Transmission Lines- Part Idghosh/tx-lines.pdf · propagation constant of the transmission line I Now, from the general expression V 1(x;t) = V+e x cos( x !t), at a particular location

Transmission Lines- Part I

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 I 1 / 30

Outline

I Motivation of the use of transmission lines

I Voltage and current analysis

I Wave propagation on transmission lines

I Transmission line parameters and characteristic impedance

I Reflection coefficient and impedance transformation

I Voltage and current maxima/minima, and VSWR

I Developing the Smith Chart


Difference between Low and High Frequency CircuitsI At low frequencies (say, up to a few MHz), the size of most circuits and circuit

elements are negligible compared to the wavelength of the signalI For time-varying voltage/current, most circuit laws- Ohm’s Law, Kirchhoff’s Voltage

and Current Laws- assume that at a given instant, the voltage/current across thelength of the circuit remains constant

I This assumption does not hold as the frequency increases. Consider ahigh-frequency signal travelling from a source to a load through a transmission lineof length l

lVL

VSLoad

I At a given instant of time, the source sees voltage VS and the load sees voltage VL,which are different. The time taken for VS to appear at the load end is equal to the

propagation time i.e. tp =lv

, where v is the wave velocity

I Applying circuital laws at high frequencies for transmission lines is therefore, not

suitable. If the length of the line l ≤ λ

20, it can be assumed that there is negligible

change of V or I along l . Practically, l ≤ λ

10is a more commonly used convention


Analysis of Transmission Lines using Circuit LawsI If circuital laws are not valid for transmission lines, then how should the analysis be

done?I Solution: consider an infinitesimally small length of the line, ∆x where ∆x � λ and

it can be assumed that the V and I do not change for ∆x at a given instant of timeI Let us represent this section by standard circuit elements. Since the line is made of

a large number of such small sections, we will represent the circuit using distributedelements (i.e. per unit length quantities) as

R- resistance per unit length (due to resistance of the conducting lines)L- inductance per unit length (self inductance of the line)G- conductance per unit length (due to loss in the dielectric between the lines)C- capacitance per unit length (due to the gap between the two lines)

R∆x L∆x

G∆xC∆x

∆x

V(x) V(x+∆x)

I(x) I(x+∆x)

I The voltages V (x) and V (x + ∆x), and currents I(x) and I(x + ∆x) can beexpressed using Kirchhoff’s laws


Voltage and Current Analysis on Transmission Line

I The voltages and currents can then be related using KVL as

V (x)− I(x)(R + jωL)∆x − V (x + ∆x) = 0

∴V (x + ∆x)− V (x)

∆x= −I(x)(R + jωL)

I In the limiting case ∆x → 0, we can rewrite as

dVdx

= −(R + jωL)I (1)

I Similarly, the currents can be related using KCL as

dIdx

= −(G + jωC)V (2)

where V and I are functions of x . The above equations are the simplified form of theTelegraphers’ Equations

I The above two differential equations are inter-related. Differentiating each equation

with x again and substituting the known expressions fordVdx

anddIdx

we obtain


Solutions to Telegraphers’ Equations

d2Vdx2 = γ2V (3)

d2Idx2 = γ2I (4)

I Where γ2 = (R + jωL)(G + jωC). The above is a pair of 2nd order homogeneousdifferential equations, whose solutions are of the form

V (x) = V+e−γx + V−eγx (5)

I(x) = I+e−γx + I−eγx (6)I In practice, the voltage and current vary with position as well as time. If the input

voltage is sinusoidal, V (x , t) = V (x)ejωt . Thus,

V (x , t) = V+e−γx ejωt + V−eγx ejωt (7)

I(x , t) = I+e−γx ejωt + I−eγx ejωt (8)I In general, if R, L,G,C 6= 0, then γ is a complex quantity. Representing, γ = α+ jβ,

the above solutions are rewritten as

V (x , t) = V+e−αx e−j(βx−ωt) + V−eαx ej(βx+ωt) (9)

I(x , t) = I+e−αx e−j(βx−ωt) + I−eαx ej(βx+ωt) (10)Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part I 6 / 30

Wave Phenomena on Transmission LineI This gives some interesting results. As an example, let’s look at V (x , t). Consider

the first term of V (x , t), the real part of which indicates the voltage along thex-direction. Thus,

V1(x , t) = V+e−αx cos(βx − ωt)I The term V+e−αx , if α > 0, indicates a decaying magnitude along the +ve x

direction, i.e. attenuation. Thus, α is called attenuation constant of the lineI Consider, without loss of generality, α = 0 and V+ as a real quantity. Thus,

V1(x , t) = |V+| cos(βx − ωt)I Consider three different time instants t = t1, t2, t3 and t3 > t2 > t1. Then, V1 at these

time instants as a function of x is shown belowVx

Vx

Vx

t = t1

t = t2

t = t3

x

The point Vx moves forwardas time increases. Thus, it isan indication of a wave movingtowards the positive x direction


Wave Phenomena on Transmission Line (cont’d..)

I Similarly, holding above assumptions, V2(x , t) = |V−| cos(βx + ωt) indicates awave travelling in the -ve x direction with time

I This indicates that in every transmission line, there are two wave components: onetravelling in the +ve x direction (forward) and the other in the -ve x direction(reverse)

I In general, the amplitude of the forward wave decreases with increase in x and thatof the reverse wave increases with increase in x

I It was assumed that γ = α+ jβ. If γ = 0, then the V (x , t) solution reduces in termsof the input ejωt only, which does not indicate a propagating wave with time

I The γ term is thus essential to denote wave propagation, hence γ is termed thepropagation constant of the transmission line

I Now, from the general expression V1(x , t) = V+e−αx cos(βx − ωt), at a particularlocation x on the line, the magnitude of V1(x , t) varies with time t . The term βxdenotes the phase of V1 at the point x . Thus, β is termed phase constant

I α determines the attenuation of the wave along the line, and it is preferred thatideally, for zero loss along the line, α = 0

I Can β ever be zero? Why/why not?


More on β and αI For a time-varying wave, βx denotes the phase at a given location. Physically on

the line, the variation in x denotes the phase change

I Assume the wavelength of the propagating wave is λ. As x varies from any pointx = x0 to x = x0 + λ, the effective phase change must be 360◦ or 2π radians

∴ β(x0 + λ)− βx0 = 2π ⇒ β =2πλ

(11)

I Unit of β ≡ radians/m

I The term αx is dimensionless (why?). Thus, the unit of alpha should be m−1

I As α denotes the change in the wave magnitude envelope along the line, the ratioof the change in voltage per unit length is expressed as Neper/m (or Np/m)

I α may also be expressed in Decibels/m or dB/m. Assume α = 1 Np/m. Define the‘‘effective travel distance’’ as the distance after which the wave magnitude becomes1/e times the initial value

∴ 20 log10

(Vfinal

Vinitial

)= 20 log10 e−1 = 20 log10 e−αx ⇒ x = 1

∴ 1 Np/m = 20 log10 e = 8.68 dB/m


Characteristic Impedance of a Transmission LineI Let’s go back to the positional solutions for V and I i.e.

V (x) = V+e−γx + V−eγx

I(x) = I+e−γx + I−eγx

I As per Telegraphers’ equations,dVdx

= −(R + jωL)I

∴ −γV+e−γx + γV−eγx = −(R + jωL)(I+e−γx + I−eγx ) (12)

∴ −V+e−γx + V−eγx = −R + jωLγ

(I+e−γx + I−eγx ) (13)

I Now,R + jωL

γ=

√R + jωLG + jωC

= Z0

I Z0 has the dimensions of impedance, and is governed by the primary distributedcharacteristics of the line (R, L,G,C), and frequency (ω = 2πf ). Z0 is termedcharacteristic impedance of the line

I Comparing exponents e−γx and eγx on both sides, we get

V+

I+= Z0

V−

I−= −Z0 (14)

I What does negative impedance −Z0 indicate?Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part I 10 / 30

Effect of Load and Reflection on a Transmission LineI So far, the discussion has been largely limited to the line in general. But the load at

the end of the line plays an important role in determining V , I along the line

I It is thus beneficial to shift the origin from the source to the load. We define l fromload to source i.e. l = −x . At the load, l = 0

I The solutions for V and I now become

V (l) = V+eγl + V−e−γl (15)

I(l) =V+

Z0eγl − V−

Z0e−γl (16)

I V+eγl and V−e−γl denote forward and backward waves, respectively. Define aquantity called reflection coefficient along the line as the ratio of the waves i.e.

Γ(l) =V−e−γl

V+eγl =V−

V+e−2γl (17)

I At the load, l = 0. Γ(0) = ΓL =V−

V+. ΓL is called the load reflection coefficient

I The impedance along any point on the line is given as

Z (l) =V (l)I(l)

= Z01 + ΓLe−2γl

1− ΓLe−2γl (18)


Load and Line Reflection Coefficient

I At the load i.e. l = 0, Z (0) = ZL =1 + ΓL

1− ΓL

I Thus, ΓL =ZL − Z0

ZL + Z0. Clearly, there will be no reflection if ZL = Z0

I Maximum possible magnitude of ΓL is 1 (unless the load has unstable activeelements)

I For finite Z0,|ΓL| = 1 is possible for three kinds of loads. Two of which are an opencircuit (ZL =∞) and a short circuit (ZL = 0). What is the third?

I The reflection coefficient at any point on the line at a distance l from the load is

given as Γ(l) =V−

V+e−2γl = ΓLe−2γl

I For a lossless line, α = 0. Thus, Γ(l) = ΓLe−j2βl


Transformation of Impedance along the Line

I It was shown that Z (l) = Z01 + ΓLe−2γl

1− ΓLe−2γl

I Substituting ΓL =ZL − Z0

ZL + Z0, and hyperbolic sine and cosine functions,

Z (l) = Z0ZL cosh γl + Z0 sinh γlZ0 cosh γl + ZL sinh γl

(19)

I This is known as the impedance transformation relationship, which shows thatthe load impedance is transformed to some other impedance along the length of theline l

I For a lossless line, α = 0⇒ γ = jβ. In such a case the impedance transformationrelation becomes

Z (l) = Z0ZL cosβl + jZ0 sinβlZ0 cosβl + jZL sinβl

(20)

I Rearranging, we get

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

(21)

I This shows that if the impedance at a point on the line l1 is known, the impedance atany other point l2 can be calculated from this relation


Lossless Transmission Lines

I The lossy elements in a transmission line are R and G (line ohmic loss anddielectric loss)

I If R,G→ 0, then the line becomes lossless i.e.

γ =√

(R + jωL)(G + jωC) = jω√

LC (22)

I Since γ is now purely imaginary, α = 0, which is consistent with earlier discussion

I ∴ β = ω√

LC. Since β =2πλ

,

2πλ

= 2πf√

LC ⇒ v = λf =1√LC

(23)

where v is the wave velocity, also known as phase velocityI The phase constant β is determined by the transmission line elements L and C,

which determine the wavelength of the wave in the medium

I For a lossless line, Z0 =

√LC

, which is a real quantity

I Complex Z0 indicates a lossy line, but does real Z0 guarantee that the line islossless?


Back to Impedance Transformation

I It is seen that the load (ZL) or line impedances (Z (l)) by themselves are notsignificant; a lot depends on the characteristic impedance Z0

I Define the normalized impedance (denoted in lower case) w.r.t. Z0 as

z(l) =Z (l)Z0

or zL =ZL

Z0(24)

I In terms of the normalized impedances, the impedance transformation relation for alossless line becomes

z(l) =zL + j tanβl1 + jzL tanβl

(25)

I If the impedance is z1 at length l1, then the impedance z2 at l2 = l1 + λ/4 is

z2 =zL + j tanβ(l1 + λ/4)

1 + jzL tanβ(l1 + λ/4)=

zL − j cotβl11− jzL cotβl1

=1 + jzL tanβl1zL + j tanβl1

=1z1

I Result 1: The normalized impedance inverts itself along a line, after every λ/4distance


More Results on Impedance Transformation

I If the impedance is z1 at length l1, then the impedance z2 at l2 = l1 + λ/2 is

z2 =zL + j tanβ(l1 + λ/2)

1 + jzL tanβ(l1 + λ/2)=

zL + j tanβl11 + jzL tanβl1

= z1

I Result 2: The normalized (and also, the actual) impedance repeats itself along aline, after every λ/2 distance

I If ZL = Z0, then the normalized load impedance zL = 1. In such a case, theimpedance along any point on the line is

z(l) =1 + j tanβl1 + j tanβl

= 1

I Result 3: If the load and characteristic impedances are equal, then the impedanceat all point on the line is equal to the characteristic impedance Z0

I We can then redefine characteristic impedance Z0 as the impedance seen at theinput of the transmission line when the line is terminated at the other end also byZL = Z0

I Such a case is called the matched condition of a transmission line and is highlydesirable as it prevents any reflections along the line


Voltage and Current Maxima/Minima on Transmission LineI We have already derived the solutions for V and I as

V (l) = V+eγl + V−e−γl

I(l) =V+

Z0eγl − V−

Z0e−γl

I Rewriting in terms of ΓL and V+, for a lossless line, (now on, the assumption willbe a lossless line unless mentioned otherwise)

V (l) = V+ejβl (1 + ΓLe−j2βl ) (26)

I(l) =V+

Z0ejβl (1− ΓLe−j2βl ) (27)

I ΓL is in general, a complex quantity which can be represented as |ΓL|ejφL . Thus,

V (l) = V+ejβl (1 + |ΓL|ej(φL−2βl)) (28)

I(l) =V+

Z0ejβl (1− |ΓL|ej(φL−2βl)) (29)

I V (l) is maximum when (φL − 2βl) is an even multiple of π, and I(l) is maximumwhen (φL − 2βl) is an odd multiple of π. Conversely, with the minimas

I It is interesting to note that the maximas of the voltage and current of the standingwave are out of phase by 180◦ or π radians


The Concept of Standing Waves on a Lossless LineI Standing wave is defined as one whose magnitude does not change at certain

points on a line, while changing at other points. Let us see howI The component of the forward wave along the line is Vf = V+ cos(βx − ωt) and the

reverse wave Vr = V− cos(βx + ωt) (temporarily switching back to x from l)I The net voltage along the line is the sum of the forward and reverse waves i.e.

Vt = V+ cos(βx − ωt) + V− cos(βx + ωt)

I Consider a simple case of an open circuit load, where ΓL =V−

V+= 1. Thus,

Vt = V+[cos(βx − ωt) + cos(βx + ωt)] (30)

= 2V+ cos(βx) cos(ωt) (31)I The spatial term cos(βx) and the time varying term cos(ωt) are now separated. For

simplicity, assuming that V+ is real, here is what Vt looks like

0 λ/4 λ/2 3λ/4

λ/4-

λ/2-

3λ/4-

t = 0

t = 1/2f0

t = 1/8f0

t = 1/4f0

|2V+|


Standing Waves and SWR on a Lossless LineI Looking at Vt , it is seen that the magnitudes at points x = ±λ/4,±3λ/4 etc. are

always zero. These points are called nodes. At pointsx = 0,±λ/2 etc. they reachmaxima depending on the time. These points are called antinodes

I The variation of Vt with time gives an indication of a ‘‘stationary’’ wave vibrating atfixed locations, hence is called a standing wave. Ex: do a similar analysis for ashort load and calculate the node and antinode positions along the line

I For an arbitrary load with 0 < ΓL < 1, the analysis is more complex with thisapproach. If the load is matched to Z0, then ΓL = 0 hence there is no standingwave; the wave always travels to the load

I Going back to V (l), it was established that

V (l) = V+ejβl (1 + |ΓL|ej(φL−2βl))

I Thus,

|V (l)|max = |V+|(1 + |ΓL|) and |V (l)|min = |V+|(1− |ΓL|) (32)

∴ Define voltage standing wave ratio (VSWR) ρ =|V (l)|max

|V (l)|min=

1 + |ΓL|1− |ΓL|

(33)

I e.g. for the open load, |Vmax | = 2|V+| and |Vmin| = 0 (remember that this is inabsolute magnitude terms), thus ρ =∞. You will find a similar result for short load


Line Impedance Transformation and VSWRI VSWR is always a real number. Like reflection coefficient Γ, it also indicates

matching on the line. The complex Γ changes as a function of length l , but theVSWR is constant along the line. For any arbitrary load, 1 ≤ ρ ≤ ∞

I For a matched load ZL = Z0, there is no standing wave because of no reflections.Hence, VSWR ρ = 1

I Along with V (l), it was established that I(l) =V+

Z0ejβl (1− |ΓL|ej(φL−2βl))

I Thus,

|I(l)|max =|V+|Z0

(1 + |ΓL|) and |I(l)|min =|V+|Z0

(1− |ΓL|) (34)

∴|V (l)|max

|I(l)|max=|V (l)|min

|I(l)|min= Z0 (35)

I As the impedance Z (l) changes along the line, the maxima and minima are thus,

Zmax =|Vmax ||Imin|

and Zmin =|Vmin||Imax |

(36)

i.e. Zmax = ρZ0 and Zmin =Z0

ρ(37)

I Thus if the VSWR is known, the maximum and minimum impedances, and thelocation of the voltage/current maxima and minima can also be found out


Voltage Maxima/Minima seen Graphically

I We know that V (l) = V+ejβl (1 + |ΓL|ej(φL−2βl))

I Consider the term 1 + |ΓL|ej(φL−2βl). It is the sum of a constant with a phasor withmagnitude |ΓL| and phase φL − 2βl .

I As one moves away from the load (i.e. increasing l), the phase term becomes morenegative, so the phasor rotates clockwise

1

|ΓL|φL-2βl Moving towards

source0

1 + |ΓL|ej(φL- 2

βl)

Voltage minimum Voltage maximum

I Voltage maxima occur at φL − 2βl = 0, 2π, 4π (all even multiples of π)

I Voltage minima occur at φL − 2βl = π, 3π, 5π (all odd multiples of π)

I It was discussed that the VSWR ρ is determined by |ΓL|. Thus the locus of ΓL

tracing a circle is called a constant VSWR circle, as ρ is fixed along the circle


Transformation Between Impedance and Reflection Coefficient

I Recall that for a normalized impedance z, the reflection coefficient Γ =z − 1z + 1

I Both numerator and denominator are linear functions of z, hence this is a bilineartransformation function

I The bilinear transformation is a one-to-one mapping, hence for every value of z,there exists a unique value of Γ

I Consider impedance z = r + jx . Consider r ≥ 0 (ignore unstable systems likeoscillators). Then on the complex Cartesian z-plane, the first and fourth quadrants(the right side of the jx axis shows the area covering all possible impedances

I Consider a polar representation of the reflection coefficient Γ = |Γ|ejφ. Now, thefollowing relation

|Γ|ejφ =r + jx − 1r + jx + 1

(38)

shows that all loads are mapped into a circle of radius |Γ| and phase angle φ on thepolar plane.

I This means all loads are mapped within a circle whose maximum radius is 1


Pictorial Representation of the Bilinear Transformation

r

jx

Re(Γ)

Im(Γ)

Open loadShort load

Matched

load

|Γ| = 1

I Purely reactive loads (±jx) get mapped to the periphery of the unit circleI Inductive impedances are mapped to the upper half, and capacitive ones are

mapped to the lower half peripheryI r + jx and r − jx are mapped to the interior of the upper and lower half of the unit

circle, respectivelyI Purely resistive loads lie on the real Γ axisI This polar form of representation of impedances on the complex Γ plane is the

foundation of the Smith ChartDebapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part I 23 / 30

Developing the Smith ChartI The Smith chart is a graphical tool for transmission line calculations. Rather than

using multiple tedious equations, one can use the Smith chart to do the jobI At the first look, a Smith chart looks quite complex. In reality, it is nothing but a

slightly different coordinate systemI The Smith chart consists of loci of normalized resistances and reactances mapped

to the complex Γ planeI The previous graph shows the readings of the points in the circle in terms of Γ; a

Smith chart has readings of normalized resistance and reactance

I In terms of Γ, the normalized impedance z =1 + Γ

1− ΓI If an impedance r + jx is transformed to Γ = u + jv , then

r + jx =1 + u + jv1− u − jv

Normalizing this, we obtain

r + jx =(1 + u + jv)(1− u + jv)

(1− u)2 + v2

I Simplifying and equating the corresponding real and imaginary parts on both sides,

r =1− u2 − v2

(1− u)2 + v2 and x =2v

(1− u)2 + v2


Constant Resistance Solution

I On the complex Γ plane, the resistance must be expressed as functions of the axesu and v

I r =1− u2 − v2

(1− u)2 + v2 simplifies to

(r + 1)u2 − 2ru + (r + 1)v2 + r − 1 = 0

I Dividing throughout by (r − 1) and by completing the square, we obtain(u − r

r + 1

)2

+ v2 =

(1

r + 1

)2

(39)

I The above equation is that of a circle with center at

(r

r + 1, 0

)and radius

1r + 1

I There is a unique circle for every distinct value of r . Hence this set of solutions isknown as a constant resistance solution

I Let us study what these circles look like


Resistance CirclesI Shown below are the constant resistance circles for r = 0, 1, 3, 7

r = 0

r = 1

r = 3

r = 7

(-1,0) (0,0) (0.5,0)

(0.75,0)

(1,0)

I As r increases, the centers of the circle shift towards the +ve u direction, with v = 0at all times

I The radii of the circles decrease as r increases. All r circles touch one another at(1,0), which indicates r =∞ or an open circuit, where the r circle is negligibly small


Constant Reactance Solution

I Like the case for r , the reactance must also be expressed as functions of u and v

I x =2v

(1− u)2 + v2 simplifies to

(1− u)2 + xv2 = 2v

I Dividing by x and completing the square, we get

(u − 1)2 +

(v − 1

x

)2

=

(1x

)2

(40)

I The above equation is that of a circle with center at

(1,

1x

)and radius

1x

I There is a unique circle for every distinct value of x . Hence this set of solutions isknown as a constant reactance solution

I Let us study what these circles look like


Reactance CirclesI Shown below are the constant reactance circles for r = 0, 1, 2, 4, 8

x = 1

x = 4

x = 8

x = 2

x = 0

x = -2

x = -4

x = -8

x = -1

u = 1

(1,1)

(1,0.5)

(1,-0.5)

(1,-1)

I As x increases, the centers of the circle shift towards the point (1,0), with centersaligned with u = 1

I The radii of the circles decrease as x increases. All x circles touch one another at(1,0), which indicates x = ±∞, where the x circle is negligibly small


The Complete Smith ChartI To complete the Smith Chart, combine the r and x circles. We are only interested in

those circles which lie inside r = 0 circleI This means part of all x circles lie outside the area of interest

12

0.5

Inductive reactance

Capacitive reactance

Movement

towards

source

ResistanceMovement

towards

load

j0.5

j1

j2

j0.5

j1

j2

I It was designed by an American engineer Philip H. Smith


References

I Electromagnetic Waves by R. K. ShevgaonkarI Microwave Engineering by D. M. PozarI Electromagnetic Waves and Radiating Systems by Jordan and BalmainI OCW EECS, MIT


Documents

Transmission Lines- Part Idghosh/tx-lines.pdf · propagation constant of the transmission line I Now, from the general expression V 1(x;t) = V+e x cos( x !t), at a particular location