Frequency Response Analysis MCQ Quiz - Objective Question with Answer for Frequency Response Analysis - Download Free PDF

Explanation:

Bode Amplitude Plot for a First-Order Lowpass System

Definition: A first-order lowpass system is a type of linear system that allows low-frequency signals to pass through while attenuating high-frequency signals. The system is characterized by a single pole in its transfer function, and the Bode plot is used to graphically represent the frequency response of the system.

The Bode amplitude plot typically consists of two regions:

• A flat region at low frequencies (where the system behaves as a constant gain).
• A sloped region at high frequencies (where the system attenuates the signal).

Working Principle: The transfer function of a first-order lowpass system can be expressed mathematically as:

H(s) = K / (1 + s/ω_c)

Where:

• K is the DC gain of the system.
• s is the Laplace variable.
• ω_c is the cutoff angular frequency of the system.

The frequency response of this transfer function can be evaluated by substituting s = jω, where ω is the frequency in radians per second. The magnitude of the frequency response is given by:

|H(jω)| = K / √(1 + (ω/ω_c)²)

Bode Amplitude Plot Analysis:

• Low Frequencies: When ω << ω_c, the term (ω/ω_c)² becomes negligible, and the magnitude of the frequency response approximates to |H(jω)| ≈ K. On the Bode plot, this appears as a horizontal line parallel to the frequency axis with a constant gain in decibels.
• High Frequencies: When ω >> ω_c, the term (ω/ω_c)² dominates, and the magnitude of the frequency response approximates to |H(jω)| ≈ K / (ω/ω_c). In decibels, this corresponds to a slope of -20 dB/decade, indicating that the amplitude decreases by 20 dB for every tenfold increase in frequency.

Correct Option Analysis:

The correct option is:

Option 2: Line with slope -20 dB/decade.

This option correctly describes the behavior of the Bode amplitude plot of a first-order lowpass system. At frequencies much greater than the cutoff frequency, the amplitude decreases at a rate of -20 dB/decade, which is characteristic of a first-order system.

Additional Information

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

Option 1: Line with slope -40 dB/decade.

This description is incorrect. A slope of -40 dB/decade is characteristic of a second-order system, not a first-order system. In a second-order system, the amplitude decreases at a faster rate (two poles contribute to the attenuation), whereas in a first-order system, the slope is limited to -20 dB/decade.

Option 3: Line with slope +20 dB/decade.

This option is incorrect because it represents a system where the amplitude increases with frequency, which is not the behavior of a first-order lowpass system. A first-order lowpass system attenuates high-frequency signals rather than amplifying them.

Option 4: Straight line parallel to the frequency axis.

This option is partially correct but incomplete. While the amplitude plot of a first-order lowpass system is indeed a straight line parallel to the frequency axis at low frequencies, this description fails to account for the sloped region at higher frequencies where the amplitude decreases at a rate of -20 dB/decade.

Option 5: None of the above.

This option is incorrect because Option 2 accurately describes the behavior of the Bode amplitude plot for a first-order lowpass system.

Conclusion:

Understanding the frequency response and Bode plot characteristics of a first-order lowpass system is crucial for analyzing its behavior in various applications. The amplitude plot consists of a flat region at low frequencies and a sloped region with a slope of -20 dB/decade at high frequencies. This characteristic distinguishes first-order systems from higher-order systems, which exhibit steeper slopes in their amplitude plots.

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

The correct answer is: 1) Magnitude versus phase angle plot

Explanation:

A Bode plot is the most common tool for frequency response analysis. It consists of two graphs:

• Magnitude (in dB) vs. frequency (log scale)
• Phase angle vs. frequency (log scale)

Nyquist Plot 
A Nyquist plot represents the frequency response of a system in the complex plane, plotting real vs. imaginary parts of the transfer function as frequency varies.

Polar Plot 
A polar plot (similar to a Nyquist plot) displays magnitude and phase in polar coordinates, where the radial distance represents magnitude and the angle represents phase.

Magnitude vs. Phase Angle Plot 
While this plot shows magnitude and phase, it does not explicitly include frequency, making it unsuitable for frequency response analysis. Instead, it is more commonly used in Nichols charts (a variation for stability analysis).

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.

Explanation:

In control systems engineering, the phase margin is a measure of the stability of a system. It is defined as the amount of additional phase lag required to bring the system to the verge of instability. The phase margin is typically measured at the gain crossover frequency, which is the frequency where the magnitude of the open-loop transfer function is unity (0 dB).

The gain crossover frequency (ω_gc) is determined from the Bode plot, which is a graph of the frequency response of the system. The Bode plot consists of two separate plots: the magnitude plot and the phase plot. The phase margin is found using the phase plot at the gain crossover frequency.

Mathematically, the phase margin (PM) is given by:

PM = 180° + Φ(ω_gc)

where Φ(ω_gc) is the phase angle of the open-loop transfer function at the gain crossover frequency.

Given in the problem:

The phase at the gain crossover frequency (Φ(ω_gc)) is -125°.

Substituting the given value into the phase margin formula:

PM = 180° + (-125°)

Simplifying this:

PM = 180° - 125°

Therefore:

PM = 55°

The correct phase margin of the system is 55°, which corresponds to Option 1.

Important Information

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

Option 2: -55°

This option is incorrect because the phase margin cannot be negative. A negative phase margin would indicate that the system is unstable. In this problem, the correct calculation shows a positive phase margin of 55°.

Option 3: -125°

This option is the phase angle at the gain crossover frequency, not the phase margin. The phase margin is calculated by adding the phase angle to 180°. Therefore, -125° is not the correct phase margin.

Option 4: 125°

This option is incorrect because it represents the absolute value of the phase angle subtracted from 180°, which does not align with the phase margin calculation process. The correct phase margin is found by subtracting the absolute phase angle (125°) from 180°, resulting in 55°.

Conclusion:

In summary, the phase margin is a critical parameter in assessing the stability of a control system. It is determined by evaluating the phase angle at the gain crossover frequency and adjusting it relative to 180°. For the given problem, the phase margin of the system is 55°, indicating that the system has a stable margin before reaching instability. Understanding this concept is essential for designing and analyzing stable control systems.

Concept:

Minimum phase system: It is a system in which poles and zeros will not lie on the right side of the s-plane.  In particular, zeros will not lie on the right side of the s-plane.

For a minimum phase system,

\(\mathop {\lim }\limits_{\omega \to \infty } \angle G\left( s \right)H\left( s \right) = \left( {P - Z} \right)\left( { - 90^\circ } \right)\)

Where P & Z are finite no. of poles and zeros of G(s)H(s)

Non-minimum phase system: It is a system in which some of the poles and zeros may lie on the right side of the s-plane. In particular, zeros lie on the right side of the s-plane.

Stable system: A system is said to be stable if all the poles lie on the left side of the s-plane.

Application:

\(G\left( s \right) = \frac{{\left( {s - 1} \right)}}{{\left( {s + 2} \right)\left( {s + 3} \right)}}\)

As one zero lies in the right side of the s-plane, it is a non-minimum phase transfer function.

As there no poles on the right side of the s-plane, it is a stable system.

Explanation:

Nyquist plot:

1. Nyquist plots are an extension of polar plots for finding the stability of the closed-loop control systems. This is done by varying ω from −∞ to ∞, i.e. Nyquist plots are used to draw the complete frequency response of the open-loop transfer function.
2. In practice, it is not enough that a system is stable. There must also be some margins of stability that describe how stable the system is and its robustness. 
3. There are many ways to express this, but one of the most common is the use of gain and phase margins, inspired by Nyquist’s stability criterion. 
4. An increase in controller gain simply expands the Nyquist plot radially and an increase in the phase of the controller twists the Nyquist plot.
5. Hence from the Nyquist plot, we can easily pick off the amount of gain or phase that can be added without causing the system to become unstable. Hence option (3) is the correct answer.

​Important Points​

Method of drawing Nyquist plot:

• Locate the poles and zeros of the open-loop transfer function G(s)H(s) in the ‘s’ plane.
• Draw the polar plot by varying ω from zero to infinity.
• Draw the mirror image of the above polar plot for values of ω ranging from −∞ to zero.
• The number of infinite radii half circles will be equal to the number of poles at the origin.
• The infinite radius half-circle will start at the point where the mirror image of the polar plot ends. And this infinite radius half-circle will end at the point where the polar plot starts.

DC Gain:

The DC gain is the ratio of the magnitude of the steady-state step response to the magnitude of step input.

DC Gain of a system is the gain at the steady-state which is at t tending to infinity i.e., s tending to zero.

DC gain is nothing but the error coefficients.

For type 0 system: \({K_P} = \mathop {\lim }\limits_{s \to 0} G\left( s \right)\)

For type 1 system: \({K_v} = \mathop {\lim }\limits_{s \to 0} sG\left( s \right)\)

For type 2 system: \({K_a} = \mathop {\lim }\limits_{s \to 0} {s^2}G\left( s \right)\)

Analysis:

Given: 

CLTF = (S + 4) / (S2 + 7S + 13)

\(\frac{G(S)}{1+G(s) H(s)} = \frac{(s+4)}{s^2+7s+13}\)

\(\frac{G(S)}{1+G(s) H(s)-G(s)} = \frac{(s+4)}{s^2+7s+13-s-4}\)

H(s) = 1

\(G(s) = \frac{(s+4)}{s^2+6s+9}\)

DC gain = \( \mathop {\lim }\limits_{s \to 0} G\left( s \right)\)

= 4/9

Concept:

Each zero at origin gives a phase shift of + 90°

For N zeros at origin gives a phase shift of + 90N°

Each pole at origin gives a phase shift of - 90°

For N poles at origin gives a phase shift of - 90N°

Calculation:

The given transfer function = 1/s²

Number of poles at origin = 2

Phase shift = 2 × (- 90°) = - 180°

Additional Information

Minimum phase system: It is a system in which poles and zeros will not lie on the right side of the s-plane.

For a minimum phase system:

\(\mathop {\lim }\limits_{\omega \to \infty } \angle G\left( s \right)H\left( s \right) = \left( {P - Z} \right)\left( { - 90^\circ } \right)\)

Where P & Z are finite no. of poles and zeros of G(s)H(s)

Non-minimum phase system: It is a system in which some of the poles and zeros may lie on the right side of the s-plane.

In particular, zeros lie on the right side of the s-plane.

All pass system:  An all-pass network is the network that is the combination of the minimum and non-minimum phase systems and have the unity magnitude for all

frequencies and imparts only a 180-degree phase shift. The pole and zero of an all-pass filter are at the same distance from the origin. 

\(\;\sin \omega t \to \boxed{G\left( s \right)} \to Y\left( s \right) = \left| {G\left( s \right)} \right| \cdot \sin \left( {\omega t + \angle G\left( s \right)} \right)\)

Output will be zero when

|G(s)| = 0

Put s = jω

\(\left| {\frac{{\left( { - {\omega ^2} + 9} \right)\left( {j\omega + 2} \right)}}{{\left( {j\omega + 1} \right)\left( {j\omega + 3} \right)\left( {j\omega + 4} \right)}}} \right| = 0\)

at         ω = 3   , |G(jω)| = 0

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

Concept:

\(G.M. = {\frac{1}{{\left| {G\left( {j{\rm{\omega }}} \right)H\left( {j{\rm{\omega }}} \right)} \right|}}_{{\rm{\omega \;}} = {\rm{\;\omega pc}}}}\)

\(So,\;G.M. = \frac{1}{{Gain}}\)

Conclusion:

It is given that Gain is doubled, i.e. (G)_New = 2(G)_Old

\(\therefore {\left( {G.M} \right)_{New}} = \frac{1}{2}{\left( {G.M.} \right)_{Old}}\)

Explanation:

1. Gain margin (GM): The gain margin of the system defines by how much the system gain can be increased so that the system moves on the edge of stability.

2. It is determined from the gain at the phase cross-over frequency.

\(GM = \frac{1}{{{{\left| {G\left( {j\omega } \right)H\left( {j\omega } \right)} \right|}_{\omega = {\omega _{pc}}}}}}\)

3. Phase crossover frequency (ωpc): It is the frequency at which the phase angle of G(s) H(s) is -180°.

\(\angle G\left( {j\omega } \right)H\left( {j\omega } \right){|_{\omega = {\omega _{pc}}}} = - 180^\circ \)

4. Phase margin (PM): The phase margin of the system defines by how much the phase of the system can increase to make the system unstable.

\(PM = 180^\circ + \angle G\left( {j\omega } \right)H\left( {j\omega } \right){|_{\omega = {\omega _{gc}}}}\)

5. It is determined from the phase at the gain cross over frequency.

6. Gain crossover frequency (ωgc): It is the frequency at which the magnitude of G(s) H(s) is unity.

\({\left| {G\left( {j\omega } \right)H\left( {j\omega } \right)} \right|_{\omega = {\omega _{gc}}}} = 1\)

7. So it is important to note that these margins of stability are valid for open-loop stable systems only.

8. A large gain margin or a large phase margin indicates a very stable feedback system but usually a very sluggish response.

9. Hence a Gain margin close to unity (or 0 in dB) or a phase margin close to zero corresponds to a highly oscillatory system.

Hence option (3) is the correct answer.

Gain margin and phase margin are frequently used for frequency response specifications by designers. Usually a Gain margin of about 6 dB or a Phase margin of 30 - 35° results in a reasonably good degree of relative stability.

Important Points

• If both GM and PM are positive, the system is stable (ωgc < ωpc)
• If both GM and PM are negative, the system is unstable (ωgc > ωpc)
• If both GM and PM are zero, the system is just stable (ωgc = ωpc)

The type of system is the number of poles present at the origin in the open-loop transfer function.

In Bode plot,

• The initial slope for each pole at origin is -20 dB/decade or -6 dB/octave.
• The initial slope for each zero at origin is +20 dB/decade or +6 dB/octave.
• The initial slope is zero if there are no poles or zeros at the origin.

The initial shape of the Bode plot gives an indication of the type of the system.