Distortionless Line MCQ Quiz - Objective Question with Answer for Distortionless Line - Download Free PDF

A system is said to be distortion less if its group delay is constant.

Explanation:

• Group Delay: Group delay is the time delay experienced by the envelope of a signal as it passes through a system. It represents how much the different frequency components of a signal are delayed relative to each other.
• Distortionless Transmission: For a system to be distortionless, all frequency components of the input signal must be delayed by the same amount. This ensures that the shape of the signal's envelope remains unchanged as it passes through the system.

Key Points:

• Constant Group Delay: If the group delay is constant across all frequencies, it means that the same amount delays all frequency components. This results in a distortion less output signal.
• Variable Group Delay: If the group delay varies with frequency, different frequency components will experience different delays. This leads to distortion, such as changes in the shape of the signal's envelope.

In summary: A constant group delay is a crucial condition for distortion-less transmission, ensuring that the output signal retains the same shape as the input signal.

Concept:

For a lossless line:

R = G = 0

A distortion less line satisfies the following condition:

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

Calculation:

Given:

R = 1 ohm/m, L = 200 nH/m, G = 300 μS/m, C = 60 pF, 

R ≠ G ≠ 0

Hence, it is not Lossless line

\(\frac{R}{L}=\frac{1}{200\times10^{-9}}=5\times10^6\) ----(1)

\(\frac{G}{C}=\frac{300\times10^{-6}}{60\times10^{-12}}=5\times10^6\)  -----(2)

From equation 1 and equation 2, we can say that the line is distortionless.

Concept:

For a distortionless transmission line:

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

The characteristic impedance is given by:

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

Also, the attenuation constant is given by:

\(α=\sqrt{RG}\)    ---(1)

Calculation:

Given R = 0.05 Ω/m, Z₀ = 50 Ω  

\(50=\sqrt\frac{R}{G}\)   ---(2)

Multiplying Equation (1) and (2), we get:

50α = R

\(\alpha=\frac{0.05}{50}=0.001~Np/m\)

In a distortion less transmission line R/L = G/C

Propagation constant \(\gamma = \sqrt {RG} + j\omega \sqrt {LC} \;\left( {is\;of\;form\;\alpha + j\beta } \right)\)

Hence when R and C are doubled L and G must also be doubled to keep line distortion less hence α and β and doubled.

A transmission line is distortionless if:

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

Rewriting this we have:

LG = RC

Note:

If \(\frac{{\rm{R}}}{{\rm{L}}} = \frac{{\rm{G}}}{{\rm{C}}}\) then phase velocity is given by 

\({v_p} = \frac{\omega }{\beta } = \frac{1}{{\sqrt {LC} }}\)

the phase velocity is independent of frequency hence the transmission line is distortionless.

Concept:

For a distortionless transmission line:

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

The characteristic impedance is given by:

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

Also, the attenuation constant is given by:

\(α=\sqrt{RG}\)    ---(1)

Calculation:

Given R = 0.05 Ω/m, Z₀ = 50 Ω  

\(50=\sqrt\frac{R}{G}\)   ---(2)

Multiplying Equation (1) and (2), we get:

50α = R

\(\alpha=\frac{0.05}{50}=0.001~Np/m\)

Concept:

For distortionless transmission line, we have

\(LG = RC\)

Characteristic impedance:

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

Attenuation constant:

 \(\alpha = \sqrt {RG} \)      --- (2)

Calculation:

From equation (1) and (2), we get

\(\alpha = \sqrt R \frac{{\sqrt R }}{{{Z_o}}} = \frac{R}{{{Z_o}}}\;\)

Putting the values of R and Z₀, we get

\( \Rightarrow \alpha = \frac{{0.1}}{{50}} = 0.002\;Np/m\)

Concept:

For a lossless line:

R = G = 0

A distortion less line satisfies the following condition:

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

Calculation:

Given:

R = 1 ohm/m, L = 200 nH/m, G = 300 μS/m, C = 60 pF, 

R ≠ G ≠ 0

Hence, it is not Lossless line

\(\frac{R}{L}=\frac{1}{200\times10^{-9}}=5\times10^6\) ----(1)

\(\frac{G}{C}=\frac{300\times10^{-6}}{60\times10^{-12}}=5\times10^6\)  -----(2)

From equation 1 and equation 2, we can say that the line is distortionless.

A transmission line is distortionless if:

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

Rewriting this we have:

LG = RC

Note:

If \(\frac{{\rm{R}}}{{\rm{L}}} = \frac{{\rm{G}}}{{\rm{C}}}\) then phase velocity is given by 

\({v_p} = \frac{\omega }{\beta } = \frac{1}{{\sqrt {LC} }}\)

the phase velocity is independent of frequency hence the transmission line is distortionless.

Concept:

For a distortionless transmission line:

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

The characteristic impedance is given by:

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

Also, the attenuation constant is given by:

\(α=\sqrt{RG}\)    ---(1)

Calculation:

Given R = 0.05 Ω/m, Z₀ = 50 Ω  

\(50=\sqrt\frac{R}{G}\)   ---(2)

Multiplying Equation (1) and (2), we get:

50α = R

\(\alpha=\frac{0.05}{50}=0.001~Np/m\)

Concept:

For distortionless transmission line, we have

\(LG = RC\)

Characteristic impedance:

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

Attenuation constant:

 \(\alpha = \sqrt {RG} \)      --- (2)

Calculation:

From equation (1) and (2), we get

\(\alpha = \sqrt R \frac{{\sqrt R }}{{{Z_o}}} = \frac{R}{{{Z_o}}}\;\)

Putting the values of R and Z₀, we get

\( \Rightarrow \alpha = \frac{{0.1}}{{50}} = 0.002\;Np/m\)

Concept:

For distortionless transmission line, we have

\(LG = RC\)

Characteristic impedance:

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

Calculation:

With R = 0.25 Ω/m, L = 0.5 μH/m, C = 200 pF/m, and G = 100 μS/m, the characteristic impedance will be:

\({{\rm{Z}}_0} = \sqrt {\frac{{\rm{L}}}{{\rm{C}}}} = \sqrt {\frac{{0.5 \times {{10}^6}}}{{200 \times {{10}^{ - 12}}}}} \)

Z₀ = 50 Ω 

Concept:

1. If voltage maxima or minima occur at the load then the load impedance is purely resistive.

2. Voltage standing wave ratio: It is the ratio of maximum voltage to minimum voltage.

If the voltage minima occur at load ZL < Z0 then

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

If the voltage maxima occur at load ZL > Z0 then

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

Analysis:

In a transmission line, Maxima and Minima are separated by λ/4. As minima occur at a distance of λ/4 from the load.

Hence there is a maxima formed at the load.

Now, VSWR = 2

\( \Rightarrow \left| \Gamma \right| = {{VSWR - 1} \over {VSWR + 1}} = {1 \over 3}\)

For lossless line when maxima or minima is formed at the load, then the phase of Γ is zero.

Thus \(\Gamma = + {1 \over 3}\ or - {1 \over 3}\)

For maxima at the load \({Z_L} > {Z_o}\) and Γ is +ve and

For minima formed at the load \(Z_L and Γ is –ve.

Thus, \(\Gamma = {1 \over 3} = {{{Z_L} - {Z_o}} \over {{Z_L} + {Z_o}}}\)

Z_L = 2Z₀ = 100 Ω 

Concept:

For a lossless line:

R = G = 0

A distortion less line satisfies the following condition:

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

Calculation:

Given:

R = 1 ohm/m, L = 200 nH/m, G = 300 μS/m, C = 60 pF, 

R ≠ G ≠ 0

Hence, it is not Lossless line

\(\frac{R}{L}=\frac{1}{200\times10^{-9}}=5\times10^6\) ----(1)

\(\frac{G}{C}=\frac{300\times10^{-6}}{60\times10^{-12}}=5\times10^6\)  -----(2)

From equation 1 and equation 2, we can say that the line is distortionless.

In a distortion less transmission line R/L = G/C

Propagation constant \(\gamma = \sqrt {RG} + j\omega \sqrt {LC} \;\left( {is\;of\;form\;\alpha + j\beta } \right)\)

Hence when R and C are doubled L and G must also be doubled to keep line distortion less hence α and β and doubled.