Bode Plot MCQ Quiz - Objective Question with Answer for Bode Plot - Download Free PDF

Last updated on Apr 3, 2025

Latest Bode Plot MCQ Objective Questions

Bode Plot Question 1:

For a control system having fourteen poles and two zeros, the slope in its magnitude plot for high frequency asymptote will be :

  1. −40 dB/decade
  2. −240 dB/decade
  3. −280 dB/decade
  4. −320 dB/decade

Answer (Detailed Solution Below)

Option 2 : −240 dB/decade

Bode Plot Question 1 Detailed Solution

Explanation:

To determine the slope in the magnitude plot for the high frequency asymptote of a control system, we need to understand the relationship between the number of poles and zeros in the system.

In control systems, the Bode plot is a graphical representation of a system's frequency response. The slope of the magnitude plot in the high frequency region is influenced by the number of poles and zeros. Specifically, each pole contributes a slope of -20 dB/decade, and each zero contributes a slope of +20 dB/decade.

Given:

  • Number of poles (P" id="MathJax-Element-893-Frame" role="presentation" style="position: relative;" tabindex="0">P ) = 14
  • Number of zeros (Z" id="MathJax-Element-894-Frame" role="presentation" style="position: relative;" tabindex="0">Z ) = 2

The formula to determine the overall slope in the high frequency asymptote is:

Slope (dB/decade) = -20 × (Number of poles - Number of zeros)

Substituting the given values:

Slope (dB/decade) = -20 × (14 - 2)

Slope (dB/decade) = -20 × 12

Slope (dB/decade) = -240 dB/decade

Therefore, the correct option is:

Option 2: -240 dB/decade

Bode Plot Question 2:

6 dB corresponds to a frequency response with magnitude equal to:

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

Answer (Detailed Solution Below)

Option 1 : 2

Bode Plot Question 2 Detailed Solution

Explanation:

Decibel (dB) and Frequency Response

Definition: The decibel (dB) is a logarithmic unit used to express the ratio of two values, commonly used in acoustics and electronics to describe the gain or loss in a system. The frequency response of a system shows how the amplitude of the output signal varies with frequency.

Understanding 6 dB: A 6 dB increase corresponds to a doubling of the amplitude of the signal. This is because the decibel scale is logarithmic. Specifically, a gain of 6 dB means that the output signal's amplitude is twice that of the input signal. The formula for converting a magnitude ratio to decibels is:

dB = 20 * log10(magnitude ratio)

To find the magnitude ratio that corresponds to 6 dB, we rearrange the formula:

6 dB = 20 * log10(magnitude ratio)

Dividing both sides by 20 gives:

0.3 = log10(magnitude ratio)

To solve for the magnitude ratio, we take the antilogarithm (base 10) of both sides:

magnitude ratio = 100.3

Using a calculator, we find:

magnitude ratio ≈ 2

Therefore, a 6 dB increase corresponds to a magnitude ratio of approximately 2.

Correct Option Analysis:

The correct option is:

Option 1: 2

This option correctly describes the magnitude ratio corresponding to a 6 dB increase in the frequency response. As calculated, a 6 dB gain means that the output signal is twice the amplitude of the input signal, which matches the magnitude ratio of 2.

Additional Information

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

Option 2: 3

This option is incorrect. A 6 dB increase corresponds to a magnitude ratio of 2, not 3. Using the formula mentioned earlier, a 3 dB increase corresponds to a doubling of power, but not of magnitude. For magnitude, a 6 dB increase is required to double it.

Option 3: 1

This option is incorrect. A magnitude ratio of 1 means no change in amplitude, corresponding to 0 dB. Therefore, it does not represent a 6 dB increase.

Option 4: 0

This option is also incorrect. A magnitude ratio of 0 would imply no output, which is not possible in the context of a 6 dB increase. The magnitude ratio corresponding to a 6 dB increase is 2, as calculated.

Conclusion:

Understanding the relationship between decibels and magnitude ratios is crucial in fields such as acoustics and electronics. A 6 dB increase corresponds to a doubling of the magnitude of the signal, leading to a magnitude ratio of 2. This logarithmic scale allows for a more manageable representation of large changes in signal strength. Thus, the correct answer to the question is option 1, which accurately reflects the magnitude ratio corresponding to a 6 dB increase in frequency response.

Bode Plot Question 3:

In a Bode plot, how is the order of a system determined from the phase plot? 

  1. By the location of the magnitude peak 
  2. By examining the gain margin
  3. By counting the number of phase crossings 
  4. By analysing the slope of the phase plot

Answer (Detailed Solution Below)

Option 3 : By counting the number of phase crossings 

Bode Plot Question 3 Detailed Solution

The correct answer is: Option 3) By counting the number of phase crossings

Concept

A Bode plot is a graphical representation of a linear, time-invariant system transfer function. It consists of two plots: one for magnitude (in dB) versus frequency and one for phase (in degrees) versus frequency. The phase plot is particularly useful in determining the stability and response characteristics of the system.

In a Bode plot, the order of a system can be determined from the phase plot by counting the number of phase crossings. The number of times the phase plot crosses -180 degrees is indicative of the order of the system. Each crossing typically represents a pole or a zero in the system's transfer function, which helps in determining the overall order.

Bode Plot Question 4:

In a Bode plot, what is the significance of the gain crossover frequency? 

  1. It indicates the unity gain frequency
  2. It is the frequency at which phase margin is minimum. 
  3. It corresponds to the phase crossover frequency
  4. It represents the damping ratio

Answer (Detailed Solution Below)

Option 1 : It indicates the unity gain frequency

Bode Plot Question 4 Detailed Solution

Concept

A Bode plot is a graphical representation of a linear, time-invariant system transfer function. It consists of two plots: one for magnitude (gain) and one for phase, both plotted against frequency.

The gain crossover frequency is a key parameter in control system analysis and design. It is defined as the frequency at which the magnitude of the open-loop transfer function is equal to one (0 dB).

The gain crossover frequency is significant because it is the point where the system's open-loop gain is unity (1 or 0 dB). At this frequency, the phase margin can be assessed to determine the stability of the system.

Additional Information 
2) It is the frequency at which the phase margin is minimum.

This statement is incorrect. The gain crossover frequency is not necessarily the point where the phase margin is minimum. The phase margin is the difference between the phase of the open-loop transfer function and -180 degrees at the gain crossover frequency.

3) It corresponds to the phase crossover frequency.

This statement is incorrect. The phase crossover frequency is the frequency at which the phase of the open-loop transfer function is -180 degrees. It is different from the gain crossover frequency, which is the frequency at which the magnitude is 0 dB.

4) It represents the damping ratio.

This statement is incorrect. The damping ratio is a measure of how oscillations in a system decay after a disturbance. It is not directly related to the gain crossover frequency.

Bode Plot Question 5:

Bode plot of a stable system is shown below.

The transfer function of the system is -

qImage67811fc6e19ef6d5ac90cb94

  1. \(G(s)=\frac{100}{s+10}\)
  2. \(G(s)=\frac{10}{s+10}\)
  3. \(G(s)=\frac{10}{s+1}\)
  4. \(G(s)=\frac{10}{s+100}\)

Answer (Detailed Solution Below)

Option 1 : \(G(s)=\frac{100}{s+10}\)

Bode Plot Question 5 Detailed Solution

Transfer function from Bode Plot

The bode plot of a transfer function is plotted between the magnitude of the system's transfer function in decibels (dB) and angular frequency which is in log scale.

A pole in a bode plot provides a slope of -20dB/decade while a zero provides a slope of 20dB/decade.

A transfer function is of the form:

\(G(s)={k\over 1+{s\over ω_c}}\)

Here, ωc is the angular frequency. It is the frequency where the bode plot changes its slope due to a finite pole or zero.

Calculation

qImage67811fc6e19ef6d5ac90cb94

In the given figure, we are changing log ω into 'ω' by:

log ω = 0 → ω = 100 = 1

log ω = 1 → ω = 101 = 10

So, ωc = 10 rad/sec

The value of k is obtained by the low-frequency region.

20 log k = 20 

log k = 1

k = 101 = 10

The transfer function is calculated as:

\(G(s)={10\over 1+{s\over 10}}\)

\(G(s)=\frac{100}{s+10}\)

Top Bode Plot MCQ Objective Questions

The asymptotic Bode magnitude plot of a minimum phase transfer function G(s) is shown below.

D234

Consider the following two statements.

Statement I: Transfer function G(s) has three poles and one zero.

Statement II: At very high frequency (ω→∞), the phase angle \(\angle G\left( {j\omega } \right) = - \frac{{3\pi }}{2}\).

Which one of the following options is correct?

  1. Statement I is true and statement II is false.
  2. Statement I is false and statement II is true.
  3. Both the statements are true.
  4. Both the statements are false.

Answer (Detailed Solution Below)

Option 2 : Statement I is false and statement II is true.

Bode Plot Question 6 Detailed Solution

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Form the given bode plot, we can write the transfer function as follows 

\(G\left( s \right) = \frac{k}{{s\left( {1 + s} \right)\left( {1 + \frac{s}{{20}}} \right)}}\)

It has 3 poles and no zeros,

So, statement I is false

At ω → ∞, the phase angle

\(G\left( {j\omega } \right) = - \frac{\pi }{2} - \frac{\pi }{2} - \frac{\pi }{2} = - \frac{{3\pi }}{2}\)

So, statement II is true

The bode magnitude plot for the transfer function \(\frac{{{V_0}(s)}}{{{V_i}(s)}}\) of the circuit is as shown. The value of R is ______ Ω. (Round off to 2 decimal places.)

F1 Shraddha Koda 20.02.2021 D10

F1 Shraddha Koda 20.02.2021 D11

Answer (Detailed Solution Below) 0.09 - 0.11

Bode Plot Question 7 Detailed Solution

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Given circuit,

quesImage5954

The transfer function for series RLC circuit by taking output voltage acroos the capacitor is given by,

\(T\left( s \right) = \frac{{{V_O}\left( s \right)}}{{{V_i}\left( s \right)}} = \frac{1}{{LC{s^2} + RCs + 1}}\)

Put, s = jω

\( \frac{{{V_O}\left( s \right)}}{{{V_i}\left( s \right)}} = \frac{1}{{ - {\omega ^2}LC + j\omega RC + 1}}\)

\( \frac{{{V_O}\left( s \right)}}{{{V_i}\left( s \right)}} = \frac{1}{{ - \left( {{{2000}^2} \times 1m \times 250\mu } \right) + j\left( {250\mu } \right)R + 1}}\)

\( \frac{{{V_O}\left( s \right)}}{{{V_i}\left( s \right)}} = \frac{1}{{0.5R}}\)

From the plot,

20 log|Mr| = 26 dB

|Mr| = 19.95

19.95 × 0.5R = 1

R = 0.1 Ω

Alternate Method

Resonant Peak formula,

\({M_r} = \frac{1}{{2\zeta \sqrt {1 - {\zeta ^2}} }}\)

Mr = 19.95

ζ = 0.025

For RLC series circuit,

\(\zeta = \frac{R}{2}\sqrt {\frac{C}{L}} \)

\(0.025 = \frac{R}{2}\sqrt {\frac{{250\mu }}{{1m}}} \)

R = 0.1 Ω

When the gain margin is positive and phase margin is negative, the system is:

  1. unstable
  2. highly stable
  3. oscillatory
  4. stable

Answer (Detailed Solution Below)

Option 1 : unstable

Bode Plot Question 8 Detailed Solution

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Phase Cross over Frequency:

  • The frequency at which the phase plot is having the phase of -180° is known as phase cross-over frequency.
  • It is denoted ωpc .
  • The unit of phase cross over frequency is rad/sec

 

Gain Cross over Frequency:

  • The frequency at which the magnitude plot is having the magnitude of zero dB is known as gain cross over frequency.
  • It is denoted by ωgc.
  • The unit of gain cross over frequency is rad/sec

 

Gain Margin:

Gain margin GM is defined as the negative of the magnitude in dB, at phase cross over frequency, i.e.

\(GM = 20\log \left( {\frac{1}{{{M_{pc}}}}} \right) = 20\log {M_{pc}}\)

Mpc is the magnitude at phase cross over frequency. The unit of gain margin (GM) is dB.

Phase Margin

The phase margin of a system is defined as:

PM = 180° + ϕgc

The stability of the control system is based on the relation between gain margin and phase margin as:

Gain Margin (GM) Phase Margin (PM) Nature
Positive Positive Stable
Zero Zero

Marginally
Stable

Negative Negative Unstable
Positive Negative Untable


F2 S.B Madhu 07.05.20 D18

F2 S.B Madhu 07.05.20 D19

The transfer function of a system is \(\frac{{10\left( {1 + 0.2s} \right)}}{{\left( {1 + 0.5s} \right)}}\)

The corner frequencies will be

  1. -0.2 and -0.5
  2. 5 and 2
  3. -5 and -2
  4. 0.2 and 0.5

Answer (Detailed Solution Below)

Option 2 : 5 and 2

Bode Plot Question 9 Detailed Solution

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Concept:

Bode plot transfer function is represented in standard time constant form as \(T\left( s \right) = \frac{{k\left( {\frac{s}{{{\omega _{{c_1}}}}} + 1} \right) \ldots }}{{\left( {\frac{s}{{{\omega _{{c_2}}}}} + 1} \right)\left( {\frac{s}{{{\omega _{{c_3}}}}} + 1} \right) \ldots }}\)

ωc1, ωc2, … are corner frequencies.

In a Bode magnitude plot,

  • For a pole at the origin, the initial slope is -20 dB/decade
  • For a zero at the origin, the initial slope is 20 dB/decade
  • The slope of magnitude plot changes at each corner frequency
  • The corner frequency associated with poles causes a slope of -20 dB/decade
  • The corner frequency associated with poles causes a slope of -20 dB/decade
  • The final slope of Bode magnitude plot = (Z – P) × 20 dB/decade


Where Z is the number zeros and P is the number of poles

Application:

The given transfer function is \(\frac{{10\left( {1 + 0.2s} \right)}}{{\left( {1 + 0.5s} \right)}}\)

\( = \frac{{10\left( {1 + \frac{s}{5}} \right)}}{{\left( {1 + \frac{s}{2}} \right)}}\)

By comparing with the standard transfer function, corner frequencies are

ω1 = 2, ω2 = 5

Slope of the asymptote in Bode plot for a second-order system is:

  1. 18 dB per octave
  2. 6 dB per octave
  3. 3 dB per octave
  4. 12 dB per octave

Answer (Detailed Solution Below)

Option 4 : 12 dB per octave

Bode Plot Question 10 Detailed Solution

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Concept:

Bode plot transfer function is represented in standard time constant form as \(T\left( s \right) = \frac{{k\left( {\frac{s}{{{\omega _{{c_1}}}}} + 1} \right) \ldots }}{{\left( {\frac{s}{{{\omega _{{c_2}}}}} + 1} \right)\left( {\frac{s}{{{\omega _{{c_3}}}}} + 1} \right) \ldots }}\)

ωc1, ωc2, … are corner frequencies.

In a Bode magnitude plot,

  • For a pole at the origin, the initial slope is -20 dB/decade
  • For a zero at the origin, the initial slope is 20 dB/decade
  • The slope of magnitude plot changes at each corner frequency
  • The corner frequency associated with poles causes a slope of -20 dB/decade
  • The corner frequency associated with poles causes a slope of -20 dB/decade
  • The final slope of Bode magnitude plot = (Z – P) × 20 dB/decade


Where Z is the number zeros and P is the number of poles

Application:

The given system is second order system. So, the number of poles is 2.

The slope of the asymptote = 2 × 20 dB/decade = 40 dB/decade

20 bB/decade = 6 dB/octave

Therefore, the slope of the asymptote = 12 dB/octave

The Bode magnitude plot of a first order stable system is constant with frequency. The asymptotic value of the high frequency phase, for the system, is −180°. This system has

F1 RaviR Madhuri 05.03.2022 D9

  1. one LHP pole and one RHP zero at the same frequency.
  2. one LHP pole and one LHP zero at the same frequency
  3. two LHP poles and one RHP zero
  4. two RHP poles and one LHP zero

Answer (Detailed Solution Below)

Option 1 : one LHP pole and one RHP zero at the same frequency.

Bode Plot Question 11 Detailed Solution

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Concept:

From the given bode plot, it is clear that magnitude is constant at all frequencies.

F1 RaviR Madhuri 05.03.2022 D9

  • Also, it is given a first-order system, hence there exists one finite pole.
  • It can be an all-pass system.
  • The transfer function of the all-pass system is

 

\(\rm T.F = \frac{1-S}{1 + S}\)

Phase angle of T.F is

ϕ = - tan-1ω - tan-1ω

At ω → ∞, ϕ = -180°

It is given in the figure also, at high-frequency phase is equal to -180°.

Therefore, it is an all-pass system having one pole at left and one zero at right at the same frequency.

Therefore, the correct option is (a).

The figure below shows the Bode magnitude and phase plots of a stable transfer function

\({\rm{G}}\left( {\rm{s}} \right) = \frac{{{{\rm{n}}_0}}}{{{{\rm{s}}^3} + {{\rm{d}}_2}{{\rm{s}}^2} + {{\rm{d}}_1}{\rm{s}} + {{\rm{d}}_0}}}\)

1104

Consider the negative unity feedback configuration with gain in the feedforward path. The closed loop is stable for 𝑘 < 𝑘0. The maximum value of 𝑘0 is ______.

Answer (Detailed Solution Below) 0.1

Bode Plot Question 12 Detailed Solution

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According to figure

Gain = 20 dB (at phase cross over frequency)

Gian margin = -20 dB

System make to marginally stable G.M = 0 or Gain = 0 dB

i.e 20 log K = -20

log K = -1

K = 10-1 = 0.1

So for K < 0.1 system will be stable

A system is _______ when gain margin is positive whereas the phase margin is negative.

  1. Stable
  2. Unstable
  3. Probabilistic
  4. Undetermined

Answer (Detailed Solution Below)

Option 2 : Unstable

Bode Plot Question 13 Detailed Solution

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Phase Cross over Frequency:

  • The frequency at which the phase plot is having the phase of -180° is known as phase cross-over frequency.
  • It is denoted ωpc .
  • The unit of phase cross over frequency is rad/sec

 

Gain Cross over Frequency:

  • The frequency at which the magnitude plot is having the magnitude of zero dB is known as gain cross over frequency.
  • It is denoted by ωgc.
  • The unit of gain cross over frequency is rad/sec

 

Gain Margin:

Gain margin GM is defined as the negative of the magnitude in dB, at phase cross over frequency, i.e.

\(GM = 20\log \left( {\frac{1}{{{M_{pc}}}}} \right) = 20\log {M_{pc}}\)

Mpc is the magnitude at phase cross over frequency. The unit of gain margin (GM) is dB.

Phase Margin

The phase margin of a system is defined as:

PM = 180° + ϕgc

The stability of the control system is based on the relation between gain margin and phase margin as:

Gain Margin (GM) Phase Margin (PM) Nature
Positive Positive Stable
Zero Zero

Marginally
Stable

Negative Negative Unstable
Positive Negative Unstable


F2 S.B Madhu 07.05.20 D18

F2 S.B Madhu 07.05.20 D19

Find out the slope of bode magnitude plot for a 3rd order all-zero system

  1. −40 dB/decade
  2. 40 dB/decade
  3. −60 dB/decade
  4. 60 dB/decade

Answer (Detailed Solution Below)

Option 4 : 60 dB/decade

Bode Plot Question 14 Detailed Solution

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Concept:

Bode plot transfer function is represented in standard time constant form as

\(T\left( s \right) = \frac{{k\left( {\frac{s}{{{ω _{{c_1}}}}} + 1} \right) \ldots }}{{\left( {\frac{s}{{{ω _{{c_2}}}}} + 1} \right)\left( {\frac{s}{{{ω _{{c_3}}}}} + 1} \right) \ldots }}\)

ωc1, ωc2, … are corner frequencies.

In a Bode magnitude plot,

  • For a pole at the origin, the initial slope is -20 dB/decade
  • For a zero at the origin, the initial slope is 20 dB/decade
  • The slope of magnitude plot changes at each corner frequency
  • The corner frequency associated with poles causes a slope of -20 dB/decade
  • The corner frequency associated with poles causes a slope of -20 dB/decade
  • The final slope of Bode magnitude plot = (Z – P) × 20 dB/decade


Where Z is the number of zeros and P is the number of poles.

Calculation:

Each pole adds a -20 dB/dec slope and each zero adds a +20 db/dec slope

Hence the overall slope at high frequency is given by:

Slope at high freq. = (-20 × No. of poles + 20 × No. of zeros) dB/dec

for P = 0 and Z = 3

⇒ Slope at high freq. = (-20 × 0 + 20 × 3) dB/dec

⇒ Slope at high freq. = 60 dB/decade

If the given system is connected to a unity negative feedback system, the steady-state error of a closed-loop system to a ramp input is;

F1 Neha Ravi 08.05.21 D9

  1. 0.01
  2. 1
  3. 0.5
  4. 0.2

Answer (Detailed Solution Below)

Option 3 : 0.5

Bode Plot Question 15 Detailed Solution

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Concept- 

For a unity feedback system with an open-loop transfer function G(s), the steady-state errors can be found identify the system type and using the respective formula:

For system type 0 : \(ess = \frac{1}{{1 + {K_p}}}\)

For system type 1 : \(ess = \frac{1}{{{K_v}}}\)

For system type 2 : \(ess = \frac{1}{{{K_a}}}\)

By identifying the system type from the open-loop Bode plot, the steady-state error can be easily found as follows-

F3 Neha B Shraddha 05.05.2021. D 2

 

F3 Neha B Shraddha 05.05.2021. D 3

 

F3 Neha B Shraddha 05.05.2021. D 4

From the given bode plot initial slope = \(\frac{{M\left( {j{\omega _2}} \right) - M\left( {j{\omega _1}} \right)}}{{\log {\omega _2} - \log {\omega _1}}}\)

\(slope = \frac{{ - 6.02}}{{\log 2 - \log 1}} = - 20\) dB / dec

So one pole present at the origin

Since it is a type 1 system so it will intersect Real Axis at kv

Kv = 2

\(ess = \frac{1}{{{K_v}}} = \frac{1}{2} = 0.5\)

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