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

u(t) = sin 2t 

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

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

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

The steady-state response is:

$Y(s)=Lt\underset{s=0}{\to }sY(s)$

$Y(s)=Lt\underset{s=0}{\to }\frac{s}{s+3}\times \frac{2}{s^2+4}$

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

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

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

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: 

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

$\sin \omega tu\left(t\right)\iff \frac{\omega }{s^2+{\omega }^2}$

$\sinh \omega tu\left(t\right)\iff \frac{\omega }{s^2-{\omega }^2}$

$e^{-at}\sin \omega tu\left(t\right)\iff \frac{\omega }{{\left(s+a\right)}^2+{\omega }^2}$

$e^{-at}\sinh \omega tu\left(t\right)\iff \frac{\omega }{{\left(s+a\right)}^2-{\omega }^2}$

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: $K_P=\underset{s\to 0}{lim}G\left(s\right)$

For type 1 system: $K_v=\underset{s\to 0}{lim}sG\left(s\right)$

For type 2 system: $K_a=\underset{s\to 0}{lim}{s}^{2}G\left(s\right)$

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 = $\underset{s\to 0}{lim}G\left(s\right)$

= 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: $K_P=\underset{s\to 0}{lim}G\left(s\right)$

For type 1 system: $K_v=\underset{s\to 0}{lim}sG\left(s\right)$

For type 2 system: $K_a=\underset{s\to 0}{lim}{s}^{2}G\left(s\right)$

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 = $\underset{s\to 0}{lim}G\left(s\right)$

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

= - 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)

$H(j\omega )=\frac{Y(j\omega )}{X(j\omega )}$

Calculation:

$H(j\omega )=\frac{j\omega }{j\omega +2}$

r(t) = cos (2t)

ω = 2

$H(j\omega )=\frac{j2}{j2+2}$

$|H(j\omega )|=\frac{1}{\sqrt{2}}$

∠ H(jω) = 90° - 45°

= +45° 

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

Concept:

The final value theorem is given as:

$f\left(t\right){|}_{t=\mathrm{\infty }}=\underset{s\to 0}{lim}\left[sF\left(s\right)\right]$

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

$=\underset{s\to 0}{\mathrm{l}\mathrm{t}}s\frac{10}{s\left(s^2+s+100\sqrt{2}\right)}$

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

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: 

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

$\sin \omega tu\left(t\right)\iff \frac{\omega }{s^2+{\omega }^2}$

$\sinh \omega tu\left(t\right)\iff \frac{\omega }{s^2-{\omega }^2}$

$e^{-at}\sin \omega tu\left(t\right)\iff \frac{\omega }{{\left(s+a\right)}^2+{\omega }^2}$

$e^{-at}\sinh \omega tu\left(t\right)\iff \frac{\omega }{{\left(s+a\right)}^2-{\omega }^2}$

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

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

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

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

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