The _______ causes a real power loss in the conductor of the transmission lines.

Option 4 : series resistance

A transmission line consists of four parameters that directly affect its ability to transfer power efficiently.

**Shunt conductance:**

- This parameter is associated with the
**dielectric losses**. - Conductance from
**line to line**or a**line to ground**accounts for losses that occur due to the**leakage current**at the**cable insulation**and the**insulators**between overhead lines.

**Series resistance:**

- The primary source of
**real power losses**incurred in a transmission system is due to the**resistance of the conductors.** - Power loss is directly proportional to the square of the RMS current traveling through the line.

**Capacitance and Inductance:**

- Power loss cannot occur due to the capacitance or inductance present in the transmission line.
- It can only occur due to the resistance in the transmission line.
- The capacitance and inductance present in the line however can
**trap energy**in their**electric fields**and**magnetic fields**and**effects the power transfer capability of the line.**

Option 1 : Inductive

__Concept__:

The input impedance of a transmission line is given by:

\({Z_{in}} = {Z_o}\frac{{\left( {{Z_L} + j{Z_0}tanβ l} \right)}}{{\left( {{Z_0} + j{Z_L}tanβ l} \right)}}\) ---(1)

Z0 = Characteristic impedance

ZL = Load impedance

__Application__:

For l = λ/8

\(β l=\frac{2\pi}{\lambda}\times \frac{\lambda}{8}\)

\(\beta l=\frac{\pi}{4}\)

Putting this Equation (1), we get:

\({Z_{in}} = {Z_o}\frac{{\left( {{Z_L} + j{Z_0}} \right)}}{{\left( {{Z_0} + j{Z_L}} \right)}}\)

For short circuit load (ZL= 0), the input impedance becomes:

\({Z_{in}} = {Z_o}\frac{{\left( {{0} + j{Z_0}} \right)}}{{\left( {{Z_0} + j{0}} \right)}}\)

\({Z_{in}} = j{Z_o}\)

Since Impedance has a positive imaginary part, the transmission line behaves as an inductive Transmission Line.

The magnitude of the input impedance of a λ/8 lossless 50 Ω transmission line terminated with 25 Ω is:

Option 1 : 50 Ω

__Concept:__

For a lossless transmission line, the input impedance is given by:

\({Z_{in}} = {Z_0}\left[ {\frac{{{Z_L} + j{Z_0}\tan β l}}{{{Z_0} + j{Z_L}\tan β l}}} \right]\)

1) For the capacitive nature of impedance, the input impedance should be in the form -jb (b > 0)

2) For inductive impedance, the input impedance should be in the form jb (b > 0)

**Calculation:**

Given:

l = λ/8, Z_{0} = 50Ω, Z_{l} = 25Ω

\({Z_{in}} = {Z_0}\left[ {\frac{{{Z_L} + j{Z_0}\tan β l}}{{{Z_0} + j{Z_L}\tan β l}}} \right]\)

\(βl = \frac{2\pi}{\lambda}{l} =\frac{\pi}{4}\)

On solving we'll get:

|Z_{in}| = 50Ω

Option 2 : shunt capacitance

**Capacitance in Transmission line:**

- Capacitance in a transmission line results due to the potential difference between the conductors.
**The conductors of the transmission line act as a parallel plate of the capacitor and the air is just like a dielectric medium between them.**- The conductors get charged in the same way as the parallel plates of a capacitor.
- The capacitance between two parallel conductors depends on the size and the spacing between the conductors.
- The capacitance of a line gives rise to the leading current between the conductors.
- It depends on the length of the conductor.
**The capacitance of the line is proportional to the length of the transmission line.** - The capacitance effect is negligible on the performance of short (having a length less than 80 km) and low voltage transmission line.
- In the case of high voltage and long lines, it is considered as one of the most important parameters.
**Therefore a long transmission line has a considerable shunt capacitance**

Option 3 : \(\dfrac{1}{\log_e\left(\dfrac{D}{r}\right)}\)

**Capacitance of a Single-Phase Two-wire Line:**

Consider a single-phase overhead transmission line consisting of two parallel conductors A and B spaced **D meters** apart in the air.

Suppose that radius of each conductor is** r meters**.

Let their respective charge be + Q and − Q coulombs per metre length.

The total potential difference. between **conductor A** and neutral “infinite” plane is,

\(V_A=\int^\infty_r{\frac{Q}{2\pi x\epsilon_0}}+\int^\infty_D{\frac{-Q}{2\pi x\epsilon_0}}\)

or, \(V_A=\frac{Q}{2\pi \epsilon_0}[{log_e\frac{\infty}{r}}-log_e\frac{\infty}{D}]\)

or, \(V_A=\frac{Q}{2\pi \epsilon_0}[{log_e\frac{D}{r}}]\)

Similarly, total potential difference. between conductor B and neutral “infinite” plane is,

\(V_B=\frac{-Q}{2\pi \epsilon_0}[{log_e\frac{D}{r}}]\)

Both these potentials are with respect to the same neutral plane.

Since the unlike charges attract each other, the potential difference between the conductors is,

V_{AB} = 2V_{A} = \(\frac{2Q}{2\pi \epsilon_0}[{log_e\frac{D}{r}}]\)

We know that the capacitance is the ratio of charge to the voltage (C =Q/V),

Hence,

C_{AB} = \(\frac{Q}{V_{AB}}=\frac{Q}{\frac{2Q}{2\pi \epsilon_0}[{log_e\frac{D}{r}}]}\)

or, C_{AB} = \(\frac{\pi\epsilon_0}{[{log_e\frac{D}{r}}]}\)

Hence, When two conductors between each of radius r are at a distance D, the capacitance between the two is Proportional to \(\dfrac{1}{\log_e\left(\dfrac{D}{r}\right)}\)

Option 2 : \(\left[ {\begin{array}{*{20}{c}} 73&88\\ 33&40 \end{array}} \right]\)

**Concept:**

In cascaded connection transmission parameters get multiplied.

**Calculation:**

\(\left[ {\begin{array}{*{20}{c}} A&B\\ C&D \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {{A_1}}&{{B_1}}\\ {{C_1}}&{{D_1}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{A_2}}&{{B_2}}\\ {{C_2}}&{{D_2}} \end{array}} \right]\)

\( = \left[ {\begin{array}{*{20}{c}} 7&8\\ 3&4 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 7&8\\ 3&4 \end{array}} \right]\)

\( = \left[ {\begin{array}{*{20}{c}} {73}&{88}\\ {33}&{40} \end{array}} \right]\)

A lossless matching circuit is shown in the figure

The values satisfying the matching condition are:

Option 1 : X_{S1} = -j25.1, X_{P} = +j100, X_{S2} = -j50

__Concept:__

**Matching Network:**

It is also called an impedance transformer is used to create matched impedance between a source and load. Between a power amplifier and antenna. For matching,

Source impedance = load impedance

__Solution:__

Given Z_{S} = 50 Ω

Choose option (a)

Let \(50 = {X_{{S_2}}} + \frac{{{X_P}\left( { - 125.0 + 125.1 + 100} \right)}}{{\left( {{X_P} + 125.1 - 125.1 + 100} \right)}}\)

\(50 = {X_{{S_2}}} + \frac{{{X_P} \times 100}}{{{X_P} + 100}}\)

Put X_{P} = j 100

\(50 = {X_{{S_2}}} + \frac{{j \times 100 \times 100}}{{j100 + 100}}\)

\(50 = {X_{{S_2}}} + \frac{{j100}}{{j + 1}}\)

\({X_{{S_2}}} = 50 - \frac{{j100}}{{j + 1}} = \frac{{50j + 50 - j100}}{{j + 1}} = \frac{{ - j50 + 50}}{{j + 1}} = - j50\)

So \({X_{{s_1}}} = - j25.1,\) X_{P} = j100, \({X_{{S_2}}} = - j50\)

Correct choice is option (a)

**More information:**

For matching Z_{L} = Z_{0} and Reflection coefficient

\({\rm{\Gamma }} = \frac{{{Z_L} - {Z_0}}}{{{Z_L} + {Z_0}}}\)

So Γ = 0 For impedance matching

Option 1 : 240 Ω

__Concept: __

- Standing wave Ratio (
**SWR**) defines mismatch on the line. - It is also the measure of the deviation of impedances or current or voltages from their central values.
- \(VSWR = \frac{{{V_{\max }}}}{{{V_{\min }}}}\)
- The value of VSWR varies from 1 to ∞
**(1 ≤ VSWR < ∞)**

__Calculation: __

Given:

SWR = 4, Z_{0} = 60 ohm

\(SWR = \frac{{{Z_L}}}{{{Z_0}}}\)

Z_{L} = SWR x Z_{0}

Z_{L} = 4 x 60 = **240 ohm**

__Important Points__

- Γ is the amplitude ratio of reflected voltage to the forward voltage.
- It also indicates a mismatch between the expected impedance and the actual impedance at that point.
- \({\rm{\Gamma }} = \frac{{{Z_L} - {Z_0}}}{{{Z_L} + {Z_0}}}\)

Where,

Z_{L} = Load Resistance

Z_{0} = Characteristic Resistance

The value of Γ varies between **0 and 1**

** Γ = 1 when there is complete reflection**.

\(SWR = \frac{{1 + {\bf{\Gamma }}}}{{1 - {\bf{\Gamma }}}}\)

Option 2 : 6.25

__Concept:__

The wavelength of the microstrip line is defined as:

\(\lambda = \frac{c}{{\sqrt {{\epsilon_r}} f}}\)

ϵ_{r}: relative permittivity / dielectric constant

f: frequency of operation

c: speed of light

__Calculation:__

Given wavelength is 12 mm and frequency is 10 GHz

\(12 \times {10^{ - 3}} = \frac{{3 \times {{10}^8}}}{{\sqrt {{\epsilon_r}} \times 10 \times {{10}^9}}}\)

\(\sqrt {{\epsilon_r}} = \frac{{3 \times {{10}^8}}}{{12 \times {{10}^{ - 3}} \times {{10}^{10}}}}\)

\(\sqrt {{\epsilon_r}} = \frac{1}{4} \times \frac{{{{10}^8}}}{{{{10}^7}}}\)

\({\epsilon_r} = \frac{{100}}{{16}}\)

**ϵ _{r} = 6.25**

Option 3 : is inversely proportional to the square root of dielectric constant of insulation between conductors

Velocity factor

- The velocity of light and all other electromagnetic waves depends on the medium through which they travel.
- It is very nearly 3 x 10
^{8}m/s in a vacuum and slower in all other media.

The velocity of light in a medium is given by

\(v = \frac{{{v_c}}}{{\sqrt k }}\)

v: velocity of the wave in the medium

vc: velocity of the wave in free space

k: dielectric constant

The velocity factor of a dielectric substance, and thus of a cable, is the velocity reduction ratio and is therefore given by

\(vf = \frac{1}{{\sqrt k }}\)

Hence, the Velocity factor of the transmission line is inversely proportional to the square root of the dielectric constant of insulation between conductors

The dielectric constants of materials commonly used in transmission lines range from about 1.2 to 2.8, giving corresponding velocity factors from 0.9 to 0.6

**NOTE:**

since ν = fλ and f is constant, the wavelength λ is also reduced by a ratio equal to the velocity factor.

**Example**

If a section of 300-Ω twin lead has a velocity factor of 0.82, the speed of energy transferred is 18 per cent slower than in a vacuum.

__Calculation:__

Given dielectric constant is 2.

The velocity factor of the line is

\(vf = \frac{1}{{\sqrt 2 }}\)

vf = 0.707

vf = 70.7 %

Two coherent microwave power sources of same frequency f, each generating P Watts of average power, are combined using a four-port network in the following manner:

If the S-parameter of the 4-port network is:

\(\left[ {\begin{array}{*{20}{c}} 0&{\frac{j}{{\sqrt 2 }}}&{\frac{{ - 1}}{{\sqrt 2 }}}&0\\ {\frac{j}{{\sqrt 2 }}}&0&0&{\frac{{ - 1}}{{\sqrt 2 }}}\\ {\frac{{ - 1}}{{\sqrt 2 }}}&0&0&{\frac{j}{{\sqrt 2 }}}\\ 0&{\frac{{ - 1}}{{\sqrt 2 }}}&{\frac{j}{{\sqrt 2 }}}&0 \end{array}} \right]\)

What should be the phase difference between the inputs to maximize the output at port 2 and what is the maximum power?

Option 1 : 90°, 2P

The input signal at port 1 is a_{1} = √P Volts

The input signal at port 4 is:

a_{4}= √P∡α Volts \( = \sqrt P {e^{j\alpha }}\) Volts

Output at port 2:

\( = {S_{21}}{a_1} + {S_{24}}\;{a_4} = \frac{j}{{\sqrt 2 }}\sqrt P - \frac{1}{{\sqrt 2 }}\sqrt P {e^{j\alpha }}\)

\( = \sqrt {\frac{P}{2}} \left( {{e^{ - \frac{{j\pi }}{2}}} + {e^{ - j\pi }}{e^{j\alpha }}} \right) = \sqrt {\frac{P}{2}} \left( {{e^{ - \frac{{j\pi }}{2}}} + {e^{ - j\left( {\pi - \alpha } \right)}}} \right)\)

To maximize output at Port2:

π – α = π/2

\(\alpha = \frac{\pi }{2} = {90^o}\)

Output at Port 2 will be:

\(\sqrt {\frac{P}{2}} \left( {{e^{ - \frac{{j\pi }}{2}}} + {e^{ - \frac{{j\pi }}{2})}}} \right) = j\sqrt {2P} \;V\)

Output Power at Port 2 will be:

\({\left| {j\sqrt {2P} } \right|^2} = 2P\;W\)

For a parallel RC circuit as shown in figure, R = 20 Ω & C = 0.9 pF, which is to be matched to 50 Ω over a bandwidth,

\(\int^{\infty}_0 \ln \frac {1}{|Γ|} dw \le \frac {\pi}{R\; C}\)

The circuit has to be matched over 6 GHz to 18 GHz. What is the very best Γ that can be achieved:

Option 3 : 0.1

__Matching Network:__

It is also called an impedance transformer is used to create matched impedance between a source and load.

Between a power amplifier and antenna. For matching,

Source impedance = load impedance

i.e Z_{0} = Z_{L} which given Zero reflection coefficient.

For RC network phase velocity and impedance of transmission line can be changed by capacitive loading

Therefore 0.1 sec is the time required for the capacitor to be fully charged itself.

Option (C) correct choice.

Option 3 : ∞

**Concept:**

- The load impedance, ZL at the end of the transmission line must match its characteristic impedance, Z0 Otherwise there will be reflections from the transmission line’s end.
- A quarter-wave transformer is a component that can be inserted between the transmission line and the load to match the load impedance ZL to the transmission line’s characteristic impedance Z0.
- The input impedance of a quarter-wave transformer is given as:

\({{\rm{Z}}_{{\rm{in}}}} = \frac{{{\rm{Z}}_0^2}}{{{{\rm{Z}}_{\rm{L}}}}}\;\)

**Analysis:**

For a λ/4 line:

Input impedance = \({{\rm{Z}}_{{\rm{in}}}} = \frac{{{\rm{Z}}_0^2}}{{{{\rm{Z}}_{\rm{L}}}}}\;\)

Where Z_{0} is the characteristic impedance.

Now, if the line is shorted i.e Z_{L} = 0

Z_{in} = ∞

Option 2 :

__Concept:__

The General RLC equivalent diagram of any transmission line(called lumped circuit diagrams) is shown below:

Where R is the resistance of conductor used in the transmission line (Ideally it is 0)

G is the conductance of the dielectric media (Ideally = 0)

__Analysis:__

We are given that the dielectric media is of Teflon with ϵ_{r} = 2.1 and tan δ = 0

i.e. tan δ = \(\frac{\sigma }{\omega \epsilon }=0\)

So, σ = 0 which indicates that G = 0.

G = 0 in the equivalent lumped diagram will be represented as shown:

Hence Option (2) is Correct.

Option 2 : L = 250 μH/m, C = 0.1 μF/m, R = 100 Ω/m, G = 0.04 S/m

**Concept:**

If the characteristic impedance is purely real and the propagation constant is complex, the transmission line is distortionless and for this line \({\rm{\alpha }} = \sqrt {{\rm{RG}}}\) and \({\rm{\beta }} = {\rm{\omega }}\sqrt {{\rm{LC}}}\). and \({{\rm{Z}}_{\rm{o}}} = \sqrt {\frac{{\rm{R}}}{{\rm{G}}}}\)

**Application: **

As \({\rm{\alpha }} = \sqrt {{\rm{RG}}}\) and \({{\rm{Z}}_{\rm{o}}} = \sqrt {\frac{{\rm{R}}}{{\rm{G}}}}\)

\(\begin{array}{l} \Rightarrow {\rm{\alpha }}{{\rm{Z}}_{\rm{o}}} = {\rm{R}}\\ {{\rm{Z}}_{\rm{o}}}{\rm{R}} = 2 \times 50 = 100{\rm{\;\Omega }}/{\rm{m}} \end{array}\)

\({\rm{G}} = \frac{{\rm{R}}}{{{\rm{Z}}_0^2}} = \frac{{100}}{{2500}} = 0.04\frac{{\rm{S}}}{{\rm{m}}}\) . We see only option b fits these two requirements. Confirming \({\rm{\beta }}\) from option b we see,

\(\begin{array}{l} {\rm{\omega }}\sqrt {{\rm{LC}}} = {10^6}\sqrt {250 \times {{10}^{ - 6}} \times 0.1 \times {{10}^{ - 6}}} \\ = {10^6} \times {10^{ - 6}} \times 5\\ = 5 = {\rm{\beta }} \end{array}\)

Thus, option b is true