EXHIBIT 1018
`
` 1
`
`EXHIBIT 1018
`
`1
`
`

`

`Radio Receiver Design
`
`2
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`This page intentionally left blank
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`

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`Radio Receiver Design
`
`Kevin McClaning
`Tom Vito
`
`ican
`
`PUBL BLE
`
`Noble Publishing Corporation
`Atlanta, GA
`
`4
`
`

`

`Library of Congress Cataloging-in-Publication Data
`
`McClaning, Kevin, 1959-
`Radio receiver design / Kevin McClaning, Tom Vito.
`p. cm.
`Includes bibliographical references and index.
`ISBN 1-884932-07-X
`
`1. Radio--Receivers and reception--Design and construction.I. Vito, Tom, 1953-TI.
`Title.
`
`00-061271
`
`TK6563 .M38 2000
`621.384'18--de21
`
`
`
`Copyright 2000 by Noble Publishing Corporation.
`
`All rights reserved. No part of this book may be reproduced in any form or by any
`meanswithout prior written permission of the publisher.
`
`Printed in the Unites States of America
`
`ISBN 1-884932-07-X
`
`5
`
`

`

`Library of Congress Cataloging-in-Publication Data
`
`McClaning, Kevin, 1959-
`Radio receiver design / Kevin McClaning, Tom Vito.
`p. cm.
`Includes bibliographical references and index.
`ISBN 1-884932-07-X
`
`1. Radio--Receivers and reception--Design and construction.I. Vito, Tom, 1953-TI.
`Title.
`
`00-061271
`
`TK6563 .M38 2000
`621.384'18--de21
`
`
`
`Copyright 2000 by Noble Publishing Corporation.
`
`All rights reserved. No part of this book may be reproduced in any form or by any
`meanswithout prior written permission of the publisher.
`
`Printed in the Unites States of America
`
`ISBN 1-884932-07-X
`
`6
`
`

`

`30 | RADIO RECEIVER DESIGN
`
`Transmission Line Models
`
`Figure 1-7 shows a useful electrical model for transmission line analy-
`sis. The combination of V, and #, could be an antenna,signal generator, dig-
`ital logic gate or any other signal source. The resistor (R,) represents the
`signal load that could be a receiver, amplifier, and so on.
`
`
`
`Signal Generator or Antenna
`
`Figure 1-7 Transmission line analysis model.
`
`There are always two elements in a transmission line. In Figure 1-7,
`the center conductor carries the signal, and the second outer conductor
`serves as both a signal ground and a shielding mechanism. Other physical
`arrangements are possible.
`Figure 1-8(a) shows an equivalent circuit of a transmission line. Each
`marked-off section in Figure 1-8(a) is an infinitesimal piece of the line. The
`shunt capacitors (A C) represent the capacitance present between thesig-
`nal conductor and ground. The shuntresistors (A B) represent signal loss
`through the insulation separating the two wires (dielectric loss). The series
`inductance (A L) and the series resistance (A 2) model the parasitic induc-
`tance and resistance of the center conductor. Any transmission line can be
`modeled as an infinite number of the marked-off equivalent circuits in
`series. As the numberof sections increases, the values of AC, AB, AL and
`A R approachzero.
`
`7
`
`

`

`INTRODUCTION |
`
`314
`
`(a)
`
`AL AR AL AR.
`
`AC AB
`
`AC AB
`
`AC AB
`
`AC AB
`
`AC: Parallel Capacitance
`
`AB: Parallel Conductance
`
`AL: Series Inductance
`
`AR: Series Resistance
`
`(b)
`
`AL
`
`AL
`
`AL
`
`2
`
`AL
`
`rtftftgqft"
`
`Figure 1-8 Transmission line equivalent circuits (a) lossy (b) lossless.
`
`Physical cables always exhibit signal loss if they are not superconducting.
`In Figure 1-8(a), the signal loss has been modeled as series (AR) and shuntresis-
`tors (A B). Removingthe resistors makestheline lossless [see Figure 1-8(b)].
`
`Characteristic Impedance
`
`One of the main properties of a transmission line is its characteristic
`impedance (Z,). When we terminate a transmission line with a resistance
`whose value equals the cable’s characteristic impedance, signals propagating
`down the cable are completely absorbed by the load resistor. When the load
`value of the load resistor does not equal the valueof the cable’s characteris-
`tic impedance, signals propagating down will reflect off of the load resistor
`and travel back up the cable, in the direction of the source (see Figure 1-9).
`
`8
`
`

`

`32 | RADIO RECEIVER DESIGN
`
`Matched Cable
`
`Z, = Transmission Line
`Characteristic impedence
`
`TEN LA
`(nothing) =
`
`A wave propagating down a matched cable will not reflect off
`of the load resistor R,
`
`Unmatched Cable
`
`Z,) = Transmission Line
`har:
`istic
`
`Im
`
`A wave propagating down a matched cable will reflect off of
`the load resistor R,
`
`Figure 1-9 Signals propagating on a transmission line under matched
`and mismatched conditions.
`
`These signal reflections have profound effects on the performance of the
`system as will be shown. A transmission line’s characteristic impedance is a
`function of the physical geometry of the cable and the materials used to build
`it. According to Figure 1-8(b), the characteristic impedance of the cable is
`
`AL
`Zo =,i-S
`AC
`[Efanit length
`C/unit length
`
`__
`
`1.76
`
`Notice that the unit length quantity of Equation 1.76 drops outofthe rela-
`tion, which allows us to measure the inductance and capacitance of any
`random length of cable and use those numbersdirectly in Equation 1.76.
`
`9
`
`

`

`INTRODUCTION | 33
`
`Propagation Velocity
`
`Waves travel slower in a transmission line than in free space. The prop-
`agation velocity is v. Referring to Figure 1-8(b), we can show that the prop-
`agation velocity in a low-loss transmission line is
`
`10
`
`10
`
`

`

`34 | RADIO RECEIVER DESIGN
`
`
`1
`
`y=
`
`ALAC
`
`1
`
`
`J@/unitlength)(C/unitlength)
`
`1.78
`
`Unlike Equation 1.79, the cable length does not cancel out in this expres-
`sion. If we want the propagation velocity to be in meters per second, we
`have to express A L in henries/meter and AC in farads/meter.
`
`1.79
`(meters|
`1
`ve
`
`J(henries / meter)(farads / meter)\second
`
`The following example shows how to measure A L and AC (see Figure 1-10).
`
`Example 1.15 — Measuring the Propagation Velocity
`We selected a 6-foot long RG-223 cable and performed the following
`measurements:
`
`¢ We open-circuited the far end of the cable [as in Figure 1-10(a)] and
`measured 62.5 pF of capacitance at the near end.
`
`¢ We short-circuited the far end of the cable [as in Figure 1-10(b)] and-
`measured 177 nH of inductance at the near end.
`
`What is the velocity factor of the cable?
`
`Solution —
`Equation 1.82 demands that the electrical measurements be expressed as
`a per meter quantity. We know AC = 62.5 pF/6 foot = 34.2 pF/meter. Likewise,
`we know AL = 177 nH/6 foot = 96.8 nH/meter. Equation 1.82 produces
`
`1
`
`(96.8. 10°°\(34.2. 10”)
`
`~550-10° meters
`second
`
`1.80
`
`Thevelocity factor is the ratio of the propagation velocity in the cable to
`the propagation velocity in free space, so the velocity factor of the trans-
`missionline is
`
`11
`
`11
`
`

`

`INTRODUCTION | 35
`
`1.81
`
`Velocity Factor = es€
`_ 550-10°
`~ 3-108
`= 0.545
`
`Velocity Factor
`
`Waves usually move slower in a transmission line than in free space.
`The velocity factor is the ratio of the propagation velocity of a wave in a
`transmission line to the wave’s velocity in free space or
`
`Velocity Factor =v = "<i
`c
`
`1.82
`
`where
`vy = the velocity factor of the transmission line (0 < vy < 1),
`v = the propagation velocity in the transmission line,
`ec = the speedoflight.
`
`Example 1-16 — Measuring Velocity Factor
`We launched an 8 nsec long pulse into a 6-foot long piece of RG-223
`coaxial cable. We left the far end of the cable open-circuited and the return
`pulse came back 19.1 nsec after we launched thefirst pulse into the cable.
`Whatis the velocity factor of this piece of cable?
`
`Solution —
`The 19.1 nsec between pulsesis the time it takes for the pulse to trav-
`el down the cable and return. The one-waytravel time is 19.1/2 = 9.55 nsec.
`Our pulse traveled 6 feet in 9.55 nsec. Using Equation 1.72, the pulse would
`travel the6 feet in 6 nsecif it were travelling in free space. The velocity fac-
`tor of the cable is
`
`Velocity Factor = vy = orn
`
`nsec
`
`.55
`
`1.83
`
`= 0.63
`
`Since waves move slower than the speed of light inside the transmis-
`sion line, the wavelength inside the transmission line (the guide wave-
`length), will be shorter than the wavelength in free space. The wavelength
`inside of a transmission line is related to the wavelength in free space
`throughthe velocity factor.
`
`12
`
`12
`
`

`

`36 | RADIO RECEIVER DESIGN
`
`VelocityFactor =v, = stsi
`
`0
`
`1.84
`
`where
`v; = the velocity factor of the transmissionline (0 < vy < 1),
`4, = the wavelength in the transmission line (g = guided wavelength),
`Ag = the wavelength in free space.
`
`Equation 1.73 relates 4, to the frequency.
`
`Transmission Lines and Pulsed Input Signals
`
`Figure 1-11 shows a transmission line experiment. A lossless transmis-
`sion line (whose characteristic impedanceis Z,) is being fed with a very nar-
`row pulse. We will plot V,,,, the voltage at the input of the transmissionline,
`over time. Wewill plot data for various values of the load resistor R;.
`Pulse
`Generator
`
`rs ®
`
`2,
`
`Vs
`42V
`
`(a) V¢
`
`(time)
`
`“wv 5
`
`(b) R, = co
`Cl #1V
`
`ty
`(time)
`
`Vin
`
`Vin
`ay
`
`Vin
`
`
`
`(d)R, = 2,
`(NoPulse)
`+ a
`(ume)
`ts
`1
`Figure 1-11 Transmission line under pulsed input conditions with var-
`tous loads.
`
`ty
`(c)R, =0
`
`(time)
`
`-1V
`
`13
`
`13
`
`

`

`INTRODUCTION | 37
`
`Open-Circuit Line
`Note that we have set Rg, the resistor in series with V,, equal to the
`characteristic impedanceof the cable. Thefirst plot [Figure 1-11(a)] shows
`the source voltage V,. We are generating a 2-volt pulse and observe the same
`waveform regardless of the value of FR.
`Figure 1-11(b) showsV,,, when the load is an open circuit (A; = «). The
`magnitude of the pulse has changed:it is now only 1volt — half of what the
`generator produced. We will also observe a second pulse at V;,. The second
`pulse has the same magnitude as thefirst pulse but appears at a later time
`(tz) from the initial pulse. We can explain the second pulse by considering
`the transmission line as a simple time delay. The first pulse was generated
`by the source. The pulse travels through the cable until it encounters the
`open circuit at the load end. Similar to a voice echoingoff of a cliff or a radar
`pulse bouncingoff of an airplane, the pulse reflects off of the discontinuity and
`travels back in the direction it came. The secondpulseat V,,, is travelling back
`from the load end of the cable toward the generator. The time(¢,) is the time
`it takes the pulse to travel down the cable and back again. This is one tech-
`nique we can use in the lab to determine the cable’s propagation velocity.
`Whenthe pulse first enters the transmission line and before the return
`pulse has had time to return from the opencircuit, the transmissionline is a
`resistor whose valueis its characteristic impedance. The change in the mag-
`nitude of the pulse whenit enters the cable can be easily explained. Since we
`set the source resistor (Rg) equal to the cable’s characteristic impedance Z,,
`we have a voltage divider whose output voltage is half the input voltage.
`
`Short-Circuit Line
`Figure 1-11(c) shows V,,, if we short-circuit the load end of the trans-
`mission line. Everything is the same except for one minor detail — the
`return pulse has the same magnitude, but it is opposite in polarity. The
`short circuit at the end of the cable causes the pulse to come back upside-
`down or inverted.
`
`Zy Terminated Line
`Figure 1-11(d) shows a terminated cable with a resistor whose value
`equals the characteristic impedance of the cable. No pulse returns
`because the matched load resistor absorbsall of the energy in the pulse.
`This is usually a desirable situation and one of the reasons to keep sys-
`tem impedances matched.
`
`Reflection Coefficient
`
`Open and short circuits are extreme conditions. What happens if we
`place a variable resistor at the load end of the cable? In general, the mag-
`
`14
`
`14
`
`

`

`38 | RADIO RECEIVER DESIGN
`
`nitude of the reflected pulse will be
`
`Vireftected = PVincident
`
`1.85
`
`where
`p is the reflection coefficient (p can be a complex numberif the load
`impedanceis complex).
`Vincident iS the voltage incident on the load. In other words, the magni-
`tude of the pulse we sent into the transmission line.
`Vincident travels from the source to the load.
`Vneflected is the magnitudeofthe voltage reflected off of the load and back
`into the transmission line. Vpoectea travels from the load to the source.
`
`We can show
`
`p=—— -lspstl
`
`1.86
`
`Pulse
`>
`: Generator
`:
`
`Rs
`
`s
`;
`
`Z)
`
`Strikes the load
`
`generates
`
`a reflection
`
`a
`
`and
`
`Launched by —3 Thi
`Pulse Generator
`Vincident
`
`—_ >
`
`Source
`
`<_
`Backto the <_
`Veehected = PVincident
`
`Figure 1-12 Transmission line reflection coefficient.
`
`Figure 1-13 showsa graphofthe load resistance vs. the reflection coef-
`ficient for a 50-ohm cable. At low values of R,, the reflection coefficient
`approaches —1. This means that mostof the voltage we send down thecable
`comesback (although its magnitude is reversed). When R, is very large, p
`
`15
`
`15
`
`

`

`INTRODUCTION | 39
`
`approaches +1. Note that everything that is sent down into the cable is
`reflected but the magnitudeis not affected.
`When the value of the load resistor equals the transmission line’s charac-
`teristic impedance, the reflection coefficient equals zero. None of the energy
`we send into the cable returns;all of it is dissipated in the load resistor.
`
`Reflection Coefficient vs. Load Resistance
`
`
`
`ReflectionCoefficient
`
`
`
`
`10
`
`10
`
`z
`
`10
`
`10
`
`Load Resistance (Ohms)
`
`Figure 1-13 Transmission line reflection coefficient us. load resistance (50-
`ohm system).
`
`Transmitting Systems
`
`In a transmitting system, a high-power amplifier feeds a cable. The
`cable, in turn, feeds an antenna. Weare interested in sending powerinto
`an antenna and emitting that powerinto free space. If the antennais not
`matched to the cable or the cable is not matched to the power amplifier,
`some of the energy that is being sent into the cable will bounce off the
`antenna, return and not radiate into space.
`
`Receiving Systems
`
`If we are interested in receiving signals, we normally place an antenna
`feeding a cable which then feeds a receiver. Any mismatch between the
`antenna and the cable or between the cable and the receiver will cause
`
`16
`
`16
`
`

`

`40 | RADIO RECEIVER DESIGN
`
`energy to reflect off of the mismatch. Signal energy will be lost. and the
`
`receiving system will not be as sensitive as it could be.
`
`17
`
`17
`
`

`

`INTRODUCTION | 44
`
`18
`
`18
`
`

`

`42 | RADIO RECEIVER DESIGN
`
`19
`
`19
`
`

`

`INTRODUCTION | 43
`
`Complex Reflection Coefficient
`
`So far we have examined resistive cable terminations and the astute
`reader may wonder what happensif we decided to terminate a cable with
`an inductor or capacitor. The answeris that everything we have discussed
`
`20
`
`20
`
`

`

`44 | RADIO RECEIVER DESIGN
`
`will still apply with the exception that the reflection coefficient will be a
`complex numberand have a magnitude and a phase angle. With a resistive
`load, the reflection coefficient is always real. With a complex load, the time-
`domain examples are harder to visualize because of the frequency-depen-
`dent nature of the load.
`For a complex load, Z; = Rz, + jX,, the reflection coefficientis
`
`p=|p|20,
`_2,-%4
`Zp +Zy
`
`1.100
`
`where
`Z, = R, + JX, = the complex load (resistance and reactance),
`p = the complex reflection coefficient,
`|p| = the magnitude of the complex reflection coefficient,
`6, = the angle of the complex reflection coefficient.
`
`A complex p means that the incident wave suffers a phase change as well
`as a magnitude change when it encounters the mismatch.All the equations
`concerning return loss will also apply for the complex reflection coefficient.
`The magnitudeis specified wherever it was required.If it is not specified,
`we use the complexreflection coefficient, which yields a complex result.
`
`Example 1.19 — Reflection Coefficient and Reactive Terminations
`Figure 1-16 shows a transmission line with a reactive termination.
`Find the reflection coefficient, return loss and VSWR for this load. Assume
`a 50-ohm system and that the frequency of operation is 4 GHz.
`
`Z,= 502
`
`102
`
`1*
`
`Figure 1-16 Transmission line with a complex load.
`
`Solution —
`The reactance of this load at 4 GHz is 10 -/13.3 © or 16.6 4 53°. Using
`Equation 1.100 with Z,; = 10 -j13.3 Q and Z, = 50 Q produces
`
`21
`
`21
`
`

`

`INTRODUCTION | 45
`
`Mismatch Loss
`
`The voltage reflecting off of a mismatched load at the end of a trans-
`mission line represents powerloss. If the load were perfectly matched to
`the transmission line, we would transfer all of the available incident signal
`powerto the load resistor. But since the load is not matched to the trans-
`mission line’s characteristic impedance,all the available poweris not trans-
`ferred to the load. This mismatch loss is defined as
`
`
`
`Mismatch Loss =FavaitablePrettected~
`
`
`P,Available
`1 . 104
`= &Delivered
`P,Available
`
`where
`Psvailable = the power that is delivered to the matched load,
`Poetivered = the power that is delivered to the unmatched load,
`Prefected = the power thatis reflected off of the unmatched load.
`
`22
`
`22
`
`

`

`46 | RADIO RECEIVER DESIGN
`
`We know
`
`and
`
`Py, a Vincident
`
`Vi
`
`ie
`
`0
`
`1.105
`
`1.106
`_ Vitehected
`P
`Reflected~~"ay0
`
`2
`
`2
`
`/
`
`EY
`
`Combining Equations 1.104 through 1.106 with Equation 1.85 produces
`Mismatch Lossy, a= Vincident/Zo — Vteflected iZo
`/
`2
`Vincident / Zo
`2
`2
`= Vincident ~ Viteflected
`2
`Vincident
`~ Vincident ~ (|pL [Vincident}
`Vincident
`
`2
`
`1.107
`
`=1-|p;)"
`
`Mismatch loss is yet another way of describing the relationship
`between the characteristic impedance of a transmission line andits load.
`Mismatchlossis specified in decibels,
`Mismatch Loss,,24.4B = 10log(1 - \px|
`
`1.108
`
`Mismatchloss is only relevant in connection with an antenna. In a receiv-
`ing situation, the mismatch loss of the antenna represents signal loss that
`affects the system noise floor and minimum detectable signal. In the trans-
`mitting case, the mismatch loss represents signal energy that is not deliv-
`ered to the antenna and so cannot be radiated into free space.
`
`Source and Load Mismatches
`We can experience mismatches at both the source and load ends of a
`transmission line. In a source mismatch, i.e., when the source impedance
`(Rg) is not equal to the characteristic impedanceof its accompanying trans-
`mission line, the maximum amount of power is not transferred into the
`line. The mismatch loss due to source mismatch is
`
`23
`
`23
`
`

`

`INTRODUCTION | 47
`
`.
`Mismatch LoSSgource,dB =1- ip1
`
`2
`
`1.109
`
`or
`
`Mismatch Lossgource,an = 10log(1 - lesl’)
`
`1.110
`
`24
`
`24
`
`

`

`48 | RADIO RECEIVER DESIGN
`
`d.
`
`e.
`
`f.
`
`Mismatch Lossqp = 10log(1-|9\*)
`,
`= 10log{1-|0.339)*)
`
`=-0.5 dB
`
`Mismatch Lossgp = 1Olog(1 - lol)
`F
`= 10log(1-|0.667)")
`
`=-2.6 dB
`
`Mismatch Lossy = 10log(1 = lel”)
`= 10log(1 — |0.855)")
`
`=-5.7 dB
`
`1.114
`
`1.115
`
`1.116
`
`Thesign of the reflection coefficient is irrelevant as far as the mismatch
`loss is concerned.
`
`Transmission Lines and Sine-Wave Input Signals
`
`We havediscussed the behavior of pulses as they move up and down the
`transmission line and when they encounter impedance discontinuities.
`Since a pulse consists of a series of sine waves, we argue that the sameeffects
`occur when we launch a continuous RF carrier (i.e., a sine wave) down a
`transmission line. Reflections still occur as the wave encounters mismatched
`impedances. However, because of the continuous nature of the sine wave, the
`effects are not as intuitive as in the pulsed case.
`Figure 1-17 shows an example where the voltage generator V, is a
`sine- wave source that turns on at time t = 0. The system is completely
`mismatched (i.e., Rs # Zy) # R,). When we turn the signal generator on at
`t = 0, the sine wavefirst encounters the mismatch between the signal
`generator’s source impedance and the impedance of the transmission
`line. This effect acts to reduce the voltage incident on the line. After the
`sine wave enters the transmission line, it travels down the line until it
`encounters the load resistor. The load absorbs someof the signal energy
`and reflects the rest according to the equation
`
`25
`
`25
`
`

`

`INTRODUCTION | 49
`
`Vreflected = PVincident
`
`Lil?
`
`Zo
`
`
`|S
`
`p
`
`L
`
`t
`
`V, = 0 fort<0
`sin (@,t) for t > 0
`
`R,#Z,#R,
`
`Rg
`
`< llO)-~
`
`Figure 1-17 Mismatched transmission line with sinusoidal inputvoltage.
`
`The energy that reflects off of the load travels back along the line
`toward the source. If p is complex, the sine wave experiences a phase shift
`as well as a magnitude change. At the source endof the line, the signal from
`the load splits into two pieces. Some of the energy is absorbed by the source
`while the rest is reflected and sent back again toward the load. This process
`of absorption and reflection at each end of the transmission line continues
`until eventually the transients recede and the line reaches a steady-state
`condition. The voltage present at any particular point on the transmission
`line is the sum ofthe initial incident wave andall of the reflections.
`The mathematical expression describing this process involves an infi-
`nite series. To simplify the analysis, we will examine a transmission line
`with an arbitrary load resistance but with a matched source resistance (see
`Figure 1-18). When Ag = Zo, the reflections caused by a mismatch at the
`source end of the cable are eliminated. The voltage at any physical point on
`the line is due to the incident wave and onereflection from the load. Any
`signals travelling from the load to the source are absorbed entirely by the
`source, because, by matching the source to the line, we have set the source
`reflection coefficient to zero.
`
`26
`
`26
`
`

`

`50 | RADIO RECEIVER DESIGN
`
`Sine Wave
`' Source
`Rg =Zy |
`
`Z,
`
`Incident
`sine wave
`V.
`_I : \Y travelling
`
`
`:
`:
`toward
`:
`the load
`
` —
`
`V.= pV,
`
`—_:
`
`(NIN, __
`
`Reflected
`
`gine wave
`:
`bouncing
`-
`: Off load and
`:
`travelling
`‘ back toward
`generator
`
`Signal reflected from the load is
`absorbed completely by the
`matched R,
`Figure 1-18 Mismatched transmission line showing incident andreflect-
`ed voltages.
`
`Voltage Minimums and Maximums
`
`Figures 1-19 and 1-20 show the magnitudes of the sine waves that can be
`observed along a length of transmission lines for several load mismatch condi-
`tions. Figure 1-19 describes loads that are greater than Z, (or 50 ohms, in this
`case); Figure 1-20 describes the situation when the loads are less than 50 ohms.
`
`27
`
`27
`
`

`

`INTRODUCTION | 51
`
`Voltage vs. Physical Position (High-Z Load)
`—
`r
`“T
`T
`
`T
`
`T
`
`
`
`
`
`— VSWR= 10 (R,=500 9)
`VSWR= 5 (R,=250 0)
`— VSWR = 2 (R,=100 0)
`— - VSWR =1 (R,=50.9)
`
`2.5}
`
`
`
`
`
`
`
`
`
`RelativeAmplitude a
`
`0.5
`
`
` 1.0
`
`L
`L
`1
`
`
`1.0
`0.9
`08
`0.7
`0.4
`0.6
`0.5
`Wavelengths From Load
`
`1
`0.3
`
`1
`0.2
`
`L
`0.1
`
`
`Load
`
`Figure 1-19 Voltage standing wave ratio (VSWR). The magnitude of the
`voltages present on a mismatched transmission line when Z, > Zo.
`
`Voltage vs. Physical Position (Lo-Z Load)
`
`VSWR= 10 (R,=5 ®)
`VSWR= 5 (R,=10 9)
`VSWR= 2 (R,=25 2)
`VSWR= 1 (R,=50 9)
`
`0.9
`
`0.8
`
`0.7
`
`0.4
`0.5
`0.6
`Wavelengths From Load
`
`0.3
`
`0.2
`
`0.1
`
`Load
`
`Figure 1-20 Voltage standing wave ratio (VSWR). The magnitude of the
`voltages present on a mismatched transmission line when Z, < Zo.
`
`28
`
`28
`
`

`

`52 | RADIO RECEIVER DESIGN
`
`29
`
`29
`
`

`

`INTRODUCTION | 53
`
`e.
`f.
`
`0.90cos(wt +45),
`1,12cos(wt + dg).
`
`where
`«w = 2nf = the frequency of operation,
`¢,, = a phase constant depending upon the frequency and the distance
`from the sourceto load.
`
`Given a line with a mismatched load, we will define the highest voltage pre-
`sent on the line asV,,,,, and the lowest voltage present on thelineasV,,,;,, oF,
`
`Largest Voltage on the T- Line = V,,,, cos(at + @,)
`Smallest Voltage on the T- Line = V,,;,, cos(a@ +2)
`
`1.118
`
`Note that V,,,,, and V,,;, change with the value of the load resistor. When
`the line is matched (R; = Z,), then V4, = Vain. When the line is not
`matched, V0, # Vinin. AS the load’s reflection coefficient increases (i.e., as
`the match gets worse), V,,,,, and V,,;, become increasingly different.
`
`Voltage Standing Wave Ratio (VSWR)
`
`The characteristic of voltage minimums and voltage maximums along
`a mismatched transmission lineis referred to as the voltage standing wave
`ratio (VSWR) of the line. VSWR is defined as
`
`VSWR = “me. >4
`Vv.min
`
`L119
`
`VSWR is a common measure of impedance, similar to the reflection coeffi-
`cient (p). The concept of VSWR was developed from the observations of
`physical transmission lines and from T-line theory. In the past, VSWR was
`a very common unit of impedance measurement because it was relatively
`easy to measure with simple equipment and without disturbing the system
`under test. RF engineers rarely measure VSWRdirectly anymore but many
`vendors still specify it as a measure of how well their systems are matched.
`
`VSWR Relationships
`A little algebra with V,,,., Vinin, Ri, Zp and p;,, reveals the following
`relationships:
`
`VSWR =
`
`
`1+|p;|
`1-|p;
`
`1< VSWR <<
`
`1.120
`
`30
`
`30
`
`

`

`54 | RADIO RECEIVER DESIGN
`
`and
`
`|o| = VSWR-1
`
`VSWR +1
`
`1.121
`
`The absolute value in these two equations contains two separate values of
`reflection coefficient, i.e., two values of load resistance, which can produce
`the same value for VSWR. Twovalues of R, will produce the same VSWR
`on a transmissionline if the two valuessatisfy the following equation:
`
`Z
`1.122
`
`
`R,=—
`R, =Z,(VSWR
`
`L=Zol ) or Ry VSWR
`
`Figure 1-21 showsa plot of VSWR vs.load resistance for a 50-ohm sys-
`tem. When the matchis perfect (2; = 50 9), the VSWR equals 1. Theclos-
`er the VSWR is to 1, the better the match. VSWR ranges from 1 (a perfect
`match) to infinity (a short or open circuit).
`
`VSWRvs. Load Resistance (4 = 50 Q)
`
`T VSWR
`
`
`
`T
`
`a
`
`0
`
`Load Resistance (Ohms)
`
`Figure 1-21 Voltage standing wave ratio (VSWR) vs. load resistance for
`Zo = 50.
`
`31
`
`31
`
`

`

`INTRODUCTION | 55
`
`War Story — VSWR and Microwave Ovens
`If we stretch our imagination, we can think of a microwave oven as a
`transmission line system. The energy source is the magnetron tube inside
`the microwave. The transmission line is the cavity where weplace the food
`to be heated; the load is the water in the food.
`As long as we have a load on the transmissionline,i.e., as long as we
`have food in the oven, energy transfers from the magnetron tube through
`the waveguide and into the food. We have a low VSWR inside of the
`microwave oven and all is well. But if we operate the oven without an ade-
`quate load, the oven runs under high VSWR conditions. There will be areas
`of “high voltages”(large electric fields) inside the cavity that will stress the
`magnetron tube. Also, since the energy from the magnetron meets a load
`with a high VSWR (or equivalently, a high reflection coefficient), all of the
`energy sent out by the magnetron will bounceoff of the poor load and trav-
`el back into the tube. The magnetron nowhas to dissipate a large amount
`of heat and which maycauseit to fail.
`Even when we operate a microwave oven with food in the cavity, the load
`is still poorly characterized. In other words, if we think of a microwave oven
`and its load as a transmission line with a source and a load, the load is not
`exactly Zp. After all, what is the impedance of a bologna sandwich? Thesys-
`tem still exhibits a nonunity VSWR. From a practical point of view, this effect
`causes “hot” and “cold” spots in the oven. The “hot” spots are areas of high
`field energy; the “cold” spots are areas of low field energy. Food thatlies in a
`hot spot heats quickly, and food in a cold spot heats slowly or not at all.
`To combatthis problem, manufacturers often arrange to rotate the food
`in the oven. Since the rotating food alternately passes through hot and cold
`spots in the cavity, the food gets heated more evenly. Another solution is to
`install stirrers in the cavities. A stirrer is a metal propeller-like structure
`that rotates slowly in the cavity. It reflects the energy from the magnetron
`in different directions as it moves. This changesthe positions of the hot and
`a cold spot in the oven and results in more evenly heated food.
`
`War Story — VSWR and Screen Room Testing
`Screen rooms are commonly used to test antennas or to measure the
`electromagnetic radiation emanating from a component. Such tests are
`necessary to make sure pieces of equipment will not interfere with one
`another. Frequently, sensitive electronic tests are performed here. A screen
`room usually takes the form of a large metal room that shields the inside
`from the ambient electromagnetic fields on the outside of the room.It has
`metal walls, doors and ceilings with the appropriate fixtures to supply
`power and air into the room.
`After constructing a screen room we need to determine how muchshield-
`ing the room supplies from outside interference. There are many suchtests
`
`32
`
`32
`
`

`

`56 | RADIO RECEIVER DESIGN
`
`to verify the performance. In orderto test a screen room, the operator often
`places a transmitter inside the room, closes all the doors, then checks for
`radiation from the transmitter on the outside of the room.
`As it was the case in the microwave oven example, we haveto be con-
`cerned with the load that the transmitter experiences and with the VSWR
`inside the room. In an empty metal room,the transmitter effectively has no
`load, and wewill experience a high VSWR condition inside the screen room.
`High VSWR causeslarge electromagnetic fields in some areas of the room
`and small electromagnetic fields in other areas of the room. If an area of
`small electromagnetic field happens to fall in an area where the screen
`room is leaky, then we may not detect the leak. If the area of strong elec-
`tromagneticfield falls in the neighborhood of a good joint, we may detect
`that the joint is leaky due to the excess field placed across the joint by the
`high VSWR condition in the room.
`Commercial gear avoids this problem by sweeping the transmitter slight-
`ly in frequency. The positions of the voltage maximums and minimums on a
`transmission line depend upon the operating frequency. When we change the
`frequency, we change the position of the voltage maxima and minima to
`obtain a morerealistic measurementof the attenuation of the screen room.
`
`(= R=4
`
`Z
`
`Measure V,,,,cos(2 ft)
`max
`at this point on the line
`
`Measure V_,cos(2 f,t)
`at this point on theline
`
`Figure 1-22 Measuring VSWR on a transmission line.
`
`33
`
`33
`
`

`

`INTRODUCTION | 57
`
`Figures 1-19 and 1-20 show another feature of the voltage peaks and
`valleys. As Figure 1-22 indicates, the distance between two adjacent voltage
`maximumsor two adjacent voltage minimum is 4,/2 where Ag is the wave-
`length in the transmission line or the guide wavelength. Also, the distance
`between a voltage minimum andits closest voltage maximum is A,/4.
`
`Return Loss
`
`Return loss is a measure of how well a device is matched to a trans-
`mission line. We have mentioned launching pulses down a transmission
`line. The load absorbed some of the pulse energy and some of the energy
`was reflected back toward the source. Return loss characterizes this ener-
`gy loss in a very direct manner (see Figure 1-23).
`
`
`
`> PpIncident
`
`
`
`>
`
`Radiate a known amountof powerinto
`the cable. That poweris incident on the
`load
`
`<_ Measure the powerwhich returns from
`
`the load
`
`Figure 1-23 Measuring return loss.
`
`Whena knownvoltageis incident on a load (Vincent), We can imply that
`a known amountof poweris incident on the load (Piygigent). Similarly, when
`voltage is reflected from a load (Vpenecteg), that voltage represents power
`that is reflected off of the load (Ppepestea)- This applies with continuoussig-
`nals as well as pulses. Return loss is defined as
`
`Return Loss,p = 10log,ists <0dB
`
`Incident
`
`1.123
`
`Return loss is measured directly by launching a known amount of power
`into a transmission line (Pygent) and measuring the amount of powerthat
`reflects off of a load (Pprepected): Return loss measures the power lost when a
`signal is launched into a transmission line.
`
`34
`
`34
`
`

`

`58 | RADIO RECEIVER DESIGN
`
`
`Return Lossy, = 20log Vinehected
`
`Incident
`
`
`
`= 20]og|p}
`
`= 20log 41-2
`Zi +Zo
`
`
`
`1.124
`
`Figure 1-24 showsa plot of return loss vs. load resistance for a 50- ohm
`system. Whenthe load resistance is very small or very large, most of the power
`welaunch into the transmissionline reflects off of the load. Since Propected =
`Pincidents the return loss is near 0 dB. This is a poorly matched condition.
`Whentheload resistance is approximately equal to the system charac-
`teristic impedance (Z, = 50
`in this case), most of the power we launch
`into the line is absorbed by the load andvery little returns. The return loss
`is a large negative number, which indicates a good match.
`
`Return Loss vs. Load Resistance (2, = 50.9)
`
`
`
`ReturnLoss(dB)
`
`
`
`
`
`0 oO
`10
`
`1
`
`10
`
`z
`
`2
`
`10'
`
`3
`
`10
`
`Figure 1-24 Return loss vs. load resistance for a 50-ohm system.
`
`Transmission Line Summary
`
`VSWR,reflection coefficient and return loss all measure one parameter:
`how well the load is matched to the system’s characteristic impedance. In the
`examples, we have used 50 ohms as Zp, but other common characteristic
`impedances are 23 ohms, 75 ohms, 93 to 125 ohms and 300 ohms.
`
`35
`
`35
`
`

`

`INTRODUCTION | 59
`
`War Story — Reflection Coefficients and Speech Synthesis
`The useofreflection coefficients is not limited to transmission lines as
`the following example from sp