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

$\left[\begin{array}{l}{V}_1^{-}\\ V_2^{-}\end{array}\right]=\left[\begin{array}{ll}{S}_{11}& S_{12}\\ S_{21}& S_{22}\end{array}\right]\left[\begin{array}{l}{V}_1^{+}\\ V_2^{+}\end{array}\right]$       ....(1)

To be calculated,

$\begin{array}{rl}& \left[\begin{array}{c}{V}_{1N}^{-}\\ V_{2N}^{-}\end{array}\right]=\left[\begin{array}{ll}{S}_{11N}& S_{12N}\\ S_{21N}& S_{22N}\end{array}\right]\left[\begin{array}{c}{V}_{1N}^{+}\\ V_{2N}^{+}\end{array}\right]\end{array}$

$V_1^{+}=V_{1N}^{+}{e}^{-j{\theta }_1}$

$V_{1N}^{-}=V_1^{-}{e}^{-j{\theta }_1}=V_1^{-}=V_{1N}^{-}{e}^{j{\theta }_1}$

$V_2^{+}=V_{2N}^{+}{e}^{-j{\theta }_2}$

$V_{2N}^{-}=V_2^{-}{e}^{-j{\theta }_2}=V_2^{-}=V_{2N}^{-}{e}^{j{\theta }_2}$

Feeding the data in equation (i),

$\left[\begin{array}{c}{V}_{1N}^{-}\\ V_{2N}^{-}\end{array}\right]=\left[\begin{array}{cc}{S}_{11N}{e}^{-j\left(2{\theta }_1\right)}& S_{12N}{e}^{-j({\theta }_1+{\theta }_2)}\\ S_{21N}{e}^{-j({\theta }_1+{\theta }_2)}& S_{22N}{e}^{-j2\left({\theta }_2\right)}\end{array}\right]$

Hence, the correct option is (A).

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:

           ${\mathrm{Z}}_{\mathrm{i}\mathrm{n}}=\frac{{\mathrm{Z}}_0^2}{{\mathrm{Z}}_{\mathrm{L}}}$

Analysis:

For a λ/4 line:

Input impedance = ${\mathrm{Z}}_{\mathrm{i}\mathrm{n}}=\frac{{\mathrm{Z}}_0^2}{{\mathrm{Z}}_{\mathrm{L}}}$

Where Z₀ is the characteristic impedance. 

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

Z_in = ∞ 

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 %

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 2^nd relation:

$Z_0=\sqrt{\frac{L}{C}}$

$L=Z_0^{2}C$

= (50²) (0.1 × 10^-9)

L = 2.5 × 10^-7 H/m

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:

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

l = distance from the load

β = Phase constant

ΓL = Reflection coefficient at the load calculated as:

${\Gamma }_L=\frac{Z_L-Z_0}{Z_L+Z_0}$

The maximum and minimum voltage is given by:

$V_{max}=\left|V^{+}\right|\left(1+\left|{\Gamma }_{\mathrm{L}}\right|\right)$

$V_{min}=\left|V^{+}\right|\left(1-\left|{\Gamma }_{\mathrm{L}}\right|\right)$

$VSWR=\frac{V_{max}}{V_{min}}=\frac{1+{\Gamma }_L}{1-{\Gamma }_L}$  ---(1)

Calculation:

Given,

$VSWR=5$

Using Equation (1), we get:

$5=\frac{1+{\Gamma }_L}{1-{\Gamma }_L}$

${\mathrm{\Gamma }}_L=2/3$

Concept:

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

$Z_{in}=Z_0\left[\frac{Z_L+j{Z}_0\tan \beta l}{Z_0+j{Z}_L\tan \beta l}\right]$

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, Z_l = 25Ω​​​​

$Z_{in}=Z_0\left[\frac{Z_L+j{Z}_0\tan \beta l}{Z_0+j{Z}_L\tan \beta l}\right]$

$\beta l=\frac{2\pi }{\lambda }l=\frac{\pi }{4}$

On solving we'll get:

Z_in = 50Ω 

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 = Z₀ and Reflection coefficient

$\mathrm{\Gamma }=\frac{Z_L-Z_0}{Z_L+Z_0}$

So Γ = 0 For impedance matching

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.

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

$SWR=\frac{Z_L}{Z_0}$

Z_L = SWR x Z₀

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.
• $\mathrm{\Gamma }=\frac{Z_L-Z_0}{Z_L+Z_0}$

Where,

Z_L = Load Resistance

Z₀ = Characteristic Resistance

The value of Γ varies between 0 and 1

 Γ = 1 when there is complete reflection.

​$SWR=\frac{1+\mathbf{\Gamma }}{1-\mathbf{\Gamma }}$

Concept:

The wavelength of the microstrip line is defined as:

$\lambda =\frac{c}{\sqrt{{ϵ}_r}f}$

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

$12\times {10}^{-3}=\frac{3\times {10}^8}{\sqrt{{ϵ}_r}\times 10\times {10}^9}$

$\sqrt{{ϵ}_r}=\frac{3\times {10}^8}{12\times {10}^{-3}\times {10}^{10}}$

$\sqrt{{ϵ}_r}=\frac{1}{4}\times \frac{{10}^8}{{10}^7}$

${ϵ}_r=\frac{100}{16}$

ϵ_r = 6.25