Transmission Lines- Part II

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 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

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

 Power Delivered to Load of a Lossless Transmission Line
I Using the standard expression in terms of the complex voltage and current, the
power at any point l along the line is

1 1 V + jβl
P(l) = Re(VI ∗ ) = Re{[V + ejβl (1 + ΓL e−j2βl )][ e (1 − ΓL e−j2βl )]∗ } (1)
2 2 Z0
I At the load, l = 0. Therefore, the load power is

1 V+
P(0) = PL = Re{[V + (1 + ΓL )][ (1 − ΓL )]∗ } (2)
2 Z0
1 |V + |2
= Re(1 + ΓL − Γ∗L − |ΓL |2 ) (3)
2 Z0
I The term ΓL − Γ∗L is purely imaginary. Thus, simplifying, we get

1 |V + |2
PL = (1 − |ΓL |2 ) (4)
2 Z0
I The same expression can be derived using an alternate approach: first calculating
the power incident on the load, and then subtracting the reflected power from the
load. Try this out, you should see the same result!

Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part II 3 / 30

 Power at any point on a Lossless Transmission Line
I The instantaneous power at any point l along the line is
1 1 V + jβl
P(l) = V (l)I(l)∗ = {[V + ejβl (1 + ΓL e−j2βl )][ e (1 − ΓL e−j2βl )]∗ }
2 2 Z0
1 |V + |2
= (1 + ΓL e−j2βl )(1 − Γ∗L ej2βl )
2 Z0
1 |V + |2
= (1 + ΓL e−j2βl − Γ∗L ej2βl − |ΓL |2 )
2 Z0
I Writing ΓL = |ΓL |ejφ and Γ∗L = |ΓL |e−jφ , we get
1 |V + |2
P(l) = (1 + |ΓL |ej(φ−2βl) − |ΓL |e−j(φ−2βl) − |ΓL |2 )
2 Z0
I Expanding the complex exponentials using Euler’s entity and simplifying, we obtain
1 |V + |2 1 |V + |2
P(l)real = (1 − |ΓL |2 ) and P(l)imag = (−2 sin(φ − 2βl)) (5)
2 Z0 2 Z0
I It is interesting to note that the real portion of P(l) is equal to the power delivered to
the 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’’ in
the line due to variation of V , I with position, which varies the electric and magnetic
fields
Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part II 4 / 30
 Calculating V + from Transmission Line Measurements
I In most of our analyses for voltage, current, reflection coefficient and power, the
quantity V + is assumed to be known
I 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 to
physically 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
RS

VS Z0 ZL

ZX L
ZL − Z0
I The voltage looking into the line VX = V + ejβL (1 + ΓL e−j2βL ), & ΓL =
ZL + Z0
ZX
I If the impedance looking into the line is transformed to ZX , then VX = VS
ZX + RS
I Equating the two and simplifying, we get
VS ZX e−jβL
V+ = (6)
(RS + ZX )(1 + ΓL e−j2|ΓL |βL )
Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part II 5 / 30
 Back to the Smith Chart
In Transmission Lines-I, the development of the Smith chart was discussed. Now, some
analyses are discussed
I The center of the Smith chart denotes z = 1 + j0, i.e. a matched impedance and

this is the center of all VSWR circles

I The normalized load zL may lie on any of the VSWR circles. The corresponding
reflection coefficient ΓL may be measured by mapping the radius of the VSWR
circle on the scale below the chart
I Movement on the VSWR circle in an anti-clockwise direction indicates movement
away from load. On this trajectory, the maximum, minimum line voltage points away
from the load, and impedance at any point on the line may be found
A ≡ zL
L2 1
B≡
zL
A
L1 C≡ρ
1
D C D≡
ρ
λ
L1 ≡
B 4
L2 ≡ distance to nearest voltage maximum
L3
L3 ≡ distance to nearest voltage minimum
Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part II 6 / 30
 Calculating Transformed Impedance along the Line
I 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 = ZZL
0
I Next, mark the zL on the Smith chart by locating the intersection of the correct r and
x 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 is
expressed in terms of the wavelength λ, and distance is marked on the Smith chart
in 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 the
correct r and x circles
l1

jx1
Move CW z1
zL
r1

Constant ρ
circle

Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part II 7 / 30

 Smith Chart with Admittance
I 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 admittance
I Consider the admittance Y at any point on a line with characteristic admittance Y0 .
Y 1
The normalized admittance y = = g + jb =
Y0 r + jx
I Substituting in the expression for reflection coefficient Γ, we obtain
1−y y −1 z −1
Γ= = ∠180◦ = (7)
1+y y +1 z +1
I It is seen that Γ in terms of both y and z are identical, except with a 180◦ phase
difference. Effectively, the Smith chart may be rotated by 180◦ as well
-jx = jb

r=g

Short Open

jx = -jb
Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part II 8 / 30
 Using the Impedance Smith Chart for Admittance
I It seems that by simply inverting the Smith chart, one can use it for
admittance-based calculations. But there is a simpler way
I Recall that Smith chart is a coordinate system on a complex Γ plane where the axes
are defined as Γ = u + jv
I Rather than inverting the Smith chart, we can invert the u and v axes and use the
impedance Smith chart coordinates as admittance
I 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 towards
source/load
Capacitive susceptance jb

u Conductance g

Inductive susceptance -jb

jv
Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part II 9 / 30
 Transmission Line Applications- Impedance Matching
I One of the most crucial considerations in transmission lines is the impedance
matching between the source, line and the load. Mismatch between these
impedances 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 line
is 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 is
called stub matching. Shown below are some examples of stubs with
transmission lines
Shunt short stub Shunt open stub

Z0 ZL Z0 ZL Z0 ZL Z0 ZL

Series open stub Series short stub

Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part II 10 / 30

 Impedance Matching using a Single Stub
I 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 short
stub
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 the
main line (over length ls ) should be y2 = −jb2
I The effective admittance seen by the line is then yeff = y1 + y2 = 1. This means the
line sees the transformed impedance equal to Z0 , i.e. line and load are matched
y1

ds

Z0 ZL

ls
yeff
y2

Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part II 11 / 30

 Single Stub Matching using Smith Chart
I 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 ls
I 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 of
the 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

jb
yL y1

g=1

ls

-jb
Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part II 12 / 30
 Some Important Points about Single Stub Matching
I 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 how
large a stub can be accommodated in your system
I Exercise: in a similar manner, work out the procedure to design a single stub
matching network, but with an open-terminated shunt stub
I If a series stub matching is required, we have to use the Smith chart as an
impedance chart
I The stub matching technique works only for a single frequency. This is decided by
the distances ds and ls which are expressed in terms of λ
I Transmission lines fabricated on a two-layer printed circuit board (PCB) are called
microstrip lines. Stubs can be easily realized using microstrip technique. Shown
below is a short circuited stub along a line

Main Line Stub

Via hole
to ground

Ground

Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part II 13 / 30

 Determining the Type of Load Connected to a Transmission Line
I 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 at
any point on the line
I We look at the variation of the voltage standing wave right from the load position
onwards. For e.g. the following two standing wave patterns (SWP) correspond to a
capacitive load, and a resistive load with 0 < ZL < Z0 , respectively
|V|
Vmax
SWP-1 Capacitive load
l
Vmax
SWP-2 Resistive load
Vmin with 0 < ZL < Z0
l
I With the help of the Smith chart, this can be easily understood
Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part II 14 / 30
 Determining the Type of Load Using Smith Chart
I 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 minima
can only be on the jx = 0 line i.e. the real axis of the Smith chart. Also, the voltage
minimum is non-zero. This means the ZL lies between a short circuit, and Z0
SWP-1 SWP-2

ZL
V minima
comes first
& is zero
V minima
exists at ZL
ZL & is non-zero

I Exercise: Work out the standing wave patterns for (i) a purely inductive load (ii) an
open load (iii) a short load (iv) an R + jX load with X > 0 (v) the same load with
R ≈ Z0 and X  R
Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part II 15 / 30
 Transmission Line Applications- Realization of Circuit Elements
I 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

Lp Lp
2 2
Cp
I As the frequency increases, these parasitic effects become more dominant.
Generally, most through-hole 2-lead capacitors and inductors do not work reliably
beyond 100−150 MHz
I Inductor and capacitors of a particular value, however, can be realized at a
particular frequency, using lossless transmission lines
I The first starting point is the impedance transformation relation i.e.
ZL + jZ0 tan βl
Z (l) = Z0 (8)
Z0 + jZL tan βl
Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part II 16 / 30
 Realization of L and C using Open and Short-ended Terminated Lines
I 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 tan βl (9)
π λ
I Thus, if 0 ≤ βl ≤ i.e. 0 ≤ l ≤ , then the magnitude of Z (l) is positive, which
2 4
indicates inductive reactance. Thus,
Z0 tan βl = ωL (10)
Z0 tan βl
∴L = (11)
ω
λ λ
I Likewise, if ≤ l ≤ , the Z (l) is negative and it indicates capacitive reactance.
4 2
Thus,
1
Z0 tan βl = (12)
ωC
1
∴C = (13)
Z0 ω tan βl
I Exercise: Derive the conditions and expressions for realizing L and C using an
open-ended transmission line
Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part II 17 / 30
 Inductive and Capacitive Behaviour of Short-ended Transmission Line
I The inductive or capacitive behaviour of a transmission line is easily understood
using the Smith chart
I Start from the short circuit point on the Smith chart, and move clockwise towards
source
+jX

Short
B A

-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
λ
reactance. Inductors and capacitors repeat after every movement (i.e. one
2
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 Lines
I We have seen that for a short-terminated line, the impedance at any point is given
by Z = jZ0 tan βl
I Graphically, the magnitude of Z as a function of the line length l looks like
Z

λ λ 3λ λ l
4 2 4

λ
I At odd multiples of , the impedance peaks up to ∞. This denotes parallel L − C
4
resonance (equivalent to admittance minima)
λ
I At even multiples of , the impedance reaches zero. This denotes series L − C
4
resonance
λ λ
4 2

Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part II 19 / 30

 The Practical Scenario- Low Loss Transmission Line
I So far, almost all analyses and discussions have been about lossless lines.
Perfectly lossless lines, however, exist only in textbooks and not in practice! There
is 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 the
α term, i.e. γ = α + jβ, and assume that α  β. This is called a low-loss
transmission 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)
tanh αl + j tan βl
= Z0 (15)
1 + j tanh αl tan βl
λ
I As α → 0 for a low-loss line, the term tanh αl ≈ αl. Therefore, as l → ,
4
Z0
Z = (16)
αl
I Thus, in practice Z is not infinite, but is of a very large value, as the term αl is quite
small. Similarly, one can prove that for a quarter-wavelength open-loaded low-loss
line, the input impedance is
Z = Z0 αl (17)
Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part II 20 / 30
 Q of a Resonant L − C using Transmission Line
I 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 be
obtained at f0 (corresponding to the physical λ0 /4 of the line)
I Using this resonant frequency f0 , and the 3 dB impedance variation around the
maxima, the Q of the line can be obtained. However, a less complicated approach
is 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 I
magnitudes are
V = V + [ejβl − e−jβl ] = 2V + sin βl = V0 sin βl (18)
+ +
V 2V V0
I= [ejβl + e−jβl ] = cos βl = cos βl (19)
Z0 Z0 Z0
I In terms of the line parameters, if it is assumed that R, G are negligible, then the
energy stored along the λ/4 section of the line is
Z λ/4 Z λ/4
1 1
E= C V 2 dl + L I 2 dl (20)
2 0 2 0
Z λ/4 Z λ/4 " #2
1 1 V0
= C [V0 sin βl]2 dl + L cos βl dl (21)
2 0 2 0 Z0

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

 Q of a Resonant L − C using Transmission Line (cont’d..)
I The energy stored E can be simplified to
Z λ/4 " # Z λ/4 2 " #
1 2 1 − cos 2βl 1 V0 1 + cos 2βl
E= C V0 dl + L dl (22)
2 0 2 2 0 Z02 2
1 λ 1 LV02 λ
CV02 +
= (23)
4 4 4 Z02 4
p
I Since Z0 ≈ L/C, the two terms are identical, and this simplifies to
1 λ
CV02 E= (24)
2 4
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
V02 V02
PL = = (25)
Z0 /(αl) Z0 /(αλ/4)
E
I By definition, Q = 2πf0 , where f0 is the frequency of operation. Therefore, this
PL
simplifies to
β
Q= (26)
2α
Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part II 22 / 30
 More About Low Loss Transmission Lines
I 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, the
propagation constant γ becomes
p
γ = (R + jωL)(G + jωC) (27)
v ! !
u
u R G
= tjωL 1 − j jωC 1 − j (28)
ωL ωC
!1/2 !1/2
√ R G
= jω LC 1 − j 1−j (29)
ωL ωC
I As R  jωL and G  jωC, expanding the root terms using Power series and
ignoring the 2nd and higher order terms,
!
√ R G
γ = jω LC 1 − j −j (30)
ωL ωC
√
r r
C L
=R +G + jω LC (31)
L C
R √
I We see α = + GZ0 . Note that β = ω LC is not affected by non-zero R, G
Z0
Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part II 23 / 30
 Lossy Transmission Lines

I So far, the discussion has largely been about lossless or low loss lines where
α  β. But it should be known as to what happens when α becomes significantly
larger than β
I In terms of the line parameters, R  jωL and G  jωC. The characteristic
impedance then becomes r
R
Z0 = (32)
G
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
p √
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 no
propagation 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 is
concerned

Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part II 24 / 30

 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 and
reverse waves thus exist near the load
V − e−αl
I Thus, the reflection coefficient at any point on the line is Γ = = ΓL e−2αl
V + eαl
I It is interesting to note that |Γ| decreases exponentially as one moves away from
the load. Close to the source, the |Γ| ≈ 0, which means that the source sees a
nearly matched line
I The VSWR along the line is

1 + |Γ| 1 + |ΓL |e−2αl

ρ= = (35)
1 − |Γ| 1 − |ΓL |e−2αl
I Thus, as one moves away from the load, ρ decreases and converges to 1

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

 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 a
VSWR circle looks like for a lossy line

zL

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

Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part II 26 / 30

 Measurement of Characteristic Impedance of a Line
I As the characteristic impedance Z0 is a distributed parameter, it is not possible to
physically measure it using, say, a multimeter
I However, one can measure the impedance at any point along the line, from which
the Z0 may be calculated
I 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 the
impedance at the other end. It is given by
Zoc = Z0 coth γl (37)
I The line characteristic impedance is then simply,
p
Z0 = Zsc Zoc (38)
I Zsc and Zoc can be measured using an instrument called Vector Network
Analyzer (VNA)
Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part II 27 / 30
 Measurement of Propagation Constant of a Line
p
I From the previous analysis, if the line has length l, then tanh γl = Zsc /Zoc = A
γl −γl
e −e
∴ =A (39)
eγl + e−γl
1+A
∴e2γl = (40)
1−A
1+A
∴e2αl ej2βl = (41)
1−A
I α can be obtained by equating the real portions of the above polar equation i.e.
1+A
e2αl = (42)
1−A

1 1+A
⇒α= ln (43)
2l 1−A
I The estimation of β, however, is tricky, as the standing wave characteristics repeat
every 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,
" #
1 1+A
β= ∠ ± 2nπ (44)
2l 1−A
Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part II 28 / 30
 Measurement of Propagation Constant of a Line
I 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 ,
" #
1 1+A
β1 = ∠ ± 2nπ (45)
2l 1−A
I At f2 , " #
1 1+A
β2 = ∠ ± 2(n + 1)π (46)
2l 1−A
I Subtracting, we obtain
π
β2 − β1 = (47)
l
2πf2 2πf1 π
∴ − = (48)
v v l
∴Wave velocity v = 2l(f2 − f1 ) (49)
2πf
I By earlier definition, β = . Therefore,
v
πf
β= (50)
l(f2 − f1 )
Debapratim Ghosh (Dept. of EE, IIT Bombay) Transmission Lines- Part II 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 Microwaves 101, IEEE MTT-S

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