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

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

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 = 10^0.3

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.

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.

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.

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

Calculation

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, ω_c = 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:

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

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

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

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

Given circuit,

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|M_r| = 26 dB

|M_r| = 19.95

19.95 × 0.5R = 1

R = 0.1 Ω

Alternate Method

Resonant Peak formula,

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

M_r = 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

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:

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

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

Concept:

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

​

• 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).

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

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:

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

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-

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 k_v

K_v = 2

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