Peak Overshoot MCQ Quiz in తెలుగు - Objective Question with Answer for Peak Overshoot - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Concept:

Time-domain specification (or) transient response parameters:

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.

Peak Time (tp):

It is the time taken by the response to reach the maximum value.

\({\text{}}{t_p} = \frac{\pi }{{{\omega _d}}}\)

If ξ increases, rise time, peak time increases and peak overshoot decreases.

Calculation:

Given peak overshoot as:

\({{\rm{M}}_{\rm{p}}} = {\rm{\;}}30{\rm{\% \;}} = {\rm{\;}}{{\rm{e}}^{ - \frac{{{\rm{ξ \pi }}}}{{\sqrt {1 - {{\rm{ξ }}^2}} }}}} \times 100\)

\( \ln \left( {\frac{{30}}{{100}}} \right) = \frac{{ - {\rm{ξ \pi }}}}{{\sqrt {1 - {{\rm{ξ }}^2}} }}\)

ζ = 0.3578

Peak Overshoot occurs at peak time:

\({\text{}}{t_p} = \frac{\pi }{{{\omega _d}}}\)

On solving we'll get:

t_p = 0.336 sec

The tolerance band is 5%

Settling time t_s = 0.5 sec

\({t_s} = \frac{3}{{\xi {\omega _n}}} = 0.5 \Rightarrow \xi {\omega _n} = 6\)         ---(1)

Delay time t_d = 0.142 sec

Delay time \( = \frac{{1 + 0.7\;\xi }}{{{\omega _n}}} = 0.142\) 

⇒ (1 + 0.7 ξ) = ω_n 0.142

\( \Rightarrow {\omega _n} = \frac{{1 + .7\xi }}{{0.142}}\)      ---(2)

From the equations (1) and (2),

\(\xi \left( {\frac{{1 + 0.7\;\xi }}{{0.142}}} \right) = 6\) 

⇒ 0.7 ξ² + ξ = 0.852

⇒ 0.7 ξ² + ξ – 0.852 = 0

⇒ ξ² + 1.43 ξ – 1.22 = 0

\(\xi = \frac{{ - 1.43 \pm \sqrt {{{\left( {1.43} \right)}^2} - 4 \times 1\left( { - 1.22} \right)} }}{{2.1}}\)  

\(= \frac{{ - 1.43 \pm 2.63}}{2} = 0.6,\; - 2.03\) 

As damping ratio cannot be negative hence

⇒ ξ = 0.6

\( \Rightarrow {\omega _n} = \frac{6}{{0.6}} = 10\;rad/sec\) 

Now, maximum overshoot: 

Maximum % over-shoot will be:

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

\( \Rightarrow \% \;{M_p} = \left( {\frac{{{e^{ - 1 \times 3.14 \times 0.6}}}}{{\sqrt {1 - {{\left( {0.6} \right)}^2}} }}} \right) \times 100\% \) 

= (e^-2.355) × 100%

% M_p = 9.48 %

We redraw the given network in s-domain as:

The transfer function of the system is:

\(\frac{{{V_0}\left( s \right)}}{{{V_i}\left( s \right)}} = \frac{{\frac{1}{{CS}}}}{{R + LS + \frac{1}{{CS}}}}\) 

\( = \frac{{\left( {\frac{1}{{LC}}} \right)}}{{{s^2} + \frac{R}{L}s + \frac{1}{{LC}}}}\) 

Given percentage overshoot is:

% M_p = 20%

\({M_p} = {e^{\frac{{ - \xi \pi }}{{\sqrt {1 - \xi } }}}} = \frac{{20}}{{100}}\) 

ξ = 0.456

Also, from the transfer function, the characteristic equation is given as:

\({s^2} + \frac{R}{L}s + \frac{1}{{LC}} = 0\) 

Comparing it with \({s^2} + 2\xi {\omega _n}s + \omega _n^2 = 0\), we get:

\({\omega _n} = \frac{1}{{\sqrt {LC} }}\) 

And \(\xi = \frac{{R/L}}{{2{\omega _n}}} = \frac{R}{{2L{\omega _n}}}\) 

 = 2ξLω_n

\( = 2 \times 0.456 \times 1 \times \frac{1}{{\sqrt {1 \times {{10}^{ - 6}}} }}\) 

= 912 Ω

Peak overshoot,

 \(= {e^{ - \frac{{n\zeta \pi }}{{\sqrt {1 - {\zeta ^2}} }}}} \times 100\)

For 3rd peak overshoot n = 5

\(\begin{array}{l} = {e^{ - \frac{{5\zeta \pi }}{{\sqrt {1 - {\zeta ^2}} }}}} \times 100 = 5.08\\ = {e^{ - \frac{{5\zeta \pi }}{{\sqrt {1 - {\zeta ^2}} }}}} = 0.0508\\ \frac{{{\zeta ^2}}}{{1 - {\zeta ^2}}} = 0.0359 \to \zeta = 0.1864 \end{array}\)

Location of dominant pole  \(= - \zeta {\omega _n} + j{\omega _n}\sqrt {1 - {\zeta ^2}}\)

\(= - 0.7455 + j3.93\)

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).

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)

Overshoot-

Moving parts of instruments have mass and thus possess inertia.

When an input is applied to instruments, the pointer does not immediately come to rest at its steady-state (or final deflected) position but goes beyond it or in other words overshoots its steady position.

The overshoot is evaluated as the maximum amount by which the moving system moves beyond the steady-state position.

Percentage error-

Percent error is the difference between the estimated value and the actual value in comparison to the actual value and is expressed as a percentage.

In other words, the percent error is the relative error multiplied by 100.

\(\% Error - \left| {\frac{{T - E}}{T}} \right| \cdot 100\)

Here,

T = True or Actual value

E = Estimated value

Calculation:

a) Overshoot = 4.18 – 4.02

= 0.16 A

% Overshoot = \(\frac{{0.16}}{{4.02}} \times 100 = 3.98\% \)

b) Here T = 4.0 Amp, E = 4.02 Amp

% error = \(\frac{{4.02 - 4}}{4} \times 100 = 0.5\% \)

Concept:-

⇒ Integral Controller:-

• It is use to improve steady state characteristics of sysem.
• When Integral gain increases then steady state error (e_ss) decreases also stability of system reduces.

⇒ Derivative controller:-

• It is use to improve transient state characteristics of system.
• When gain of derivative controller improves then damping ratio increases & hence peak overshoot of system reduces.

i. e. ξ ↑, % M_p ↓, \(\left\{ {\% \;{M_p} = {e^{\frac{{ - {\xi _H}}}{{\sqrt {1 - {\xi ^2}} }}}}} \right\}\)

Conclusion:-

Since both the statement is true without any interconnection, we will select option (2)

Note:-

⇒ What is use of controller in control system?

\({M_r} = 1.4 = \frac{1}{{2\zeta \sqrt {1 - {\zeta ^2}} }}\)

ζ = 0.387

Maximum overshoot mp

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

= 0.2675 × 100

= 26.75%

OLTF = G_c(s) G(s)

\(= \left( {{k_1} + \frac{1}{s}} \right)\left( {\frac{{500}}{{\left( {s + 5} \right)}}} \right)\) 

\(\frac{{500\left( {1 + {k_1}s} \right)}}{{s\left( {s + 51} \right)}} = T\left( s \right)\) 

\(CLTF = \frac{{T\left( s \right)}}{{1 + T\left( s \right)H\left( s \right)}}\) 

\(= \frac{{500\left( {1 + {k_1}s} \right)}}{{s\left( {s + 5} \right) + 500\left( {1 + {k_1}s} \right)}}\) 

\(CLTF = \frac{{500\left( {1 + {k_1}s} \right)}}{{s\left( {s + 5} \right) + 500 + 500\;{k_1}s}}\) 

\(CLTF = \frac{{500 + 500\;{k_1}s}}{{{s^2} + \left( {5 + 500\;{k_1}} \right)s + 500}}\) 

\(\omega _n^2 = 500\) 

\({\omega _n} = \sqrt {500} = 10\sqrt 5\) 

\(2\xi {\omega _n} = 5 + 500\;{k_1}\) 

\(\xi {\omega _n} = 2.5 + 250\;{k_1}\) 

\(\xi = \frac{{2.5}}{{10\sqrt 5 }} + \frac{{250}}{{10\sqrt 5 }}{k_1}\)       ____(1)

Peak overshoot:

\({e^{ - \pi \cot \theta }} = 0.2\) 

π cot θ = 1.61

\(\cot \theta = \frac{{1.61}}{{3.14}} = 0.512\) 

\(\frac{\xi }{{\sqrt {1 - {\xi ^2}} }} = 0.512\) 

ξ = 0.458

substitute in 1

\(0.458 = \frac{{2.5}}{{10\sqrt 5 }} + \frac{{250\;{k_1}}}{{10\sqrt 5 }}\)