956 15. THE SMITH CHART, IMPEDANCE MATCHING. AND IRANSMISSION LINE CIRCUITS

214¿+jX¡

FIGURE 15.7

15.2 THE SMITH

A simple Eansmission line used ro introduce the Smith chart.

CHART'

To better understand the Smith chart and to gain some insight in is use, we will':build" a Smith chart, gradually, based on the definitions of the reflection coeffi-cient. Then; efter all aspects of iüe cha;t are understood, we will use the chart in anumber of examples to show its utility. In the process, we will also define a numberof transmission line circuits for which the Smiü chart is commonly used. Considerthe circuit in Figure 15.1. The line impedance is real and equals Z6,but the loadis a complex impedance Zt : Rt * jXt, where R¿ is the load resistance and X¿the load reactance. The reflection coefñcient (see Eqs. (14.91) and (14.92)) maybe written in one of two forms. The first is a rectangular form (that is, written incomplex variables):

n__Zt-h_(h-h)+jXt n.IL: zL+ñ: úTá+ñ:

r ,* i r ¡ (15.1)

The reflection coefficienr is not modified by normalizing the nominator anddenominator by Zs:

(zL-zúlh :\!,,3 !!*i!:,3 :\, ,2!i: : r,*jr¡ (r5.2)' L : (zL + hyh -

Gu,^T r) +jwT - e ¡ 1¡ *¡*

: t

To obtain this resulg we substin¡ted r : Ry/Z¡ and x : XtlZo as the normalizedresistance and reactance- For much of the re¡nainder of this chapter, we will dropthe specific notation for load pardy to simpli$' norarion but mosdy because themagnitude of the reflection coefñcient remains constant along the line and, therefore,the results we obtain apply equally well for any impedance on rhe line (see Figuref5.2). In the latter case, the generalized reflection coefficient is obtained and thiscan be written in exacdy the same form as Eq. (15.1) or (15.2) by replacing Zr v¡tútZ(z). Equation (15.2) defines a complex plane fsr the reflection coefEcient as shownin Figure 15.3a. Any normalized impedance (load impedance or line impedance) isrepresented by a point on this diagram.

The second form of the reflection coefÉcient is the polar form. This may bewritten as

17: lfleior : l/-l(cos 0r * j sin d¡) (15.3)

tThe Smitl chart "r'as

incoduced by Phillip H. Smith inJanuary 1939. Snüúr deveiopcd rhc ch¿n as anaid in calculetion and c¿lled it a "trarxmission line calbulator." Over half a cenn¡¡y later, the chart is asuseñ¡l as ever as e standard tool in analysis either in its printed form, slide-rule form, or, more recentl¡as computer programs and instrument displays.

15.2. THE SMITH CHART 9$7

T+ h l lzm.t l

____l/ZunÁ.me+jXmc

FIGURE l5.2 Use of an equivalent transmission line to dqscribe the line impedance ¿t a distance a fior¡the load.

'illa-la

)erlerradXrrayin

.1)

Lnd

: .2)

'nd.opüe,f€r

trehisithwn)is

be

.3)

:s ¿srdy,

I

I-=-1

G=l

r,(0,0)

rr=-1

(f , - l ) r*r , : : - ( / : ,+t)f ¡ r*( f , - l )x:- f ¡

t

,r +t

-l n(0,0)

\_,

nn

b.

FIGURE 15.3 The complex plane ieprese¡c¿tion of the reflection coefficient. (a) In rectangular fonn.

(b) tn polar iorm.

where d¡ is the phase angle of the load reflection coefficient as discussed in Sectio¡"r,14.7.t. For a given magnitude of the reflection coefiñcient, the phase angle definesa point on the circle of radius l&1. Thus, since lf¿l S 1, only that section of the

rectangular diagram enclosed by the circle ofradius I is used, as shown in Figurc15.3b. The polar form is more convenient to use ihan tle rectangriar form but we

will, for the moment, retain both.We now go. back to the rectangular representation and calculate the real and

imaginary parts of the reflection coefficient in terms of the normalized impedance.The starting point is Eq. (15.2):

(r- l )+jxf, * if¡" r ' ( r+l)* j*

Cross-multiplying gives

? + Dr, - f¡x *jf¡(r + l) +jrcr, : (r - L) +jx

(15.4)

(1 5.s)

Separating the real and imaginary parts and rearranging terns' we get two equations:

(1s.ó)(1s.7)

We now write two equations: one for r and one for r, by first eliminating r and then,separatel¡ r. First, we eliminate r by substituting from Eq. (15.7) into Eq. (15.5)-

958 I5. THE SMITH CHARI IMPEDANCE MATCHING. AND IRANSMISSION LINE CIRCUITS

After rearranging terms, this gives

r j@ + r) -2r . r + r |@+ l) : | - r 'Diüding by the conunon term (r a l),

pz- 2f , r t r2- l - r' r ( r+l) t ' i - ( r+t)

(1s.8)

(1s.e)

Adding r2/(r + l)2 to both sides of the equation and rearranging tenns, we get

/ * 12(n-, ' ) -+r l\ ' ( r+l) / ' - ' ( r*r)2 (15.10)

Repeating the process starting with Eqs. (l¡.ó) and (15.s) and eüminating r fromboth equations, we get an equation in tirms'of r alone:

(15.r1) FIGintr

lothEq. (15.10) and Eq. (t5.l l) describe circles in the complex /- plane. Equation(15.10) is the equation of a circle, with its center ar f, : r/(i + l), i. : 0 and radiusI/(r ¡ 1)- The center of any circle is on the real axis and can be anywhere betweenf, :0 for r :0 to ,ll. : I for r --+ @. For example,for r: 1, ú\. center of thecircle is

^t r, -- 0.5 and its radius equals 0.5. A number of these circles are drawn

in Figure 15.4a. The larger the normalized resistance, the smaller the circle. Alicircles pass through f, : 1,1r. :0.

From Eq. (15.11), we obtain yetanother set of circles forr. Sincer can be positiveor negadve, üe circles are centered ¿t + - Lo l¡ : l/x for positive values áfr andñ r, - l, f¡ : -r/x for r negative. These circles ¿re shcwn in Figure r5.4b fora number of values of the normalized reactance r. Figure l5.s shows the r and r

I;=0J'1=l

l-.-lfl'=O

IF1Il=0

n=0Jl-l

FIGURE 15.4 The basic components of the Smith chan. (a) Circles of constant values of r. (b) Circles ofconstent values of¡ or -*.

tr.

e, _ t¡2* (" _;)' : ('1)'

t52. IHESMI]|CHART {t5iÍ}

r)

rnts)nrenil

te

rd)rx

FIGURE 15'5 The Smith chart. A norlaliryd impedance is_ a point on the Smith chart defined hy tlreintersection of a circle of constant no¡malized resistnce t and e circle of consEnt norrn¿lized reactnncrr r.

circlSon$ef plane,tn¡ncarcdatthecirclel/-l: l.ThisisthebasicSmithcfrp¡.:.A.number of properties of dre two sets of circles are immédiately apparent:(f) The circles are loci of constant r or constanc r.(2) r and r circles are always orthogonal to each other.(3) There is an infinite number of circles for r and for r.(4) All circles pass through the point f, : l, L : 0.(5) The circles for ¡ and -.r are images of each other, reflected about üe real ¿xis.(6) Ttre center of the chart is at 1l. : 0, 1]. : 0.

-q0 The intersection of the r circles with the real axis, for r : r0 and r : llro, occrlr

' I at points q¡mmetric about the center of the chart (^|-, = 0, /l : 0).

^8)

The intersections of the r circles with the outer circle ([/-l : l) fior r - 16 an{i( r - l/xo occur ar points syrnmerically opposite each other.

(9) The intersecdon of any r circle with anyr circle gives a normalized impecla.ncc,point.

The chart as described above is an_impedance chart since we defined all points interms of normalized impedance. We will see how to use the chart as an admitancechart later.

In addition to the properties of the r and x circles given above, we note rhefollowing:

(1) The point [. : l,.C : 0 (rightrnost point in Figure 15.5) represenrs r : oo,r : oo. This is the impedance of an op"n tt"rrr-ission line. This point istherefore the open ciratit point.

(2) IF

diamericallyopposite point, at f" : -l,C : 0 represen6 r : 0, r : 0-This is the impedance of a short circuit and is called the'sbart ciranit point-

(3) The outer circlb represents l^|- | : l. The center of the diagram r€presenrslf | : 0. Any circle centered at the center of the diagram (4 j O, 1]. _' 0) wirlr

.l0

cf

960 I5. THE SMITH CHART. IMPEDANCE MATCHING. AND TRANSMISSION ttNE CIRCUITS

radius ¿ is a circle on which the magnitude of the reflection coefficient is con-stant, l/- | : ¿. Moreover, if we take the intersection between any r and r circles,the distance between thispoint to the center of the diagram is tire magnitude ofthe reflection coefficient for this normalized impedanci. A circle drañ throughthis point represents the reflecdon coefficient at different locations on the liiefor this normalized load impedance. The intersection of the reflection coef6-cient circle with r and r circles represents line impedances at various locations.These aspece of the use of transmission lines ari shown in Figure 11.5. Forexample, point A represenrs an impedance ra * jr¿ and point.á r"pr"r.rrt *impedance rB * jrn, but the magnitude of the reflection cóefñcient is the same.This will later be used to calculate the üne impedance as well as voltages andcurrents on the line.

(4) Aty point on the-chart represents a normalized impedance, say, z:r*jr.The adminance of this point is y : (r - jxl(r2 + rt¡. The ,drrri*rr". po'irrtcorresponding to an impedance point lies on the reflection coefi6cient circli thatpasses tlSough the impedance poing diametrically opposite to the impedancepoint. Thus, if we mark a normalized impedance as z-and draw the reflectioncoefficient circle ürough point z, ¡¡t¡s sirslipasses through the admittence point! = l/2. The admittance pointy is found by passing a line through

" *i th"

center of the diagram. The intersection of thii line with the reflectiJn coefficientcircle is pointy. These steps are shown in Figure 15.6a. These considerationswill later be used to calculate admitances instead of impedances.

The smittr chart also provides for calculation of phase angles and lengths oftransmission lines. For this purpose, the Smith chart ii eqüppgi with a nuriber ofscales, marked on tlre outer periphery of the diagram. Thise are defined as follows:

(1) For a given impedance, a point on the chart is found. The distance frorn thecenter of the chart to the point is the magnitude of the line reflection coefficien¿If the line connecting the center of theihart with the impedance point is con-tinued until it intersected the outer (1- : 1) circle, the lot¿tion ofintersectí6ngi_ves-the phase angle of the reflection coefficient in degrees. This is the first setof values given on tre circumference of the smith chai and is shown in Figure

FIGURE I 5.6 (a) Normalized impedance, reflection coefficient and normalized admirrance. (b) Indicationofphase angle ofthe reflection coef6cient on the Smith char¿

FIGop(

:on-:les,le ofrughline

rcffi-ions.. Forts an.ame.¡ end

* jx.

loint: thatlancection¡ointI thecienttions

hs ofer oflows:

: the:ient.con-:tionIt Set

gure

15.2. THE SMITH CHART 961

15.6b. Note that the open ci¡cuit point has zero phase angle (f : +l) and theshort circuit point has either a 180e or - 180' phase angle. The difference is i¡rthe sign of the imaginary paft of the load impedance Oelow or above the realaxis). Intermediate points will vary in phase depending on the distance from theload. For example, for pointl in Figure 15.6b, the phase angle of the reflectioncoefficient is 104", whereas for point B it is -120" .

(2) we recall that the disance between a point of maximum voltage and a pointof minimum voltage was found tobe ),/4 in section 14.7.3. tn panicutar, ttreimpedance of a shorted transmission line changes from zero to iofitt;ty (orneg-ative infinity) if we move a distance L/4 froÁ the shor¿ Thus, thé distanlebetween the short circuit and open circuit points is ),/4. This fact is indicatedon the outer circle of the chart, starting at the short circuit point- Since theshort (or any other load) can be anywhere on a line, we may wisir to move eithertoward the generator or toward the load to evaluare the line behavior. Thesetwo possibilities are indicated with arrows showing the diréction toward loadand toward generator @gure 15.7). Although the disunce is marked from the

¡hon-- circuil point, the distance is always relarive: If a point is given at anylocation on the charg movement on the chart a distance t74 represéna half rh;circr¡mference of the chart.

(3) The direction toward the generator is the cloch¡¡ise direction. If we wish tocalculate the line impedance starting from the load, we move in the clockwisedirection toward the geíerator. If, on the other hand, we wish to calculate the line

1nn9d""ce sarting from the generator going toward the load or, starting at theload and going away from t}re generator, we must move in the counterclóckwisedirection and use the appropriate distance charrs (see Figure 15.7).

(a) The whole smith chart encompasses one-half wavelength. This, of course, isdue to the fact that ¿ll conditions orr lines repeat at intervais oi lJ2 regardless ofloading or any other effect that may happen on the line. If we need io analyzelines longer that),/2, we simplymove around the chartas manyhalf-wavelensrhs

phase angle

FIGURE 15.7 Directions on the Smith charg and indication of SWR. The distance between shon andopen circuit points is 1/4.

¡tion

off l

962 15. THE SMITH CHARI IMPEDANCE MATCHING, AND TRANSMISSION LINE CIRCUITS

as_are necessary. Only the remainder length (length beyond any integer numbersof half-wavelengths) need be nalfzed.

The Smith chartalso allows for the calculation of sandingwave ratios. The standingweve ratio is calculated from the reflection coefficient as

swR: I + l/-l1- l / - l (15.12)

We note that the circle of radius lfl intersects the positive real axis ^ttc

:0. At thispoint, the normalized impedance is equal to r and the reflection coef6cient is givenas /- : (r - l)/(r * 1). Substituting this into the relation for SWR, we ger

SWR: T (15.13)

Thus, the standing wave ratio equals the value of normalized resistance at the lo-cation of intersection of the reflection coefficient circle and the positive real axis.From property (7, above, the intersection of the reflection coefficient circle withthe negative real axis is at a point 1/r. Thus, this point gives the value I/SWR. Therwo points are given for the reflecrion coefficient in Figure 15.7.

Now that we discussed the individual pars making up the Smith chart and havediscussed its use for individual calculations on a-simple line, it is time to put it alltogether. The result is the Smith chart sholm in Figure 15.8. You will immediatelyrecognize the r and r circles as well as tJre scales discussed. There are, however,a number of other scales given at the bottom of the chart as well as a number ofindications on the chart itself which we have not discussed. These have to do withlosses on the line (which we have neglected) and t}re use of the chart as an admittancerather than impedance chart (which we will take up later). This chart is availablecommercially as well as in the form of computer programs. Any values that fallbetween given circles are found by interpolation.

Although the chart is relatively simple, it contains considerable information andcan be used in many different ways and for purposes other than transmission lines.To see how the chart is used, we will discuss nexr a number of applicatiorrs of theSmith chart to design of transmission lines. Because the chart givés numerical data,the examples must also be nurneric, but, in general, the equations in the previouschapter can also be used for this purpose. The main difference in rhe Smith chartsolution and the analytic soludon is that the Smith chartuses normalized impedances,whereas in analydc calculations, we tend to use the actual values of the impedance.AIso, because it is a graphical chart, the results are approximate and depend on ourability to accurately read üe values off the chart.

w'EXAMPLE 15.1 Calculat ion of l ine cond¡t¡ons

A long line with characteristic impedance Zo : 50 f2, operates at I GHz. Thespeed of propagation on the line is r and load impedance is 75 +/i00 e. Fintl:

(a) The reflection coefificient at the load.

O) The reflection coefi6cient at a distance of 20 m from the load toward thegenerator.

¡¡0.9

01 0.6 @03 0¿ 0.1

0.9 0.E 0.7 0.6 60.9 03 0x

ling

.12)

thisiven

r . r3)e lo-axis.withThe

haveit allrtely3ver':r ofwithtnceablefall

andnes.- the

lata,ioushart.ces,nce.our

15.2. IHESMÍMCHART sCIr3

t . l t2t31¡ lÁtr2 t a5 ro¡ ' -0.a 0f0t l t5 I

0-a 03 0¿oá 05 0{

FIGURE 15.8 The comolete Smith chan.

Input impedance at 20 m from the load.

'T.L^ - .^-J:- -^;^ ^-

FL; l : -^l l tL )L¿¡¡u¡¡ lE wdvL ¡4uv v¡¡ urw ¡¡rr ! .

Locations of the 6rst voluge ma:qimum and first voltage minimum from theload.

t5 tó {s lF€ü2Cl0

3u¡lú 2 , tUMÚ

lhe(c)

/J\\u,,

7 {")_/,

the

:--L

IIlII964 I5. THE SMITH CHART, IMPEDANCE MATCHING, AND TRANSMISSION IINE CIRCUITS

FIGURE 15.9 The Smith ch¿rt for Example 15.1.

Solution.

(a) (l) Normalize the load impedance: z¡ : (75 + jl0})/50: 1.5 +12. Enterthis on the Smith chart at the intersection of the resistance circle equalto 1.5 and reactance circle equal to 2. This is point P2 in Figure 15.9.

(2) With center at origin (point P¡), draw a circle rhat passes through pointPz. This circle is t}re reflection coefficient circle and gives lf I anywhereon the line. Measure the length of the radius (distance beween P1 andP2) and divide by the radius of the Smith chart's outer circle. This givesthe magnitude of the reflection coefficient" In this cese, lf l :0.645.Note: The radius of the Smith chart should be equal to l, but to facil-itate reading, the size is often different; thus, the need to calculate themagnitude of rhe reflection coefificient.

15.2. THE SMÍIII CfIART 965

rterlual

9.

)intere¡ndves45.cil-the

(3) Draw e straighr line betrreen Pr and p2 and exrend it to the peripheryof the chaft, ro p-o1nt Pj_. The angle (in degrees, on the periphery) isthe phase ahgle of the reflection coefficient at the load. In this cas", it is37'. Alternativel¡ read the "wavelength toward generatoro circle. Thisis'equal to 0.198 at point Pi. To celcilate the an"gle, mbtrect this va¡rr:from the value on the real axis (open circuit point), and multipty by 4n:(0.25 - 0.198) x 4n :0.208rr radians or 37.. Thus, the amwér to (a) is

F1 : ll-¿lsis, ; 0.645efl.208n _ 0.645 t37"

(b) To calculate the reflection coefificient at 20 m from the load, moving tower:dthe generator, we first calculate the wavelength bec-ause dre chart can onlyaccommodate wavelengths :

. c 3x108

i : *n :o' ' [m]

si¡ce the circumference of the smith chan represenrs 0.51 (or 0.15 m),the 20 m dist¿nce represent (20/0.15) : L33.3334 half-uavelengths. Thus,we move around the reflection coefficient circle towerd dre generator 133times, starting at P2- This puts us exacdy where we srarred (at point p2). Theremainder is one-third of a half-wavelength or ),/6.

We now move from point P2 along the reflection coefficient circle, a distanceof 0.167 wavelengths to¡¡¡ard the generator to point P3. connecting this pointwith the center of the éhart and with the circumference gives the intersectionwith the reflecrion ioefficie¡t circle at P¡ end with the-circumftrence at pj.This point gives the phase angle of the reflection coefficientes -8J" Thus thereflection coefficient at 20 m from the load is

f=0.6451-83"

(c) The input impedance 20 m frorn the load is represented at point pl. Thenorrrralized input impedance is

z( l :20m):0.46 - j1.02

Multiplying by the characteristic line impedance (fr : 50 O), we get the actualline impedance as

Z(l = 20m):23 - j | r [o:

(d) The refl ection coefficient circle. intersects the real axis.at pointPa. At this poingr : 4.76:, This is the standing w¿ve r¿tio: SWR= 4.76. At point p5 (on theother side of the reflection coefficientcircle) r : I/SWR : 0.21. At point p4,the line impedance is real and maximum and equals !rr, : fux4.76 - Z:S O.At poitrt P5, the impedance is minimum and red and equals {,6o : fr/4.76 :10.5 s¿.

(e) Location of maximum voltage is on the real axis at the same point whereswR : 4.7ó since, at this point, the line impedance is maxi¡nrm (and real).Thus, moving from point P2 to the positive rial axis, we reeclr avoltage max-imum: The distance is the difference in wavelengths be¡reen point p* and

I

VII

I

L-

I

966 15. THE sMrTH CHARI TMpEDANCE MATcHTNG, AND TRANsMrssroN LINE ctRCUtTS

point P2 or /nrax : 0.25¡. - 0.198¡' : 0.0521' from the load. The voltage min-imum is a quarter-wavelength tway (where I/SWR : 0.21) at point P5 or/^in : 0.302¡, from the load. In terms of actual distance the first maximumoccurs ata distance of 0.052 x 0.3 :0.015ó m, or 15.ó mm from the load.The first minimum occurs at 0.302 x 0.3 : 0.0906 m or 90.ó mm from theload.

15.3 THE SMITH CHART AS ANADMITTANCE CHART

We mentioned earlier that the Smith chart may be used as an admittance chart. InFigure 15.óa, we showed that for any given normaüzed impedance, the admittanceis found by locating the normalized impedance point z : r * jr on the S¡irithchart, drawing the reflection coefficient circle, and then drarving a straight line thacpasses through the impedance poing the center of the chart, and then intersects thereflection coefficient circle, again, on a point diametrically opposite the impedancepoint, at pointy. This point represents the normalized admittance of the load. Anynormalized impedance may be converted into is equivalent admittance using thissimple step.

In addrtion to this, we note that an infinite normaiized impedance (open circuitpoint on the impsdance Smith chart) represents infinite admittance. Similarly, theshort circuit point on the impedance Smith chart represents zero admittance on theadmittance Smith chan.

The admittanee may be w¡itten in terms of the impedance at point z as

(15.14)

Since we use the same charq the constant-resistance circles now become constant-conductance circles and the constant-reactance circles become constant susceptancecircles. All other aspects of the chart, including phase angles, distances, etc. remainunchanged.

-l'he use of the Smith chart as an admitance chart is shown in Figure 15.10, incompárison with the impedance chart.

V EXAMPLE 15.2

A load, such as an antenna, of impedance Z¡ :50-i 100 Q is connected to a losslesstransmission line with characteristic impedance h : L0O Q. The line operates at300 MHz and the speed of propagation on the line is 0.8c.

(a) Calculate the input admitance a distance 2.5 m from the load.

(b) Calculate the input impedance a distance 2.5 m from the load.

(c) Suppose the load is shorted accidentally. What is the input adrnittance at thesame point?

rn-or

.um)ad.the

t,hancemiththats theenceAny'this

rcuit; ther the

r.14)

tant-ancenain

0, itt

;less:s at

I5.3. THESMITHCHARTASANADM|ITANCTCHART 9$7

OC (infinite ímpedance)IOC (infuite adnittance)l

FIGURE 15.10 Relations between the impedance and admittance Smith chare. Descriptiom in squarebrackets are for the admittance chaft.

Solution. To calculate the inprrt admittance, w'e first cdculate the wavelength onthe line. The load is then located on the impedance chart and the admitdnce isfound on the reflection coefficient circle. Then, we move towerd the generaror adisance 2.5 m (in wavelengths, ofcourse) to fnd the normalized input idmittence.The admittance is found by multiplying *ith the characteristic admitance of the!ry. rhg input impedance'can be found from the input admi$arce.by finding thediametrically opposite point on the reflection coefif,cient circle.

(a) The normalized load impedance is

50 - t100zL: -1ft- :0.5 - j l

This is marked on the chart as point P2 in Figure rs.li. The reflectioncoefficient circle is drawn around point P1, g'irh a radius.equal to the dis-tance between P2 and P1. The admittance point is P3. The normalized loadadmittance is

yr:0.4 + j0.8

The wavelength on the line is i. : 0.8c/f : 2.4 x 10813 x lOs : 0.8 m.The given disance represenrs 2.5/0.8 :3.L25 wavelengths. To 6nd the inputadminance, we move from the load adminance point toward the generarora distance of 0.125r" (the three wavelengt]rs mean simply m ing six timesaround the chart ro get 19 the initial point). Moving from point.pi a disance0.1?5]" bringsus topointP{ (0.114¡"+0.125}, :0.2391). Connectingthispointwith Pt intersects the reflection coefficient circle at point Pa. The nonnalizedinput line admittance is

la:4.0 + j l .Os

The innut line adrnittance is úe nc:malized inn-'rt line rrlmitto..o .h^,.o_ -__ ---r__ úrrs. e¡ ! ¿ú¡uL¡ '&L_ óuvvÉ

multiplied by the char¿cteristic line admittance, which equals 0.01:Ya:0.04 +10.0105 [t/f,¿]

: the

968 I5. THE SMITH CHARI IMPEDANCE MATCHING, AND TRANSMISSION LINE CIRCUITS

FIGURE 15.11 Use of the Smith chart as an adminance charr @xampte 15.2).

The normalized input impedance is found by locating point P5, which is thediametrically opposite point to Pa, on the reflection coefficienr circle. The nor-malized line impedance at this point is 0.2I =i0.06. The line impedance is foundby multiplying this normalized impedance by the characteristic impedance ofthe line:

z;" :23 - j6 ta l

I f the lorr l is shorte. l tho l^ .á i - - - . ln-^- ic -qr¡ ^- l

t l . . ! i - . "J- ie^-.^- i "r ¡ r ¡ l aurruLLd¡tLL ¡J

infinite. This is represented at point Po, on the admittance chart. From here,we move 0.125 wavelengths toward the generator on the outer circle, since forshorted loads, l/- | : 1. This point is shown as P7. The normalized input line

o)

1?s

,s

G)

t_

15.4. IMPEDANCEMATCHINGANDTHESMITHCIIART 96(}

admittance is -71. The line admittance is, therefore, i0.01 Qine impedanc,rtisl l00, at point Ps).

15.4 IMPEDANCE MATCHING ANDTHE SMITH CHART

15.4.1 lmpedanceMatching

When connecting a transmission line to a generator, a load, or another transmisslt,¡¡lirie, the impedances are, in general, mismatched and the result is a reflection coeffi-cient at the load, generator, or discontinuity, which, in turn, generetes stending wavr,s

on the line. The effect of this reflection was disorssed at some length in Chapter14. It is often necessary to match a transmission line to a load or.to a generator, forthe purpose of eliminating standing waves bn the line. Similarl¡ if a discontinuity

exists, such as the connection,of an unmatched line section, it is often necessary toeüminate this mismatch before the line can be used. The result of mismatch on aline can be disasuous: Large amounts of'reactive power may travel along the lincwhich can easily damage circuitry, especially generators.

A transmission line is matched to a load if the load impedance is equal to thecharacteristic impedance. Similarly, if the line impedance is equal to the generatorimpedance, the two are matched. To match a load to a line (cr a generator for thatmatter), a matching network is connected between the line and the load, as shov'¡¡in Figure L5.12.

The location of the matching network depends on the application. If we wish toredtrce the standing rvaves on the line, the matching ner*'ork should be located asclosely as possible to the mismatched impedance. If, however, the. line can operatewith standing wav€s, then a more convenient location, at some distance away, canbe found. The latter approach is possible since all conditions on the line repeat atintervals of )J2. Thus, if a matching network has been designed to be located ata given point on the line, the network can now t¡e moved a distance ]./2'withoutaffecting the line conditions.

There are two types of impedance matching necworks that are particulady useful.One is the so-called stub matching, which makes use of properties of shorted (or

open) uansmission lines. In this type of networh the impedance on the line is alteredbyconnecting shorted or open transmission lines in parallel or in series with the line

t¡I

nor-rundce of

; the

'ee ishere,:e fort line

a. b. c.

FIGUR€ 1 5.12 Matching netwórks at (a) generator side; (b) load side; (c) arbitrary location on the lirrc.

¡i

I

970 15. THE SMITH CHART. IMPEDANCE MATCH¡NG, AND IRANSMISSION LINE CIRCUITS

to adjust the impedance. The second method of impedance matching is based on theproperties of transformers. In effecg we build a transformer which then can matchtwo impedances in a manner similar to that discussed in section 10.7.

The following sections discuss these methods and develop the relations requiredto design

-matching nerworks. we use the Smith chart in the design of matching

networks for two reasons: First, in many cases, the design is greatly simplified by th!use of the Smith chart. Second, and more importantly, t¡re Smitfr chart is rouúnelyused for this type of application.

15.4.2 Stub MatchingThe idea of stub matching is to connect open or short circuited secrions of trans-mission lines, either in parallel or in series with the transmission line as shown inFigure 15.13. The impedance of the srub anüor location on the line are chgsensuch that the combined impedance of line and stubs is equal to the characteristicimpedance of üe line. Although, in orinciple, matching ro any line impedance ispossible (other than a purely reactive line), because shonád transmission lines have apurelyimaginaryimpedance, stubs are normallyused to match loads to transmissionlines with resistive characteristic impedances; that is, the impedance of the shrb, orstubs, is chosen such that it cancels the reactive part of the line impedance.

ConsiderfirstthematchingnetworkinFiguré l5.13a.fusumin¡gacharacteristicimpedance Ze (or admiaance l¡) and a line

"d-itt"rr.. yo*jBo,af a?istance /r from

the load, the two can be matched by adding a stub in p"r"ll"l, at distance d1 fromthe load, such that the admittance of the stub is -.78¡ . The distance /1 defines theimaginarypart of the line admittance from Eq. (14.102). /1 is trren that length of theshorted transmission line stub that cancels thé imaginary part of the line aimittanceat the location of the srub. The choice of /¡ and ¿i ir

"bi unique, but any practical

combination that satisfies the above conditions can be used.Akhough a single srub may be used ro match any load (except for a purely imagi-

nary-load) to any line which has real characteristic impedance, sómetimes the physilalconditions of the line do not allow perfect matching with a single stub becaúse ofphysical constraints. In such cases, two stubs, at wó fixed locaüons, mav be used.This method is similar to the single stub method, bur now we

-ort design Áe lengths

\ and 12 while /1, and d.2 are fixed as shown in Figure 15.l3b.In the series matching method in Figure l5.l3c, the idea is the same as in single

stub matching: W'e must choose a srub length /1 and place it a distance /1 from úre

¡s

Xr

br.+Xr

FIGURE 15.13 (a) Single stub matching. (b) Double stub matching. (c) Series scub matching.

the.tch

rredúngthe

nely

rs.4. TMpEDANcEMATCHINGANDTHESMITHcHART 97X

load so that the sum of the line impedance at that point with that of the stub equalsZo.

To summarize, in the single stub matching method, we bhooie the length ar-rrjposition of the stub. In the double stub matching method, we choose the lengths oftwo stubs while their positions are fixed and often prescribed by the deüce bein¡i.matched. It is also possible to match loads and other devices by more tltan two stub:',but we will not discuss these here.

15.4.2.1 Single Stub Matching

The idea of single stub matching relies on the fact that the line impedance variesalong the line and a parallel or series stub changes only the reacrive part of th_eline impedance. To see how this is accomplished, consider a load impedance Z¡ :

h * jXt connected on a line of characteristic impedance Zs. For the load to bematched, its impedance must be changed so that Z'r: Zs. This is done as follows:

(l) Move along the line from the load (Figore 15.13a) and find a point at whichZ(z) : Zo + jX(z). Note th¿t Zx d,oes not have to be real, but in most cases, ifwill be.

(2) At this point (a distance fi from the load), connect a shorted or opened trans-mission line of a length { such that the termjX(z) cancels. As a result, the linesees a total impedance equal to 26 and the new load (which now is the wholcline section to the right of the location of the snrb) is matched.

These steps are implemented with the use of the Smith chart with the foilowingdifferences:

(1) The impedance is first normalized to conform r¡'ith the requirements of theSrnith chart.

(2) If the stub is connected in parallel (Figure l5.l3a), it is easier to workwithadrnittances. Therefore, the normalized load admittance is first located on thechart.

(3) If the stub is connected in series (Fig,r.e 15.13c), it is easier to work withnormalized impedances.

The stubs will be assumed to have the same characteristic impedance as the line,but this is not a necessary condition. The following two examples show the stepsand details involved in parallel and series single stub matching.

V EXAMPLE 15.3 Application: Single parallel stub matching at anantenna

An antenna operates at a wavelength of 2 m and is designed with an impedance of75 Q: However, because of mistakes in design, the antenna is badly mismatched.The measured impedance afterinstallation is l5 +160 O. The antenna is connecredto e75 O line as shown in Figure 15.14. Calculate:

(a) The required shorted stub and its location on the line to match the antennato the line. The line and stub have the same characteristic impedance.

:ans-minLOS€ll

risticrce isiave a

ssionrb, or

risticfromfrom:s iherf the:ance:tical

nagl-¡sical

se ofused.rgths

- !t rIt.JX

ingleo the

i

972 I5. THE SMITH CHART. IMPEDANCE MATCHING,,AND TRANSMISSION LINE CIRCUITS

15+i60 Cr

FIGUREline.

15.14 Mismatched antennaconnected to a line and a stub designed to match the antenna to the

(b) The shortest required open circuit stub that will accomplish the same purposeas the short circuit stub in (a).

Solution. Firsg we find a location on the line at which the real part of the liíeadmittance is equal to the characteristic admittance of the line; that is, flrl.d Z(d)such that Y(d) : Yo +jB(d). Now, we connect a shorted stub in parallel with theline at üis point and of a length such thet the imaginary part of the line admitanceis canceled. The open circuit stub in (b) is placed at the same location and is lengthis that of the short circuit stub +¡./4.

(a) Itt rhis case, it is simpler to use the Smith chan as an admitance chart. To doso, we first calculate the normalized load impedance:

rs + j60t )

zL: :0.2 +10.8

\.I/e mark this point as P2 on the Smith chart in Figure 15.15, usingüe chart as an irrpedance chart. The reflection coef6cient¡i¡cle is nowdrawn around the center of the chart, with the radius equal to the distancebetween P2 and P1.

fb find the load admittance, we draw a suaight line from P2, throughP1 and extend this line to the periphery of the chart. The line intersectsthe reflection coef6cient circle at point P3. This point is the normalizedload admittance:

J7:0.3 - i l . l8

As we move around the reflection coef6cient circle, the line admitancechanges. To match the load, we must find the location at which the realpart of the line admittance equals the characteristic admittance. Since weare working with normalized admittances, this happens when Re[¡] ;1. This happens at the locations at which the reflection coefñcient circleintersects the circle g : 1. The two possible points *e Pa and P5. Theline admitance at these points is

At P+,

la: | + j2.6

( t )

(2)

(3)

T-

1,sI

15.4. IMPEDANCE MATCHING AND THE SMITH CHART 9V3

' '

FIGURE 15.15 Smith chart for Example 15.3.

at ?5,

yb: l_ j2.6

Each one of.these points provides one possible solution.

(4) Solution No. 1 : Point Pa. The distance /1 for this solution is the distanceuaveled from point P3 to point P4, on the reflection coefficient circle.The distance in wavelengths is the difference in reading:s berween pointPi and Pi moving from Pj to Pi toward the generator. First, we movea distance of 0.5), - 0.358¡" : 0.14D, up to the short circuit poinuThen,'te mowe an addi¡onal 0.198¡, to point Pf. The total distance isdu :0.142), + 0.198¡. : 0.341".

The normalized line susceptance at this point is 2.6. The stub must,therefore, have a normalized susceptance. of -2.ó. This point is shown

r the

)ose

line'(di,, thelncergth

¡do

,uBh;ectsized

ulcerealewe

i) :i¡cleThe

l

I:

smgnowlnce

¡I,

974 I5. THF SMITH CHARI, IMPEDANCE MAÍCHING, AND TRANSMISSION LINE CIRCUITS

as point P'i-The length of the stub is the disance from the open circuitpoint P,. (inñnite adminance) to point P{ (moüng toward the generator).This is /r, : 0.J081. - 0.25.1,'- 0.058¡,. Thus, the first possible solutionis

dto :0.34),, /ro : 0:058,1"

Since we know the wavelength Q, : 2 m), we can write the solution inactual lengths:

du:0.68, lv :0.116 [m]

(5) Solution No. 2. Point P5. The disance d1 at this point is the distancebetween points P3 and &.Ag"io, we move a distance of 0.142). up tothe shoit circuit point and then a disance of 0.i02), from the sñortgircuit point to point P5. Thus, /1¿ - Q.g4¡. The lirte susceptance etP5 is -2.6. The stub susceptance must be *2.ó. This is marked as pointP'i.The distance from P* to pointPf, moving toward the generator, ish :0.25], + 0.I92L :0.44D,. The second soludon is therefore

t

du:0.444L, ht:0.442) ' or / r¿:0.888m, /r¿ :0.884m

(b) Because an open line behaves as a shorted line ar a distance of ),/4 from theshort, the lines in (a) can be replaced by open circuit lines by either shortening.the stubs bv )'/4 or lengthening them by L/+.Thking in each case the shorcesipossible stub length (length ening l¡ and sirorténi ng hú, the solutions for opencircuit stubs are

du :0.34X: 0.ó8 [m],dru = 0.444)': 0.888 [n ],

ho :0.442),: 0.884 tm]/l¿ : 0.0581. : 0.116 [m]

of half-wavelengths ro /¡ or /¡ or

ho:0.308), : 0.ó1ó [*]Iu, :0.192) ' : 0.384 [*]

a EXERCTSE 1s.r

Suppose that in Example 15.3, it is not phpically possible ro connecr the stub ateither location found. The nearest location at which a stub may be connected isI m from the load.

(a) What are the solutions for d1 and I]

(b) fue these solutions unique?

Answer. 1 m:0.5),. The solutions are:

(")

db: Q.444 + 0.5)¡" : 1.888 [*],úb: (0.34 + 0.5)¡, : l .ó8 [m],

(b) No. The addition of any integer numberboth are also acceptable solutions.

/

)

I

crrltror).tion

v EXAMPLE 15.4 Application: series stub matching at an antenna

Consider again the uañsmission line and load in F-xample 15.3. The load has al

impedanceá f 15 +i6O e and the line impedance is 75O, as shown in Figury 15.14.

Há*eo.r, now it is required to match the load using a shorted,-series stub similar

to that strown in Figuie 15.13c. Calculate the required-t"ryt t!f"d circuit st¡rl'

and its distance from the load to match ihe antenna to the line. The line and smb

have the same characteristic impedance-

Solution. The solutio'n is similar to that in Example 15.3. To match the load, v"c

seek a location /1 and a stub length /1 as shourn in Figure l5'13c' Since the stub''c

reactance is in series with the line impedanc e at d1, the sum of the line impedanee

and stub reactance must be equal to the line resistance. Therefora we should now

use the Smith chart as an impedance chart. we move a disance /1, from the load

at which the normalized line impedance is z¡(d.¡) -- | * jx. Then, we fnd a stub

f""g,t /1 such that zr(/1) : -jx.th. ro* of the two gives the correct match at /r-

(1) The normalized load impedance is z¿ :0.2 +i0.8. This is marked at point

Pz. (Figure 15.16).

(2) Now, we move toward the generatoi orr the reflection coefficient circle until

we intersect the r : 1 circlé at points Pt and P+. Connection of Pl to P3 andp1 to pa and extending the linei to the circumference gives points P\ nd P'*-

Each of these is a possible solution'

(3) Solution No. 1: The distanóe between Pi and fi i^t^th" fiist Possible solution

for /1. In this case, ri'e moved a distance tlv :0'198)' - 0'109)' : 0'089;"

The normalized line readance at point fu is j2.6. The stub length must be such

that its normalized input impedance is -j2.6.This requr¡ed im- pedance is marked

as point Pi. The distance bet*een the shórt circuit point P,, t!! li ,noving tow-ard

áá g.rr.ri.or is the stub length necessary. This distance is 0.3081.- Thus' the first

solution (with ¡' = 2 m) is

15.4. IMPEDANCEMATCHINGANDTHESMITHCIIART 975

dv:0.089) ' : 0.178 [m], /ro :0.308). :0.616 [m]

a¡n

tflce

ptorgfteatoint'r, is

]4m

theúngtest

Pen

batdis

(4) Solution No. 2: This occurs at point P+. The distance /1 now is the distance

ben¡¡een point Pi and P', or d15 :0'302;' - 0'109)' : 0'193)''

The normalized line reactance at pointPa is -12.ó. The stub normalized impedance

mustbe +72.ó. This impedance is marked T point P{- The disence /1¿ is the distance

bet*een ihe short circrút point to point P{: lp : 0.192},. The second solution is

therefore

d16:0.193L = 0.386 [m], l ¡ :0 '192] ' :0 '384 [- ]

Either solution is correcg but perhaps in practical teÍns, the closest stub to the

load (solution No. 1) may be chosen'

15.4.2.2 Double Stub Matching

fu mentioned earlier, double snrb matching takes a different approach than single

,*ü -o"t

i"g. There are now rrryo stubs ,i fit d locations /t .and /2, as shown iir

976 I5. THE SMITH CHART, IMPEDANCE MATCHING. AND TRANSMISSION LINE CIRCUITS

FIGURE 15.16 Smith chrt for Example 15.4.

Figure 15.13b. Matching is achieved by adjusting the two stub lengths \ and l2.Tosee how tiis is accomplished, it is best to look at the process in reverse. Suppose thatwe have already accomplished matching. From the results for single stub matching,we know that when the load is matched, we must be on a point on the unit circle(g = 1). In fact, we know that there will usually be two points at which matching canbe accomplished, but, for clarity, only point P1 is shown in Figure 15.17. The pointshown represents the load impedance at a distance ú * dz from the load. Now, wemust move toward the load a distance d2. For any of the points on tlte unit circle, thismeans rnoving on its reflection coefficient circle. The locus of all points on the unitcircle, moved toward the load a distance /2, is a shifted unit circle, as shown in Figure15. 17. This shifted unit circle represents the equivalent load impedance ar a distancefi from the load (this equivalent load impedance is due to the line impedance andthe stub at this point). Point Pi is the equivalent impedance at the location of stub (2)corresponding to the matched point P1. Stub (1) only adds a suscept¿nce to the line

a\f"\,

/.J

!

,

15.4. IMPEDANCE MATCHING AND IHE sMITH CH¡fiT s7'ir

H-Át _te

FIGURE 15.17 Smithchan for Example 15.4.

admittance. Therefore, to get to the load admittance point, we must first removethis susceptance by moving along the circles of constant conductance. This bringsus to point P{' marked on the chan in Figure lí.l7.In addition we must move edistance /1 from Pl toward the load (not shou'n on the chart). Note, also, that thedifference in susceptance betrieen points Pi and Pi' is the susceptance stub (l) mustadd to the line while the susceptance of stub (2) is the imaginary partof dre admittanceat point P1.

Of course, when matching a load, we will staft with the load impedance, but theabove process is more instructive because it explains the need for the shifted unitni-^ lonnA

' r r l 'at t t r p^n-eih"dnn nf-o^h <¡ 'k ic Tn oF(^n t ' , , mctco¡r thot thonrtmnso

! ¡ er^!s5

of the firsr stub (the stub closer to the load) is to modift the line susceptance so thatthe second stub can then take the line admittance to the unit cirde. The followingtwo examples show the steps and the details of double snrb matching.

i 4.Torse thattchingt circleing cane pointlow, we:le, thishe unitFigurelistancerce andstub (2)üe line

978 15. THE SM|TH CHART. |MPEDANCE MATCH|NG, AND TRANSM|SS|0N L|NE c|RcU|Ts

Zrl50+i225 Q

point P3 and isyr :0.62 -i0.92.

In preparation for the calculation of the stubs, we draw the two unit circles'

tre unit circle for stub (2) is the g : 1 circle of the chart' The unit circle for

srub (1) is the same circle, shifted toward the load a distance of 0.1tr, as shown

in Figure 15.19.

Now, we add the stub at the load. The stub's impedance is purely lmagular-y:Therefore, it can only change the susceptance of the combined stub and load

while the conductance remains the same. To find the combined adminance

on the unit circle for stub (1), we move on the constant-conductance circie,

sBrring from Pl (load admittance). This path is shown Gfrtli*l in Figure

15.1g."The path intersects unit circle (1) at two points, marked Pa and Ps. The

FIGURE 15.1S A load impedance matched to the line with two stubs.

V EXAMPLE 15.5 Double stub matching

A line with characteristic impedance 7'o : 300 Q and load impe'dance Zr : I50 +

j225 Ois given. Design a dóuble stub matching network such that the two stubs

are 0.1). apart as shown in Figure 15.18.

Solution. After calculating the normaüzed load impedance, we draw the reflec-

tion coefficient circle and en¿ Ae normalized load admittance since, as with the

single stub, the Smith chart is used as an admittance chart. In the single stub cese,

m"ichirrg óonsisted of finding the intersection of the reflection coefñcient circle

with thei : 1 circle. The same principle is used here, but the actual matchingis at

the secorid stub from the load (stub (2)) since we want to metch the load to the line.

Thus, stub (1) is represented by its own unit circle, which is shifted a distance 0'11'

ao*"rd rh. load. Now, we start at the load (P3) and move from the admittance point

on the constant conductance circle at the load until we intersect the unit circle for

stub (1). The intersection points represent the reflection coefficients of the com-

bined load and stub (1). The combined impedance oí the load and stub represent

a neq modified load with a stub a distance 0.1). away, toward the generator' Tlús

modiáed 1¡ne is a load with a single stub; therefore, its treatrnent is the same as

for the single stub matching in thát the length of the stub is chosen to cancel the

susceptanc; for each ofthe two stubs possible at the load. Stubs (1) and (2) refer to

the notation used in Figures 15.13b and 15.18, wittr stub (l) closer to the load'

(1) The normalized load impedance (without stubs) is zr : 0.5 *10.75 and is

shown at point P2 in Figure 15.19. The normalized load admittance is at

a)

(3)

I5.4. IMPEDANCE MATCHTNG AND IHE SMNH CHART i r l : i

\-\

o'¿¿1, '

3 '

. \\

FIGURE tS.l9 Smith chart for the line in Figure l5.lg.

admiftances at p4 and p5 are

lPc = 0.62 *fl.195 yps :0.62 + jZ.6

In moving from the load admittance point p3 ro poinls pa andp5, üe changc inadmittance is only due to .h. ,,rr".f."',.i. "o.rt

iu,r..d by;; ¿itolt ".t'g th.load adr.qittancé iom the ,dmin"íces-;;;;;

stub (r) ;;.;";;;.; ;;;ijü,,*P"iff,:;:Lr"|' gives the susceprar¡oc

At Pa:

! t , : yp4 - lL - 0.62 +j0. tg5 - 0.62 +jo.gz: i | . l15

L

i0+¡tubs

'flec-r thecase,:ircle: isatline.0. l rninte for

':om-:sentThis.le asI theter to,ad.

ad isis at

-'cles.e for¡own

nary.load

anceircle,gureThe

(4)

980 I5. THE SMITH CHART, IMPEDANCE MATCHING, AND TRANSMISSION LINE CIRCUITS

l' 0pq or yp5)

ll9ylE 1 5'20 -

The equivalent condition at the location of stub (1) after the load and stub (2) in Figure15.18 were taken into accounL

AtP5:

ytb : lps - lL:0.62 + j2.6 - 0.62 + j0.g2 : j3.52 ,

These twovalués a¡eshown atpoints{ and P(. Thepossible stub lengths are fouñdby-moling from the short circuit admittan"e pointipr,¡ toward trr."g.rr.rat*, topoine P{- and Pf. For point P4, the o¡,r""pt tt"" of the stub must be r.írs. Sartingat P* and *gl"q, in rurn, ,o p9Tl.pí

^Áa e; (alwap roward .t

" g.""r"*1lgi*,

the rwo possible lengths for stub (1): '

lu :0.25), + 0.133,1, : 0.3g3). btei¡ht :0.25), + 0.20ó), : 0.456), (at pi)

(5) Now, we consider the admittances at P4 and P5 as rhe new load admittances asshown in Figure rs.20. From here on, w: trear rhe problem as a single stub*'l"hg for each of these impedances and with the distance between load andstub (2) ho|" and equal to 0.1)..IMe startwithye+ an,i use Figure 15.2i, o¡rwhich the unit circle has been marked. We draw íhe reflection óefficient círclefor the admitunceypa. fu we move on the reflection coefficient

"i."t., ri"*i.rf*

P4, toward the generatoq and mbve 0.11., we intersect the unit circle at poir,t"p6.Although rve .r,t the unit circle at another poirrt, q¡ *etrically located about the

. real axis, this intersection cannot be used iinc. the stub musibe a distance 0.1i"from the load. At P6, t}le line adminance is I + j0.55. Thus, the srub must haveadmittance -j0-55 so that the line susceptance ií.^.r.eled ,, ,h. lo.*rio" ,f,*U(2)' The latter is marked as point Pi The stub length that will accomplish this isthe distance between the short circriit point and r¿lrhis is

lzo :0.4D, _ 0.2f l : 0.17),

Similarly, for point P5, we draw the reflection coefficient circle and move 0.1),toward the generatoq to point p7. The rine admittance ar p7 is l - j3.4. Therequired srub admittance is *Jr3 .4, which is marked as point pi . Ther*ú t"r,g.h i,the distance berween po, and,this point:

lu :0.25), + 0.204X :0.454,),

Therefore, the two possible solutions are

/r = 0.3831, tz :0.17), or l t :0.456A., tz :0.454X

i

I, ll !:I

¡l :1rr iiJ . . -

'undr, totingjves

rSure

ls as

;tuband.onrclegatFo.üe

t . tL

Iave;tub

is is

). 1;.fhehis

L

I5.4. IMPEDANCE MATCHING AND THE SMITH CHART 981

, '2'

FIGURE 15.21 Smith chart for the line in Figure 15.20'

Y EXAMPLE 15.6

A uansmission line and load are given in Figure 15.22a.It is required to calculate

the lengths of the snrbs só that the load is matched to the line'

Solution The steps in the sqlution are as follov's:

(1) First, we normalize the load impedance:

Zr $Q-t- ;7\- , : -

Zo 50

982 15. THE SMITH CHARI IMPEDANCE MATCHING, AND IRANSMISSION LINE CIRCUITS

FIGURE 15'22 (a) The double srub matching network for Example 1s.6. (b) Equivalent netr¡¡ork afterthe load impedance has been moved to the loladon of stub (t).

F|GURE l5'23 Smith chan for.the configuration in Figure 15.22a. Calculation of rhe equivalent loadimpedance at the location of stub (l).

¿bl

efter

\t \

,É\; tiEl- l

J9l*/

I

: load

(2)

I5.4. IMPEDANCEMATCHINGANDTHESMTTHCHART'S#3

This is marked as point P2 in Figure 15.23. The reflection coefiñcient cir¡:i':cen now be d¡awn. The point opposite P2 is P3. This is the load admitta¡:,.,:.r

lL:0.32 - j0.4

The line admitance at the location of the first stub is found by moving ii'r.r::rpoint P3 toward the generator a distance of 0.2).. This brings us to poi*r: ;':.¡ .The admittance at P4 (without the stub) is

!p+ :0.57, +i0.98

that is, the normaüzed line impedance at th€ location of the second stub, bc:.: ,;: ,.:the stub is added, is

{: zii5: 0.45 - j0.75

Note This is at point P5, which is the opposite point to point P4. Since P4 re¡r -resents the normaüzed admitance, P5 represents the norrnalized impedancar"Now the line and stubs eppeer as in Figure 15.22b. The new load admittan,: ¡yi'is marked as Pa in Figure 15.24.

The distance between tt¡e two stubs is 8.35.1,. Stub (2) is calculated to match Li-,'line. Stub (1) must be calculated for a unit circle (g : l) that has been mov(:' ito$'ard the load a disance of 8.35.¡,. Figure 15.24 shows the actual unit cir:;:,and the shifted circle after moving it 8.351. toward the load (that is, from snrt¡(1) to stub (2)). Note that this is the same as moüng the circle 0.35). towardthe load.

Now, we move on the conducance circle that passes through poinc F.1(g : 0.57) until the shifted unit circle is intersected at points P5 and P6. T'htg:0.57 circle is shown as a gray line.

The admitances at point P5 and F6 are

lps :0.57 *70.025, lpe :0.57 - j1.48

To find the length of stub (1), we argue as ̂ tollows: Moving from Pa to P5 íirP6,we have changed the imaginary part of admittance only. This change is

From P+ to Ps:

!1, : lps - !p+ = (0.57 +i0.025) - (0.57 +70.98) : -j0.955

From Pa to P6:

yLh : yp6 - !p+ : Q.57 -iL.+$ - (0.57 +70.98) : -j2.46

The admitances/ta andyr, are the admittances ¿dded to the load by the twopossible choices for stub (1). The two admitt¡nces required are shown as pointsPi anóP{. Thus, the length of stub (1) is the distance between the short circuitpoint and Pi or P'i. Foryro(Ps), the input stub admittance must be equal to-j0.955.W'e move from the infinite admitance point (point Po, on t*re chart),tcward thb generator, on the outer circle of the Smith chart up to point Pi-The toal distance traveled is the length of the snrb:

lv :0.378), - 0.2i l ' : 0.128)"

(3)

(4)

984 15. THE SMITH CHART, IMPEDANCE MATCHING. AND TRANSMISSION LINE CIRCUITS

FIGURE 15.24 Smith chait for Example 15.6 (continuation).

Similarly, the stub admimance for point Pa is lt¡ : -j2.46.The stub length isthe distance between Po, anó P'/:

lv = 0.31)" - 0.25I = 0.06¡. '

(5) For each one of these solutions, we have an equivalent admimance point: P5and Po. The problem now is that of an equivalent admittancey ps or lpe , and tsingle stub a distance 8.35L toward the generator. To avoid confusion, we usethe new chart in Figure 15.25. Points Ps and P6 as well as the unit circle forstub (2) are shown. We now draw the reflection coefñcient circles for eachof these two admittance points starting with point P5. From P5, w€ rDov€0.351 toward the generator. This intersects the unit circle at point Pf. Theline admittance at this point @efore connecting stub (2)) is I -70.5ó. The

L

f

!

s

I5.4. IMPEDANCE MATCHING AND THE SMITH CHARÍ 985

FIGURE 15.25 Smith chert for Example 15.6 (continuation).

admittance of the stub must be +j0.56, a value shown at point P7. The lenpi:of stub (2) corresponding to the point is the disance between Po, and P7,moving toward the generator:

lzo :0.25), + 0.081 : 0.3311

Starting with point P6 and moüng 0.3 5), toward the generator, we reach pointPf . The üne admittance at this point is L + j2.2. The stub admittance must be-j2.2, shown at point Pe. The stub length is therefore

12¡ = 0.32D' - 0.25¡, :0.07il\.

The two possible solutions are tierefore

lv -- 0.L2$', ht :0.06], arrd lzo:0.331)', lzb :0.07Dt

lw

his

P5rdauseforachove[hefhe

986 I5. THE SM{TH CHART. IMPEDANCE MATCHING, AND TRANSMISSION UNE CIRCUITS

7,=150+j225 tL

FIGURE 15.26

I EXERCISE 15.2

In Figure l5.26,the load impedance is 0.21 from the first stub (sub (1)) and thedistance bétween the two stubs is 0.1i.. Calculate the lengths of the ¡¡¡o stubs tomatch the load to the line.

Answer. Iu : 0.256)', lzo :0.102.1, or lp - 0.42Or, lu : 0.461),

15.5 OUARTER-WAVELENGTH TRANSFORMERMATCHING

I. t -

Stub matching, in effect, is capable of removing a mismatch for any load (except apurely reactive load), but it is not an impedance transformer. If different lines mustbe nratched, a transmission line transfonner can be used, as inl!'igure 15.27.

From Eq. (14.102), the line impedance Z¡o of a lossless üansmission üne ofcharacteristic impedance Zs, at a disunce 26, from the load may be viewed as theinput impedance of the line section between zs and the load:

(1 5.1 5)

Now; suppose rve chose a transmission line section. with characteristic impedance2,, ant it so it is ,1,/4 long, and connect it to a load impedance Z¡. Setting zo : ),/4and pzs : B)"/4 : (2ilL)Q,/4) : n/2 and replacing Zrby Zt znd h by Z' it Eq.

l l

r l

v o fZt cos 0zo + jZo sin Pze]or" :"0@

A quarter-wavelength transformer loc¿ted at distence / from loed.FIGURE 15.27

rd therbs to

xcepc a35 mustt.üne of

I as the

(1 5. r5)

pedance

o : )r/4I, in Eq.

1,rl-r-

I 5.5. OUARTER-WAVEI,"ENGTH TRANSFORM ER MATCHING

(15.16), we get for the input impedance of the ]./4 section

987

(1s.16)

(15.1e)

v ,lzt cos i +izt sn ilan:4@

swR: i+ifii

_zlZt

Referring now to Figure 15.27, where Zt is the line impedance at a distancc

/ from the load, we get the condition for matching using the quarter-wavelength

transformer shown:

(1s.17)

Thus, two different transmission lines or eny two impedances may be matched,provided a transformer of proper characteristic impedance Zt can be found. Thequarter-wavelength transformer is normally connected at a point of maximum orminimum voltage since the line impedance is real at that point. The line impedanceat a point of minimum volage is

(15.18)

where Zo is the characteristic impedance cf the line, and the sandingwave retio onthe line is given as

zhor:M

The location of the minimum volage on the line for a general load is at a disance(see Section 14.7.3)

r ¡ ./ - io: ; (0r+r)+ni

( r 5.2 1)

(1s.20).

from the load, where n is eny integer, including zero. For a resistively loaded line,the location of minimum voitage is either at the lo¿d (if R¿ < 4) ot at a distance

^J4 (if fu , h). Thus, the transformer can be located at any of the points iq Eq.

(15.20). If the characteristic line impedance is Zs, the characteristic impedance of atransforrnet located at a point of minimum volage must be

tSWR

Similarly, if the transformer is located at a point of maximum volage (by movingit a quaiter-wavelength in either direction of any of the poins in Eq. (15-20)), thecharacteristic impedance of the transformer for matching is

4: ^JSWR

(rs.22)

How can we use the Smith chart to design a quafter wavelength transformer and,

therefore, match two lines or a line and a load? First, we note that two Parametersare imporant in this design. The first is the sanding wave ratio SWR The second is

the location of the minimum (or maximum) voitage on the iine. For any given load,

these ¿re obainable from the Smith chart. Once the SWR andlocation of minimum

or maximum are found, the transformer impedance is found from Eq. (15.21) or

TJI

Ii¿-

FIGURE 15'28 A quarter wavelength tansformer used to match two different transmission lines.

988 i5. THE SMITH CHART. IMPEDANCE MAICHING, AND TRANSMISSION TINE CIRCUITS

(15.22),.depending on where the transformer is placed. The following examplesdiscuss the design sequence.

v EXAMPLE r s.7 Application: Matching of two different linesA student has found out trat he/she is out of money and ca.not pay his cable TVbill- Helshe decides to cancel the service and go u"ct to the old i"Lnop anrenna.F:y:":t 9" ry input is 75 o, while the catle coming down from rhe antennais 300 Q. Design a matching nenvork to match the mó hnes assuming that theantenna is matched to the 300 o line and the TV is matched to the is o ürr".Whcre should the matching network be placed?

solution. A quarter-wavelength transformer can be used, although, because TVreception is in a range of frequencies, the l¡nes will only be rnatched ít th" frequencyat which the transformer is exactry one-quarrer wavelengh. The characteri.sticimpedance of the transfotm.. *rrsi be

z,:,/ffi: .',f5 " 3oo: l5o tol

The.transform.I Try be placed anywhere ben¡¡een the antenna and rv because

one line is matched to the TV and the second to the antenn" ,rrá, th"r"fore, theimpedance anywhere on.each line equals i,, characteristic impedance. The only11nog"t point is that the line betwien'the transformer and the Tv must be a75 o line, and between the transformer and the anrenna the line

-"st be a 300 eline @igure 15.28).

Y EXAMPLE 1S.B Matching a load to a l ineA load_ zr : 45 - j6a e is connected to a line with characteristic impedance!0.:

50 Q. Design a cuarter-wavelength transformer to match the ¡oad to the line.It is required to connect the transforirer as close to the load

"r p"siUf". Find therequired characteristic impedance of the transformer and its location.

solution. There are rwo methods ro solve this problem. The most obvious is toy_s_e Eq9. (15.15) th¡ough (15.22). The second is to use the Smith chart instead.We will do both, starting with the Smith chart method.

, ).14 -antenna

TVG75c¿ o_

L/4 transfortner150 c¿

I300 c)75fJ

IIll

amples

15.5. OUARÍER.WAVEI.TNGTH IRANSFORüER MATCHING 9SS

. FIGURE 15.29 Smith chan for Example 15.8.

Method A: The Smith chart

(1) Find the normalized impedance of the load and mark it on an impedance Smithchart. The normalized load impedanceis z¡ : (45 -j6AV50 :0.9 -j't .2 (pointP2 in Figure 15.29).

(2) Find the first extremum in impedance from the load (minimum or maximum).This is done by moving on the reflection coefiñcient circle, toward the genera-tor, from point P2, until the real axis of the chart is met. This happens at point &ar a point of ¡¡rini¡tru¡i¡ impe.lance and is a distance of 0.5), - 0.3-?6)' : 0.164.7.,Éom the load. At point P3, the value on the axis is I/SWR : 0.3. Thus,SWR: 3.334.

¡le TVtfenna.

:rtennarat the.2 line.

.seTV

luency:eristic

ecausere, thee onlyi tbea300 f¿

,dancee line.rd the

.s rs tostead.

15. THE SMITH CHARI TMPEDANCE MATCHING, AND TRANSMI'SION LINE CIRCUITS

(3) From Eq' (15.2r), the characteristic impedance of the transformer is

z' : zol #: 5o/bJ :27 '39 ' tolMethod B: Direct calculationmagninrde, and is phase angle:

Firsg we find the load reflection coefficient, irs

, , _Zt-h 4s - j6o-so -s - i6o' r : / ] f i : 4sA+so= f f i :o '2475 - io-47s

lf¿l = {(ur+?sy + A.4?sy: 0.53ó0t: tan-r

-J0'475o.Ñ:-62'48" -> h=- l ¡9 [ rad]

The st¿nding wave ratio is

swR= l+lr¿l l+0.536t- lr lT I - oJ36

:3'3lo

The location of the firstminimum from the load(n= 0 in Eq. (f S.20)) is, l ld-i" -

* (ez * n) -

fif_t.oo * n) _0.1ó3.¡.

The transformer's intrinsic impedance is

z,:4l #.: ro/r# :27.48 tolNote that the rwo solutions are not identicar although trrey are close. This, ofcourse' is due to the nature ofthe chart The precisián depends on accurary ofreading rhe values on the chart. Much

"f ,rú, ;id;"I;;;"i;ffi"h computerizedSmith charts since these chars use *"""ao"l mathematicar rerations invorved.

3 EXERCISE 15.3

Find the location ""d.:h:r1:yistic impedance.of trre quafter-wavelength trans-

3::l I Example 15.g if ,'" o"nrFo.*.r is connected ,t th" first vortagemaxlmum.

Answer. d - 0.4l4J.r, Zt:-90.97 tC2]

15.6 EXPERIMENTS

990

l

I

fTll:t,l^9t-ottstr¿tes: Matching using shorted stubs). Thke a 2-3 m¡Éngril 0i a JUU 52 transmission iine: you can use an antenna dor¡¡n wire (wo_conductor flat cable) or a- simpre two-conductor üre ,rr"h ",

th" Jr., osed ro connectspeakers. Leave one end opán ""¿ "on

,."ith" oth., "J;;;ü

to the antenna

ls. RMEWOUtSnONS 991

)f)fd

lugs on your TV while the external antenna is connected as well, and tune a low-frequency channel (VHF channel 2 has a frequen a¡ of 5 440 lvtrIachannel 3, 60-{¡6MHz, channel4,6046 MHz). with a n..á1., slort the line by piercing th-oghthe insulation at different locations on the wire. Note the locations of the"short thatproduce the best and worst receptions. Measur. O. Jirtarr"e benyeen *o p."to

"rratwo minima in reception. What can you say about these locations? Can you relatethis distance with the frequency received? iYot r Do not perform thir

"tp..irrr".rton amplified antenna systems or cable TV connectiot r, orriy on TVs widr portable,passive antemas.

Experiment 2 @emonstrates: Matching using open stubs). Repeat experiment Iby cutting small sections from the free."¿ of O"-ttansmission line. Cutlnlv aboucI cm at a time and make sure tltat the conductors ere not shorted after ortting.

. REVIEW OUESTIONS

t. What is a Smith chart? Describe its important features.2. The_upper pan of the Smith chart represenrc (mark correct answer):

(a) Impedances with positive imaginary tenns or admitances wiú positive imaginaryteEns.

(b) Impedances with positile imaginary tenns or admittances with negative imaginaryterfns.

(c) Impedances üth neg"ative imaginary teflns or admitances with negative imaginaryterlns.

3. What is rhe reflection coefficient ¡t the center of rhe Smith chan?+. Standing wave ratios are marked on the Smith chart on the line of real normalizedimpedance. 7F.

5. What do infinite and zero impedances represent on the Smith chan? Where are thesepoints?

7: Describe the properties of the Smith chart when used as admitance charr In particular,discuss shon and open circuit points, relation to the impedance charg etc7. Define the conditions for matching on a transmission üne.8. what is a matching network? what purpose does it serve end how does it do so?9. Is conjugate matching also possible on a tr¿nsmission line? Explain.10. Does matching on a line ensure tlat more power is trensferred to the load than in theunmatched case? Explain.

t t. How is stub matching accomplished? wh¿t is the basic principle involved?L2. Does srub matching guarantee:

(a) Maximum transfer of power to the load?(b) More power transferred to the load than in rhe unmatched case? Explain.

13. Single stub matching is essentially a quescion of coirnecdng the correct reacrance on$e

ling "j.ú. lplt"Priate location. C¿n this be accomplished usúg capacitors or inductors

insrcad oí iines? Explain.

L4. can a single stub march all lines to any load? If nog what are the exceptions?15. why does it sornetimes become necess¿ry to use more than one srub for matching?

n)-rt:?

992 ls. rHE sMltH CHARL TMpEDANcE MATCHTNG, AND TRANsMrssroN uNE crRcurTs

16' A stub may elso be connected in series with the line. Discuss why this method is norused as often as parallel stubs.

17 ' In double.stub, matcfrilg both iles are placed at fixed locations. Matching is achievedsolely by changing the stubs lengths. ZF. Explain.18' Double stub matching can match any line to any load. T/F.If nog what are theexceptions?

19' Single or double stub matching is frequency dependent. What happens if the frequencychanges?

20' Can we use more than two stubs for matching? [f yes, explain how this might work21. Describe the basics of quarter_wavelength transformer.22' Does a quarter-wavelength transformer ensu¡e maximum power transfer? Explain.23' Cable TVcompanies.allow a single connection per zubscriber. How can they tell if thesubscriber connects more than one TV s.t to the sa*e line?

O PROBLEMS

I

General design using the Smith chart

!5't' -Line properties using the Smith ch¿rt. A long line with characteristic impedance

Zo-: 100 Q operates at I GHz. The speed of propagation on the üne is c ¿nd load impedanceis 260 +it80 Q. Find:

(a) The reflection coefficient at the load.(b) The reflecdon coefficient at a distance of 20 m from the load toward the generator.(c) Standing wave rado.(d) Input impedance at 20 m from the load.(e) Location of the firstvoltage maximum and first voltage minimum from the load.

ls'?' Calculation of voltage/current along tr¿nsmission lines. A rransmission tinewith a characteristic impedance of 100 Q and aload of 50 -750 e is connected to a matched

lt::r:::|1le line is very long and the voltage measured aithe road is 50 v carcurate using

[ne Jmltn crrart:(a) Maximum voltage on the line (magnitude only).(b) Minirnum voltage on the line (m"gnir,rde onryj.(c) Location of maxima and minima olvoltage or,-.h" lirr" (starring from the load).

15.3- Impedance of composite rine. A uansmission line is made of two segments, eacht.m lols-Gigure 15.30). Calculate the inputimpedance of the combin.Jii.r" urirrg a Smithchart if the speed of propagation on rine 1i¡ ir l "

108 m/s ""¿

." ri"" ól i x lOs m./s. Thelines operare ar 300 MHz.

Zz=LN {l

L-

FIGURE 15.30

Zrá0+j5O Q

ved

the

ency

k

l .

if the

:danceedance

rerator.

re load.

.on linenatchedte using

e load).

rts, eacha Smith/s. The

IIlI, t

15. PROBLEMS ss"?

FIGURE 15.31

1 5.4. SWR on line. The transmission line in Figure 15.3 I is given. A gen eralload (Z¡ :Zo + ¡Xt¡ is connected as shown in Figure 15.3 I . The shoned section is made of a differenrline with a different characteristic impedance Z¡.

(a) fusuming the generator is matched, calcul¿te the sanding wave ratio on the line.(b) What must be the length of tlle shorted line to ensure m¿rchhg of the lo*d

(no reflection). A¡e there any other conditions that must be satisfied for this ir¡happen?

15.5. Line properties. A lossless tr¿nsmission line has characteristic impedance 26 .-300 Q, is 6.1 wavelengths long, and is terminated in a load impedance Zr:35 +j25 Q. Fin.ei:

(a) The input impedance on the line.(b) The standing wave ratio on the main line.(c) If the load current is I d calculate the input power ro the line.

15.6. Line properties. A lossless transmission line has characteristic impedance 2650 Q and its input impedance is 50 - j25 A. The line operates at a wavelength of 0.45 m ais 3.85 m long. Calculate:

(a) The load impedance con¡ected to the line.(b) The location of the voltage minima ¿nd maxima on the line, starring from the

load.(c) The reflection coef6cient at the load (magrritude and angle) and the standinl;

wave refic on the iine.

L5.7. Application: Design of transmission lines. It is reqüred ro design a load of 75 -750 Q to simulate a device operating at 100 MHz. It is proposed using a secrion of a 50 Q li¡r¡;and connecdng to its end a lumped resisrance R. The linet phase velocity is c/3.

(a) Calculate the length of line and the required resisance R drat will accomplisfrthis.

(b) Is the solution unique? Explain and 6nd all possible solutions if the solution isnot unique.

15.8. Line properties using the Smith chart. An unknown load is connected to a 75 f)lossless transmission line. To find the load, two measurements are performed: (l) The locationof the 6rst voltage minimum is found et 0.18¡. from the load. (2) The SWR is measured ¿s2.5. Find using the Smith chart:

(a) The load impedance.O) The load reflection coef6cient (magnitude and angle).

Stub matching

15.9. iltatching with shorted/opcn loads. The uansmission i.ine necwork in Figurc15.32 is given. The shoned transmission line and the open transmission line are part ofthe network- Show that no stub network will match the rwo line sectiom to the main line.

tirl, l

i994 I5. THE SMITH CHARI,IMPEDANCE MATCHING. AND IRANSMISSION UNE CIRCUITS

Ztz=50+j 50 dl

FIGURE 15.32

lLlO' Appücation: Series stub-matching. Atransmissionline ofcharacteristicimpedanceZo : 50 Q is loaded with en impedan ce Z¡' J 100 t7g0 A @igure f 5J3). Á open transmis_sion lineis connected in series with the line as shown. The opeiline has the s,¿me characteristicimrredalce. Finq rh: length oftre. open line and the rocadán (d.; ;Á; load) it should beinserted to metch the load to the lini.

t f l sdr l l Ar l l :¿i l v

-

Z¿=100+j80 Cl

FIGURE I5.33

15.11. Application: singte ¡ru! matching. A transmission line is loaded as in Figure

15'3-4 Il the wavelength:".q" fine equars ] m, find a shorted p"oriJr*u (ocation andtength ot stub) ptaced to the left of points A-A to match the load to the line.

FIGURE 15.34

1.5'12' Application: Series stub matching. A load is connecred ro a Eansmission line asshown in Figure 15.35- k is required.o *r.Ih rhe load to the line (whiÁ;;;, characteristic

:::."j:::-.Ltj,?] ,T11" tocation and.length of a stub to *"t"ith" r¡r,.lr," rtub is open

ra. ^ ^5s^v ^J.JJ.

!f.11. Application: single stub matching. A 75 o rvcable is used to connecr to a TVThe load is matched to the line (Figure l5.3Za). A second rV must be connected r 0 m from

idence;mis-risticld b€

gure: and

re asistic)pen

TViom

-t-II.l

IIIIII

I

ls. PRoBt¡Ms 995

2é0+Í70 A

FIGURE 15.35

FIGURE 15.36

úe first fi again with a marched section of dre same cable (trigure 15.36b). fuzuming rkephase velocity on the line is r/2, calculate:

(a) 'ihe reflection coefñcient at the loc¿tion ofconnection ofthe two lines.(b) The stending weve ratio on the main line.(c) Design a single stub flocation to the left of the discontinuity and length) ro march

the line for chennel 3 et ó3 MHz. Use the same linc impedance for the srubs.(d) For the design in (c) calculate the reflection coeflicient to the left of rhe stub for

ch¿nnel 2 (57 MHz). What is your conclusion from dris calculation ¿s far as stubmatching ecross a range offrequencies?

15.14. Application: Double stub matching. fivo stubs ere used on a transmission lineas shown in Figure 15.37. Calculete sa¡b lengths d¡ and d2'w match the load to the line. [sthis arrangement of stubs a good arrangement? Why?

Zt5A+jñ0 {2

.* 0.251. *^-0.251.

stub2 - stub

FIGURE 15.37

996 15. IHE SMITH CHART. IMPEDANCE MATCHING, AND TRANSMISSION LINE CIRCUITS

f 5,.15. Application: Double stub,matching. An ántennahas animpedance ofó8*7100 A.The antenna needs to be connected to a 75, Q line. Beceuse rhe antenna goes on a mast, thedesign engineer decided to fabric¿te a marching secrion as shown in ñigure ls.3g. Thematching-section is then hoisted and connected to the anrenna during installarion. At therequired frequency, the secdon is 0.1], ¡nd two sn¡bs are made of the same line. Calculate drelengths of the stubs, if the antenna is connected

^tA - A.

75fJ 68+jl00A

FIGURE 15.38

Transformer match¡ng

15-ló. Application: V4 tr¿nsformer. Show that two lines with any characteristic (real)impedances 21 and, 22 may be matched with, a quarter-wavelettgth lioe. what is thecharacteristic impedance of the matching section?

15.17. Application: 3W4 tr¿nsformer. A lossless transmission line with charecteristicimpedance Z0 Eansfers power to a load Zt (real). To match the line, a matching section isconnected as shown in Figure 15.39. At what distence / from the load

-,rst th. line be

connected (minimum disance) anC whet ¡nust the cheracteristic impedance of the matchingsection be?

FIGURE I5.39

f 5.18. Application: V4 tr¿nsformer. A tr¡nsmission line is given as shown in Figure15.40. If the characteristic impedance of the quarter-wavelength transformer must be-real,find the location of the transformer (disunce / in the figure) and the intrinsic impedance oithe uansf<rrmer Zr.

15.19. Application: V4 tr¿nsformer. A ¡wo-wire transmission line has characterisdcimpedance of 300 o and connects to an entenna. The line is long, the antenna has an impedanceof 200 a and operares at a wavelength of 3.8 m. To matcñ the line and load,

" qo.*.r-

)0s¿.:, theThet thee the

real)the

istic,n is:b.ring

ureeal,eof

strcnceler-

15. PROBIEMS e&"

ZzáO+j50 {l

HGURE 15.¿[0

wavelength trarúformer is connecrcd on the line, but dre location ¿t which the transformermay be connected is l0 m from the antenftr or larger. Calculate:

(a) The closest location at which the transformer may be onnected.(b) For the result in (a), celculate the characteristic impedence ofthe transforrner

section.(c) Calculate the standing wave ratios on the sections of line between the transfor¡ner

and antenn¿ and between tansformer end generator.

ttl

:¡

rl

C H P T E R , I

J

The Smith Chart,lmpedance Matching, andTransmission Li ne Circuits

I

;

I

15.1 INTROD{JCTIQN

A look back at much of wh¿t we did.with.tr¿nsmission lines reveals i:trf:*: r¡|:tjffi #'i":ilH;ti;;;"hdo"'i"h'."'"dú:,':li::::?:T*T"T::to calculate the line

reflection coefñcient waswave ratio. Voltage, clurent' and power

imTédtnce-ñilto cacoefficienq in turn, was

ffi ::l'f '¿'ff J:rffi ffiH;;;ü;"'"'t"'""v'q"¡-"r":*:Í^:*::T:

approximations involved'

]if:: l'iTi"llfi"T;; io,, *"v;',o reJtlperhary'widr some fondness' the

complicated ""L,rt"tiorirl1'¡]"f,l[;ir.din addition to dre useofcomplexvariables,

rhe use of trigonom.J;;;;;b"lic functions. Thus, the following proposition:

Build a graphical chart (or an equivalent computer program):tiy:;f *presenting

the refl ection .o"mti"'li"' ;;ü; b; ; *pti""tti i" io*t genera] fashion and vou

haveasimplemethodofdesigningtransmissionlinecircuitswithouttheneedtoperform rather tedious;;b;i;ft.*: This has been accomplished in a rather general

iool called the Smiü Oft"t,. itt. Srnith chart is a chart óf normelized impedances

(or admittanc.s) in tf'" "Á"ttio"

coefñcient plane' fu sudl it allows calculations

of all param",.r, ,.h.Jii" o*r*i*ion lines as w.ell as impedances in open space'

circuits; and the rrn". e,r".tl"ñ;;¡"rrrh chan is rather ol4 it is a common design

toolinelectromagnetics.somemeasuringinsmrmencsuchasnetworkanalr¿ersactually use a Smith;h".;;;ttpl"y.orrái,io.rs on lines and network' Naturall¡

anv chart can also be implemented i'' " tot"pt'ttr Pro8ranu 11j,*.?**

chart has'

ill#il. n.ri ""¿.r!ád

how it work before we can use it either on paPer or

on the screen. ^

a"*p"r;^Ji*irj, chan can then be used to analvze conditions

on lines. The examples provided here are tJ;;;;;ggralhi€al tools and a printed

Smith chart, rather than the computer p'ogt"*' to !tiptt"ti'" the techniques and

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