Transmission Lines- Part II - IIT Bombaydghosh/tx-lines-2.pdf · Back to the Smith Chart In Transmission Lines-I, the development of the Smith chart was discussed. Now, some analyses

Transmission Lines- Part II

Debapratim Ghosh

Electronic Systems GroupDepartment of Electrical Engineering

Indian Institute of Technology Bombay

e-mail: [email protected]

Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part II 1 / 30

Outline

I Power delivered to transmission line load

I Transmission line calculations using Smith chart

I Using the Smith chart with admittance

I Transmission line applications- Single stub impedance matching

I Transmission line applications- Determining the load type

I Transmission line applications- Realization of circuit elements

I Low loss and lossy transmission line

I Transmission line measurements


Power Delivered to Load of a Lossless Transmission Line

I Using the standard expression in terms of the complex voltage and current, thepower at any point l along the line is

P(l) =12

Re(VI∗) =12

Re{[V +ejβl (1 + ΓLe−j2βl )][V +

Z0ejβl (1− ΓLe−j2βl )]∗} (1)

I At the load, l = 0. Therefore, the load power is

P(0) = PL =12

Re{[V +(1 + ΓL)][V +

Z0(1− ΓL)]∗} (2)

=12|V +|2

Z0Re(1 + ΓL − Γ∗L − |ΓL|2) (3)

I The term ΓL − Γ∗L is purely imaginary. Thus, simplifying, we get

PL =12|V +|2

Z0(1− |ΓL|2) (4)

I The same expression can be derived using an alternate approach: first calculatingthe power incident on the load, and then subtracting the reflected power from theload. Try this out, you should see the same result!


Power at any point on a Lossless Transmission LineI The instantaneous power at any point l along the line is

P(l) =12

V (l)I(l)∗ =12{[V +ejβl (1 + ΓLe−j2βl )][

V +

Z0ejβl (1− ΓLe−j2βl )]∗}

=12|V +|2

Z0(1 + ΓLe−j2βl )(1− Γ∗Lej2βl )

=12|V +|2

Z0(1 + ΓLe−j2βl − Γ∗Lej2βl − |ΓL|2)

I Writing ΓL = |ΓL|ejφ and Γ∗L = |ΓL|e−jφ, we get

P(l) =12|V +|2

Z0(1 + |ΓL|ej(φ−2βl) − |ΓL|e−j(φ−2βl) − |ΓL|2)

I Expanding the complex exponentials using Euler’s entity and simplifying, we obtain

P(l)real =12|V +|2

Z0(1− |ΓL|2) and P(l)imag =

12|V +|2

Z0(−2 sin(φ− 2βl)) (5)

I It is interesting to note that the real portion of P(l) is equal to the power delivered tothe load. Since the line is lossless, the entire power is sent to the load

I The imaginary portion of P(l) is dependent on l and denotes the energy ‘‘stored’’ inthe line due to variation of V , I with position, which varies the electric and magneticfields


Calculating V + from Transmission Line MeasurementsI In most of our analyses for voltage, current, reflection coefficient and power, the

quantity V + is assumed to be knownI V + is the voltage across the load when it is matched to the line characteristic

impedance. In practice, this condition need not be met always, and it is not easy tophysically connect a voltage probe across the load and expect reliable reading

I V + is estimated using the source parameters and line parameters, which are known.Consider a source-load arrangement through a transmission line as shown below

Z0 ZL

RS

VS

LZX

I The voltage looking into the line VX = V +ejβL(1 + ΓLe−j2βL), & ΓL =ZL − Z0

ZL + Z0

I If the impedance looking into the line is transformed to ZX , then VX = VSZX

ZX + RSI Equating the two and simplifying, we get

V + =VSZX e−jβL

(RS + ZX )(1 + ΓLe−j2|ΓL|βL)(6)


Back to the Smith ChartIn Transmission Lines-I, the development of the Smith chart was discussed. Now, someanalyses are discussed

I The center of the Smith chart denotes z = 1 + j0, i.e. a matched impedance andthis is the center of all VSWR circles

I The normalized load zL may lie on any of the VSWR circles. The correspondingreflection coefficient ΓL may be measured by mapping the radius of the VSWRcircle on the scale below the chart

I Movement on the VSWR circle in an anti-clockwise direction indicates movementaway from load. On this trajectory, the maximum, minimum line voltage points awayfrom the load, and impedance at any point on the line may be found

A

B

CD

L1

L2

L3

A ≡ zL

B ≡ 1zL

C ≡ ρ

D ≡ 1ρ

L1 ≡λ

4L2 ≡ distance to nearest voltage maximumL3 ≡ distance to nearest voltage minimum


Calculating Transformed Impedance along the LineI Suppose we have a load ZL connected to a transmission line of characteristic

impedance Z0. The first step is to get the normalized impedance i.e. zL = ZLZ0

I Next, mark the zL on the Smith chart by locating the intersection of the correct r andx circles. Draw a constant ρ circle through zL

I Now, if we wish to find the impedance z1 at a distance l1 from the load. Usually, l1 isexpressed in terms of the wavelength λ, and distance is marked on the Smith chartin terms of λ as well

I Move clockwise (away from load) along the ρ circle over the distance required(equivalent to angle 2βl). On that point, read the r and jx values by identifying thecorrect r and x circles

jx1

r1

zL

z1Move CW

l1

Constant ρ

circle


Smith Chart with AdmittanceI Often it is easier to work with admittance, rather than impedances (e.g. parallel

loads). Let us see how the Smith chart changes when working with admittanceI Consider the admittance Y at any point on a line with characteristic admittance Y0.

The normalized admittance y =YY0

= g + jb =1

r + jxI Substituting in the expression for reflection coefficient Γ, we obtain

Γ =1− y1 + y

=y − 1y + 1

∠180◦ =z − 1z + 1

(7)

I It is seen that Γ in terms of both y and z are identical, except with a 180◦ phasedifference. Effectively, the Smith chart may be rotated by 180◦ as well

-jx = jb

jx = -jb

OpenShort

r = g


Using the Impedance Smith Chart for AdmittanceI It seems that by simply inverting the Smith chart, one can use it for

admittance-based calculations. But there is a simpler wayI Recall that Smith chart is a coordinate system on a complex Γ plane where the axes

are defined as Γ = u + jvI Rather than inverting the Smith chart, we can invert the u and v axes and use the

impedance Smith chart coordinates as admittanceI The only change is that the phase of Γ must be measured using the inverted axes

as a reference. There is no change in the direction of movement towardssource/load

Capacitive susceptance jb

Inductive susceptance -jb

Conductance gu

jvDebapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part II 9 / 30

Transmission Line Applications- Impedance MatchingI One of the most crucial considerations in transmission lines is the impedance

matching between the source, line and the load. Mismatch between theseimpedances result in reflections, which reduce power delivered to the load

I Suppose a line of characteristic impedance Z0 is terminated with an impedance ZL,where ZL 6= Z0. Here, impedance matching needs to be done

I A classic technique involves using another transmission line of impedance Z0,connected to the main transmission line in series or shunt fashion. This second lineis usually terminated on the other end by an open or a short circuit

I This second line is known as a stub, and this impedance matching technique iscalled stub matching. Shown below are some examples of stubs withtransmission lines

ZL ZL ZL ZL

Series open stub Series short stub

Shunt short stub Shunt open stub

Z0Z0Z0Z0


Impedance Matching using a Single StubI Consider the example of matching an arbitrary load impedance zL = r + jx

(normalized to the line impedance). Let us match this to z0 = 1 using a shunt shortstub

I As we want to use shunt stub, it is better to use admittance rather than impedances.Let yL = g + jb. Transforming this admittance to the point of the stub connection, i.e.after length ds, the admittance should be y1 = 1 + jb1

I The transformed admittance from the short-end of the stub to the connection on themain line (over length ls) should be y2 = −jb2

I The effective admittance seen by the line is then yeff = y1 + y2 = 1. This means theline sees the transformed impedance equal to Z0, i.e. line and load are matched

ds

ls

ZLZ0

y1

y2

yeff


Single Stub Matching using Smith ChartI The Smith chart is a useful tool for matching calculations. For the shunt short stub,

we will use it as an admittance chart. The unknowns to be calculated are ds and lsI First, mark the normalized load admittance yL. On the VSWR circle, move towards

the source up to the admittance point y1 = 1 + jb. b is be the intersection point ofthe VSWR circle and the g = 1 circle

I The measured distance from yL to y1 is equivalent to ds

I Now comes the calculations for the shunt stub. Corresponding to jb, mark the point−jb on the periphery of the chart. This corresponds to y2. Move towards load (i.e.anti-clockwise) to the short circuit point. This distance gives us ls

ds

ls

yL y1

-jb

jb

g = 1


Some Important Points about Single Stub MatchingI Rather than choosing y1 = 1 + jb, one can also choose y1 = 1− jb. In that case,

y2 = jb. Both are acceptable solutions. You can choose either, depending on howlarge a stub can be accommodated in your system

I Exercise: in a similar manner, work out the procedure to design a single stubmatching network, but with an open-terminated shunt stub

I If a series stub matching is required, we have to use the Smith chart as animpedance chart

I The stub matching technique works only for a single frequency. This is decided bythe distances ds and ls which are expressed in terms of λ

I Transmission lines fabricated on a two-layer printed circuit board (PCB) are calledmicrostrip lines. Stubs can be easily realized using microstrip technique. Shownbelow is a short circuited stub along a line

Main Line StubVia hole

to ground

Ground


Determining the Type of Load Connected to a Transmission LineI Often, it is not easy to directly measure the impedance of a load connected to a

transmission line (e.g. if it is soldered or mechanically affixed)I In such a case, measuring the standing wave patterns and VSWR provides some

interesting information about the load.I There exists a measurement system known as a slotted line, wherein a movable

voltage probe is connected to a transmission line, and a reading can be obtained atany point on the line

I We look at the variation of the voltage standing wave right from the load positiononwards. For e.g. the following two standing wave patterns (SWP) correspond to acapacitive load, and a resistive load with 0 < ZL < Z0, respectively

Vmax

Vmax

Vmin

l

l

|V|

Capacitive load

Resistive load

SWP-1

SWP-2with 0 < ZL < Z0

I With the help of the Smith chart, this can be easily understood


Determining the Type of Load Using Smith ChartI In SWP-1 , while moving towards source, we first see a voltage minima. This

means on a Smith chart, the load impedance lies somewhere in the lower half. Also,the voltage minimum is 0, thus Re(ZL) = 0. This can only mean a capacitive load

I In SWP-2, at the load, a voltage minima exists. Now, voltage maxima and minimacan only be on the jx = 0 line i.e. the real axis of the Smith chart. Also, the voltageminimum is non-zero. This means the ZL lies between a short circuit, and Z0

ZL

V minima

comes first

& is zero

ZL

V minima

exists at ZL

& is non-zero

SWP-1 SWP-2

I Exercise: Work out the standing wave patterns for (i) a purely inductive load (ii) anopen load (iii) a short load (iv) an R + jX load with X > 0 (v) the same load withR ≈ Z0 and X � R


Transmission Line Applications- Realization of Circuit ElementsI As the frequency increases, ordinary discrete inductors and capacitors behave in a

different manner. The inter-coil capacitance (Cp) of inductor, and lead inductance(Lp) of capacitor start becoming significant

L C

L C

Cp

Lp

2

Lp

2

I As the frequency increases, these parasitic effects become more dominant.Generally, most through-hole 2-lead capacitors and inductors do not work reliablybeyond 100−150 MHz

I Inductor and capacitors of a particular value, however, can be realized at aparticular frequency, using lossless transmission lines

I The first starting point is the impedance transformation relation i.e.

Z (l) = Z0ZL + jZ0 tanβlZ0 + jZL tanβl

(8)


Realization of L and C using Open and Short-ended Terminated LinesI Suppose we have a lossless transmission line terminated with a short circuit. Then,

the impedance along the line Z (l) is given as

Z (l) = jZ0 tanβl (9)

I Thus, if 0 ≤ βl ≤ π

2i.e. 0 ≤ l ≤ λ

4, then the magnitude of Z (l) is positive, which

indicates inductive reactance. Thus,

Z0 tanβl = ωL (10)

∴L =Z0 tanβl

ω(11)

I Likewise, ifλ

4≤ l ≤ λ

2, the Z (l) is negative and it indicates capacitive reactance.

Thus,

Z0 tanβl =1ωC

(12)

∴C =1

Z0ω tanβl(13)

I Exercise: Derive the conditions and expressions for realizing L and C using anopen-ended transmission line


Inductive and Capacitive Behaviour of Short-ended Transmission LineI The inductive or capacitive behaviour of a transmission line is easily understood

using the Smith chartI Start from the short circuit point on the Smith chart, and move clockwise towards

source

Short

AB

+jX

-jX

I As one moves on the upper part of r = 0 circle of the Smith chart, it indicates +jX(inductor) and movement on the lower part of r = 0 circle indicates −jX (capacitor)

I Movement along this trajectory periodically results in inductive and capacitive

reactance. Inductors and capacitors repeat after everyλ

2movement (i.e. one

complete trajectory of the r = 0 circle)Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part II 18 / 30

Realization of Resonant L − C Circuits using Transmission LinesI We have seen that for a short-terminated line, the impedance at any point is given

by Z = jZ0 tanβlI Graphically, the magnitude of Z as a function of the line length l looks like

Z

lλ 3λ

4 4

λ λ

2

I At odd multiples ofλ

4, the impedance peaks up to∞. This denotes parallel L− C

resonance (equivalent to admittance minima)

I At even multiples ofλ

4, the impedance reaches zero. This denotes series L− C

resonanceλ

4

λ

2


The Practical Scenario- Low Loss Transmission LineI So far, almost all analyses and discussions have been about lossless lines.

Perfectly lossless lines, however, exist only in textbooks and not in practice! Thereis always some non-zero loss in a line, no matter how small

I Thus, in the first approximation of a propagation constant, we must re-introduce theα term, i.e. γ = α + jβ, and assume that α� β. This is called a low-losstransmission line

I We can now study the effect of α on the realization of L and C using low-loss lines.For a short-terminated quarter-wavelength line, the impedance now becomes

Z (l) = Z0 tanh γl = Z0 tanh(α + jβ)l (14)

= Z0tanhαl + j tanβl

1 + j tanhαl tanβl(15)

I As α→ 0 for a low-loss line, the term tanhαl ≈ αl . Therefore, as l → λ

4,

Z =Z0

αl(16)

I Thus, in practice Z is not infinite, but is of a very large value, as the term αl is quitesmall. Similarly, one can prove that for a quarter-wavelength open-loaded low-lossline, the input impedance is

Z = Z0αl (17)Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part II 20 / 30

Q of a Resonant L − C using Transmission LineI For the quarter-wavelength short ended line, the maximum impedance was shown

to be Z0/(αl). As the signal frequency is varied, the maximum impedance will beobtained at f0 (corresponding to the physical λ0/4 of the line)

I Using this resonant frequency f0, and the 3 dB impedance variation around themaxima, the Q of the line can be obtained. However, a less complicated approachis using the voltage and current along the line, and for a this short-terminated line

I Assuming that γ ≈ jβ, work out for yourself that for this line, the V and Imagnitudes are

V = V +[ejβl − e−jβl ] = 2V + sinβl = V0 sinβl (18)

I =V +

Z0[ejβl + e−jβl ] =

2V +

Z0cosβl =

V0

Z0cosβl (19)

I In terms of the line parameters, if it is assumed that R,G are negligible, then theenergy stored along the λ/4 section of the line is

E =12

C∫ λ/4

0V 2dl +

12

L∫ λ/4

0I2dl (20)

=12

C∫ λ/4

0[V0 sinβl]2dl +

12

L∫ λ/4

0

[V0

Z0cosβl

]2

dl (21)


Q of a Resonant L − C using Transmission Line (cont’d..)I The energy stored E can be simplified to

E =12

C∫ λ/4

0V 2

0

[1− cos 2βl

2

]dl +

12

L∫ λ/4

0

V 20

Z 20

[1 + cos 2βl

2

]dl (22)

=14

CV 20λ

4+

14

LV 20

Z 20

λ

4(23)

I Since Z0 ≈√

L/C, the two terms are identical, and this simplifies to

E =12

CV 20λ

4(24)

I The power lost in the circuit is dependent on the impedance seen by the source i.e.Z0/(αl). Therefore, the power lost is

PL =V 2

0

Z0/(αl)=

V 20

Z0/(αλ/4)(25)

I By definition, Q = 2πf0EPL

, where f0 is the frequency of operation. Therefore, this

simplifies to

Q =β

2α(26)


More About Low Loss Transmission LinesI We have already discussed that a low loss line has α� β. In terms of the per-unit

length line parameters, this implies R � jωL and G� jωC. Therefore, thepropagation constant γ becomes

γ =√

(R + jωL)(G + jωC) (27)

=

√√√√jωL

(1− j

RωL

)jωC

(1− j

GωC

)(28)

= jω√

LC

(1− j

RωL

)1/2(1− j

GωC

)1/2

(29)

I As R � jωL and G� jωC, expanding the root terms using Power series andignoring the 2nd and higher order terms,

γ = jω√

LC

(1− j

RωL− j

GωC

)(30)

= R

√CL

+ G

√LC

+ jω√

LC (31)

I We see α =RZ0

+ GZ0. Note that β = ω√

LC is not affected by non-zero R,G


Lossy Transmission Lines

I So far, the discussion has largely been about lossless or low loss lines whereα� β. But it should be known as to what happens when α becomes significantlylarger than β

I In terms of the line parameters, R � jωL and G� jωC. The characteristicimpedance then becomes

Z0 =

√RG

(32)

I It is interesting to note that the Z0 of a lossy line is real, just like a lossless line.Thus, it cannot be said if a line is lossy or lossless just because Z0 is real. It can,however, be said that a line with complex impedance is moderately lossy

I The propagation constant γ of a lossy line is

γ =√

(R + jωL)(G + jωC) ≈√

RG (33)

I It is seen that here, γ is real. α is finite and β is negligible. Thus, there is nopropagation of the wave as such, as the power would be dissipated in the line itself

I Thus, a lossy line, not surprisingly, is useless as far as delivering power to a load isconcerned


Reflection Coefficient and VSWR Along a Lossy Line

I The voltage at any point on a lossy line is given as

V (l) = V +eαl + V−e−αl (34)

I Close to the load, αl → 0, and the term jβl becomes significant. Forward andreverse waves thus exist near the load

I Thus, the reflection coefficient at any point on the line is Γ =V−e−αl

V +eαl = ΓLe−2αl

I It is interesting to note that |Γ| decreases exponentially as one moves away fromthe load. Close to the source, the |Γ| ≈ 0, which means that the source sees anearly matched line

I The VSWR along the line is

ρ =1 + |Γ|1− |Γ| =

1 + |ΓL|e−2αl

1− |ΓL|e−2αl (35)

I Thus, as one moves away from the load, ρ decreases and converges to 1


VSWR Circles for Lossy Transmission Line

As VSWR is a function of line length, there is no concept of a ‘constant’ VSWR circle.VSWR may be aprroximated as a piecewise constant function along the l . This is what aVSWR circle looks like for a lossy line

zL

The spiral indicates reducing reflection coefficient and VSWR moving closer to 1, as onemoves away from the load


Measurement of Characteristic Impedance of a LineI As the characteristic impedance Z0 is a distributed parameter, it is not possible to

physically measure it using, say, a multimeterI However, one can measure the impedance at any point along the line, from which

the Z0 may be calculatedI Suppose we are given a line of length l with standard connectors affixed at either

end. How can its Z0 be measured?I Step 1: Connect one end of the line to a short circuit load, and then measure the

impedance at the other end. It is given by

Zsc = Z0 tanh γl (36)

I Step 2: Replace the short load by an open circuit load, and then measure theimpedance at the other end. It is given by

Zoc = Z0 coth γl (37)

I The line characteristic impedance is then simply,

Z0 =√

ZscZoc (38)

I Zsc and Zoc can be measured using an instrument called Vector NetworkAnalyzer (VNA)


Measurement of Propagation Constant of a LineI From the previous analysis, if the line has length l , then tanh γl =

√Zsc/Zoc = A

∴eγl − e−γl

eγl + e−γl = A (39)

∴e2γl =1 + A1− A

(40)

∴e2αlej2βl =1 + A1− A

(41)

I α can be obtained by equating the real portions of the above polar equation i.e.

e2αl =

∣∣∣∣∣1 + A1− A

∣∣∣∣∣ (42)

⇒ α =12l

ln

∣∣∣∣∣1 + A1− A

∣∣∣∣∣ (43)

I The estimation of β, however, is tricky, as the standing wave characteristics repeatevery λ/2 distance along the line (equivalent to an integral phase multiple of 2π).For a line, it is difficult to estimate the no. of λ/2 sections. Thus,

β =12l

[∠

1 + A1− A

± 2nπ

](44)


Measurement of Propagation Constant of a LineI The ambiguity in the estimation of β can be removed using analysis at two

successive frequencies f1 and f2 which have identical sets of (Zsc ,Zoc). At f1,

β1 =12l

[∠

1 + A1− A

± 2nπ

](45)

I At f2,

β2 =12l

[∠

1 + A1− A

± 2(n + 1)π

](46)

I Subtracting, we obtain

β2 − β1 =π

l(47)

∴2πf2

v− 2πf1

v=π

l(48)

∴Wave velocity v = 2l(f2 − f1) (49)

I By earlier definition, β =2πfv

. Therefore,

β =πf

l(f2 − f1)(50)


References

I Electromagnetic Waves by R. K. ShevgaonkarI Microwave Engineering by D. M. PozarI Electromagnetic Waves and Radiating Systems by Jordan and BalmainI Microwaves 101, IEEE MTT-S


Documents

Transmission Lines- Part II - IIT Bombaydghosh/tx-lines-2.pdf · Back to the Smith Chart In Transmission Lines-I, the development of the Smith chart was discussed. Now, some analyses