Transmission Linesranga/courses/en4620_2008/... · transmission line with known termination. In...

Transmission Lines

Ranga Rodrigo

January 27, 2009

Antennas and Propagation: Transmission Lines 1/72

1 Standing Waves

2 Smith Chart

3 Impedance MatchingSeries Reactive MatchingShunt Reactive Matching

Antennas and Propagation: Transmission Lines Outline 2/72

Transmission Line ModelZ0

R: series loss resistance (copper losses).G: shunt loss conductance (losses in dielectric).L: series inductance representing energystorage within the line.C : shunt capacitance representing energystorage within the line.

Antennas and Propagation: Transmission Lines Summary of last Week’s Lecture 3/72

If an infinitely long pair of wires were consideredand the voltage and current were somehowmeasured at uniform spaced points along theline, then

I1= V2

I2= ·· · = Vk

Ik= constant = Z0Ω. (1)

This is termed the characteristic impedance ofthe line and is denoted by Z0.

Characteristic Impedance

R + jωL

G + jωC

Ω . (2)

At very low frequencies:

Ω. (3)

For high frequencies (ωL À R and ωC ÀG):

Ω (4)

Propagation Constantd 2V

d x2= (R + jωL)(G + jωC )V. (5)

Equation 5 can be written as

d x2= γ2V , (6)

where γ=√(R + jωL)(G + jωC ) is termed the

propagation constant. This is usually expressed as

γ=α+ jβ, (7)

where α represents the attenuation per unit lengthand β the phase shift per unit length.

Equation 6 has a solution of the form

V (x) = Ae−γx +Beγx . (8)

This suggests that the line will contain two waves,one traveling in the positive x-direction (e−γx) and theother traveling in the negative x-direction (eγx).

Sending-End Impedance

IincIref

1+e−2γl ZT −Z0ZT +Z0

1−e−2γl ZT −Z0ZT +Z0

= ZT (1+e−2γl )+Z0(1−e−2γl )

ZT (1−e−2γl )+Z0(1+e−2γl ). (9)

Z0= ZT cosh(γl )+Z0 sinh(γl )

ZT sinh(γl )+Z0 cosh(γl ). (10)

+ tanh(γl )

1+ ZTZ0

tanh(γl ). (11)

Short Circuit Termination

If a short circuit is used as the load termination, thenthe sending-end impedance becomes,

ZS|SC = Z0 tanh(γl ), (12)

and, neglecting line losses for short lengths we get

ZS|SC = Z0 tanh( jβl ) = j Z0 tan(βl ). (13)

Open Circuit Termination

When the termination is open circuit, then thesending-end impedance becomes,

ZS|OC = Z0

tanh(γl ), (14)

and, neglecting line losses for short lengths we get

ZS|OC = − j Z0

tanh( jβl )=− j Z0 cot(βl ). (15)

βl = 2πlλ

− tan(βl )

π 3π

2π 5π

Normalized input resistance versus βl for ashort-circuited, lossless, transmission line.Antennas and Propagation: Transmission Lines Summary of last Week’s Lecture 12/72

βl = 2πlλ

−cot(βl )

π 3π

2π 5π

Normalized input resistance versus βl for aopen-circuited, lossless, transmission line.Antennas and Propagation: Transmission Lines Summary of last Week’s Lecture 13/72

Standing WavesFor any transmission line, a sinusoidal signal ofappropriate frequency introduced at the sendingend by a generator will propagate along thelength of the transmission line.If the line has infinite length, then the signalnever reaches the end of the line.If the signal is viewed at some distance downthe line away from the generator, then it willappear to have the same frequency, but willexhibit smaller peak to peak voltage swing thanat the generator.The signal, therefore, has been attenuated bythe line losses due to the conductor resistanceand dielectric imperfections.Antennas and Propagation: Transmission Lines Standing Waves 15/72

ReflectionIf the line is lossless, then the signal viewed atsome remote point will be identical to that at thegenerator but time delayed by an amountdependant on the position.When a sinusoidal signal reaches the open endof a section of a lossless transmission line, itcan dissipate no energy.This means that all the energy propagatingalong the line in the forward direction (incident)will be reflected completely, on reaching theopen circuit termination.The reflected wave (backward wave) must besuch that the total current at the open circuit iszero.Antennas and Propagation: Transmission Lines Standing Waves 16/72

ReflectionAs the reflected signal travels back along theline toward the generator, it reinforces theincident waveform at certain points formingmaxima (nodes).Similarly, it can cancel the incident waveform atcertain other points producing minima(antinodes).In an open circuit line, node voltage points willoccur at the same position as the antinodecurrent points.These waves do not represent traveling waves.They are standing waves, implying that there isno net power flow from generator to the(open-circuit) load.Antennas and Propagation: Transmission Lines Standing Waves 17/72

A point one-quarter wavelength away from ashort circuit will have voltage and currentmagnitudes equivalent to those obtained for anopen circuit.For a lossless line the peak value of the standingwave envelope is twice that of the incident wave.There are nulls as well due to the completecancelation.For lossy (practical) line, the peak will be lessthan twice that of the incident wave, andcomplete cancelation rarely results.

Antennas and Propagation: Transmission Lines Standing Waves 18/72

Voltage Standing Wave RatioNode to antinode voltage ratio is infinity for theideal case.Voltage standing wave ratio (VSWR) or simplystanding wave ratio is defined as

VSWR= Vmax

= |Vinc|+ |Vref||Vinc|− |Vref|

|Vinc| is the peak value of the incident voltagewave, similarly the others. Here the voltagevalues are peak values. RMS values can beused as well.

Reflection Coefficient

IincIref

Itrans

Considering the current continuity at the boundaryA A′, we get

Z0− Vref

Z0= Vtrans

ZT. (17)

For voltage continuity across the boundary

Vinc+Vref =Vtrans. (18)Antennas and Propagation: Transmission Lines Standing Waves 20/72

From Equations 17 and 18 we can obtain thereflection coefficient Γ.

Γ= Vref

Vinc= ZT −Z0

ZT +Z0. (19)

An alternative symbol for the reflection coefficient isρ. Sometimes ρ is used to represent |Γ|. FromEquation 16

VSWR=1+

∣∣∣VincVref

∣∣∣1−

∣∣∣VincVref

∣∣∣ = 1+|Γ|1−|Γ| . (20)

Therefore,

|Γ| =∣∣∣∣Vinc

∣∣∣∣= VSWR−1

VSWR+1. (21)

ExampleA certain transmission line with a characteristicimpedance of 50 Ω is required to deliver 1 kW over ashort distance to a load. The cable can withstand amaximum of 250 V. Determine

1 the maximum VSWR, and2 amount of power that must be provided by the

generator.

Solution

We can assume that the line is lossless. Powerdissipated by the load PL is

PL = Pinc−Pref.

PL =V 2

Z0− V 2

Z0(Vinc−Vref)(Vinc+Vref),

= 1000 W.

(Vinc−Vref)(Vinc+Vref) = 50000.

Because RMS voltage in the line must be less than250 V,

(Vinc+Vref)|max < 250.

(Vinc−Vref) > 200.

VSWR= |Vinc|+ |Vref||Vinc|− |Vref|

VSWR< 250

200= 1.25.

We can show that

Pinc= 1−

(VSWR−1

VSWR+1

For this value of VSWR, 98.8% of the incident poweris delivered to the load. Therefore, the generatormust be capable of supplying at least 1012 W andmust be matched to the line. In normal design,VSWR is kept below 1.1.

Sending-End Reflection Coefficient

ZTZ0, γ

Γin = ZS−Z0ZS+Z0

ΓL = ZT −Z0ZT +Z0

Rearranging Equation 9,

1+e−2γl ZT −Z0ZT +Z0

1−e−2γl ZT −Z0ZT +Z0

we can obtainΓin = ΓLe−2γl . (22)

Impedance MatchingOne of the major tasks in transmission linecircuit design is impedance matching, and Smithchart is a graphical procedure for solvingimpedance transformation problems.A mismatched transmission line (a lineterminated in an impedance other than itscharacteristic impedance) would reflect some ofthe incident wave back along the line toward thegenerator.This interaction of the forward waves andreflected waves give rise to a resultant waveformwith nodes and antinodes at fixed points alonglength of the transmission line.This way, standing waves are formed.Antennas and Propagation: Transmission Lines Smith Chart 28/72

Effects of MismatchAdverse Effects

Energy loss resulting in reduced systemperformance.Frequency pulling of the signal generator.Possible transmitter (and other circuitry)damage.

Desirable EffectCarefully controlled mismatch can improvesystem performance by improving the noisecharacteristics.

Antennas and Propagation: Transmission Lines Smith Chart 29/72

Controlling Impedance Mismatch

There are two requirements:1 A methodical scheme for determining the

degree of mismatch between the source and theload impedance.

2 A method whereby the degree of mismatch canbe varied in a known manner.

Smith Chart

Matching requires a way of handling theexpressions governing the impedancetransformation that occurs along a length oftransmission line with known termination.In 1939, P. H. Smith, an engineer with the BellTelephone Laboratories, developed a graphicalpresentation of transmission line data.He found that for passive loads (0 < |Γ| < 1) thegraphical representation is bounded.Smith chart is a plot of the voltage reflection.

All possible values of impedance andadmittance can be can be plotted on the chart.An easy method of converting impedances toadmittances is available.Smith chart provides a simple graphical methodfor determining the impedance transformationsdue to a length of transmission line.

Normalized Resistance Loci

Γ= ZL −Z0

ZL +Z0(23)

LetΓ= u + j v and

Z0= Zn = Rn + j Xn.

u + j v = (Rn −1)+ j Xn

(Rn +1)+ j Xn. (24)

Equating real and imaginary parts gives

Rn(u −1)−Xn v =−(u +1),

Rn v +Xn(u −1) =−v.(25)

Xn = Rn(u −1)+ (u +1)

v. (26)

Substituting 26 in 25,

Rn v + [Rn(u −1)+ (u +1)](u −1)

v=−v. (27)

v2(Rn +1)−2uRn +u2(Rn +1) = 1−Rn.

v2 − 2uRn

1+Rn+u2 = 1−Rn

v2 +(u − Rn

− R2n

(1+Rn)2= 1−Rn

1+Rn. (29)

Finally, the desired relationship is

v2 +(u − Rn

(1+Rn)2. (30)

When plotted on the u − v plane, in Cartesiancoordinates, Equation 30,

v2 +(u − Rn

(1+Rn)2,

represents a circle with center

u = Rn

Rn +1and v = 0,

and radius1

Rn +1.

For different values of Rn, we can generate a familyof circles. They represent normalized resistance loci.

v Normalized Resistance

Normalized Reactance LociWe can repeat the above process for the normalizedreactive element Xn.

(u −1)2 +(

v − 1

. (31)

This equation too results in a family of circles, thistime, centered at

u = 1 and v = 1

and with radius1

v Normalized Reactance

For any passive impedance, Re(Z ) ≥ 0 and z0

real, |Γ| ≤ 1. Thus all possible values of Γ can beplotted on this polar chart having a maximumradius value of unity.It can accommodate reflection coefficient valuesfor all impedances from short (Γ= 1∠180) to anopen (Γ= 1∠0)

Constant VSWR CirclesSmith chart is really a plot of the reflection coefficientΓ= u + j v . Constant VSWR circles are actually |Γ| =constant circles. From Equation 20 we have therelation

VSWR= 1+|Γ|1−|Γ| .

We can, therefore, relate |Γ| and VSWR to plot thecircles.

For a matched system, the reflection coefficientwill be zero and VSWR will be unity, the centerof the Smith chart.For al reflection coefficient of one, the VSWRwill be infinity and will overlay the contour forzero resistance.

∞Constant VSWR

vNormalized resistanceNormalized reactance

Constant VSWR

Points and Properties of Smith ChartAt point u = 1, v = 0 normalized resistance andreactance terms are large: position for an opencircuit.At point u = 0, v = 1 the value is 0+ j 1: positionfor perfect inductive reactance.At point u = 0, v =−1 the value is 0− j 1: positionfor perfect capacitive reactance.At point u =−1, v = 0 zero resistance and zeroreactance contours intersect: position for theshort circuit.Point u = 0, v = 0, the center of the Smith chart,represents the normalized impedance 1+ j 0,which under normal circumstances, representsthe transmission line characteristic impedance.Antennas and Propagation: Transmission Lines Smith Chart 44/72

Point Comment Representationu = 1,v = 0

normalized resis-tance and reactanceterms are large

open circuit

u = 0,v = 1

0+ j 1 perfect inductivereactance

u = 0,v =−1

0− j 1 perfect capaci-tive reactance

u =−1,v = 0

zero resistance andzero reactance con-tours intersect

short circuit

u = 0,v = 0

center of the Smithchart, normalizedimpedance 1+ j 0

characteristicimpedance

ExampleConsider a load impedance ZL = 17.7+ j 11.8Ωconnected to a 50Ω line of length l =λ/6. Plot ΓL,and graphically obtain Γin. Compare the result withthe computed value.

Solution

Form Equation 18

ΓL = 17.7+ j 11.8−50

17.7+ j 11.8+50= 0.5∠150.

We notice that if the line is lossless α= 0, and γ= jβ.So, 2βl = 2×2π/λ× (λ/6) = 2π/3 = 120. We can useEquation 22 to obtain the input reflection coefficient.

Γin = 0.5∠150−120 = 0.5∠30.

The same result can be graphically obtained.Because |ΓL| = |Γin| for a lossless line, we merelyhave to rotate clockwise from ΓL an angle 2βl (120).

Γ= Vref

Vinc= ZT −Z0

ZT +Z0.

Γin = ΓLe−2γl .

Normalized resistanceNormalized reactance

Constant Γ

+150 +30

Impedance Transformation UsingSmith Chart

The above example showed us that thegraphical solution of the reflection coefficienttransformation is simple.However, we had to compute ΓL using theimpedances. It is possible to solve impedancetransformations completely graphically.We just need to replace every point on the polarreflection coefficient chart by it normalizedequivalent impedance.We can do this by using the following equations:

Z0= Zn = 1+Γ

1−Γ , andY

Y0= Yn = 1−Γ

1+Γ (32)Antennas and Propagation: Transmission Lines Smith Chart 49/72

ExamplePlot the following on the Smith chart.

1 Γ1 = 1∠180

2 Γ2 = 1∠0

3 Γ3 = 0.5∠180

4 Γ4 = 0.5∠0

5 Γ5 = 0.5∠90

6 Γ6 = 0.5∠−90

+ j 0.2

+ j 0.4

− j 4

− j 2

− j 1

− j .4

− j .2

311 30

Constant VSWR

Γ1 : Z = 0 ∞= Z : Γ2Γ3 : Z = 1/3 Γ4 : Z = 3

Γ5 : Z = 0.6+ j 0.8

Γ6 : Z = 0.6− j 0.8

Fortunately, we do not have to use theequations as, on the Smith chart, the nor-malized impedance grid has been superim-posed.

ExampleA 5.2 cm length of lossless 100 Ω line is terminatedin a load impedance ZL = 30+ j 50Ω.

1 Calculate ΓL and SWR along the line.2 Determine the impedance and admittance at the

input and at a point 2.0 cm from the load end.The signal frequency is 750 MHz.

Solution

ZL = 30+ j 50, and Z0 = 100. Therefore, ZLn = 0.3+ j 0.5.

+ j 0.2

+ j 0.4

− j 4

− j 2

− j 1

− j .4

− j .2

311 30

Constant VSWR+ j 0.5

ZLnZLn

+ j 0.5

123.42

Z1n = Rn = 4.23

|Γ| = 0.6176

1 Γ= 0.617∠123.42. The constant |Γ| circle is theSWR circle. We obtain 4.23 as the SWR valuefrom the point of intersection between the SWRcircle and the right half of the resistance axis.This point is marked as Z1n = Rn. At this pointRn = 1+|Γ|

1−|Γ| .2 λ= 3×108/750×106 = 40 cm. In order to find the

impedance at the input, we must travel5.2/40 = 0.130λ toward the generator, along aconstant VSWR circle.

+ j 0.2

+ j 0.4

− j 4

− j 2

− j 1

− j .4

− j .2

311 30

Constant VSWR+ j 0.5

ZLnZLn

+ j 0.5

123.42,0.0786λ

Z1n = Rn = 4.23

|Γ| = 0.6176

1 We read off the chart that Zinn = 2+ j 2.Therefore, Zin = 200+ j 200.

2 Admittance points are directly opposite (across(1+ j 0)) in the Smith chart. We read off the chartthat YLn = 0.88− j 1.47.

3 Do the other calculations on your own.

Some Impedance MatchingTechniques

There are two common types of impedancematching:Conjugate Matching: The matching of a load

impedance to a generator formaximum transfer of power.

Z0 Matching: The matching of a load impedanceto a transmission line to eliminatewave reflection at the load.

Antennas and Propagation: Transmission Lines Impedance Matching 58/72

Conjugate Matching

Maximum power is delivered to a load when ZL

is set equal to the complex conjugate of thegenerator impedance ZG . That isZL = Z ∗

G = RG − j XG .In situations where the load impedance is notadjustable, a matching network may be placedbetween the generator and the fixed load.

Z0 MatchingThis type matches a load impedance to thecharacteristic impedance of a transmission line,i.e., ZL = Z0.In this case ΓL = 0 and hence SWR along theline is unity.If ZL 6= Z0, a matching network may be used toeliminate the standing waves on the line.With ZL connected to the transmission linethrough the matching network, Z0 matchingrequires that the input impedance of the networkequal to Z0 or Zinn = Zin/Z0 = 1+ j 0. For a Smithchart normalized to Z0, this means that Zinn

must be at the center of the chart.

Γ= Vref

Vinc= ZT −Z0

ZT +Z0

VSWR= 1+|Γ|1−|Γ|

For a well-designed source, ZG = Z0, thecharacteristic impedance of its output line.With Z0 real, matching the load to the line(ZL = Z0) results in a conjugate match betweenthe generator and the load.

Series Reactive MatchingExampleFind the reactance X in the following figure toachieve matching.

25+ j 30

To generatorZ0 = 50Ω Z01 = Z0

Matching network

+ j 0.2

+ j 0.4

− j 4

− j 2

− j 1

− j .4

− j .2

311 30

Constant VSWR

ZLn2.87

61.090.065λ

Effect of adding j X n =− j 1.1045

−61.09

0.235λ

Effect of adding j X n = j 1.1045

If we assume that the design frequency is 2000MHz and the wavelength within the line is equalto the free space wavelength, we find thatl = 0.065×λ= 0.065×15 cm. l = 0.975 cm. This isif we pick the point E , the first intersection pointbetween the constant VSWR circle and Rn = 1circle.If we pick point A, we get l = 0.212×15 = 3.525cm. Since ωL/Z0 = 1.1, we have ωL = 55Ω. Thewe get L = 55/2π f = 4.38 nH.With the matching network, Zinn = 1+ j 0 and thusthe SWR on the 50Ω line to the left of Zin is unity.

The matching network that we designed will notperfectly match at frequencies other than the designfrequency. This is because both l/λ0 and ωL/Z0 arefrequency dependant. λ0 is the free spacewavelength.

Freq.(MHz)

λ0 (cm) l/λ0ωlZ0

ZAn Zin SWR

1800 16.67 0.212 0.99 1.4− j 1.25 1.4− j 0.26 1.52000 15.00 0.235 1.10 1.0− j 1.10 1.00 1.002200 13.64 0.259 1.21 0.7− j 0.95 0.7+ j 0.26 1.65

Shunt Reactive MatchingExampleFind the susceptance B in the following figure toachieve matching.

ZL = 20j BTo generator

Z0 = 50Ω Z01 = Z0

Matching network

Smith chart solution must be carried out inadmittances.ZLn = 50/Z0 = .40Ω.We can find YLn by rotating the ZLn point halfwayaround the chart on its SWR circle.From the chart, we get YLn = 2.5+ j 0.Since matching requires Yinn = 1, the length l ischosen so that the real part of YAn is equal tounity.As before, there are two possibilities since theSWR circle intersects the Gn = 1 circle at twopoints, YAn = 1+ j 0.95 and YAn = 1− j 0.95.

+ j 0.2

+ j 0.4

− j 4

− j 2

− j 1

− j .4

− j .2

311 30

Constant VSWR

YLn = 2.5+ j 0 0

YAn = 1− j 0.95

0.09λ

−64.623

Effect of adding j X n = j 0.95

For l = 0.09λ, YAn = 1− j 0.95, and with Y0 = 0.01 S,YA = 0.02− j 0.19 S.This means that a shunt capacitance ofadmittance value + j 0.019 S is required to cancelthe inductive effect of YA.Thus adding a capacitive susceptanceB =ωC = 0.019 S at plane A results in Yin = 0.02 Sor Yinn = 1.0, a matched condition.Short-circuit or open-circuit stubs may be usedto obtain the required susceptance.It is best to use a high impedance shorted linefor inductive susceptance and a low impedanceopen-circuited line for capacitive susceptance.

Shunt Reactive Matching with aShorted Stub

shortls

Shunt Reactive Matching with aOpen-Circuited Stub

openls

Stub Tuners

We can do stub tuning using either series orshunt stubs.However, series and shunt reactance matchingtechniques usually result in low SWR valuesover narrow range (typically, less than 20%).Improved broadband performance can often berealized by utilizing series and parallel resonantcircuits.

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