Transmission Line Parameters MCQ Quiz in मल्याळम - Objective Question with Answer for Transmission Line Parameters - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Last updated on Mar 9, 2025

നേടുക Transmission Line Parameters ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Transmission Line Parameters MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

Latest Transmission Line Parameters MCQ Objective Questions

Top Transmission Line Parameters MCQ Objective Questions

Transmission Line Parameters Question 1:

The S-parameters of a two port network is given as

\(\rm [S]=\begin{bmatrix}S_{11}&S_{12}\\\ S_{21}&S_{22}\end{bmatrix}\)

with reference to 𝑍0. Two lossless transmission line sections of electrical lengths 𝜃1 = 𝛽𝑙1 and 𝜃2 = 𝛽𝑙2 are added to the input and output ports for measurement purposes, respectively. The S-parameters [𝑆′] of the resultant two port network is

F1 Engineering Arbaz 27-12-23 D11

  1. \(\rm \begin{bmatrix}S_{11}e^{-j2\theta_1}&S_{12}e^{-j(\theta_1+\theta_2)}\\\ S_{21}e^{-j(\theta_1+\theta_2)}&S_{22}e^{-j2\theta_2}\end{bmatrix}\)
  2. \(\rm \begin{bmatrix}S_{11}e^{j2\theta_1}&S_{12}e^{-j(\theta_1+\theta_2)}\\\ S_{21}e^{-j(\theta_1+\theta_2)}&S_{22}e^{j2\theta_2}\end{bmatrix}\)
  3. \(\rm \begin{bmatrix}S_{11}e^{j2\theta_1}&S_{12}e^{j(\theta_1+\theta_2)}\\\ S_{21}e^{j(\theta_1+\theta_2)}&S_{22}e^{j2\theta_2}\end{bmatrix}\)
  4. \(\rm \begin{bmatrix}S_{11}e^{-j2\theta_1}&S_{12}e^{j(\theta_1+\theta_2)}\\\ S_{21}e^{j(\theta_1+\theta_2)}&S_{22}e^{-j2\theta_2}\end{bmatrix}\)

Answer (Detailed Solution Below)

Option 1 : \(\rm \begin{bmatrix}S_{11}e^{-j2\theta_1}&S_{12}e^{-j(\theta_1+\theta_2)}\\\ S_{21}e^{-j(\theta_1+\theta_2)}&S_{22}e^{-j2\theta_2}\end{bmatrix}\)

Transmission Line Parameters Question 1 Detailed Solution

F1 Engineering Arbaz 27-12-23 D12

\(\left[\begin{array}{l} V_1^{-} \\ V_2^{-} \end{array}\right]=\left[\begin{array}{ll} S_{11} & S_{12} \\ S_{21} & S_{22} \end{array}\right]\left[\begin{array}{l} V_1^{+} \\ V_2^{+} \end{array}\right] \)       ....(1)

To be calculated,

\(\begin{aligned} & {\left[\begin{array}{l} V_{1 N}^{-} \\ V_{2 N}^{-} \end{array}\right]=\left[\begin{array}{ll} S_{11 N} & S_{12 N} \\ S_{21 N} & S_{22 N} \end{array}\right]\left[\begin{array}{l} V_{1 N}^{+} \\ V_{2 N}^{+} \end{array}\right]} \\ \end{aligned}\)

\(V_1^{+}=V_{1 N}^{+} e^{-j \theta_1} \)

\(V_{1 N}^{-}=V_1^{-} e^{-j \theta_1}=V_1^{-}=V_{1 N}^{-} e^{j \theta_1} \)

\(V_2^{+}=V_{2 N}^{+} e^{-j \theta_2} \)

\(V_{2 N}^{-}=V_2^{-} e^{-j \theta_2}=V_2^{-}=V_{2 N}^{-} e^{j \theta_2}\)

Feeding the data in equation (i),

\(\left[\begin{array}{c} V_{1 N}^{-} \\ V_{2 N}^{-} \end{array}\right]=\left[\begin{array}{cc} S_{11 N} e^{-j\left(2 \theta_1\right)} & S_{12 N} e^{-j\left(\theta_1+\theta_2\right)} \\ S_{21 N} e^{-j\left(\theta_1+\theta_2\right)} & S_{22 N} e^{-j 2\left(\theta_2\right)} \end{array}\right]\)

Hence, the correct option is (A).

Transmission Line Parameters Question 2:

A λ/4 line, shorted at one end presents impedance at the other end equal to

  1. Z0
  2. √ 2 Z0
  3. ∞ 
  4. 0

Answer (Detailed Solution Below)

Option 3 : ∞ 

Transmission Line Parameters Question 2 Detailed Solution

Concept:

  • The load impedance, Zat 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 Z0 is the characteristic impedance. 

Now, if the line is shorted i.e ZL = 0

Zin∞ 

Transmission Line Parameters Question 3:

Velocity factor of transmission line

  1. is directly proportional to the dielectric constant of insulation between conductors
  2. is inversely proportional to the dielectric constant of insulation between conductors
  3. is inversely proportional to the square root of dielectric constant of insulation between conductors
  4. does not depend on dielectric constant of insulation between conductors

Answer (Detailed Solution Below)

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

Transmission Line Parameters Question 3 Detailed Solution

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 108 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 %

Transmission Line Parameters Question 4:

The attenuation constant for a 50 Ω distortionless line is 0.01 dB/m and its capacitance is 0.1 nF/m. The value of inductance/meter is _____ (μH/m)

Answer (Detailed Solution Below) 0.25

Transmission Line Parameters Question 4 Detailed Solution

Concept:

For distortion less T.L.

\(\frac{R}{L} = \frac{G}{C}\)

\({Z_0} = R + j\;{X_0} = \sqrt {\frac{L}{C}} + j0\)

\(y = \propto + jB = \sqrt {\frac{C}{L}} \left( {R + j\omega L} \right)\)

From 2nd relation:

\({Z_0} = \sqrt {\frac{L}{C}}\)

\(L = Z_0^2C\)

= (502) (0.1 × 10-9)

L = 2.5 × 10-7 H/m

= 0.25 uH/m

Transmission Line Parameters Question 5:

Calculate the reflection coefficient when VSWR is 5. 

  1. 0
  2. 1
  3. 1/2 
  4. 2/3

Answer (Detailed Solution Below)

Option 4 : 2/3

Transmission Line Parameters Question 5 Detailed Solution

Concept:

The voltage standing wave ratio is defined as the ratio of the maximum voltage to the minimum voltage.

\(VSWR = \frac{{{V_{max}}}}{{{V_{min}}}} = \frac{{{I_{max}}}}{{{I_{min}}}}\)

The general expression of the voltage across a transmission line is given by:

\(V\left( l \right) = {V^ + }{e^{j\beta l}}\left( {1 + \left| {{{\rm{Γ }}_{\rm{L}}}} \right|{e^{ - j2\beta z}}} \right)\)

l = distance from the load

β = Phase constant

ΓL = Reflection coefficient at the load calculated as:

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

The maximum and minimum voltage is given by:

\({V_{max}} = \left| {{V^ + }} \right|\;\left( {1 + \left| {{{\rm{Γ }}_{\rm{L}}}} \right|} \right)\)

\({V_{min}} = \left| {{V^ + }} \right|\;\left( {1 - \left| {{{\rm{Γ }}_{\rm{L}}}} \right|} \right)\)

\(VSWR = \frac{{{V_{max}}}}{{{V_{min}}}} = \frac{{1 + {{\rm{Γ }}_L}}}{{1 - {{\rm{Γ }}_L}}}\)  ---(1)

Calculation:

Given,

\(VSWR = 5\)

Using Equation (1), we get:

\(5 = \frac{{1 + {{\rm{Γ }}_L}}}{{1 - {{\rm{Γ }}_L}}}\)

\(\Gamma_L = 2/3\)

Transmission Line Parameters Question 6:

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

F2 Shubham.B 03-11-20 Savita D 11

  1. 50 Ω
  2. 100 Ω
  3. 25 Ω
  4. 150 Ω
  5. None of the above

Answer (Detailed Solution Below)

Option 1 : 50 Ω

Transmission Line Parameters Question 6 Detailed Solution

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, Z0 = 50Ω, Zl = 25Ω​​

F2 Shubham.B 03-11-20 Savita D 11

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

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

On solving we'll get:

Zin = 50Ω 

Transmission Line Parameters Question 7:

A lossless matching circuit is shown in the figure

F1 Tapesh.S 21-01-21 Savita D13

The values satisfying the matching condition are:

  1. XS1 = -j25.1, XP = +j100, XS2 = -j50
  2. XS1 = +j25.1, XP = -j100, XS2 = -j50
  3. XS1 = -j25.1, XP = -j100, XS2 = -j50
  4. XS1 = +j25.1, XP = +j100, XS2 = +j100

Answer (Detailed Solution Below)

Option 1 : XS1 = -j25.1, XP = +j100, XS2 = -j50

Transmission Line Parameters Question 7 Detailed Solution

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

Correct choice is option (a)

More information:

For matching ZL = Z0 and Reflection coefficient

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

So Γ = 0 For impedance matching

Transmission Line Parameters Question 8:

A lossy transmission line has the propagation constant of γ = (1 + 2j) m-1 and characteristic impedance of 50 Ω at an angular frequency of 107 rad/sec. The value of inductance per unit length is _______ μH/m.

Answer (Detailed Solution Below) 9.9 - 10.1

Transmission Line Parameters Question 8 Detailed Solution

Concept:

For a lossy transmission line, if the characteristic impedance is real, then the line is considered as a distortion-less line.

Also, the characteristic impedance is given by:

\({Z_0} = \sqrt {\frac{R}{G}} = \sqrt {\frac{L}{C}} \) ---(1)

Attenuation constant is given by:

\(\alpha = \sqrt {RG} \;\)  ---(2)

Phase constant is given by:

\(\beta = \omega \sqrt {LC} \) ---(3)

Calculation:

Given:

γ = (1 + 2j)m-1, Z0 = 50 Ω, ω = 107 rad/sec

Multiplying equation (1) and (3), we get:

Z0β = ωL

\(L = \frac{{{Z_0}\beta \;}}{\omega } = \frac{{50 \times 2}}{{{{10}^7}}} = 10\;\mu H/m\)

Hence, the value of inductance is 10 μH/m.

Transmission Line Parameters Question 9:

An ideal lossless transmission line of Z0 60 Ω is connected to unknown ZL If SWR = 4. find ZL

  1. 240 Ω 
  2. 480 Ω
  3. 120 Ω
  4. 100 Ω

Answer (Detailed Solution Below)

Option 1 : 240 Ω 

Transmission Line Parameters Question 9 Detailed Solution

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, Z0 = 60 ohm

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

ZL = SWR x Z0

ZL = 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,

ZL = Load Resistance

Z0 = Characteristic Resistance

The value of Γ varies between 0 and 1

 Γ = 1 when there is complete reflection.

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

Transmission Line Parameters Question 10:

On a microstrip line, the wavelength measured is 12 mm for a 10 GHz signal. The dielectric constant of the equivalent homogeneous line is

  1. 3.5
  2. 6.25
  3. 5.5
  4. 7.0

Answer (Detailed Solution Below)

Option 2 : 6.25

Transmission Line Parameters Question 10 Detailed Solution

Concept:

The wavelength of the microstrip line is defined as:

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

ϵr: relative permittivity / dielectric constant

f: frequency of operation

c: speed of light

Calculation:

Given wavelength is 12 mm and frequency is 10 GHz

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

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

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

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

ϵr = 6.25

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