Time Domain Specifications MCQ Quiz - Objective Question with Answer for Time Domain Specifications - Download Free PDF

A second-order control system is characterized by a second-order differential equation. It typically consists of two poles and is widely used in engineering applications to model dynamic systems. When a step input is applied to such a system, its response is analyzed in terms of parameters like damping ratio (𝛿), natural frequency (𝜔n), overshoot, settling time, rise time, and oscillations.

Correct Option Analysis:

The correct option is:

Option 2: For critical damping i.e., 𝛿 = 1, there are oscillations with dying down amplitudes.

This statement is FALSE. In a critically damped system (𝛿 = 1), the system does not exhibit oscillations. Instead, it returns to equilibrium as quickly as possible without overshooting or oscillating. A critically damped system is designed to achieve the fastest response time to a step input without oscillations. Therefore, the assertion that there are "oscillations with dying down amplitudes" for 𝛿 = 1 is incorrect.

Additional Information

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

Option 1: Damping ratio 𝛿 = 0 will give sustained oscillations.

This statement is TRUE. When the damping ratio (𝛿) is zero, the system is undamped, and it will exhibit sustained oscillations. This is because there is no energy dissipation in the system, and the response oscillates indefinitely at the natural frequency (𝜔n).

Option 3: For an overdamped system i.e., 𝛿 > 1, there is no oscillation.

This statement is TRUE. In an overdamped system, the damping ratio (𝛿) is greater than 1. The system response is sluggish, and it approaches equilibrium without oscillations. Overdamping ensures that the system does not overshoot or oscillate, but it takes longer to settle compared to a critically damped system.

Option 4: Settling time is inversely related to the damping ratio.

This statement is TRUE. Settling time, which is the time required for the system to settle within a specific percentage of its final value, is inversely related to the damping ratio for underdamped systems (𝛿 < 1). As the damping ratio increases, the oscillations decrease, and the system settles faster. However, this relationship is not valid for overdamped systems (𝛿 > 1).

Concept:

The standard second-order system has the characteristic equation:

\( s^2 + 2\zeta\omega_n s + \omega_n^2 = 0 \)

The roots of this equation determine the nature of the system (overdamped, underdamped, critically damped).

Condition for Complex Conjugate Roots:

Roots are complex conjugates if the discriminant is negative:

\( (2\zeta\omega_n)^2 - 4\omega_n^2 < 0 \Rightarrow \zeta^2 < 1 \)

So, complex roots exist when:

\( 0 \leq \zeta < 1 \)

Such systems are called underdamped systems.

Therefore,  the correct answer is   \( 0 \leq \zeta < 1 \) 

Concept:

The time constant (τ) of an RC circuit is given by the formula:

τ = R_eqC

where R_eq is the equivalent resistance seen by the capacitor, and C is the capacitance.

While finding out the time constant the voltage source will be shorted.

Calculation:

Given:

Resistors: 3Ω and 2Ω will be in parallel
Capacitance: C = 10 F

Solution:

If the resistors are actually in parallel (common in RC circuits):

R_eq = (3Ω × 2Ω)/(3Ω + 2Ω) = 6/5Ω = 1.2Ω

τ = R_eqC = 1.2Ω × 10 F = 12 sec

Final Answer:

The correct time constant is 1) 12 sec when considering parallel resistors.

Concept:

\(G(s) = \frac{{C\left(s \right)}}{{R\left( s \right)}}= \frac{1}{{s^2} + 2s + 1}\)

\(C\left( s \right) = \frac{1}{{\left( {{s^2} + 2s + 1} \right)}} \times \frac{1}{s}\)

\(C\left( s \right) = \frac{1}{{s{{\left( {s + 1} \right)}^2}}}\)

\(C\left( s \right) = \frac{A}{s} + \frac{B}{{s + 1}} + \frac{C}{{{{\left( {s + 1} \right)}^2}}}\)

A = 1

B = -1

C = -1

\(C\left( s \right) = \frac{1}{s} - \frac{1}{{s + 1}} - \frac{1}{{{{\left( {s + 1} \right)}^2}}}\)

Applying inverse Laplace transform:

C(t) = 1 – e^-t – te^-t

at steady state i.e. at t → ∞

\(\mathop {{\rm{lt}}}\limits_{t \to \infty } C\left( t \right) = 1 - {e^{ - \infty }} - t{e^{ - \infty }}\)

C(t) = 1

94% of steady state \( = \frac{{94}}{{100}} \times 1\) 

= 0.94.

0.94 = 1 – e^-t – te^-t

Substitute all options

Let us substitute options

option (a) t= 5.25

1 – e^-5.25 – 5.25 e^-5.25

\(\Rightarrow 1 - \left( {\frac{{1 + 5.25}}{{{e^{5.25}}}}} \right)\)

= 0.967

option (b)

t = 4.50

\(1 - \frac{{\left( {1 + 4.50} \right)}}{{({e^{4.50}})}}\)

1 – 0.938

Concept

A second-order transfer function is represented as:

\({C(s)\over R(s)}={ω_n^2\over s^2+2ζω_ns+ω_n^2}\)

where, ω_n = Natural frequency

ζ = Damping ratio

The nature of damping with respect to the damping ratio is:

Calculation

Given, \({C(s)\over R(s)}={20\over s^2+8s+20}\)

\(ω_n^2=20\)

ω_n =4.472 rad/s

2ζω_n = 8

\(\zeta={8\over 2\times 4.472}\)

ζ = 0.894

Hence, the system is underdamped.

Concept:

Damped natural frequency of a second order system is given by

\({ω _d} = {ω _n}\sqrt {1 - {\xi ^2}} \)

Where,

ωn = Natural frequency

ζ = Damping ratio

Calculation:

Given-

ωn = 11 rad/sec

ζ = 0.6

Now, natural damped frequency can be calculated as

\({ω _d} = {11}\sqrt {1 - {(0.6)^2}} \)

ωd = 11 x 0.8

ωd = 8.8 rad/sec  

Concept:

The nature of the control system is defined on the basis of the value of the damping ratio.

Concept: 

The transfer function of the standard second-order system is:

\(TF = \frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{ω _n^2}}{{{s^2} + 2ζ {ω _n}s + ω _n^2}}\)

Characteristic equation: \({s^2} + 2ζ {ω _n} + ω _n^2 = 0\)

ζ is the damping ratio

ωn is the undamped natural frequency

\({M_p} = {e^{\frac{{ - ζ \pi }}{{\sqrt {1 - {ζ ^2}} }}}}\)   ----(1)

Calculation:

Given:

Mp = 100%

From the above equation,

\({M_p} = {e^{\frac{{ - ζ \pi }}{{\sqrt {1 - {ζ ^2}} }}}}\)

\(ln\;1 = \frac{{ - \zeta \pi }}{{\sqrt {1 - {\zeta ^2}} }}\) ; (ln 1 = 0)

So, ζ = 0

Note:

Mp is the maximum peak overshoot of the closed-loop transfer function

\(M_p \ \alpha \ \frac{1}{ζ}\)

Concept:

First-order system:

The transfer function of the standard first-order system is given by

\(TF = \frac{{k}}{{\left( {1 + τ s} \right)}}\)

Where,

τ = time constant of the system

The time constant can be defined as the negative reciprocal of the pole of the system.

Calculation:

From the given figure,

\(\frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{\frac{{3K}}{{2s + 1}}}}{{1 + \frac{{3K}}{{2s + 1}}}}\)

\( = \frac{{3K}}{{2s + 3K + 1}}\)

\( = \frac{{3K}}{{3K + 1}} \times \frac{1}{{1 + \frac{{2s}}{{3K + 1}}}}\)

From the above transfer function, the time constant is

\(T = \frac{2}{{3K + 1}}\)

The time constant of the system is less than 0.2

\( \Rightarrow \frac{2}{{3K + 1}} < 0.2\;\)

⇒ K > 3   

Concept:

The general expression of the transfer function of the standard second-order system is:

\(TF = \frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{ω _n^2}}{{{s^2} + 2ζ {ω _n}s + ω _n^2}}\)

Where,

ζ is the damping ratio

ωn is the undamped natural frequency

Characteristic equation: \({s^2} + 2ζ {ω _n} + ω _n^2 = 0\)

The roots of the characteristic equation are: 

\(- ζ {ω _n} + j{ω _n}\sqrt {1 - {ζ ^2}} = - \alpha \pm j{ω _d}\)

α is the damping factor

Calculation:

Given, the characteristic equation s2 + 6s + 25 = 0

Comparing this with the general expression of the transfer function of the second-order control system, we can write:

2 ζ ω_n = 6

⇒ ζ ω_n = 3

Additional Information

 The nature of the system is described by its ‘ζ’ value

• The transfer function for a negative closed-loop is given by:

\(T.F. = {G(s) \over 1+G(s)H(s)}\)

where G(s) = Forward path gain

H(s) = Feedback path gain

• Due to the presence of a feedback path, the gain of the transfer function decreases.
• As the gain-bandwidth product must be constant, hence bandwidth increases for decreasing value of gain.
• As the transient response of the system is improved by the use of feedback and it causes the settling time to reduce and the closed-loop system has higher bandwidth than open-loop systems and this implies an increase in the speed of response.

Concept:

Time-domain specification (or) transient response parameters:

Peak Time (tp): It is the time taken by the response to reach the maximum value.

\({\left. {\frac{{dc\left( t \right)}}{{dt}}} \right|_{t = {t_p}}} = 0,{\text{}}{t_p} = \frac{\pi }{{{ω _d}}}\)

\(As \ \omega_d=\omega_n\sqrt{ 1-\xi^2}\)

\(=\frac{\pi }{{{ω _n}√ {1 - {\zeta ^2}} }}\)

Where,

ξ = damping ratio

ω_n = natural frequency

ωd = damped frequency

Calculation:

Given:

ωn = 13 rad/sec

ξ = 0.8

\(\omega_d=13\sqrt{ 1-0.8^2}\)

ωd = 7.8 rad/sec.

\(t_p=\frac{\pi}{7.8}=0.4 \;sec\)

Hence option(4) is the correct answer.

Concept:

The transfer function of a 2^nd order is given by:

\(H(s) = {ω^2_n \over s^2+\space2ζω_ns+\space ω^2_n }\)

where, ω_n = Natural frequency

ζ = Damping ratio

Calculation:

Given, \(H(s) = {25 \over s^2\space+\space8s\space+\space 25 }\)

ω_n² = 25

ω_n = 5

2ζω_n = 8

\(ζ = {8 \over 2\times\omega_n}\)

\(ζ = {8 \over 2\times5}\)

ζ = 0.8

Concept:

The transfer function of the standard second-order system is:

\(TF = \frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{\omega _n^2}}{{{s^2} + 2\zeta {\omega _n}s + \omega _n^2}}\)

ζ is the damping ratio

ωn is the natural frequency

Characteristic equation: \({s^2} + 2\zeta {\omega _n} + \omega _n^2\)

The roots of the characteristic equation are: \( - \zeta {\omega _n} + j{\omega _n}\sqrt {1 - {\zeta ^2}} = - \alpha \pm j{\omega _d}\)

α is the damping factor

Settling time (Ts): It is the time taken by the response to reach ± 2% tolerance band as shown in the fig above.

\({e^{ - \xi {\omega _n}{t_s}}} = \pm 5\% \;\left( {or} \right) \pm 2\% \)

 \({t_s} \simeq \frac{3}{{\xi {\omega _n}}}\) for a 5% tolerance band.

\({t_s} \simeq \frac{4}{{\xi {\omega _n}}}\) for 2% tolerance band

Calculation:

\(G(s) = \frac{4}{{{s^2} + 0.4s}}\)

By comparing the above transfer function with the standard second-order system,

\(\omega _n^2 = 4 \Rightarrow {\omega _n} = 2\)

⇒ 2ζωn = 0.4 ⇒ ζ = 0.1

As the damping ratio is less than unity, the system is underdamped.

Settling time, \({t_s} = \frac{4}{{\zeta {\omega _n}}} = \frac{4}{{0.1 \times 2}} = 20\;sec\)

\(G(s) = \frac{{C\left(s \right)}}{{R\left( s \right)}}= \frac{1}{{s^2} + 2s + 1}\)

\(C\left( s \right) = \frac{1}{{\left( {{s^2} + 2s + 1} \right)}} \times \frac{1}{s}\)

\(C\left( s \right) = \frac{1}{{s{{\left( {s + 1} \right)}^2}}}\)

\(C\left( s \right) = \frac{A}{s} + \frac{B}{{s + 1}} + \frac{C}{{{{\left( {s + 1} \right)}^2}}}\)

A = 1

B = -1

C = -1

\(C\left( s \right) = \frac{1}{s} - \frac{1}{{s + 1}} - \frac{1}{{{{\left( {s + 1} \right)}^2}}}\)

Applying inverse laplace transform:

C(t) = 1 – e^-t – te^-t

at steady state i.e. at t → ∞

\(\mathop {{\rm{lt}}}\limits_{t \to \infty } C\left( t \right) = 1 - {e^{ - \infty }} - t{e^{ - \infty }}\)

C(t) = 1

94% of steady state \( = \frac{{94}}{{100}} \times 1\) 

= 0.94.

0.94 = 1 – e^-t – te^-t

Substitute all options

Let us substitute options

option (a) t= 5.25

1 – e^-5.25 – 5.25 e^-5.25

\(\Rightarrow 1 - \left( {\frac{{1 + 5.25}}{{{e^{5.25}}}}} \right)\)

= 0.967

option (b)

t = 4.50

\(1 - \frac{{\left( {1 + 4.50} \right)}}{{({e^{4.50}})}}\)

1 – 0.938