Peak Overshoot MCQ Quiz - Objective Question with Answer for Peak Overshoot - 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 maximum peak overshoot in the time domain is given by:

\(M_p=e^{-({\piζ\over \sqrt{1-ζ^2}})}\)

Resonant frequency (ω_r): It is the frequency at which maximum magnitude occurs.

For maximum frequency:

\({dM_p\over dζ }=0\)

\((-\pi)e^{-({\piζ\over \sqrt{1-ζ^2}})}[{\sqrt{1-ζ^2(1)}\space -ζ({1\over 2})({2ζ}\sqrt{1-ζ^2})\over (1-ζ^2)}]=0\)

\({\sqrt{1-ζ^2}\space ={{ζ^2}\over \sqrt{1-ζ^2}}}\)

1 - ζ² = ζ²

ζ = 0.707

Solution

Concept

For a 2nd order underdamped system, the maximum  percentage overshoot is given by

M_p%= ^{\(e ^\frac{-\pi ζ }{\sqrt {1- ζ^2}}\)} × 100 %

here ζ is the damping ratio

Calculation 

Given, M_p% =10%

⇒\(\frac{10}{100}\) = \(e ^\frac{-\pi ζ }{\sqrt {1- ζ^2}}\) 

⇒0.1=\(e ^\frac{-\pi ζ }{\sqrt {1- ζ^2}}\)

⇒ln(0.1)= \(\frac{-\pi ζ }{\sqrt {1- ζ^2}}\)

⇒-2.302= \(\frac{-\pi ζ }{\sqrt {1- ζ^2}}\)

⇒0.733=\(\frac{ζ }{\sqrt {1- ζ^2}}\)

Squaring and cross multipicating we get

1=2.8596 ζ²

ζ =0.59 ≈0.60

Hence the damping ratio is 0.6

Therefore the correct option is 2

Solution

OLTF is given by  G (s) = \(\frac{k}{(s+4)}\)

Cascade compensator GC(s) = \(\frac{s+α}{s}\)

Hence overall OLTF of the new compensated system will be

⇒G(s)GC(s) = \(\frac{k}{(s+4)}\) × \(\frac{s+α}{s}\) = \(\frac {k(s+α)}{s(s+4)}\)

The CLTF will be =\(\frac {k(s+α)}{s^2+s(4+α)+kα}\)

Given the peak overshoot M%=20%

⇒Mp%= \(e ^\frac{-\pi ζ }{\sqrt {1- ζ^2}}\) × 100 %

⇒0.2=\(e ^\frac{-\pi ζ }{\sqrt {1- ζ^2}}\)

⇒ln(0.2) = \(e ^\frac{-\pi ζ }{\sqrt {1- ζ^2}}\)

⇒0.5127 =\(\frac{ ζ }{\sqrt {1- ζ^2}}\)

Squaring and calculating we get

⇒ζ =0.456

Comparing denominator of CLTF with standard result we have

⇒s² + s(4+α) + kα = s2 + 2ζω_n s + ω_n²

We have

⇒2ζωn =(4+α) and ⇒ωn2 = kα 

⇒ω =\(\frac{(4+k)}{0.912}\)  and ⇒ωn2 = kα 

We have 2 unknowns and 1 equation ,so will use the option elimination method to calculate the value of  K and α

Checking for the options 

1. k = 16, α = 29.23 ,

putting k=16, and calculating α

ω =21.93 , ⇒α =30

2. k = 4, α = 19.23

putting k=4, and calculating α

ω =8.767, α = 19.23

3. k = 19.23, ⇒α = 4

putting k= 19.23, and calculating α

ω= 25.47 ,⇒ α =33.73

4. k = 8, α = 4

putting  k=8, and calculating α

ω =13.15 ,⇒α =21.62

Calculation:

Given 

\(\rm G_1(s)=\frac{10}{s^2+s+1} \ \ \)  &  \(\rm G_2(s)=\frac{10}{s^2+s√{10}+10} \ \ \)

Comparing standard second-order equation 

S2 + 2 ζ wn S + Wn2 = 0

Wn1 = 1,  2 ζ1 Wn1  = 1

ζ1 = 0.5

Wn2 = √ 10

2 ζ2 Wn2 = √ 10

ζ2 = 0.5

Percentage peak overshoot:

% Mp = \(\dfrac {e^{-\pi ζ }}{\sqrt {1-ζ ^2}}\ \ \)

Percentage peak overshoot is depend only on ζ 

ζ1 = ζ2 

Therefore % Mp is the same as for both y1(t) and y2(t)         (option 1 is correct)

Steady-state value: 

  \(lim _{s→ 0}~~sY(s)\ \ \)

\(\rm Y_1(s)=\frac{10}{s(s^2+s+1)} \ \ \) and  \(\rm Y_2(s)=\frac{10}{s(s^2+s√{10}+10)} \ \ \)

Steady-state value of y1(t) →  10

Steady-state value of y1(t) →  1 

Therefore the steady-state values of both systems are different          (option 2 is incorrect)

Damped frequency of oscillation (Wd): 

  \(W_d=W_n\sqrt {(1-ζ ^2}\ \ \ \ \)
\(W_{d1}=W_{n1} \sqrt {1-ζ_1 ^2}\ \ \ \ \)
\(W_{d1}=1 \sqrt {1-\dfrac {1}{4}}\ \ =\dfrac {\sqrt 3}{2}= 0.86\ \ \ \ \)
\(W_{d2}=\sqrt{10} \sqrt {1-\dfrac {1}{4}}\ \ =\sqrt {10} \times \dfrac {\sqrt 3}{2}= 2.73\ \ \ \ \)

Wd1 ≠ Wd2        (option 2 is incorrect)

Settling time (ts):

For 2% tolerance band 

\(t_s=\dfrac{4}{ζ W_n}\ \ \)

By putting values of ζ and Wn

\(t_{s1}= \dfrac {4}{\dfrac {1}{2}\times 1}=8 \ \ sec\ \ \ \ \)

\(t_{s2}= \dfrac {4}{\dfrac {1}{2}\times \sqrt {10}}=2.53\ \ sec\ \ \space \)

ts1 ≠ ts2     (option 4 is incorrect)

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

\(M_p \ \alpha \ \frac{1}{ζ}\)

Peak over-shoot \( = {{\rm{e}}^{ - \frac{{{\rm{\pi \xi }}}}{{\sqrt {1 - {{\rm{\xi }}^2}} }}}}\) 

Peak over-shoot is maximum for minimum damping ratio (ξ).

Standard second order transfer function is

\(= \frac{{{\rm{\omega }}_{\rm{n}}^2}}{{{{\rm{s}}^2} + 2{\rm{\xi }}{{\rm{\omega }}_{\rm{n}}}{\rm{s}} + {\rm{\omega }}_{\rm{n}}^2}}\) 

1) \(\frac{{100}}{{{{\rm{s}}^2} + 10{\rm{s}} + 100}}\)

\({\rm{\omega }}_{\rm{n}}^2 = 100 \Rightarrow {{\rm{\omega }}_{\rm{n}}} = 10\) 

\(2{\rm{\xi }}{{\rm{\omega }}_{\rm{n}}} = 10 \Rightarrow {\rm{\xi }} = \frac{{10}}{{2 \times 10}} = 0.5\) 

2) \(\frac{{100}}{{{{\rm{s}}^2} + 15{\rm{s}} + 100}}\)

\({\rm{\omega }}_{\rm{n}}^2 = 100 \Rightarrow {{\rm{\omega }}_{\rm{n}}} = 10\) 

\(2{\rm{\xi }}{{\rm{\omega }}_{\rm{n}}} = 15 \Rightarrow {\rm{\xi }} = \frac{{15}}{{2 \times 10}} = 0.75\) 

3) \(\frac{{100}}{{{{\rm{s}}^2} + 5{\rm{s}} + 100}}\)

\({\rm{\omega }}_{\rm{n}}^2 = 100 \Rightarrow {{\rm{\omega }}_{\rm{n}}} = 10\) 

\(2{\rm{\xi }}{{\rm{\omega }}_{\rm{n}}} = 5 \Rightarrow {\rm{\xi }} = \frac{5}{{2 \times 10}} = 0.25\)

3) \(\frac{{100}}{{{{\rm{s}}^2} + 20{\rm{s}} + 100}}\)

\({\rm{\omega }}_{\rm{n}}^2 = 100 \Rightarrow {{\rm{\omega }}_{\rm{n}}} = 10\) 

\(2{\rm{\xi }}{{\rm{\omega }}_{\rm{n}}} = 20 \Rightarrow {\rm{\xi }} = \frac{{20}}{{2 \times 10}} = 1\) 

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

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

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

\(M_p \ \alpha \ \frac{1}{ζ}\)

Calculation:

Option A:

 \(T(s)=\frac{9}{(s^2+2s+9)}\)

On Solving we'll get:

ω_n = 2

ζ = 0.33

Option B:

\(T(s)=\frac{16}{(s^2+2s+16)}\)

On Solving we'll get:

ωn = 4

ζ = 0.25

Option C:

\(T(s)=\frac{25}{(s^2+2s+25)}\)

On Solving we'll get:

ωn = 5

ζ = 0.2

Option D:

\(T(s)=\frac{36}{(s^2+2s+36)}\)

On Solving we'll get:

ωn = 6

ζ = 0.16

Thus, Option 4 has the lowest ζ, So it has maximum Overshoot.

Concept:

Peak overshoot: It denotes the normalized difference between the steady-state output to the first peak of the time response. It is given as:

\({{\rm{M}}_{\rm{p}}} = {\rm{\;}}{{\rm{e}}^{ - \frac{{{\rm{ξ \pi }}}}{{\sqrt {1 - {{\rm{ξ }}^2}} }}}} \times 100\)

where ξ : damping factor.

Calculation:

The open-loop transfer function is given as:

\({\rm{G}}\left( {\rm{s}} \right) = \frac{{\rm{K}}}{{{\rm{s}}\left( {{\rm{s}} + 2} \right)}}\)

Characteristics equation is \({{\rm{s}}^2}{\rm{\;}} + {\rm{\;}}2{\rm{s}} + {\rm{\;K\;}} = 0{\rm{\;}}\)

Compare with the standard equation we get

\(2{\rm{ξ \;}}{{\rm{\omega }}_{\rm{n}}} = 2{\rm{\;and\;\omega }}_{\rm{n}}^2{\rm{\;}} = {\rm{\;K\;}}\)

\( {\rm{\;}}{{\rm{\omega }}_{\rm{n}}}{\rm{\;}} = {\rm{\;}}\sqrt {\rm{k}}\)

\(\therefore {\rm{\;\;ξ }} = \frac{1}{{\sqrt {\rm{k}} }}\)

Given peak overshoot as:

\({{\rm{M}}_{\rm{p}}} = {\rm{\;}}10{\rm{\% \;}} = {\rm{\;}}{{\rm{e}}^{ - \frac{{{\rm{ξ \pi }}}}{{\sqrt {1 - {{\rm{ξ }}^2}} }}}} \times 100\)

\( \ln \left( {\frac{{10}}{{100}}} \right) = \frac{{ - {\rm{ξ \pi }}}}{{\sqrt {1 - {{\rm{ξ }}^2}} }}\)

\(-2.3 = \frac{{ - {\rm{ξ \pi }}}}{{\sqrt {1 - {{\rm{ξ }}^2}} }}\)

\(\frac{{{{\rm{ξ }}^2}}}{{1 - {{\rm{ξ }}^2}}} = 0.537\)

\( {\rm{ξ }} = 0.6\)

\( {\rm{K}} = \frac{1}{{{{\rm{ξ }}^2}}} = 2.86\)

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:

ζ = 0

From Equation (1);

M_p = 1 i.e.

Mp = 100%

Note:

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

\(M_p \ \alpha \ \frac{1}{ζ}\)

Solution

Concept

For a 2nd order underdamped system, the maximum  percentage overshoot is given by

M_p%= ^{\(e ^\frac{-\pi ζ }{\sqrt {1- ζ^2}}\)} × 100 %

here ζ is the damping ratio

Calculation 

Given, M_p% =10%

⇒\(\frac{10}{100}\) = \(e ^\frac{-\pi ζ }{\sqrt {1- ζ^2}}\) 

⇒0.1=\(e ^\frac{-\pi ζ }{\sqrt {1- ζ^2}}\)

⇒ln(0.1)= \(\frac{-\pi ζ }{\sqrt {1- ζ^2}}\)

⇒-2.302= \(\frac{-\pi ζ }{\sqrt {1- ζ^2}}\)

⇒0.733=\(\frac{ζ }{\sqrt {1- ζ^2}}\)

Squaring and cross multipicating we get

1=2.8596 ζ²

ζ =0.59 ≈0.60

Hence the damping ratio is 0.6

Therefore the correct option is 2

Concept:

Time response of the second-order system

The second-order system nature completely depends on ζ 

The second-order system is stable for all the positive values of ζ < ∞ and ζ > 0 because poles lie in the left half of s-plane.

The impulse response is given by:

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

The generalized response for the value of 0 < ζ < 1 is defined below;

\(c\left( t \right) = 1 - \frac{{{e^{ - \zeta {\omega _n}t}}}}{{\sqrt {1 - {\zeta ^2}} }}\sin \left( {\left( {{\omega _n}\sqrt {1 - {\zeta ^2}} } \right)t + {{\tan }^{ - 1}}\left( {\frac{{\sqrt {1 - {\zeta ^2}} }}{\zeta }} \right)} \right)\)

The relation between maximum overshoot and ζ is defined as:

\({M_p} = A{K_{CL}}{e^{ - \frac{{n\pi \zeta }}{{\sqrt {1 - {\zeta ^2}} }}}}\)

K_CL: Closed-loop gain

From the response in the concept

1^st peak occurs at n = 1

1^st undershoot occurs at n = 2

2^nd  peak occurs at n = 3

2^nd  undershoot occurs at n = 4

3^rd  peak occurs at n = 5

3^rd  undershoot occurs at n = 6

Option (D) is correct.

Concept:

Setting time: It is the time required for response to the rise and reaches to the tolerance band. It is given as:

For a 2% tolerance band:

\({t_s} = 4\tau = \frac{4}{{\xi {\omega _n}}}\)

For a 5% tolerance band:

\({t_s} = 3\tau = \frac{3}{{\xi {\omega _n}}}\)

Peak overshoot: It denotes the normalized difference between the steady-state output to the first peak of the time response. It is given as:

\(\% MP = {e^{ - \left( {\frac{{\pi \xi }}{{\sqrt {1 - {\xi ^2}} }}} \right)}} \times 100\% \)

Calculation:

Given:

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

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

⇒ Maximum setting time and no peak overshoot is obtained when ξ = ξ_max = 1

So, it is a critically damped system.

Now,

Characteristic equation: 1+G(s)H(s)= s²+2s+K

Compare it with a standard Characteristic equation: s²+2ξω_n+ω_n²

We will get,

\({\omega _n} = \sqrt K \;\;\;\& \;\;\;\;2\xi {\omega _n} = 2\)

\( \Rightarrow {\omega _n} = 1\;\;\;\;\;\;\therefore \;\xi = 1\)   

\(\Rightarrow K = 1\)

Concept:

The maximum peak overshoot in the time domain is given by:

\(M_p=e^{-({\piζ\over \sqrt{1-ζ^2}})}\)

Resonant frequency (ω_r): It is the frequency at which maximum magnitude occurs.

For maximum frequency:

\({dM_p\over dζ }=0\)

\((-\pi)e^{-({\piζ\over \sqrt{1-ζ^2}})}[{\sqrt{1-ζ^2(1)}\space -ζ({1\over 2})({2ζ}\sqrt{1-ζ^2})\over (1-ζ^2)}]=0\)

\({\sqrt{1-ζ^2}\space ={{ζ^2}\over \sqrt{1-ζ^2}}}\)

1 - ζ² = ζ²

ζ = 0.707

Solution

OLTF is given by  G (s) = \(\frac{k}{(s+4)}\)

Cascade compensator GC(s) = \(\frac{s+α}{s}\)

Hence overall OLTF of the new compensated system will be

⇒G(s)GC(s) = \(\frac{k}{(s+4)}\) × \(\frac{s+α}{s}\) = \(\frac {k(s+α)}{s(s+4)}\)

The CLTF will be =\(\frac {k(s+α)}{s^2+s(4+α)+kα}\)

Given the peak overshoot M%=20%

⇒Mp%= \(e ^\frac{-\pi ζ }{\sqrt {1- ζ^2}}\) × 100 %

⇒0.2=\(e ^\frac{-\pi ζ }{\sqrt {1- ζ^2}}\)

⇒ln(0.2) = \(e ^\frac{-\pi ζ }{\sqrt {1- ζ^2}}\)

⇒0.5127 =\(\frac{ ζ }{\sqrt {1- ζ^2}}\)

Squaring and calculating we get

⇒ζ =0.456

Comparing denominator of CLTF with standard result we have

⇒s² + s(4+α) + kα = s2 + 2ζω_n s + ω_n²

We have

⇒2ζωn =(4+α) and ⇒ωn2 = kα 

⇒ω =\(\frac{(4+k)}{0.912}\)  and ⇒ωn2 = kα 

We have 2 unknowns and 1 equation ,so will use the option elimination method to calculate the value of  K and α

Checking for the options 

1. k = 16, α = 29.23 ,

putting k=16, and calculating α

ω =21.93 , ⇒α =30

2. k = 4, α = 19.23

putting k=4, and calculating α

ω =8.767, α = 19.23

3. k = 19.23, ⇒α = 4

putting k= 19.23, and calculating α

ω= 25.47 ,⇒ α =33.73

4. k = 8, α = 4

putting  k=8, and calculating α

ω =13.15 ,⇒α =21.62