Standing Wave Ratio MCQ Quiz - Objective Question with Answer for Standing Wave Ratio - Download Free PDF

Last updated on Apr 3, 2025

Latest Standing Wave Ratio MCQ Objective Questions

Standing Wave Ratio Question 1:

A load of 3 + j4Ω is connected to a 1Ω lossless line. The magnitude of reflection coefficient at the load will be :

  1. 0.058 
  2. 0.79
  3. 0
  4. 1

Answer (Detailed Solution Below)

Option 2 : 0.79

Standing Wave Ratio Question 1 Detailed Solution

Explanation:

To find the magnitude of the reflection coefficient (Γ) at the load, we need to use the formula:

|Γ| = |(ZL - Z0) / (ZL + Z0)|

Where ZL is the load impedance and Z0 is the characteristic impedance of the lossless transmission line. Given:

  • ZL = 3 + j4Ω
  • Z0 = 1Ω

Step-by-Step Solution:

1. Calculate the numerator (ZL - Z0):

ZL - Z0 = (3 + j4) - 1 = 2 + j4

2. Calculate the denominator (ZL + Z0):

ZL + Z0 = (3 + j4) + 1 = 4 + j4

3. Find the magnitude of the numerator and the denominator:

|ZL - Z0| = |2 + j4| = √(22 + 42) = √(4 + 16) = √20 = 2√5

|ZL + Z0| = |4 + j4| = √(42 + 42) = √(16 + 16) = √32 = 4√2

4. Calculate the magnitude of the reflection coefficient:

|Γ| = |(ZL - Z0) / (ZL + Z0)| = |(2 + j4) / (4 + j4)| = (2√5) / (4√2) = (2√5) / (2√2 * 2) = √5 / (2√2) = (√5 / √2) / 2 = √(5/2) / 2

By simplifying √(5/2):

√(5/2) ≈ 1.58

Therefore:

|Γ| ≈ 1.58 / 2 ≈ 0.79

Hence, the magnitude of the reflection coefficient at the load is approximately 0.79. Therefore, the correct answer is Option 2.

Standing Wave Ratio Question 2:

A lossless transmission line is terminated in a load which reflects a part of the incident power. If VSWR is 2, the reflection coefficient will be : 

  1. 1/3
  2. 3/4 
  3. 1/2
  4. 2/3

Answer (Detailed Solution Below)

Option 1 : 1/3

Standing Wave Ratio Question 2 Detailed Solution

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Explanation:

Lossless Transmission Line and VSWR:

Definition: A lossless transmission line is a theoretical transmission line in which there is no dissipation of electrical energy as heat. This means that the conductors and the dielectric material between them are perfect, resulting in no energy loss as the signal propagates along the line.

VSWR (Voltage Standing Wave Ratio) is a measure of how efficiently RF power is transmitted from the power source, through a transmission line, into the load. It is defined as the ratio of the maximum voltage to the minimum voltage in a standing wave pattern along the transmission line.

The VSWR is given by:

VSWR = (1 + |Γ|) / (1 - |Γ|)

Where |Γ| is the magnitude of the reflection coefficient.

The reflection coefficient (Γ) represents the fraction of the incident power that is reflected back from the load. It is a measure of the impedance mismatch between the transmission line and the load.

Calculation of Reflection Coefficient:

Given that the VSWR is 2, we can use the VSWR formula to find the reflection coefficient.

VSWR = (1 + |Γ|) / (1 - |Γ|)

Substitute VSWR = 2 into the formula:

2 = (1 + |Γ|) / (1 - |Γ|)

To find |Γ|, we solve the equation step-by-step:

Step 1: Cross-multiply to eliminate the fraction:

2 * (1 - |Γ|) = 1 + |Γ|

Step 2: Distribute the 2 on the left-hand side:

2 - 2|Γ| = 1 + |Γ|

Step 3: Combine like terms to isolate |Γ|:

2 - 1 = 2|Γ| + |Γ|

1 = 3|Γ|

Step 4: Solve for |Γ|:

|Γ| = 1 / 3

Therefore, the reflection coefficient (|Γ|) is 1/3.

Correct Option Analysis:

The correct option is:

Option 1: 1/3

This option correctly represents the calculated value of the reflection coefficient when the VSWR is 2.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 2: 3/4

This option is incorrect. If we use 3/4 in the VSWR formula, we would get:

VSWR = (1 + 3/4) / (1 - 3/4) = 7 / 1 = 7

This does not match the given VSWR of 2.

Option 3: Thu Jan 02 2025 00:00:00 GMT+0530 (India Standard Time)

This option is clearly irrelevant to the calculation of the reflection coefficient.

Option 4: Mon Feb 03 2025 00:00:00 GMT+0530 (India Standard Time)

Similar to option 3, this option is irrelevant to the calculation of the reflection coefficient.

Conclusion:

Understanding the relationship between VSWR and the reflection coefficient is crucial for analyzing transmission line performance. The reflection coefficient (|Γ|) provides valuable insight into the impedance matching of the transmission line and the load. In this case, with a VSWR of 2, the reflection coefficient is accurately calculated to be 1/3, confirming that option 1 is correct.

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Standing Wave Ratio Question 3:

A transmission line having Z0 = 75 Ω is used to deliver power to 300 Ohm load. The VSWR of the circuit is

  1. 0.25
  2. 4
  3. 0.5
  4. 2
  5. 0.6

Answer (Detailed Solution Below)

Option 2 : 4

Standing Wave Ratio Question 3 Detailed Solution

Concept:

The voltage standing wave ratio is defined as the ratio of the maximum voltage (or current) to the minimum voltage (or current).

\(VSWR = \frac{{{{\rm{V}}_{{\rm{max}}}}}}{{{{\rm{V}}_{{\rm{min}}}}}} = \frac{{{{\rm{I}}_{{\rm{max}}.}}}}{{{{\rm{I}}_{{\rm{min}}.}}}}\)

VSWR is also given by:

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

Γ = Reflection coefficient, defined as:

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

ZL = Load impedance

Z0 = Characteristic Impedance

For ΓL varying from 0 to 1, VSWR varies from 1 to .

Application:

Given ZL = 300 Ω

The reflection coefficient is calculated to be:

\({\rm{Γ }} = \frac{{{300} - {75}}}{{{300} + {75}}}=0.6\)

VSWR is now calculated as:

\(VSWR = \frac{{1 + {\rm{\Gamma }}}}{{1 - {\rm{\Gamma }}}} = \frac{{1 + 0.6}}{{1 - 0.6}} =4\)

Standing Wave Ratio Question 4:

Find the SWR, if the reflection coefficient is 0.8

  1. 4.5
  2. 18
  3. 5
  4. 9

Answer (Detailed Solution Below)

Option 4 : 9

Standing Wave Ratio Question 4 Detailed Solution

Calculation:

VSWR and reflection coefficient are related by:

\(VSWR = \frac{{1 + \left| {\rm{Γ }} \right|}}{{1 - \left| {\rm{Γ }} \right|}}\)

Γ = 0.8

\(= \frac{{1 + 0.8}}{{1 - 0.8}} = \frac{1.8}{0.2} \)

= 9

26 June 1

Proof:

The voltage standing wave ratio is defined as the ratio of the maximum voltage (or current) to the minimum voltage (or current)

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

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:

\({{\rm{Γ }}_L} = \frac{{{Z_L} - {Z_0}}}{{{Z_L} + {Z_0}}}\)     -  -  - (1)

The maximum and minimum voltage is given by:

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

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

\(\therefore VSWR = \frac{{{V_{max}}}}{{{V_{min}}}} = \frac{{1 + {{\rm{Γ }}_L}}}{{1 + {{\rm{Γ }}_L}}}\)Vmax=|V+|(1+|ΓL|)" id="MathJax-Element-123-Frame" role="presentation" style="display: inline; position: relative;" tabindex="0">Vmax=|V+|(1+|ΓL|)Vmin=|V+|(1|ΓL|)" id="MathJax-Element-124-Frame" role="presentation" style="display: inline; position: relative;" tabindex="0">Vmin=|V+|(1|ΓL|)VSWR=VSWR=VmaxVmin=1+ΓL1+ΓL" id="MathJax-Element-125-Frame" role="presentation" style="display: inline; position: relative;" tabindex="0">

Standing Wave Ratio Question 5:

A 50 Ω transmission line operating at 250 MHz frequency is terminated by the series combination of a 75 Ω resistor and variable capacitor The value of Capacitor C that will produce a standing wave ratio of 5 on the line is ______.

  1. 5.9 pF
  2. 5.91 F
  3. 59 pF
  4. 59 nF

Answer (Detailed Solution Below)

Option 1 : 5.9 pF

Standing Wave Ratio Question 5 Detailed Solution

Concept:

The reflection coefficient is used to define the reflected wave with respect to the incident wave.

The reflection coefficient of the transmission line at the load is given by:

\({\rm{Γ }} = \frac{{{{\rm{Z}}_{\rm{L}}} - {Z_0}}}{{{Z_L} + {Z_0}}}\)

ZL = Load impedance

Z0 = Characteristic Impedance

Calculation:

\(\Gamma = \frac{1-5}{1+5} = \frac{-4}{6}\)

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

\(= \frac{75 - jX_c - 50}{75 - jX_c + 50}\)

\(= \frac{25 - jX_c}{125 - jX_c}\)

\(|\Gamma| = \frac{\sqrt{(25)^2 + X_c^2}}{\sqrt{(125)^2 + X_c^2}}= \frac{4}{6}\)

\((625 + X_c^2) \times 36 = 16 (15625 + X_c^2)\)

\(22500 + 36X_c^2 = 250000 + 16 X_c^2\)

\(20 X_c^2 = 227500\)

\(X_c^2 = 11375\)

Xc = 106.65 Ω

\(X_c = \dfrac{1}{\omega C} = 106.65\)

\(\omega C = \dfrac{1}{106.65}\)

\(C = \dfrac{10^{-6}}{2 \pi \times 250 \times 106.65}= 5.97 \ \rm{pF}\)

Top Standing Wave Ratio MCQ Objective Questions

If a transmission line is terminated with a resistance equal to its characteristic impedance

  1. reflection coefficient will be unity
  2. standing wave ratio will be minimum
  3. the line loss will be maximum
  4. The input impedance will be twice the terminating resistance

Answer (Detailed Solution Below)

Option 2 : standing wave ratio will be minimum

Standing Wave Ratio Question 6 Detailed Solution

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Concept:

The voltage standing wave ratio is defined as the ratio of the maximum voltage (or current) to the minimum voltage (or current).

\(VSWR = \dfrac{{{{\rm{V}}_{{\rm{max}}}}}}{{{{\rm{V}}_{{\rm{min}}}}}} = \dfrac{{{{\rm{I}}_{{\rm{max}}.}}}}{{{{\rm{I}}_{{\rm{min}}.}}}}\)

VSWR is also given by:

\(VSWR = \dfrac{{1 + {\rm{Γ }}}}{{1 - {\rm{Γ }}}}\)

Γ = Reflection coefficient, defined as:

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

ZL = Load impedance

Z0 = Characteristic Impedance

For ΓL varying from 0 to 1, VSWR varies from 1 to ∞.

Application:

Given ZL = Z0

The reflection coefficient is calculated to be:

\({\rm{Γ }} = \dfrac{{{Z_0} - {Z_0}}}{{{Z_0} + {Z_0}}}=0\)

VSWR for Γ = 0 equals 1, which is the minimum value (because it varies from 1 to ∞)

The desirable VSWR on a transmission line is

  1. 1
  2. 0
  3. 2
  4. Infinity

Answer (Detailed Solution Below)

Option 1 : 1

Standing Wave Ratio Question 7 Detailed Solution

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The reflection coefficient is used to define the reflected wave with respect to the incident wave.

The reflection coefficient of the transmission line at the load is given by:

\({\rm{Γ }} = \frac{{{{\rm{Z}}_{\rm{L}}} - {Z_0}}}{{{Z_L} + {Z_0}}}\)

ZL = Load impedance

Z0 = Characteristic Impedance

It is always desirable to have a perfectly matched load for maximum power transfer, i.e. ZL = Z0

For ZL = Z0

\({\rm{Γ }} = \frac{{{{\rm{Z}}_{\rm{0}}} - {Z_0}}}{{{Z_0} + {Z_0}}}=0\)

The Voltage Standing Wave Ratio (VSWR) is defined as:

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

The voltage standing wave ratio for a perfectly matched load (which is always desirable) is obtained by putting Γ = 0 in the above equation.

\(VSWR = \frac{{1 + {\rm{0 }}}}{{1 - {\rm{0}}}}=1\)

A transmission line having 50 Ω impedance is terminated in a load of (40 + j30) Ω. The VSWR is

  1. 1.333
  2. 0.8
  3. 1
  4. 2

Answer (Detailed Solution Below)

Option 4 : 2

Standing Wave Ratio Question 8 Detailed Solution

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Concept:

Voltage standing wave ratio (VSWR) is mathematically defined as:

\(VSWR = \frac{{\left( {1 + \left|\Gamma \right|} \right)}}{{\left( {1 - \left| \Gamma \right|} \right)}}\)   ---(1)

Γ = Reflection coefficient given by:

\(\Gamma_L= \frac{{{Z_L} - {Z_o}}}{{{Z_L} + {Z_o}}}\)

ZL = Load Impedance

Z0 = Characteristic impedance

Application:

Given: Zo = 50 Ω and ZL = 40 + j30

Reflection co-efficient will be:

\(\Gamma_L= \frac{{{Z_L} - {Z_o}}}{{{Z_L} + {Z_o}}}\)

\(\Gamma_L= \frac{{40 + j30 - 50}}{{40 + j30 + 50}}\)

\(\Gamma_L = \frac{{ - 10 + j30}}{{90 + j30}}\)

\(\left| \Gamma \right| = \sqrt {\frac{{{{\left( {10} \right)}^2} + {{\left( {30} \right)}^2}}}{{{{\left( {90} \right)}^2} + {{\left( {30} \right)}^2}}}} \)

\(\Gamma= \sqrt {\frac{{10}}{{90}}} \)

Now, the voltage standing wave ratio will be:

\(VSWR = \frac{{1 +\Gamma}}{{1 - \Gamma}}\)

\(VSWR= \frac{{1 + \frac{1}{3}}}{{1 - \frac{1}{3}}}\)

VSWR = 2

A 75 Ω transmission line is first short-terminated and the minima locations are noted. When the short is replaced by a resistive load RL, the minima locations are not altered and the VSWR is measured to be 3. The value of RL is

  1. 25 Ω 
  2. 50 Ω 
  3. 225 Ω 
  4. 250 Ω 

Answer (Detailed Solution Below)

Option 1 : 25 Ω 

Standing Wave Ratio Question 9 Detailed Solution

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Concept:

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:

Z0 = 75 Ω 

When the load is shorted the voltage minima occur at the load, and it does not change even when short is replaced by RL

\(VSWR = \frac{Z_0}{R_L}\) and RL < Z0 

VSWR = 3

\(\frac{75}{R_L} = 3\)

RL = 25 Ω 

Voltage standing wave ratio is expressed as:

  1. \(\frac{{\left| {{V_{max}}} \right|}}{{\left| {{V_{min}}} \right|}}\)
  2. |Vmax| * |Vmin|
  3. |Vmax| + |Vmin|
  4. \(\frac{{\left| {{V_{min}}} \right|}}{{\left| {{V_{max}}} \right|}}\)

Answer (Detailed Solution Below)

Option 1 : \(\frac{{\left| {{V_{max}}} \right|}}{{\left| {{V_{min}}} \right|}}\)

Standing Wave Ratio Question 10 Detailed Solution

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The voltage standing wave ratio is defined as the ratio of the maximum voltage (or current) to the minimum voltage (or current).

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

Important Derivation:

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

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

l = distance from the load

β = Phase constant

ΓL = Reflection coefficient at the load calculated as:

\({{\rm{\Gamma }}_L} = \frac{{{Z_L} - {Z_0}}}{{{Z_L} + {Z_0}}}\)     -  -  - (1)

The maximum and minimum voltage is given by:

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

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

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

Which of the following is NOT an application of SWR?

  1. Microwave-based medical applications
  2. Measuring the efficiency of a communication line
  3. Step down transformer
  4. Installation and tuning of transmitting antennas

Answer (Detailed Solution Below)

Option 3 : Step down transformer

Standing Wave Ratio Question 11 Detailed Solution

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Standing wave ratio (SWR) is a measure of impedance matching of loads to the characteristic impedance of a transmission line or waveguide.

Impedance mismatches result in standing waves along the transmission line, and SWR is defined as the ratio of the partial standing wave's amplitude at an antinode (maximum) to the amplitude at a node (minimum) along the line.

Practical applications of SWR are following:

  • SWR  have impact on the performance of microwave-based medical applications. In microwave electrosurgery an antenna that is placed directly into tissue may not always have an optimal match with the feedline resulting in an SWR. The presence of SWR can affect monitoring components used to measure power levels impacting the reliability of such measurements.
  • SWR is  the ratio between transmitted and reflected waves. A high SWR indicates poor transmission-line efficiency and reflected energy, which can damage the transmitter and decrease transmitter efficiency.
  • SWR is used as a measure of impedance matching of a load to the characteristic impedance of a transmission line carrying radio frequency (RF) signals.
  • When a transmitter is connected to an antenna by a feed line, the driving point impedance of the antenna must match the characteristic impedance of the feed line in order for the transmitter to see the impedance it was designed for.

Which of the following is NOT true about SWR?

  1.  \(VSWR = \frac{V_{max}}{V_{min}}\)
  2.  \(VSWR = \frac{V_{fwd} + V_{ref}}{V_{fwd} - V_{ref}}\)
  3. ISWR meter can measure current SWR and VSWR can measure voltage SWR
  4. Standing waves are also known as stationary waves in physics.

Answer (Detailed Solution Below)

Option 2 :  \(VSWR = \frac{V_{fwd} + V_{ref}}{V_{fwd} - V_{ref}}\)

Standing Wave Ratio Question 12 Detailed Solution

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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 < ∞)

ISWR meter can measure current SWR and VSWR can measure voltage SWR.

In physics, a standing wave, also known as a stationary wave, is a wave that oscillates in time but whose peak amplitude profile does not move in space.

Important Points

The voltage standing wave ratio is defined as the ratio of the maximum voltage (or current) to the minimum voltage (or current)

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

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

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

l = distance from the load

β = Phase constant

ΓL = Reflection coefficient at the load calculated as:

\({{\rm{\Gamma }}_L} = \frac{{{Z_L} - {Z_0}}}{{{Z_L} + {Z_0}}}\)        ---(1)

The maximum and minimum voltage is given by:

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

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

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

A two-wire transmission line terminates in a television set. The VSWR measured on the line is 5.8. The percentage of power that is reflected from the television set is ___________.

Answer (Detailed Solution Below) 48.0 - 51.0

Standing Wave Ratio Question 13 Detailed Solution

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Concept:

VSWR (Voltage standing wave Ratio) is defined as \( = \frac{{\left( {1 + \left|\Gamma \right|} \right)}}{{\left( {1 - \left| \Gamma \right|} \right)}}\) where Γ = Reflection coefficient.

Case 1:

For Γ = 1; VSWR = ∞

 All the power is reflected back to the transmission line.

Case 2:

For Γ = 0; VSWR = 1

Transmission line is perfectly matched to the Antenna and no power is reflected back to the transmission line

So, 1 ≤ VSWR < ∞ (Range)

Calculation:

Given, VSWR = 5.8

\( \Rightarrow 5.8 = \frac{{\left( {1 + \left| \Gamma \right|} \right)}}{{\left( {1 - \left| \Gamma \right|} \right)}}\)

Solving, we get

|Γ| = 0.7

\(\Gamma = \frac{{{V_ - }}}{{{V_ + }}}\)

(Where V- = Reflected voltage Amplitude at load)

(Where V+ = Forward path voltage Amplitude)

% Power reflected \(= \frac{{V_ - ^2}}{{V_ + ^2}} = \Gamma ^2 = {\left( {0.7} \right)^2} = 0.49 \times 100\) 

= 49%

Find the SWR, if the reflection coefficient is 0.8

  1. 4.5
  2. 18
  3. 5
  4. 9

Answer (Detailed Solution Below)

Option 4 : 9

Standing Wave Ratio Question 14 Detailed Solution

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Calculation:

VSWR and reflection coefficient are related by:

\(VSWR = \frac{{1 + \left| {\rm{Γ }} \right|}}{{1 - \left| {\rm{Γ }} \right|}}\)

Γ = 0.8

\(= \frac{{1 + 0.8}}{{1 - 0.8}} = \frac{1.8}{0.2} \)

= 9

26 June 1

Proof:

The voltage standing wave ratio is defined as the ratio of the maximum voltage (or current) to the minimum voltage (or current)

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

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:

\({{\rm{Γ }}_L} = \frac{{{Z_L} - {Z_0}}}{{{Z_L} + {Z_0}}}\)     -  -  - (1)

The maximum and minimum voltage is given by:

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

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

\(\therefore VSWR = \frac{{{V_{max}}}}{{{V_{min}}}} = \frac{{1 + {{\rm{Γ }}_L}}}{{1 + {{\rm{Γ }}_L}}}\)Vmax=|V+|(1+|ΓL|)" id="MathJax-Element-123-Frame" role="presentation" style="display: inline; position: relative;" tabindex="0">Vmax=|V+|(1+|ΓL|)Vmin=|V+|(1|ΓL|)" id="MathJax-Element-124-Frame" role="presentation" style="display: inline; position: relative;" tabindex="0">Vmin=|V+|(1|ΓL|)VSWR=VSWR=VmaxVmin=1+ΓL1+ΓL" id="MathJax-Element-125-Frame" role="presentation" style="display: inline; position: relative;" tabindex="0">

A transmission line having a characteristic impedance of 50 Ω has to deliver 10 kW power at 100 kHz to a load. The maximum permissible value of rms current anywhere along the line is 20 A. What is the maximum VSWR that can be tolerated on this line?

  1. 2
  2. 1
  3. 3
  4. 2.5

Answer (Detailed Solution Below)

Option 1 : 2

Standing Wave Ratio Question 15 Detailed Solution

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Concept:

\({\rm{\Gamma }}\;\left( {Reflection\;coefficient} \right) = \frac{{{Z_L} - {Z_0}}}{{{Z_L} + {Z_0}}}\)

\(VSWR = \frac{{1 + \left| {\rm{\Gamma }} \right|}}{{1 - \left| {\rm{\Gamma }} \right|}}\)

And power delivered at load = \(I_{rms}^2 \times {Z_L}\)

Calculation:

Given

P = 10k = \(I_{rms}^2\;{Z_L}\;and\;{Z_0} = 50\)

\({Z_L} = \frac{{10k}}{{I_{rms}^2}} = \frac{{10k}}{{20 \times 20}} = 25\;{\rm{\Omega }}\)

\({\rm{\Gamma }} = \frac{{{Z_L} - {Z_0}}}{{{Z_L} + {Z_0}}} = \frac{{25 - 50}}{{25 + 50}} = \frac{{ - 25}}{{75}} = \frac{{ - 1}}{3}\;\)

\(VSWR = \frac{{1 + \left| {\rm{\Gamma }} \right|}}{{1 - \left| {\rm{\Gamma }} \right|}} = \frac{{1 + \frac{1}{3}}}{{1 - \frac{1}{3}}} = 2\)
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