Transmission Lines- Part I

Debapratim Ghosh

Electronic Systems Group

Department of Electrical Engineering
Indian Institute of Technology Bombay

e-mail: dghosh@ee.iitb.ac.in

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

Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part I 2 / 30

 Difference between Low and High Frequency Circuits
I 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 signal
I 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 the
length of the circuit remains constant
I This assumption does not hold as the frequency increases. Consider a
high-frequency signal travelling from a source to a load through a transmission line
of length l
l VL
VS
Load

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
l
propagation time i.e. tp = , where v is the wave velocity
v
I Applying circuital laws at high frequencies for transmission lines is therefore, not
λ
suitable. If the length of the line l ≤ , it can be assumed that there is negligible
20
λ
change of V or I along l. Practically, l ≤ is a more commonly used convention
10
Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part I 3 / 30
 Analysis of Transmission Lines using Circuit Laws
I 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 time
I 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 distributed
elements (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)
I(x) R∆x L∆x I(x+∆x)

V(x) G∆x V(x+∆x)

C ∆x

∆x
I The voltages V (x) and V (x + ∆x), and currents I(x) and I(x + ∆x) can be
expressed using Kirchhoff’s laws
Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part I 4 / 30
 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)
∴ = −I(x)(R + jωL)
∆x
I In the limiting case ∆x → 0, we can rewrite as
dV
= −(R + jωL)I (1)
dx
I Similarly, the currents can be related using KCL as
dI
= −(G + jωC)V (2)
dx
where V and I are functions of x. The above equations are the simplified form of the
Telegraphers’ Equations
I The above two differential equations are inter-related. Differentiating each equation
dV dI
with x again and substituting the known expressions for and we obtain
dx dx

Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part I 5 / 30

 Solutions to Telegraphers’ Equations

d 2V
= γ2V (3)
dx 2
d 2I
= γ2I (4)
dx 2

I Where γ 2 = (R + jωL)(G + jωC). The above is a pair of 2nd order homogeneous

differential equations, whose solutions are of the form
V (x) = V + e−γx + V − eγx (5)
+ −γx − γx
I(x) = I e +I e (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)
+ −γx jωt − γx jωt
I(x, t) = I e e +I e e (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)
+ −αx −j(βx−ωt) − αx j(βx+ωt)
I(x, t) = I e e +I e e (10)
Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part I 6 / 30
 Wave Phenomena on Transmission Line
I 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 the
x-direction. Thus,
V1 (x, t) = V + e−αx cos(βx − ωt)
+ −αx
I The term V e , if α > 0, indicates a decaying magnitude along the +ve x
direction, i.e. attenuation. Thus, α is called attenuation constant of the line
I 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 below
Vx
t = t1

Vx
The point Vx moves forward
t = t2 as time increases. Thus, it is
an indication of a wave moving
towards the positive x direction
Vx
t = t3

x
Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part I 7 / 30
 Wave Phenomena on Transmission Line (cont’d..)
I Similarly, holding above assumptions, V2 (x, t) = |V − | cos(βx + ωt) indicates a
wave travelling in the -ve x direction with time
I This indicates that in every transmission line, there are two wave components: one
travelling 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 that
of 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 terms
of 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 the
propagation constant of the transmission line
I Now, from the general expression V1 (x, t) = V + e−αx cos(βx − ωt), at a particular
location x on the line, the magnitude of V1 (x, t) varies with time t. The term βx
denotes 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 that
ideally, for zero loss along the line, α = 0
I Can β ever be zero? Why/why not?

Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part I 8 / 30

 More on β and α
I 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 point
x = x0 to x = x0 + λ, the effective phase change must be 360◦ or 2π radians
2π
∴ β(x0 + λ) − βx0 = 2π ⇒ β = (11)
λ
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 ratio
of 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 becomes
1/e times the initial value
!
Vfinal
∴ 20 log10 = 20 log10 e−1 = 20 log10 e−αx ⇒ αx = 1
Vinitial
∴ 1 Np/m = 20 log10 e = 8.68 dB/m

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

 Characteristic Impedance of a Transmission Line
I 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
dV
I As per Telegraphers’ equations, = −(R + jωL)I
dx
∴ −γV + e−γx + γV − eγx = −(R + jωL)(I + e−γx + I − eγx ) (12)
R + jωL + −γx
∴ −V + e−γx + V − eγx = − (I e + I − eγx ) (13)
γ
r
R + jωL R + jωL
I Now, = = Z0
γ G + jωC
I Z0 has the dimensions of impedance, and is governed by the primary distributed
characteristics of the line (R, L, G, C), and frequency (ω = 2πf ). Z0 is termed
characteristic impedance of the line
I Comparing exponents e−γx and eγx on both sides, we get
V+ V−
= Z0 = −Z0 (14)
I + I−
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 Line
I 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 from
load 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)
+ −
V γl V
I(l) = e − e−γl (16)
Z0 Z0
I V + eγl and V − e−γl denote forward and backward waves, respectively. Define a
quantity called reflection coefficient along the line as the ratio of the waves i.e.
V − e−γl V−
Γ(l) = + γl
= + e−2γl (17)
V e V
V−
I At the load, l = 0. Γ(0) = ΓL = . ΓL is called the load reflection coefficient
V+
I The impedance along any point on the line is given as
V (l) 1 + ΓL e−2γl
Z (l) = = Z0 (18)
I(l) 1 − ΓL e−2γl
Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part I 11 / 30
 Load and Line Reflection Coefficient

1 + ΓL
I At the load i.e. l = 0, Z (0) = ZL =
1 − ΓL
ZL − Z0
I Thus, ΓL = . Clearly, there will be no reflection if ZL = Z0
ZL + Z0
I Maximum possible magnitude of ΓL is 1 (unless the load has unstable active
elements)
I For finite Z0 ,|ΓL | = 1 is possible for three kinds of loads. Two of which are an open
circuit (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
V−
given as Γ(l) = + e−2γl = ΓL e−2γl
V
I For a lossless line, α = 0. Thus, Γ(l) = ΓL e−j2βl

Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part I 12 / 30

 Transformation of Impedance along the Line
1 + ΓL e−2γl
I It was shown that Z (l) = Z0
1 − ΓL e−2γl
ZL − Z0
I Substituting ΓL = , and hyperbolic sine and cosine functions,
ZL + Z0
ZL cosh γl + Z0 sinh γl
Z (l) = Z0 (19)
Z0 cosh γl + ZL sinh γl
I This is known as the impedance transformation relationship, which shows that
the load impedance is transformed to some other impedance along the length of the
line l
I For a lossless line, α = 0 ⇒ γ = jβ. In such a case the impedance transformation
relation becomes
ZL cos βl + jZ0 sin βl
Z (l) = Z0 (20)
Z0 cos βl + jZL sin βl
I Rearranging, we get
ZL + jZ0 tan βl
Z (l) = Z0 (21)
Z0 + jZL tan βl
I This shows that if the impedance at a point on the line l1 is known, the impedance at
any other point l2 can be calculated from this relation
Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part I 13 / 30
 Lossless Transmission Lines
I The lossy elements in a transmission line are R and G (line ohmic loss and
dielectric loss)
I If R, G → 0, then the line becomes lossless i.e.
p √
γ = (R + jωL)(G + jωC) = jω LC (22)
I Since γ is now purely imaginary, α = 0, which is consistent with earlier discussion
√ 2π
I ∴ β = ω LC. Since β = ,
λ
2π √ 1
= 2πf LC ⇒ v = λf = √ (23)
λ LC
where v is the wave velocity, also known as phase velocity
I The phase constant β is determined by the transmission line elements L and C,
which determine the wavelength of the wave in the medium
r
L
I For a lossless line, Z0 = , which is a real quantity
C
I Complex Z0 indicates a lossy line, but does real Z0 guarantee that the line is
lossless?

Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part I 14 / 30

 Back to Impedance Transformation
I It is seen that the load (ZL ) or line impedances (Z (l)) by themselves are not
significant; a lot depends on the characteristic impedance Z0
I Define the normalized impedance (denoted in lower case) w.r.t. Z0 as
Z (l) ZL
z(l) = or zL = (24)
Z0 Z0
I In terms of the normalized impedances, the impedance transformation relation for a
lossless line becomes
zL + j tan βl
z(l) = (25)
1 + jzL tan βl
I If the impedance is z1 at length l1 , then the impedance z2 at l2 = l1 + λ/4 is

zL + j tan β(l1 + λ/4) zL − j cot βl1

z2 = =
1 + jzL tan β(l1 + λ/4) 1 − jzL cot βl1
1 + jzL tan βl1 1
= =
zL + j tan βl1 z1
I Result 1: The normalized impedance inverts itself along a line, after every λ/4
distance

Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part I 15 / 30

 More Results on Impedance Transformation
I If the impedance is z1 at length l1 , then the impedance z2 at l2 = l1 + λ/2 is

zL + j tan β(l1 + λ/2) zL + j tan βl1

z2 = = = z1
1 + jzL tan β(l1 + λ/2) 1 + jzL tan βl1
I Result 2: The normalized (and also, the actual) impedance repeats itself along a
line, after every λ/2 distance
I If ZL = Z0 , then the normalized load impedance zL = 1. In such a case, the
impedance along any point on the line is
1 + j tan βl
z(l) = =1
1 + j tan βl
I Result 3: If the load and characteristic impedances are equal, then the impedance
at 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 the
input of the transmission line when the line is terminated at the other end also by
ZL = Z0
I Such a case is called the matched condition of a transmission line and is highly
desirable as it prevents any reflections along the line

Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part I 16 / 30

 Voltage and Current Maxima/Minima on Transmission Line
I We have already derived the solutions for V and I as
V (l) = V + eγl + V − e−γl
V + γl V − −γl
I(l) = e − e
Z0 Z0
I Rewriting in terms of ΓL and V + , for a lossless line, (now on, the assumption will
be a lossless line unless mentioned otherwise)
V (l) = V + ejβl (1 + ΓL e−j2βl ) (26)
+
V jβl
I(l) = e (1 − ΓL e−j2βl ) (27)
Z0
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)
V + jβl
I(l) = e (1 − |ΓL |ej(φL −2βl) ) (29)
Z0
I V (l) is maximum when (φL − 2βl) is an even multiple of π, and I(l) is maximum
when (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 standing
wave are out of phase by 180◦ or π radians
Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part I 17 / 30
 The Concept of Standing Waves on a Lossless Line
I 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 how
I 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)
V−
I Consider a simple case of an open circuit load, where ΓL = = 1. Thus,
V+
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
|2V+|
t = 1/2f0

t = 1/4f0
- 3λ/4 - λ/4

- λ/2 0 λ/4 λ/2 3λ/4

t = 1/8f0
t=0
Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part I 18 / 30
 Standing Waves and SWR on a Lossless Line
I 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 reach
maxima depending on the time. These points are called antinodes
I The variation of Vt with time gives an indication of a ‘‘stationary’’ wave vibrating at
fixed locations, hence is called a standing wave. Ex: do a similar analysis for a
short 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 this
approach. If the load is matched to Z0 , then ΓL = 0 hence there is no standing
wave; 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)
|V (l)|max 1 + |ΓL |
∴ Define voltage standing wave ratio (VSWR) ρ = = (33)
|V (l)|min 1 − |ΓL |
I e.g. for the open load, |Vmax | = 2|V + | and |Vmin | = 0 (remember that this is in
absolute magnitude terms), thus ρ = ∞. You will find a similar result for short load
Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part I 19 / 30
 Line Impedance Transformation and VSWR
I 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 the
VSWR 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
V + jβl
I Along with V (l), it was established that I(l) = e (1 − |ΓL |ej(φL −2βl) )
Z0
I Thus,
|V + | |V + |
|I(l)|max = (1 + |ΓL |) and |I(l)|min = (1 − |ΓL |) (34)
Z0 Z0
|V (l)|max |V (l)|min
∴ = = Z0 (35)
|I(l)|max |I(l)|min
I As the impedance Z (l) changes along the line, the maxima and minima are thus,
|Vmax | |Vmin |
Zmax = and Zmin = (36)
|Imin | |Imax |
Z0
i.e. Zmax = ρZ0 and Zmin =
(37)
ρ
I Thus if the VSWR is known, the maximum and minimum impedances, and the
location of the voltage/current maxima and minima can also be found out
Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part I 20 / 30
 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 with
magnitude |ΓL | and phase φL − 2βl.
I As one moves away from the load (i.e. increasing l), the phase term becomes more
negative, so the phasor rotates clockwise
2βl)
j(φ L-
Γ |e
1+| L |ΓL|
φL-2βl Moving towards
1 source
0
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

Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part I 21 / 30

 Transformation Between Impedance and Reflection Coefficient
z −1
I Recall that for a normalized impedance z, the reflection coefficient Γ =
z +1
I Both numerator and denominator are linear functions of z, hence this is a bilinear
transformation 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 like
oscillators). 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, the
following relation
r + jx − 1
|Γ|ejφ = (38)
r + jx + 1
shows that all loads are mapped into a circle of radius |Γ| and phase angle φ on the
polar plane.
I This means all loads are mapped within a circle whose maximum radius is 1

Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part I 22 / 30

 Pictorial Representation of the Bilinear Transformation
jx
Im(Γ)
|Γ| = 1

r Re(Γ)

Matched
load
Short load Open load

I Purely reactive loads (±jx) get mapped to the periphery of the unit circle
I Inductive impedances are mapped to the upper half, and capacitive ones are
mapped to the lower half periphery
I r + jx and r − jx are mapped to the interior of the upper and lower half of the unit
circle, respectively
I Purely resistive loads lie on the real Γ axis
I This polar form of representation of impedances on the complex Γ plane is the
foundation of the Smith Chart
Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part I 23 / 30
 Developing the Smith Chart
I 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 job
I At the first look, a Smith chart looks quite complex. In reality, it is nothing but a
slightly different coordinate system
I The Smith chart consists of loci of normalized resistances and reactances mapped
to the complex Γ plane
I 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
1+Γ
I In terms of Γ, the normalized impedance z =
1−Γ
I If an impedance r + jx is transformed to Γ = u + jv , then
1 + u + jv
r + jx = Normalizing this, we obtain
1 − u − jv
(1 + u + jv )(1 − u + jv )
r + jx =
(1 − u)2 + v 2
I Simplifying and equating the corresponding real and imaginary parts on both sides,
1 − u2 − v 2 2v
r= and x=
(1 − u)2 + v 2 (1 − u)2 + v 2
Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part I 24 / 30
 Constant Resistance Solution
I On the complex Γ plane, the resistance must be expressed as functions of the axes
u and v
1 − u2 − v 2
I r= simplifies to
(1 − u)2 + v 2

(r + 1)u 2 − 2ru + (r + 1)v 2 + r − 1 = 0

I Dividing throughout by (r − 1) and by completing the square, we obtain
!2 !2
r 2 1
u− +v = (39)
r +1 r +1
!
r 1
I The above equation is that of a circle with center at , 0 and radius
r +1 r +1
I There is a unique circle for every distinct value of r . Hence this set of solutions is
known as a constant resistance solution
I Let us study what these circles look like

Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part I 25 / 30

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

r=1
r=3
r=7

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

(0.75,0)

I As r increases, the centers of the circle shift towards the +ve u direction, with v = 0
at 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
Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part I 26 / 30
 Constant Reactance Solution

I Like the case for r , the reactance must also be expressed as functions of u and v
2v
I x= simplifies to
(1 − u)2 + v 2

(1 − u)2 + xv 2 = 2v
I Dividing by x and completing the square, we get
!2 !2
2 1 1
(u − 1) + v − = (40)
x x
!
1 1
I The above equation is that of a circle with center at 1, and radius
x x
I There is a unique circle for every distinct value of x. Hence this set of solutions is
known as a constant reactance solution
I Let us study what these circles look like

Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part I 27 / 30

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

x=1

(1,1)
x=2
(1,0.5)
x=4
x=8
x=0
x = -8
x = -4
(1,-0.5)
x = -2

(1,-1)

x = -1

I As x increases, the centers of the circle shift towards the point (1,0), with centers
aligned 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
Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part I 28 / 30
 The Complete Smith Chart
I To complete the Smith Chart, combine the r and x circles. We are only interested in
those circles which lie inside r = 0 circle
I This means part of all x circles lie outside the area of interest
Inductive reactance
j1

j0.5 j2

Movement Movement
Resistance
towards 0.5 1
2 towards
load source

j0.5 j2

j1

Capacitive reactance
I It was designed by an American engineer Philip H. Smith
Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part I 29 / 30
 References

I Electromagnetic Waves by R. K. Shevgaonkar

I Microwave Engineering by D. M. Pozar
I Electromagnetic Waves and Radiating Systems by Jordan and Balmain
I OCW EECS, MIT

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