Steady State Output MCQ Quiz - Objective Question with Answer for Steady State Output - Download Free PDF

Concept

A transfer function is defined as the ratio of the Laplace transform of the output to the Laplace transform of the input by assuming initial conditions are zero.

\(TF = \frac{L[Output]}{L[Input]}\)

\(TF = \frac{{Y\left( s \right)}}{{U\left( s \right)}}\)

Calculation

Given, \(H(s)={Y(s)\over U(s)}=\frac{1}{s+3}\)

u(t) = sin 2t 

\(U(s)={2\over s^2+4}\)

\({Y(s)}=\frac{1}{s+3}\times U(s)\)

\({Y(s)}=\frac{1}{s+3}\times {2\over s^2+4}\)

The steady-state response is:

\(Y(s)=Lt\underset{s=0}{\rightarrow}s \space Y(s)\)

\(Y(s)=Lt\underset{s=0}{\rightarrow}\frac{s}{s+3} \space \times {2\over s^2+4}\)

Y(s) = 0

Concept:

A transfer function is defined as the ratio of Laplace transform of the output to the Laplace transform of the input by assuming initial conditions are zero.

TF = L[output]/L[input]

\(TF = \frac{{C\left( s \right)}}{{R\left( s \right)}}\)

For unit impulse input i.e. r(t) = δ(t)

⇒ R(s) = δ(s) = 1

Now transfer function = C(s)

Therefore, transfer function is also known as impulse response of the system.

Transfer function = L[IR]

IR = L-1 [TF]

Short-cut to convert CLTF to OLTF

CLTF = Num / Deno

OLTF = Num / (Deno - Num)

Calculation:

Given that r(t) = δ(t), c(t) = e-2t

\(\Rightarrow R\left( s \right) = 1,\;C\left( s \right) = \frac{1}{s + 2}\)

Transfer function for close loop 

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

Transfer function for open loop 

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

Given,

 \(G(S)=\frac{K(2S+1)}{S(45+1)(5+1)^2}\)

Steady state error e_ss is less than 0.1

Given ramp input r(t) = 1 + 5t

Velocity constant is calculated for ramp input

\({K_V} = \mathop {Lt}\limits_{S \to 0} S.G(S) = \mathop {Lt}\limits_{S \to 0} \frac{{S.K(2S + 1)}}{{S(4S + 1){{(S + 1)}^2}}}\)

\({K_V} = \mathop {Lt}\limits_{S \to 0} \frac{{K(2S + 1)}}{{S(4S + 1){{(S + 1)}^2}}} = K\)

Slope of ramp input, \(V =\frac{dt}{dt}=5\)

Steady state error, 

\(e_{ss}=\frac{Slope}{K_V}=\frac{5}{K}.\)

e_ss < 0.1

\(\frac{5}{K}<0.1\)

\(\frac{5}{0.1}

K > 50 

If the steady gate error is less than 0.1 then value of K is

Transfer function: It is defined as the ratio of the Laplace transform of the output variable to the Laplace transform of the input variable assuming all initial conditions to be zero.

Open-loop transfer function ⇒ \(G\left( s \right) = \frac{{2s}}{{\left( {s + 1} \right)\left( {s + 2} \right)}}\)

Nothing about feedback path, so H(s) = 1

Close-loop transfer function ⇒ \(T\left( s \right) = \frac{{G\left( s \right)}}{{1 + G\left( s \right)H\left( s \right)}}\;\)

\(T\left( s \right) = \frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{2s}}{{{s^2} + 5s + 2}}\)

Where C(s) = Output function

R(s) = Input function

Given input function is a unit step ⇒ \(R\left( s \right) = \frac{1}{s}\)

\(C\left( s \right) = T\left( s \right)R\left( s \right) = \frac{1}{s} \times \frac{{2s}}{{{s^2} + 5s + 2}} = \frac{2}{{{s^2} + 5s + 2}}\)

\(C\left( s \right) = \frac{2}{{{s^2} + 5s + 2 + \frac{{25}}{4} - \frac{{25}}{4}}}\)

\(C\left( s \right) = \frac{2}{{{{\left( {s + \frac{5}{2}} \right)}^2} - {{\left( {\sqrt {\frac{{17}}{4}} } \right)}^2}}}\)

\(C\left( s \right) = \frac{1}{{\sqrt {\frac{{17}}{4}} }}\left( {\frac{{2\sqrt {\frac{{17}}{4}} }}{{{{\left( {s + \frac{5}{2}} \right)}^2} - {{\left( {\sqrt {\frac{{17}}{4}} } \right)}^2}}}} \right)\)

Taking Inverse Laplace

\(c\left( t \right) = \frac{2}{{\sqrt {\frac{{17}}{4}} }}{e^{ - \frac{{5t}}{2}}}\sinh \left( {\sqrt {\frac{{17}}{4}} t} \right)u\left( t \right)\)

 \(c\left( t \right) = \frac{4}{{\sqrt {17} }}{e^{ - \frac{{5t}}{2}}}\sinh \left( {\frac{{\sqrt {17} }}{2}t} \right)u\left( t \right)\)

The steady-state value of c(t) for t → ∞ is zero as the value e^(-2.5t) tends to zero.

Important Points

Laplace Transform: 

\({e^{ - at}}u\left( t \right) \Leftrightarrow \frac{1}{{s + a}}\)

\(\sin \omega t\;u\left( t \right) \Leftrightarrow \frac{\omega }{{{s^2} + {\omega ^2}}}\)

\(\sinh \omega t\;u\left( t \right) \Leftrightarrow \frac{\omega }{{{s^2} - {\omega ^2}}}\)

\({e^{ - at}}\sin \omega t\;u\left( t \right) \Leftrightarrow \frac{\omega }{{{{\left( {s + a} \right)}^2} + {\omega ^2}}}\)

\({e^{ - at}}\sinh \omega t\;u\left( t \right) \Leftrightarrow \frac{\omega }{{{{\left( {s + a} \right)}^2} - {\omega ^2}}}\)

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

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

Given the transfer function, \(H\left( s \right) = \frac{{{s^2} + 100}}{{s - p}}\)

dc gain = 5

To find the DC gain, put s = 0 in the above transfer function.

\(\frac{{100}}{{ - p}} = 5 \Rightarrow p = - 20\)

Now, the transfer function becomes

\(H\left( s \right) = \frac{{{s^2} + 100}}{{s + 20}}\)

Put the value of s = jω in transfer function

\(H\left( {j\omega } \right) = \frac{{ - {\omega ^2} + 100}}{{j\omega + 20}}\)

\(\left| {H\left( {j\omega } \right)} \right| = \frac{{ - {\omega ^2} + 100}}{{\sqrt {\left( {{\omega ^2} + 400} \right)} }}\)

We need to find the frequency ω at |H(jω)| = 1

\( \Rightarrow \frac{{ - {\omega ^2} + 100}}{{\sqrt {\left( {{\omega ^2} + 400} \right)} }} = 1\)

\( \Rightarrow {\left( { - {\omega ^2} + 100} \right)^2} = {\omega ^2} + 400\)

Let ω² = t, now the above equation becomes

(100 – t)² = t + 400

⇒ 10000 + t² – 200t = t + 400

⇒ t² – 201t + 9600 = 0

⇒ t = 122.86 and 78.13

⇒ ω = 11.08 and 8.84

The smallest possible frequency = 8.84 rad/sec

Concept:

When any system is subjected to Sinusoidal input, the output is also sinusoidal having different magnitude and phase angle but same frequency ω(r/sec)

Magnitude, B = |F(s)|_{s = jω} × A

Phase, ϕ₂ = ∠ F(s)|_{s = jω} ± ϕ₁

Calculation:

\(F\left( s \right) = \frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{1}{{s + 1}}\)

Put s = jω      

\(F\left( {j\omega } \right) = \frac{{C\left( {j\omega } \right)}}{{R\left( {j\omega } \right)}} = \frac{1}{{j\omega + 1}}\)

r(t) = sin t

Here ω = 1 r/sec

A = 1, ϕ₁ = 0

Magnitude, \(\left| {F\left( {j\omega } \right)} \right| = \frac{1}{{\sqrt {{\omega ^2} + 1} }}\)

\(\left| {F\left( {j\omega } \right)} \right| = \frac{1}{{\sqrt {{1^2} + 1} }} = \frac{1}{{\sqrt 2 }}\)

Phase, \(\angle F\left( {j\omega } \right) = \frac{1}{{{{\tan }^{ - 1}}\left( {\frac{\omega }{1}} \right)}}\)

= - tan^-1(1)

\( = - 45^\circ \;\;or\;\frac{{ - \pi }}{4}\)

Output C(t) = B sin (ωt ± ϕ₂)

B = |F(jω)| × A

\( = \frac{1}{{\sqrt 2 }} \times 1 = \frac{1}{{\sqrt 2 }}\)

ϕ₂ = ∠ F(jω) ± ϕ₁

\( = \frac{{ - \pi }}{4} \pm 0 = \frac{{ - \pi }}{4}\)

C(t) = B sin [ωt₁ ± ϕ₂]

\(C\left( t \right) = \frac{1}{{\sqrt 2 }}\sin \left[ {t - \frac{\pi }{4}} \right]\)

Concept:

For an LTI system, the sinusoidal input produces a sinusoidal output of the same frequency but different amplitude and phase. The amplitude and phase depend upon the transfer function of the system.

For example:

Let, r(t) = cos(ωt) then

y(t) = a cos(ωt + ϕ)

where a = |H(jω)|, ϕ = ∠ H(jω)

\(H(jω)=\frac{Y(jω)}{X(jω)}\)

Calculation:

\(H(jω)=\frac{jω}{jω+2}\)

r(t) = cos (2t)

ω = 2

\(H(jω)=\frac{j2}{j2+2}\)

\(|H(jω)|=\frac{{1}}{{\sqrt 2}}\)

∠ H(jω) = 90° - 45°

= +45° 

\(y(t)=\frac{1}{\sqrt 2}~cos (2t + 45^{\circ})\)

Concept:

The final value theorem is given as:

\(f\left( t \right){\left. \right|_{t = \infty }} = \mathop {\lim }\limits_{s \to 0} \left[ {sF\left( s \right)} \right]\)

This theorem is only applicable when:

(I) f(t) = 0; t < 0

(II) The term [sF(s)] should have poles in the left-hand side of the s-plane.

Explanation:   

\(Y\left( s \right) = \frac{{10}}{{s\left( {{s^2} + s + 100\sqrt 2 } \right)}}\)

By using Final value theorem, the steady state value of the given system is

\(= \mathop {{\rm{lt}}}\limits_{s \to 0} s\frac{{10}}{{s\left( {{s^2} + s + 100\sqrt 2 } \right)}}\)

\(G\left( s \right) = \frac{{{e^{ - \frac{s}{{500}}}}}}{{s + 500}}\)

ϕ = ∠G(jω) at ω = 100π

\( = - {\tan ^{ - 1}}\left( {\frac{\omega }{{500}}} \right) - \frac{\omega }{{500}}\)

\( = - {\tan ^{ - 1}}\left( {\frac{{100\pi }}{{500}}} \right) - \frac{{100\pi }}{{500}}\)

\( = - {\tan ^{ - 1}}\left( {\frac{\pi }{5}} \right) - \frac{\pi }{5}\)

= - 68.12°

y(t) = A sin (100 πt - ( 68.12°))

i(t) = 2 sin (3t + π)

\(G\left( s \right) = \frac{8}{{{{\left( {s + 10} \right)}^2}}}\)

ω = 3.

Transfer function: It is defined as the ratio of the Laplace transform of the output variable to the Laplace transform of the input variable assuming all initial conditions to be zero.

Open-loop transfer function ⇒ \(G\left( s \right) = \frac{{2s}}{{\left( {s + 1} \right)\left( {s + 2} \right)}}\)

Nothing about feedback path, so H(s) = 1

Close-loop transfer function ⇒ \(T\left( s \right) = \frac{{G\left( s \right)}}{{1 + G\left( s \right)H\left( s \right)}}\;\)

\(T\left( s \right) = \frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{2s}}{{{s^2} + 5s + 2}}\)

Where C(s) = Output function

R(s) = Input function

Given input function is a unit step ⇒ \(R\left( s \right) = \frac{1}{s}\)

\(C\left( s \right) = T\left( s \right)R\left( s \right) = \frac{1}{s} \times \frac{{2s}}{{{s^2} + 5s + 2}} = \frac{2}{{{s^2} + 5s + 2}}\)

\(C\left( s \right) = \frac{2}{{{s^2} + 5s + 2 + \frac{{25}}{4} - \frac{{25}}{4}}}\)

\(C\left( s \right) = \frac{2}{{{{\left( {s + \frac{5}{2}} \right)}^2} - {{\left( {\sqrt {\frac{{17}}{4}} } \right)}^2}}}\)

\(C\left( s \right) = \frac{1}{{\sqrt {\frac{{17}}{4}} }}\left( {\frac{{2\sqrt {\frac{{17}}{4}} }}{{{{\left( {s + \frac{5}{2}} \right)}^2} - {{\left( {\sqrt {\frac{{17}}{4}} } \right)}^2}}}} \right)\)

Taking Inverse Laplace

\(c\left( t \right) = \frac{2}{{\sqrt {\frac{{17}}{4}} }}{e^{ - \frac{{5t}}{2}}}\sinh \left( {\sqrt {\frac{{17}}{4}} t} \right)u\left( t \right)\)

 \(c\left( t \right) = \frac{4}{{\sqrt {17} }}{e^{ - \frac{{5t}}{2}}}\sinh \left( {\frac{{\sqrt {17} }}{2}t} \right)u\left( t \right)\)

The steady-state value of c(t) for t → ∞ is zero as the value e^(-2.5t) tends to zero.

Important Points

Laplace Transform: 

\({e^{ - at}}u\left( t \right) \Leftrightarrow \frac{1}{{s + a}}\)

\(\sin \omega t\;u\left( t \right) \Leftrightarrow \frac{\omega }{{{s^2} + {\omega ^2}}}\)

\(\sinh \omega t\;u\left( t \right) \Leftrightarrow \frac{\omega }{{{s^2} - {\omega ^2}}}\)

\({e^{ - at}}\sin \omega t\;u\left( t \right) \Leftrightarrow \frac{\omega }{{{{\left( {s + a} \right)}^2} + {\omega ^2}}}\)

\({e^{ - at}}\sinh \omega t\;u\left( t \right) \Leftrightarrow \frac{\omega }{{{{\left( {s + a} \right)}^2} - {\omega ^2}}}\)

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

p(t) = sin (t)

Steady state output y(t) = sin (t + ϕ)

ϕ = ∠G(jω)

\(G\left( {j\omega } \right) = \frac{{j\omega - 1}}{{j\omega + 1}}\)

ϕ = -tan^-1(1) + 180 – tan^-1(1)

Given,

 \(G(S)=\frac{K(2S+1)}{S(45+1)(5+1)^2}\)

Steady state error e_ss is less than 0.1

Given ramp input r(t) = 1 + 5t

Velocity constant is calculated for ramp input

\({K_V} = \mathop {Lt}\limits_{S \to 0} S.G(S) = \mathop {Lt}\limits_{S \to 0} \frac{{S.K(2S + 1)}}{{S(4S + 1){{(S + 1)}^2}}}\)

\({K_V} = \mathop {Lt}\limits_{S \to 0} \frac{{K(2S + 1)}}{{S(4S + 1){{(S + 1)}^2}}} = K\)

Slope of ramp input, \(V =\frac{dt}{dt}=5\)

Steady state error, 

\(e_{ss}=\frac{Slope}{K_V}=\frac{5}{K}.\)

e_ss < 0.1

\(\frac{5}{K}<0.1\)

\(\frac{5}{0.1}

K > 50 

If the steady gate error is less than 0.1 then value of K is