =Y0 D D D The feedline is now matched at a single frequency The real-part condition is met by choosing the series line length d1 or d2 to make Yd1 or Yd2 respectively lie on the unit conductance circle of the Smith chart (this is always possible for a passive load)

Once on the unit conductance circle, the remaining positive or negative susceptance is cancelled by the shunt L or C negative or positive susceptance respectively

A fully analytic solution, is described in the text The key result of the analytical solution is that cos r 2ˇd ; D j j 2-95 CHAPTER 2. TRANSMISSION LINES

which ﬁnds d and

2 sin.r 2ˇd b j j s 2 D 1 2 cos r 2ˇd C j j C j j C

which ultimately ﬁnds L or C

For the upcoming example, the focus is on a Smith chart solution, as it is more straightforward and practical for working quiz and exam problems (not as accurate however)

Example 2.19: Series Line Shunt L or C Matching to a Series RC Load

Once again the load is R 25 in series with C 60 pF D D operating at 100 MHz

A Smith chart solution is formulated by following along with the tutorial of Module 2.8.

2-96 2.11. IMPEDANCE MATCHING

Figure 2.71: Lumped match Smith chart: Applet set-up

Figure 2.72: Lumped match Smith chart: Step 1

2-97 CHAPTER 2. TRANSMISSION LINES

Module 2.8 Tutorial Instructions ,color Discrete Element Matching I

Go Back Next Step Back to Start

Step 2 - Plot SWR circle with radius r = I r I - Identifynormalized load admittance 0.5 - j 0.53052 = = • ZL Y L GL + j BL 0.018817 + j 0.019965 S rL 0.45819 L -H3.8263 ° 0 YL 0.94083 + j 0.99825 0 z (d) 0.5 - j 0.53052 = = YL gL + j b l 0.94083 + j 0.99825 r(d) = 0.45819 L -113.8.263 ° • y (d) = 0.94083 + j 0.99825 (' d = 0.0 11 2� d = 0.0 rad= 0.0° ("' 0.5 11 - d = 0.5 11 2� (0.5 11 - d) = 6.2832 rad= 360.0°

Figure 2.73: Lumped match Smith chart: Step 2

Figure 2.74: Lumped match Smith chart: First solution

2-98 2.11. IMPEDANCE MATCHING

Figure 2.75: Lumped match Smith chart: Step 3

Figure 2.76: Lumped match Smith chart: Step 4

2-99 CHAPTER 2. TRANSMISSION LINES

Figure 2.77: Lumped match Smith chart:Second solution Step 3

Figure 2.78: Lumped match Smith chart:Second solution Step 3

2-100 2.11. IMPEDANCE MATCHING

Figure 2.79: Lumped match Smith chart:Second solution Step 4

Figure 2.80: Lumped match Smith chart:Summary

2-101 CHAPTER 2. TRANSMISSION LINES

Remaining questions?

2-102 Feedline M y(d2) λ/4 transformer A Feedline M A

Z0 Zin Z02 ZL A' Z Z Z Z M' 0 in L 0 L M' d2 A' (a) If ZL is real: in-series λ/4 transformer inserted at AA' 2.11. IMPEDANCE MATCHING

(d) In-parallel insertion of inductor at distance d2 Z(d) 2.11.3 Single-Stub Matching Feedline M λ/4 d A y (l ) B s 1 Feedline M d1 A Z01 Zin Z02 Z01 ZL B' M' A' Z0 Z0 ZL

(b) If ZL = complex: in-series λ/4 transformer inserted M' A' at d = dmax or d = dmin

Z0 y(d1) Feedline MA l1

Z0 Zin C Z0 ZL M' d1 A'

(c) In-parallel insertion of capacitor at distance d1 Figure(e) In-parallel2.81: Stub matchinginsertion usingof a short-circuited a shunt short circuit stub stub or an open-circuit stub (not shown). Figure 2-33 Five examples of in-series and in-parallel matching networks. In 2.50 circuit (e) considers a special case of open/short circuit single stub matching, which employs a series transmission line connected to the load with a shunt open or short circuit stub transmission line shunted across the line

The design steps are similar to the lumped element design pre- sented earlier

The main difference is the shunt L and C values are replaced with shunt open or short circuit stubs

2-103 CHAPTER 2. TRANSMISSION LINES

– For the case an open circuit stub the admittance is a pure susceptance Bs jY0 tan ˇl and the short circuit stub is D also a pure susceptance Bs jY0 cot ˇl D

The tutorial of text Module 2.9 does a great job of explaining the steps

2-104 2.11. IMPEDANCE MATCHING

Example 2.20: Series Line Single Stub Matching to a Series RC Load

Once again the load is R 25 in series with C 60 pF D D operating at 100 MHz

A Smith chart solution is formulated by following along with the tutorial of Module 2.9.

One item missing is?

Figure 2.82: Stub match Smith chart: Applet set-up

2-105 CHAPTER 2. TRANSMISSION LINES

Figure 2.83: Stub match Smith chart: Step 1

Figure 2.84: Stub match Smith chart: Step 2

2-106 2.11. IMPEDANCE MATCHING

Figure 2.85: Stub match Smith chart: First Solution

Figure 2.86: Stub match Smith chart: Step 3

2-107 CHAPTER 2. TRANSMISSION LINES

Figure 2.87: Stub match Smith chart: Step 4

Figure 2.88: Stub match Smith chart: Solution 2 Step 3

2-108 2.11. IMPEDANCE MATCHING

Figure 2.89: Stub match Smith chart: Solution 2 Step 4

Figure 2.90: Stub match Smith chart: Summary

2-109 CHAPTER 2. TRANSMISSION LINES

2.12 Transients on Transmission Lines

Transmission lines carry more than pure sinusoidal signals Many applications of transmission lines do involve sinusoidal- based signals, such communications waveforms where the amplitude, frequency, or phase of a sinusoid is a made a function of a message signal, analog or digital

The high frequency microwave/radio frequency sinsuoid serves as a carrier of the information

Still, the spectral bandwidth of these modulated carrier waveforms is small relative to the carrier frequency

True wideband signals or so-called baseband digital waveforms are clocked at 100 MHz and above, e.g., the PCB tracks forming a data bus on PC motherboard or the data bus of a ﬂexible PCB on an array of disk drives in storage rack

This is where time-domain of transmission lines comes to the stage

Time domain modeling may even resort to Laplace transform methods and will make use of simulations using LTspice in the .tran

Two useful signals are the unit step function u.t/ and a rectangular pulse of duration s

– A step signal truning on at t 0 takes the form D

V .t/ V0u.t/ D 2-110 2.12. TRANSIENTS ON TRANSMISSION LINES

– A pulse of duration s that turns on at t 0 takes the D form

V .t/ V0u.t/ V0u.t / D where in both cases u.t/ is the unit step function deﬁned as ( 1; t > 0 u.t/ D 0; t < 0

V(t) V(t)

V1(t) = V0 u(t) V0 V0

t τ t τ

V2(t) = −V0 u(t − τ)

(a) Pulse of duration τ (b) V(t) = V1(t) + V2(t)

Figure 2-39FigureArectangularpulse 2.91: FormingV (t) of duration a pulseτ can be signal represented as as the the sum differnce of two step functions between of opposit twoepolarities displaced by τ relative to each other. step functions skewed in turn-on time by s.

2.12.1 Response to a Step Function

To get started with transient modeling, consider a single section of transmission line with resistive load and source terminations

2-111 CHAPTER 2. TRANSMISSION LINES

R t = 0 g Transmission line + Vg Z0 RL −

z z = 0 z = l (a) Transmission-line circuit

+ Rg I1 + + Vg V + Z0 − 1 − Figure 2.92:(b) Circuit Equivalent model circuit for at transient t = 0+ analysis.

BigFigure change 2-40: TheAt sourcet = 0+,immediatelyafterclosingthe end is now taken as z 0 and the D loadswitch end in is thez circuitl in (a), the circuit can be represented by the equivalentD circuit in (b). Upon closing the switch (step function u.t/ turns on) incident voltage, V1C, and current, I1C, waves are launched on the line The boundary conditions dictate that only the line impedance Z0 is all that is seen by the generator at t 0, so D Vg I1C D Rg Z0 C VgZ0 V1C D RgZ0

2-112 2.12. TRANSIENTS ON TRANSMISSION LINES

– The subscript 1 on V C and I C denotes the initial voltage and current wave launched onto the line in the z direc- C tion

2.12.2 Dr. Wickert’s Archive Notes At this point we jump to some pages taken from the intro chapter to ECE 4250/5250

The title of this chapter is Transmission Theory Review, so don’t be too worried about the course number

2-113 Chapter Review of Transmission Line 1 Theory

• Transmission lines and waveguides are used to transport electromagnetic energy at microwave frequencies from one point in a system to another • The desirable features of a transmission line or waveguide are: – Single-mode propagation over a wide band of frequencies – Small attenuation • The transmission line structures of primary interest for this course are those for which the dominant mode of propagation is a transverse electromagnetic (TEM) wave • Recall that for TEM waves the components of electric and magnetic fields in the direction of wave propagation are zero • We wish to consider transmission lines which consist of two or more parallel conductors which have axial uniformity • That is to say their cross-sectional shape and electrical properties do not vary along the axis of propagation

ECE 5250/4250 Microwave Circuits 1–1 Review of Transmission Line Theory

Figure 1.1: Transmission lines with axial uniformity

• The TEM wave solution for axially uniform transmission lines can be obtained using: – Field Analysis - (obtain electric and magnetic field waves analogous to uniform plane waves) – Distributed-Circuit Analysis - (obtain voltage and current waves) • Distributed circuit analysis will be at the forefront of all analysis in this course, in particular consider Pozar1, “Modern microwave engineering involves predominantly distributed circuit analysis and design, in contrast to the waveguide and field theory orientation of earlier generations”

1.David Pozar, Microwave Engineering, 3rd edition, John Wiley, New York, 2005.

1–2 ECE 5250/4250 Microwave Circuits Distributed Circuit Analysis Distributed Circuit Analysis

Figure 1.2: Distributed element circuit model

Parameters: L – series inductance per unit length due to energy storage in the magnetic field C – shunt capacitance per unit length due to energy storage in the electric field R – series resistance per unit length due to power loss in the conductors G – shunt conductance per unit length due to power loss in the dielectric. (i.e. !!"= – j!#!#% $ 0 ) • Using KCL, KVL and letting &z ' 0 it can be shown that –(v)* z% t (i)* z% t ------= Ri)* z% t + L------(1.1) (z (t and – i z t (v)* z% t ------( )*% - = Gv)* z% t + C------(1.2) (z (t

ECE 5250/4250 Microwave Circuits 1–3 Review of Transmission Line Theory

• For now assume the line is lossless, that is R = 0 and G = 0, so we have: v (i –-----( = L---- (z (t i (v (1.3) –-----( = C----- (z (t • Now differentiate the first equation with respect to z and the second with respect to time t 2 2 v ( i ------( = –L------2 (z (t(z 2 2 (1.4) i ( v ------( - = C------(z(t (t2 • Combine the two resulting equations to get 2 2 v ( v ------( = LC------(voltage eqn.) (1.5) (z2 (t2 similarly obtain 2 2 ( i ( i ------= LC------(current eqn.) (1.6) (z2 (t2 • These are in the form of the classical one-dimensional wave equation, often seen in the form 2 2 y 1 ( y ------( = ------(1.7) 2 2 2 (z +p (t

where +p has dimension and significance of velocity

1–4 ECE 5250/4250 Microwave Circuits Distributed Circuit Analysis

• A well known solution to the wave equation is + z – z yy= ./t – ----- + y ./t + ----- (1.8) ,-,- +p +p –y+ propagates in the positive z direction – y– propagates in the negative z direction • This solution can be checked by noting that 0 0 0 y (y (x 21(y z ------( - =------=1------% x = t0----- (1.9) (z (x (z +p (x +p • A solution for the voltage wave equation is thus + z – z v)* z% t = v ./t – ----- + v ./t + ----- (1.10) ,-+p ,-+p • The current equation can be written in a similar manner, but + – it can also be written in terms of V and V since –(v (i ------= L---- (1.11) (z (t or v+ v– (i – ------( + ------( = L---- (1.12) (z (z (t

Letting x= t0 z 3 +p and using the chain rule the above equation becomes + – –1(v 1 (v (i – ------+ ------= L---- (1.13) +p (x +p (x (t

ECE 5250/4250 Microwave Circuits 1–5 Review of Transmission Line Theory

• This implies that 1 + z – z i)* z% t = ------v ./t – ----- – v ./t + ----- (1.14) ,-,- L+p +p +p or 1 + z – z i)* z% t = ----- v ./t – ----- – v ./t + ----- (1.15) ,-,- Z0 +p +p where 1 +p = ------% (velocity of propagation) LC (1.16) L Z = ----% (characteristic impedance) 0 C • At this point the general lossless line solution is incomplete. The functions v+ and v- are unknown, but must satisfy the boundary conditions imposed by a specific problem • The time domain solution for a lossless line, in particular the analysis of transients, can most effectively be handled by using Laplace transforms • If the source and load impedances are pure resistances and the source voltage consists of step functions or rectangular pulses, then time domain analysis is most convenient • In the following we will first consider resistive load and source impedances, later the analysis will be extended to complex impedance loads using Laplace transforms

1–6 ECE 5250/4250 Microwave Circuits Transient Analysis with Resistive Loads

• The Laplace transform technique will in theory allow for a generalized time-domain analysis of transmission lines • In Section 2 sinusoidal steady-state analysis will be intro- duced. This approach offers greatly reduced analysis com- plexity

Transient Analysis with Resistive Loads

Infinite Length Line

Figure 1.3: Infinite length line circuit diagram

• Assume that the line is initially uncharged, i.e. for z 4 0 and t 5 0 + z – z v)* z% t ==v ./t – ----- + v ./t + ----- 0 (1.17) ,-,- +p +p and 1 + z – z i)* z% t ==----- v ./t – ----- – v ./t + ----- 0 (1.18) ,-,- Z0 +p +p the above equations imply that + z v ./t – ----- = 0 for tz– 3 + 6 0 (1.19) ,- p +p

ECE 5250/4250 Microwave Circuits 1–7 Review of Transmission Line Theory

and – z v ./t + ----- = 0 for all t (1.20) ,- +p + Note: For the given initial conditions only v )*tz– 3 +p can exist on the line. • We thus conclude that

+ z : v)* z% t = v ./t – ----- ,-8 +p 8 z 9 for all t – ----- ; 0 (1.21) 1 + z 8 +p i)* z% t = -----v ./t – ----- Z ,-8 0 +p 7

+ • Suppose that at t = 0 a voltage source vg)*t is

applied through a source resistance Rg , at z = 0 • Apply Ohm's law at z = 0 for t > 0 and we obtain

vg)*t – v)*0% t = i)*0% t Rg (1.22) or

+ Rg + vg)*t – v )*t = ------v )*t (1.23) Zg which implies

+ Z0 v )*t = ------vg)*t (1.24) Z0 + Rg

1–8 ECE 5250/4250 Microwave Circuits Transient Analysis with Resistive Loads

• The final result is that under the infinite line length assumption for any z we can write

Z0 z v)* z% t = ------v ./t – ----- (1.25) g,- Z0 + Rg +p 1 z i)* t% z = ------v ./t – ----- (1.26) g,- Z0 + Rg +p – Note: That the infinite length of line appears as a voltage divider to the source – Voltages and currents along the line appear as replicas of

the input values except for the time delay z 3 +p

Terminated Line

Figure 1.4: Terminated line circuit diagram

• Note: As a matter of convenience the reference point z = 0 has been shifted to the load end of the line • Suppose that a wave traveling in the z+ direction is incident

upon the load, RL , at z = 0 • Thus, + z 1 + z v)* z% t ==v ./t – ----- and i)* z% t -----v ./t – ----- (1.27) ,- ,- +p Z0 +p

ECE 5250/4250 Microwave Circuits 1–9 Review of Transmission Line Theory

• By applying Ohm’s law at the load, we must have

v)*0% t = RLi)*0% t (1.28) • This condition cannot, in general, be met by the incident wave alone since

v)* z% t = Z0i)* z% t (1.29) • Since the line was initially discharged, it is reasonable to

assume that a fraction,

+ – + v )*t v )*t )*1 + < v )*t i)*0% t ==------– ------L - (1.32) Z0 Z0 Z0 • To satisfy Kirchoff’s laws,

Net load voltage 1 +

RL – Z0

1–10 ECE 5250/4250 Microwave Circuits Transient Analysis with Resistive Loads

– Note:

RL = Z0 .

General Transmission Line Problem

Figure 1.5: Circuit for general transmission line problem

• From earlier analysis, we know that for 0 55tl3 +p = Tl , : Z0 z v)* z% t = ------v ./t – ----- 8 g,- Z0 + Rg +p 8 9 0 55tTl (1.35) 1 z 8 i)* z% t = ------v ./t – ----- Z + R g,-8 0 g +p 7

• When tT= l the leading edge of vg)*t has traveled to the load end of the line (zl= )

• Assuming RL $ Z0 a reflected wave now returns to the source during the interval Tl 5 t 6 2Tl

Z0 z Z0 2l z v)* z% t = ------v ./t – ----- + ------< v ./t – ----- + ----- g,- L g,- Z0 + Rg +p Z0 + Rg +p +p

= 8 8 + 8 > 8 8 8 ? = 8 8 8 – 8 > 8 8 8 8 ? v )*tz– 3 + v )*tz+ 3 + p p (1.36)

ECE 5250/4250 Microwave Circuits 1–11 Review of Transmission Line Theory

and 1 z 1 2l z i)* z% t = ------v ./t – ----- – ------< v ./t – ----- + ----- (1.37) g,- L g,- Z0 + Rg +p Z0 + Rg +p +p – • When the load reflected wave v )*tz+ 3 +p arrives at the source (z = 0 ), a portion of it will be reflected towards the

load provided Rg $ Z0 • The reflection that takes place is independent of the source voltage • The wave traveling in the positive Z direction after

2l 3 +p = 2Tl seconds has elapsed, can be found by applying Ohm's law for z = 0 and t = 2Tl :

vg)*2Tl – v)*02% Tl = Rgi)*02% Tl (1.38) • Now substitute + – v)*02% Tl = v )*2Tl + v )*2Tl 1 + – (1.39) i)*02% Tl = -----@Av )*2Tl – v )*2Tl Z0 + and solve for v )*2Tl • The results is

+ Z0 – Rg–Z0 v )*2Tl = vg)*2Tl ------+ v )*2Tl ------Z0 + Rg Rg + Z0

= 8 8 > 8 8 ? = 8 8 > 8 8 – ? incident part of vg)*t reflected portion of v )*2Tl Z Z = v )*2T ------0 + v )*0 ------0 < < g l Z + R g Z + R L g 0 g 0 g (1.40)

1–12 ECE 5250/4250 Microwave Circuits Transient Analysis with Resistive Loads

where

Rg – Z0

Note:

2Tl 5 t 6 3Tl , thus consists of the original source signal plus a reflected component due to mismatches at both the load and source ends of the line

Z0 z z – 2l v)* z% t = ------v ./t – ----- + < v ./t + ------g,-L g,- Z0 + Rg + + p p (1.42) z + 2l + < < v ./t – ------% 2T 5 t 6 3T g L g,-l l +p and 1 z z – 2l i)* z% t = ------v ./t – ----- –< v ./t + ------g,-L g,- Z0 + Rg +p +p (1.43) z + 2l + < < v ./t – ------% 2T 5 t 6 3T g L g,-l l +p • The process of reflections occurring at both the source and load ends continues in such a way that in general over the nth

time interval, )*n – 1 Tl 55t nTl , the v)* z% t and i)* z% t solutions each require n terms involving vg)*t

ECE 5250/4250 Microwave Circuits 1–13 Review of Transmission Line Theory

Example: Consider the following circuit.

Figure 1.6: Transmission line circuit

Here we have

Z0 1 1 ------=---%

a) Find V)*0% t and I)*0% t for Tp = 1Bs

b) Find V)*0% t and I)*0% t for Tp = 6Bs • For both parts (a) and (b) the basic circuit operation is as follows: i) An 8v pulse will propagate toward the load, reaching the load in 2 ms. ii) A -8v pulse will be reflected from the load, requiring 2 ms to reach the source. iii) A -4v pulse will be reflected at the source. It will take 2 ms to reach the load. iv) The process continues. • The +z and -z direction propagating pulses can be displayed on a distance-time plot or bounce diagram (see Figure 1.7)

1–14 ECE 5250/4250 Microwave Circuits Transient Analysis with Resistive Loads

– In the bounce diagram the boundaries at z = 0 and z = 400 m are represented as surfaces with reflection coefficient of 1/2 and -1 respectively. – To obtain the voltage waveform at say z = 380 m as a function of time, you sum the contributions from the various +z and -z traveling waves – For pulses that are short in comparison with the one-way delay time of the line, only at most two wave terms need to be included at a time. – For long pulses (in the limit say a step function) all wave terms need to be included

Figure 1.7: Bounce diagram showing pulse propagation

ECE 5250/4250 Microwave Circuits 1–15 Review of Transmission Line Theory

a)Tp = 1Bs : at t = 4Bs we have v)*04% Bs ==– 4 + )*–8 –12v. 1 i)*04% Bs ==------@A– 4 – )*–8 0.08A. 50

b)Tp = 6Bs : at t = 4Bs we have v)*04% Bs ==)*84+ )*– + )*–8 –4v. 1 i)*04% Bs ==------@A)*84+ )*– – )*–8 0.24A. 50 • We can simulate this result using the Agilent Advanced Design System (ADS) software • In this example we will use ideal transmission line elements

Current measurement A short circuit means

Source I_Probe R R I_Probe1 TLIND R2 R1 TLD1 R=0 Ohm R=150 Ohm Z=50.0 Ohm VtPulse Delay=2 usec SRC1 ‘TLIND’ allows the line Vlow=0 V to be spec’d in terms of t Vhigh=32 V propagation delay Delay=0 nsec Edge=linear TRANSIENT Rise=50 nsec Fall=50 nsec Tran Width=1 usec Tran1 Period=20 usec StopTime=20.0 usec MaxTimeStep=50 nsec

Figure 1.8: Circuit schematic (Example1.dsn)

1–16 ECE 5250/4250 Microwave Circuits Transient Analysis with Resistive Loads

• The source end voltage, v)*0% t 10 Tp = 1Bs 5 (v) 0 ource

S -5

-10

-15 02 4 6 8 10 12 14 16 18 20 time, usec • The source end current entering the line, i)*0% t 0.20

Tp = 1Bs 0.15 (A) i . 0.10 1 e b ro P

_ 0.05 I

0.00

-0.05 02 4 6 8 10 12 14 16 18 20 time, usec • Modify the schematic by increasing the pulse width to 6 Bs

ECE 5250/4250 Microwave Circuits 1–17 Review of Transmission Line Theory

• The source end voltage, v)*0% t 10 Tp = 6Bs

(v) 0 ource

S -5

-10

-15 02 4 6 8 10 12 14 16 18 20 time, usec • The source end current entering the line, i)*0% t

0.25

Tp = 6Bs 0.20

(A) 0.15 i . 1 e

b 0.10 ro P _ I 0.05

0.00

-0.05 02 4 6 8 10 12 14 16 18 20 time, usec

1–18 ECE 5250/4250 Microwave Circuits CHAPTER 2. TRANSMISSION LINES

Example 2.21: LTspice Extension to ADS Example

ADS (from Keysight) is a far more complex tool than LTspice

This example repeats and then extends some of the content of the ADS example

The schematic is shown below for pulse width of 1s

Figure 2.93: LTspice schematic of a simple transient response example.

We can plot the total voltage, incident, and reﬂected voltages at z 0 D

By breaking out the incident and reﬂected voltage waves the math becomes clear

2-114 2.12. TRANSIENTS ON TRANSMISSION LINES

Total voltage: , |

𝑣𝑣 𝑧𝑧 𝑡𝑡 𝑧𝑧=0

Incident voltage: / | + 𝑣𝑣 𝑡𝑡 − 𝑧𝑧 𝑢𝑢𝑝𝑝 𝑧𝑧=0

Reflected voltage: + / | − 𝑣𝑣 𝑡𝑡 𝑧𝑧 𝑢𝑢𝑝𝑝 𝑧𝑧=0

Figure 2.94: Voltage at z 0 for a 1s pulse launched on the line. D

2-115 Transient Analysis with Resistive Loads

Transient Analysis using Laplace Transforms

Figure 1.9: Lossless line with arbitrary terminations • In the time domain we know that for a lossless line + z – z v)* z% t = v ./t – ----- + v ./t + ----- (1.45) ,-,- +p +p and 1 + z – z i)* z% t = ----- v ./t – ----- – v ./t + ----- (1.46) ,-,- Z0 +p +p + – where v )*tz– 3 +p and v )*tz+ 3 +p are determined by the boundary conditions imposed by the source and load • Laplace transform each side of the above equations with respect to the time variable, using the time shift theorem which is given by –sc LCDf)* t– c = Fs)*e (1.47) where Fs)* is the laplace transform of ft)* • The result is + –sz 3 + – sz 3 + v)* z% s = v )*s e p + v )*s e p (1.48) and

ECE 5250/4250 Microwave Circuits 1–19 Review of Transmission Line Theory

1 + –sz 3 +p – sz 3 +p i)* z% s = -----@Av )*s e – v )*s e (1.49) Z0 + + – – where v )*s = LCDv )*t and v )*s = LCDv )*t

Case 1: Matched Source

• For the special case of Zg)*s = Z0 , the source is matched to the transmission line which eliminates multiple reflections • Thus, we can write

+ Z0 1 V )*s ==Vg)*s ------Vg)*s (1.50) Z0 + Z0 2 at zl= , the incident wave is reflected with the reflection coefficient

ZL)*s – Z0

+ –sl 3 +p V)* l% s = V )*s e @A1 +

– –s2l 3 +p + V )s *=

1 –sz 3 +p –s)2lz– *3 +p V)* z% s = ---V )*s @Ae + < )*s e 2 g L (1.54) 1 –sz 3 +p 1 –s2l 3 +p sz 3 +p = ---V )*s e + ---< )*s e V )*s e 2 g 2 L g

1–20 ECE 5250/4250 Microwave Circuits Transient Analysis with Resistive Loads

• To obtain V)* z% t inverse transform: 1 z –1 v)* z% t = ---v ./t – ----- + @A< )*s V )*s (1.55) g,-L L g 2 +p t' t+ ) z – 2l *3 +p

Example:

• Let vg)*t = v0ut)* and ZL be a parallel RC connection

Figure 1.10: Parallel RC circuit

Find: v)* z% t • To begin with in the s-domain we can write 1 R------Cs R Z )*s ==------(1.56) L 1 1 RCs R + ------+ Cs and R ------– Z Z )*s – Z 1 + RCs 0 < )*s ==------L 0------L Z s Z R L)*+ 0 ------+ Z 1 + RCs 0 RZ– ------0 – s (1.57) RCZ bs– RZ+ RZ– = ------0 = ------% a = ------0% b = ------0 RZ– as+ RCZ RCZ ------0 + s 0 0 RCZ0

ECE 5250/4250 Microwave Circuits 1–21 Review of Transmission Line Theory

• Now since Vg)*s ==v0LCDut)* v0 3 s

v –sz 3 +p bs– –s)2lz– *3 +p V)* z% s = -----0 e + ------e (1.58) 2s as+ • To inverse transform first apply partial fractions to bs– K K ------= ------1 + ------2 (1.59) s)* a+ s s sa+ Clearly,

b RZ– 0 –)*ab+ –2R K1 ==------K2 ==------(1.60) a RZ+ 0 a RZ+ 0 so

v 1 –sz 3 +p V)* z% s = ----0- ---e 2 s

s)*2lz– (1.61) RZ– –------=:0 1 2R 1 +p + >9------E --- – ------E ------e ?7RZ+ 0 s RZ+ 0 sa+

and

–1 v0 z =RZ– 0 v)* z% t ==L CDV)* z% s ----- ut./– ----- + ------,-> 2 +p ?RZ+ 0

2lz– RZ+ 0 (1.62) –./t – ------,- – 2R +p RCZ0 : 2lz– ------ut./– ------e 9 ,- RZ+ 0 7 +p

1–22 ECE 5250/4250 Microwave Circuits Transient Analysis with Resistive Loads

• As a special case consider zl=

l RZ+ 0 –./t – ------,-RCZ v0 RZ– 2R +p 0 l v)* z% t = ----- 1 + ------0 – ------e ut./– ----- ,- 2 RZ+ 0 RZ+ 0 +p (1.63) RZ+ –./t – ----l------0 ,-RCZ R +p 0 l = v ------1 – e ut./– ----- 0 ,- RZ+ 0 +p

Figure 1.11: Sketch of the general v)* l% t waveform

Example: The results of an ADS simulation when using v0 = 2v, Z0 ==R 50 ohms, C = 50 pf, and Tl = 5 ns is shown below

Figure 1.12: ADS Circuit diagram with nodes numbered • The ADS schematic is shown below

ECE 5250/4250 Microwave Circuits 1–23 Review of Transmission Line Theory

Source Load R C R TLIND R2 R1 TLD1 C1 R=50 Ohm C=50 pF R=50 Ohm Z=50.0 Ohm VtPulse Delay=5 nsec SRC1 Vlow=0 V t Vhigh=2 V Delay=0 nsec Edge=linear TRANSIENT Rise=0.1 nsec Fall=0.1 nsec Tran Width=1 usec Tran1 Period=1 usec StopTime=20.0 nsec MaxTimeStep=0.1 nsec • Next we plot the source and load end waveforms

1.0 load end source end 0.8

0.6 d oa L

Source 0.4

0.2

0.0 02 4 6 8 10 12 14 16 18 20 time, nsec

1–24 ECE 5250/4250 Microwave Circuits Transient Analysis with Resistive Loads

Case 2: Mismatched Source Impedance

• For the general case where Zg)*s $ Z0 to reflections occur at the source end of the line as well as at the load end • The expression for V)* z% s now will consist of an infinite number of terms as shown below: Z 0 –sz 3 +p –s)2lz– *3 +p V)* z% s = Vg)*s ------@e +

–s)2lz+ *3 +p +

–s)4lz– *3 +p +

ZL)*s – Z0 Zg)*s – Z0

V )*s Z –sz 3 +p G n n –s)*2n l 3 + V)* z% s = ------g 0 e < )s *< )*s e p Z + Z )*s H L g 0 g n = 0 (1.66) G sz 3 +p n + 1 n –s)*2n + 2 l 3 +p + e H

ECE 5250/4250 Microwave Circuits 1–25 Review of Transmission Line Theory

Example: Consider a circuit with v0 = 2 v, Z0 = 50 ohms, Tl = 5 ns, Zg ==Rg 100 ohms, and ZL a parallel RC circuit with R = 100 ohms and C = 20 pf, as shown below in Figure 1.13.

Figure 1.13: Circuit diagram with Spice nodes indicated

• In the s-domain the solution is of the form

G n n 2 –sz 3 +p )*bs– 1 –s)*2n l 3 +p V)* z% s = --- e ------./--- e 3 H n,-3 n = 0 s)* s+ a (1.67) G n + 1 n sz 3 +p )*bs– 1 –s)*2n + 2 l 3 +p + e ------./--- e H n + 1,-3 n = 0 s)* s+ a • To inverse transform V)* z% s note that each series term consists of the product of a constant, a ratio of polynomials in s , –sI and a time shift exponential (i.e. e ) • In the time-domain each series term to within a constant is of the form

n –1=:)*bs– L ------% n = 012%%%F (1.68) >9n ?7s)* s+ a tt' – In

1–26 ECE 5250/4250 Microwave Circuits Transient Analysis with Resistive Loads

• A partial fraction expansion of the ratio of polynomials in (1.68) is n bs– K K K K ------)*= ------1 +------12 +------22 - +F + ------2n - (1.69) n 2 n s)* s+ a s sa+ )*sa+ )*sa+ where n b K = ----- (1.70) 1 n a and )*nk– n 1 d )*bs– K ==------% k 12%%%F n (1.71) 2k )*nk– ! )*nk– s ds sa= – • To obtain a partial solution for comparison with a Spice simulation we will solve (1.69) for n = 0 , 1, and 2. – Case n = 0 : 1 --- J ut)* (1.72) s – Case n = 1

bs– ba3 )*ab+ 3 a b ab+ –at ------= ------– ------J --- – ------e ut)* (1.73) s)* s+ a s sa+ a a

ECE 5250/4250 Microwave Circuits 1–27 Review of Transmission Line Theory

– n = 2 2 2 2 2 2 2 bs b a 1 b a ba a ------)*– = ------3 - + )*------– 3 - – ------)*+ 3 2 2 s)* s+ a s sa+ )*sa+ (1.74) 2 2 2 b ./b –at )*ba+ –at J ----- + KL1 – ----- e – ------te ut)* 2 2 a ,-a a • Using Mathematica the analytical solution valid for t up to 25 ns was obtained

1–28 ECE 5250/4250 Microwave Circuits Transient Analysis with Resistive Loads

• This required the use of +z and -z wave solutions from the series for n = 0 and 1 • The theoretical voltage waveforms at z = 0 and z = 1 are shown below

1.2

1.0 Load 0.8 Source 0.6

0.4 Line Voltage (v) Line Voltage 0.2

0.0 5 10 15 20 25t (ns)

!0.2

• A circuit simulation using ADS was also run, the results compare favorably as expected

ECE 5250/4250 Microwave Circuits 1–29 Review of Transmission Line Theory

Source Load R R TLIND C R2 R1 TLD1 C1 R=100 Ohm R=100 Ohm Z=50.0 Ohm C=20 pF VtStep Delay=5 nsec SRC1 Vlow=0 V t Vhigh=2 V Delay=0 nsec Rise=0.05 nsec TRANSIENT

Tran Tran1 StopTime=25.0 nsec MaxTimeStep=0.05 nsec • Plots of v)*0% t * and v)* l% t 1.2

m2 m3 1.0 Load m1

0.8 (v) Source m1 0.6 indep(m1)=1.148E-8 d plot_vs(Load, time)=0.889 oa

L m2

Source 0.4 indep(m2)=1.608E-8 plot_vs(Source, time)=0.963 0.2 m3 0.0 indep(m3)=2.443E-8 plot_vs(Load, time)=0.988

-0.2 02 4 6 8 10 12 14 16 18 20 22 24 26 time, nsec

1–30 ECE 5250/4250 Microwave Circuits CHAPTER 2. TRANSMISSION LINES

2.12.3 The Time-Domain Reﬂectometer A time-domain reﬂectometer (TDR) is an instrument that sends a step function down a transmission line

The TDR is known for having a very fast edge or step function rise time; e.g., 10’s of ps and much lower

A fundamental use of the TDR is in looking for cable faults and PCB track discontinuities

The TDR does all of its detective work from the source end of the line

Example 2.22: Generate Pulse From Step

In this example we consider a microstrip design for converting the step function output from a TDR into a rectangular pulse

Start with a LTspice simulation to see the basis for the design

Figure 2.95: LTspice design of the step to pulse converter circuit.

2-116 2.12. TRANSIENTS ON TRANSMISSION LINES

Voltage at Source End

Voltage at Junction End

Voltage at Load End

Figure 2.96: Pulse converter waveforms.

The microstrip design places two 50 stubs in parallel to form a more balanced circuit layout, that is all transmission widths are equal Ω 50

= = s 0

= s 𝑍𝑍 = s = 50 𝑇𝑇𝑠𝑠 1𝜇𝜇 𝑇𝑇1 =2𝜇𝜇50 𝑇𝑇2 = 504𝜇𝜇 𝑅𝑅𝑔𝑔 Ω + 0 0 = - 𝑍𝑍 Ω Ω 𝑍𝑍 Ω 50 50 𝐿𝐿 𝑔𝑔 𝑅𝑅 𝑉𝑉 = = 0 s Ω 𝑍𝑍 Not to scale 𝑇𝑇𝑠𝑠 1𝜇𝜇

Figure 2.97: Microstrip circuit design concept.

2-117