Time Response Analysis MCQ Quiz in தமிழ் - Objective Question with Answer for Time Response Analysis - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

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

Steady state error for different inputs is given by

Calculation:

Given:

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

It is a type 0 system, so:

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

K_p = 1 = DC Gain

Steady-state error for unit step input is given as:

\(e_{ss}=\frac{1}{1+K_p}\)

e_ss = 0.5

K_p = position error constant = \(\mathop {\lim }\limits_{s \to 0} G\left( s \right)H\left( s \right)\)

Kv = velocity error constant = \(\mathop {\lim }\limits_{s \to 0} sG\left( s \right)H\left( s \right)\)

Ka = acceleration error constant = \(\mathop {\lim }\limits_{s \to 0} {s^2}G\left( s \right)H\left( s \right)\)

Steady state error for different inputs is given by

From the above table, it is clear that for type – 1 system, a system shows zero steady-state error for step-input, finite steady-state error for Ramp-input and \(\infty \) steady-state error for parabolic-input.

As the type of the system increases, the steady-state error decreases.

The steady-state error is inversely proportional to the gain. Therefore, it can be reduced by increasing the system gain.

Concept:

Time-domain specification (or) transient response parameters:

Rise time (t­r): It is the time taken by the response to reach from 0% to 100% Generally 10% to 9% for overdamped and 5% to 95% for the critically damped system is defined.

\(c\left( t \right){\left. \right|_{t = {t_r}}} = 1 - \frac{{{e^{ - \xi {\omega _n}{t_r}}}}}{{\sqrt {1 - {\xi ^2}} }}\sin \left( {{\omega _n}{t_r} + \varphi } \right)\)

\({t_r} = \frac{{\pi - \varphi }}{{{\omega _d}}}\)

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

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

Calculation:

Rise time \(\left( {{t_r}} \right) = \frac{{\pi - \phi }}{{{\omega _d}}}\)

Peat time \(\left( {{t_p}} \right) = \frac{\pi }{{{\omega _d}}}\)

t_p = 2 t_r

\( \Rightarrow \frac{\pi }{{{\omega _d}}} = \frac{{2\left( {\pi - \phi } \right)}}{{{\omega _d}}}\)

\( \Rightarrow \frac{\pi }{2} = \pi - \phi \Rightarrow \phi = \frac{\pi }{2}\)

\(\Rightarrow {\tan ^{ - 1}}\left( {\frac{{\sqrt {1 - {\xi ^2}} }}{\xi }} \right) = \frac{\pi }{2}\)

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

⇒ ξ = 0

Therefore, the system is undamped.

Delay time (td): It is the time taken by the response to change from 0 to 50% of its final or steady-state value.

\(c\left( t \right){\left. \right|_{t = {t_d}}} = 0.5\)

\({t_d} \simeq \frac{{1 + 0.7\xi }}{{{\omega _n}}}\)

Maximum (or) Peak overshoot (Mp): It is the maximum error at the output. 

\({M_p} = c\left( {{t_p}} \right) - 1,\;{M_p} = e^ \left( {\frac{{-\xi \pi }}{{\sqrt {1 - {\xi ^2}} }}} \right)\)

\(\% {M_p} = \frac{{c\left( {{t_p}} \right) - c\left( \infty \right)}}{{c\left( \infty \right)}} \times 100\% \)

If the magnitude of the input is doubled, then the steady-state value doubles, therefore Mp doubles, but % Mp, tr, tp remains constant.

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

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

Concept:

Final value theorem:

• A final value theorem allows the time domain behavior to be directly calculated by taking a limit of a frequency domain expression
• The final value theorem states that the final value of a system can be calculated by \(f(\infty)= \mathop {\lim }\limits_{s \to 0}~ sF(s)\)  ( F(s) is the Laplace transform of the function)
• For the final value theorem to be applicable system should be stable in steady-state and for that real part of the poles should lie on the left side of s plane.

Initial value theorem:

\(C\left( 0 \right)=\mathop {\lim }\limits_{t \to 0}c\left( t \right)=\mathop {\lim }\limits_{s \to \infty}\,sC(s)\)

.

Application:

1) \(F\left( s \right)=\frac{s-2}{\left( s+1 \right)\left( s+3 \right)}\), all the poles are lying in the left half of the s plane. So, the final value theorem is applicable.

2) \(F\left( s \right)=\frac{s}{s+1}\), all the poles are lying in the left half of the s plane. So, the final value theorem is applicable.

3) \(F\left( s \right)=\frac{s+1}{\left( s+2 \right)\left( s+3 \right)}\), all the poles are lying in the left half of the s plane. So, the final value theorem is applicable.

4) \(F\left( s \right)=\frac{s}{\left( s-1 \right)\left( s+2 \right)}\), one of the poles is lying in the right half of the s plane. So, the final value theorem is not applicable.

• The LFVT can be applied to not only strictly proper functions but also to non-strictly proper functions such as improper, marginally stable, or stable functions.
• However, it is important to note that the condition for the Laplace Final Value Theorem to be applicable is that all the poles of the Laplace transform lie in the left half of the complex s-plane or on the imaginary axis.
• If the poles lie in the right half of the s-plane, the LFVT is not applicable.

Concept:

Standard second-order closed-loop transfer function is given by:

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

ξ = damping ratio

ω_n = undamped natural frequency

Analysis:

Given:

ξ = 1.25, ω_n = 200 rad/sec

DC Gain: K = 1

The closed loop transfer function will be:

\(\frac{{G\left( s \right)}}{{R\left( s \right)}} = \frac{{{{\left( {200} \right)}^2}}}{{{s^2} + 2 \times 1.25 \times 200s + {{\left( {200} \right)}^2}}}\)

\(= \frac{{40000}}{{\left( {s + 100} \right)\left( {s + 400} \right)}}\)   

For a unit step input:

\(R\left( s \right) = \frac{1}{s}.\)

∴ The response C(s) of the system for a unit step input will be:

\(C\left( s \right) = \frac{{40000}}{{s\left( {s + 100} \right)\left( {s + 400} \right)}}\)

Using partial fraction, the above expression can be written as:

\(C\left( s \right) = \frac{1}{s} - \frac{4}{{3\left( {s + 100} \right)}} + \frac{1}{{3\left( {s + 400} \right)}}\)

Taking the inverse Laplace transform on both the sides, we get:

\(c\left( t \right) = 1 - \frac{4}{3}{e^{ - 100t}} + \frac{1}{3}{e^{ - 400t}}\)

Concept:

KP = position error constant = \(\mathop {\lim }\limits_{s \to 0} G\left( s \right)H\left( s \right)\)

Kv = velocity error constant = \(\mathop {\lim }\limits_{s \to 0} sG\left( s \right)H\left( s \right)\)

Ka = acceleration error constant = \(\mathop {\lim }\limits_{s \to 0} {s^2}G\left( s \right)H\left( s \right)\)

Steady state error for different inputs is given by

From the above table, it is clear that for the type – 1 system, a system shows zero steady-state error for Unit step-input.

The correct answer is K = 100

Concept:

The formula of e_ss for such type of the system is \(E(s) = \frac{ R(s) - D(s).G_2(s)}{1+G_1(s).G_2(s)}\)

Solution:

Given

 \(G_1(s) = \frac{1}{(s + 5)}\) 

 \(G_2(s) = \frac{K}{(s + 2)}\),

R(s) is unit step input and D(s) is unit step disturbance.

Steady-state error

 \(e_{ss} = \frac{ R(s) - D(s).G_2(s)}{1+G_1(s).G_2(s)}\)

\(e_{ss} = \displaystyle\lim_{s \rightarrow 0} s \left[ \frac{1/s - 1/s \frac{K}{(s + 2)}}{1 + \frac{1}{(s + 5)} \frac{K}{(s + 2)}} \right]\)

\(\Rightarrow \displaystyle\lim_{s \rightarrow 0} s \left[ \frac{1/s - 1/s \frac{K}{(s + 2)}}{1+ \frac{K}{(s + 5)(s + 2)}} \right]\)

\(\Rightarrow \frac{10( 2 - K}{ 2 (10 + K)} = -4.45\)

 \(∴\) K =  100

The correct answer is \(\frac{10s}{s^2 +2s + 200}\)

Solution:

From the graph, we can say it is the output of a 1st order system

let's say the transfer function of an LTI system

 \(G(s) = \frac{K}{(s + a)}\)

For the unit step input-output

 \(Y(s) = \frac{K}{s (s + a)}\)

It is given that a steady-state output value is

\(\lim_{S \rightarrow 0} \frac{S.K}{S(S + a)} = 5\)

∴ \(\frac{K}{a}= 5\)

Now the value of the output

 \(Y(s) = \frac{K}{s (s + a)} = \frac{2a}{s (s + a)}\)

Apply partial fraction

 \(∴\) \(\frac{5a}{s (s + a)} = \frac{A}{s} + \frac{B}{(s + a)}\)

after solving above partial fraction we get A = 5 and B = -5.

\(∴\)  \(Y(s) = \frac{5}{s} - \frac{5}{(s + a)}\)

Now take inverse Laplace you will get

y(t) = 5 ( 1 - e^-at) u(t)

from the graph, at t = 1.15-sec output is 4.5 so put on this value in above equation.

\(∴\) y(t) = 5 ( 1 - e^-a(1.15)) = 4.5

\(∴\) a = 2

So the overall transfer function of an LTI system is

= \(\frac{10}{(s + 2)}\)

Now we have to add feedback to it, which will make the block diagram as below.

The final transfer function of the above system is

T/F = \(\frac{\frac{10}{(s + 2)}}{1 + \frac{10}{(s + 2)}.\frac{20}{s}}\) 

= \(\frac{10 s}{s^2 +2s + 200}\)

From the given response curve,

Peak over shoot, (MP) = 0.254

Peak time, tp = 3 s

\({e^{ - \frac{{\xi \pi }}{{\sqrt {1 - {\xi ^2}} }}}} = 0.254\)

⇒ ξ = 0.4

\({t_p} = \frac{\pi }{{{\omega _d}}} = \frac{\pi }{{{\omega _n}\sqrt {1 - {\xi ^2}} }} = 3\)

\( ⇒ \frac{\pi }{{{\omega _n}\sqrt {1 - {{\left( {0.4} \right)}^2}} }} = 3\)

⇒ ωn = 1.14

From the block diagram,

\(\frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{K}{{T{s^2} + s + K}}\)

\({\omega _n} = \sqrt {\frac{K}{T}} ,\;\;2\xi {\omega _n} = \frac{1}{T}\)

\(T = \frac{1}{{2\xi {\omega _n}}} = \frac{1}{{2 × 0.4 × 1.14}} = 1.09\)

\(K = \omega _n^2T = {\left( {1.14} \right)^2} × 1.09 = 1.42\)

⇒ KT = 1.09 × 1.42 = 1.5478 and K+T = 2.51

Concept:

The steady-state error for a system is defined as:

\({e_{ss}} = \mathop {\lim }\limits_{s \to 0} \frac{{sR\left( s \right)}}{{1 + G\left( s \right)H\left( s \right)}}\) 

R(s) = Input

G(s) = open loop transfer function

H(s) = feedback gain = 1 for unity feedback system

Analysis:

For a unit step input, we have:

\(R\left( s \right) = \frac{1}{s}\) 

The steady-state error becomes:

\({e_{ss}} = \mathop {\lim }\limits_{s \to 0} \frac{{s\left( {\frac{1}{s}} \right)}}{{1 + G\left( s \right)}} = 0.1\) 

1 + G(0) = 10

G(0) = 9

Again we have the input as:

r(t) = 10 [u(t) – u(t - 1)]

\(R\left( s \right) = 10\left[ {\frac{1}{s} - \frac{1}{s}{e^{ - s}}} \right]\) 

\(= 10\left( {\frac{{1 - {e^{ - s}}}}{s}} \right)\) 

∴ The steady-state error for the input will be:

\(e_{ss}' = \mathop {\lim }\limits_{s \to 0} \frac{{s \times 10\frac{{\left( {1 - {e^{ - s}}} \right)}}{s}}}{{1 + G\left( s \right)}}\) 

\(= \frac{{10\left( {1 - {e^0}} \right)}}{{1 + 9}}\) 

\(e_{ss}' = 0\)