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

Explanation:

Bode Plot Analysis for an Open Loop Transfer Function

Definition: A Bode plot is a graphical representation of the frequency response of a system. It consists of two plots: one showing the magnitude (in decibels) versus frequency, and the other showing phase versus frequency. The slope of the magnitude plot at high frequencies depends on the number of poles and zeros in the transfer function.

Problem Statement: The given open loop transfer function has 8 poles and 5 zeros. The question asks for the slope of the Bode plot at very high frequencies.

Solution:

The slope of the Bode plot at very high frequencies is determined by the difference between the number of poles (P) and zeros (Z) in the transfer function. The general formula for the slope is:

Slope (in dB/decade) = -20 × (P - Z)

Step-by-Step Calculation:

1. Given data:
   • Number of poles (P) = 8
   • Number of zeros (Z) = 5
2. Calculate the difference between the number of poles and zeros:
   • P - Z = 8 - 5 = 3
3. Apply the formula for the slope:
   • Slope = -20 × (P - Z) = -20 × 3 = -60 dB/decade

Conclusion: The slope of the Bode plot at very high frequencies is -60 dB/decade. Hence, Option 1 is correct

Concept:

The Cauchy Residue Theorem is a fundamental result in complex analysis and is used in control systems through the Nyquist criterion to relate the encirclements of a point in the complex plane to the poles and zeros of a function.

According to the theorem, for a closed contour in the complex plane enclosing singularities, we have:

\( \oint G(s) ds = 2\pi j (m - n) \)

Where,

m = number of poles of the function inside the contour

n = number of zeros of the function inside the contour

Calculation:

Given a transfer function G(s), its logarithmic derivative \(\frac{G'(s)}{G(s)}\) has poles at each zero and pole of G(s).

By applying the Cauchy Residue Theorem to this function around a closed contour, we get:

\( \oint \frac{G'(s)}{G(s)} ds = 2\pi j (m - n) \)

This gives the relationship used in the Nyquist stability criterion.

Concept:

The transfer function \( G(s) = e^{-Ts} \) represents a pure time delay of T seconds.

When analyzing this in the frequency domain for Bode plots, we substitute \( s = j\omega \):

\( G(j\omega) = e^{-j\omega T} \)

This is a complex exponential which affects only the phase of the frequency response, not the magnitude.

Magnitude:

\( |G(j\omega)| = |e^{-j\omega T}| = 1 \Rightarrow \text{Magnitude (in dB)} = 20\log_{10}(1) = 0~\text{dB} \)

So the magnitude plot is a horizontal straight line at 0 dB.

Phase:

\( \angle G(j\omega) = \text{arg}(e^{-j\omega T}) = -\omega T \)

Phase decreases linearly with frequency. Specifically, for T = 3 seconds:

\( \angle G(j\omega) = -3\omega \) radians

In degrees: \( \angle G(j\omega) = -3\omega \cdot \frac{180}{\pi} \angle G(j\omega) = -3\omega \cdot \frac{180}{\pi} \)

This is a straight line with a negative slope in the phase plot.

Given:

Transfer function: e-3s (pure delay of 3 seconds)

Conclusion:

• Magnitude plot: Constant at 0 dB (flat line)
• Phase plot: Straight line with a negative slope (due to the increasing delay with frequency)

Concept:

The Nyquist plot is used to assess the stability of a closed-loop control system using the open-loop transfer function \( G(s) \).

Given transfer function: \( G(s) = \frac{1}{s(s+2)} \)

Poles of G(s): \( s = 0 \) and \( s = -2 \) ⇒ Both in Left Half Plane (LHP), so open-loop system is stable.

According to the Nyquist criterion:

\( N = Z - P \), where:

• N = number of encirclements of point \( -1 + j0 \) (clockwise = negative, anticlockwise = positive)
• Z = number of right-half plane poles of closed-loop transfer function
• P = number of right-half plane poles of open-loop transfer function (here, P = 0)

Since the open-loop poles are in LHP, the Nyquist plot of \( G(s) \) does not encircle \( -1 + j0 \).

\( Z = 0 \Rightarrow N = 0 \)

Hence, the number of encirclements is: zero

Final Answer:  zero

Explanation:

Marginally Stable Systems

Definition: A marginally stable system is a type of dynamic system where the output oscillates indefinitely without converging to a steady-state value or diverging to infinity when subjected to certain initial conditions. In control systems, marginal stability occurs when the poles of the closed-loop transfer function lie on the imaginary axis of the s-plane but are not repeated (i.e., their multiplicity is 1).

Stability Criteria:

To understand the stability of a system, the location of the poles of its closed-loop transfer function is analyzed in the s-plane:

• If all poles are in the left half of the s-plane (real parts are negative), the system is stable.
• If any pole lies in the right half of the s-plane (real part is positive), the system is unstable.
• If poles lie on the imaginary axis (real part is zero) without repetition, the system is marginally stable.
• If poles lie on the imaginary axis with repetition (multiplicity greater than 1), the system is unstable.

Correct Option Analysis:

The correct option is:

Option 1: Marginally stable systems have closed-loop transfer functions with only imaginary axis poles of multiplicity 1.

To confirm this, let us analyze the characteristics of marginally stable systems:

1. For a system to be marginally stable, the poles of the closed-loop transfer function must lie entirely on the imaginary axis.
2. Each pole on the imaginary axis should have a multiplicity of exactly 1. This means that the poles are simple (non-repeated).
3. If any pole has a multiplicity greater than 1 on the imaginary axis, the system becomes unstable, as the response will grow indefinitely.

For example, consider a system with a transfer function:

H(s) = 1 / (s² + 1)

The poles of this system are at s = ±j. Since these poles lie on the imaginary axis and have a multiplicity of 1, the system is marginally stable. Any input to this system will result in a sustained oscillatory output without divergence or convergence.

Therefore, the correct answer is Option 1, as marginal stability requires the poles to be on the imaginary axis with a multiplicity of 1.

Additional Information

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

Option 2: Marginally stable systems have closed-loop transfer functions with only imaginary axis poles of multiplicity 3.

This option is incorrect. If a pole on the imaginary axis has a multiplicity of 3, the system becomes unstable. Higher multiplicities lead to instability because the response due to these poles grows indefinitely with time, which violates the condition of marginal stability.

Option 3: Marginally stable systems have closed-loop transfer functions with only imaginary axis poles of multiplicity 4.

This option is also incorrect. Similar to Option 2, a pole on the imaginary axis with a multiplicity of 4 would cause instability. The higher the multiplicity, the more pronounced the instability, as the system response becomes unbounded over time.

Option 4: Marginally stable systems have closed-loop transfer functions with only imaginary axis poles of multiplicity 2.

This option is incorrect. A pole on the imaginary axis with a multiplicity of 2 results in an unstable system. While the initial response may resemble marginal stability, the repeated pole causes the amplitude of oscillations to grow, leading to instability.

Conclusion:

In control systems, the concept of marginal stability is critical for analyzing systems that exhibit sustained oscillations without diverging or converging. The key characteristic of such systems is the presence of poles on the imaginary axis with a multiplicity of exactly 1. The analysis of the other options demonstrates that any pole with a multiplicity greater than 1 on the imaginary axis leads to instability, reinforcing the correctness of Option 1.

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

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

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.

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)