Step Input MCQ Quiz in தமிழ் - Objective Question with Answer for Step Input - இலவச PDF ஐப் பதிவிறக்கவும்

$G\left(s\right)=\frac{5\left(2+4s\right)}{\left(1+9s\right)\left(4+3s\right)}$

For step input, steady state error is given by

$E_{ss}=\frac{A}{1+K_p}$

$K_p=\underset{s\to 0}{\mathrm{i}\mathrm{t}}G\left(s\right)=\underset{s\to 0}{\mathrm{i}\mathrm{t}}\frac{5\left(2+4s\right)}{\left(1+9s\right)\left(4+3s\right)}=\frac{10}{4}=2.5$

$E_{ss}=\frac{1}{1+2.5}=\frac{1}{3.5}=0.286$

The simplified system rearranged to show disturbance as input and error or output is:

$e_D\left(\mathrm{\infty }\right)=-\underset{s\to 0}{lim}\frac{s{G}_2\left(s\right)D\left(s\right)}{1+G_1\left(s\right)G_2\left(s\right)}$ 

For step input $D\left(s\right)=\frac{1}{s}$

$e_D\left(\mathrm{\infty }\right)=-\underset{s\to 0}{lim}\frac{-1}{\frac{1}{G_1\left(s\right)}\underset{s\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}G\left(s\right)}$ 

Steady-state error will occur when the output variation reaches a steady state, i.e. step output.

∴ The steady-state error is calculated at ω = 0.

|M| = 0.9 → steady-state value of unit step response

e_ss = 1 – 0.9

solving 1^st loop

${\mathrm{G}}_1\left(\mathrm{s}\right)=\frac{\frac{1}{\mathrm{s}+3}}{1+\frac{{\mathrm{K}}_3}{\mathrm{s}+3}}$

${\mathrm{G}}_1\left(\mathrm{s}\right)=\frac{1}{\mathrm{s}+3+{\mathrm{k}}_3}$

Step 2

$\left(\frac{1}{\mathrm{s}+3+{\mathrm{k}}_3}\right)\left(\frac{1}{\mathrm{s}+1}\right)$

$\left(\frac{1}{{\mathrm{s}}^2+\left(4+{\mathrm{k}}_3\right)\mathrm{s}+\left(3+{\mathrm{k}}_3\right)}\right)$

Step 3 = Solving 2^nd loop

$G\left(s\right)=\frac{1}{{\mathrm{s}}^3+\left(4+{\mathrm{k}}_3\right){\mathrm{s}}^2\left(3+{\mathrm{k}}_3+{\mathrm{k}}_2\right)\mathrm{s}}$

H(s) = K₁

$\mathrm{C}\mathrm{L}\mathrm{T}\mathrm{F}=\frac{1}{{\mathrm{s}}^3+\left(4+{\mathrm{k}}_3\right){\mathrm{s}}^2+\left(3+{\mathrm{k}}_3+{\mathrm{k}}_2\right)\mathrm{s}+{\mathrm{k}}_1}$

For Non-unity feedback system

e(t) = r(t) – y(t)

$e\left(t\right)=r\left(t\right)\left[1-\frac{y\left(t\right)}{r\left(t\right)}\right]$

Laplace domain

$E\left(s\right)=SR\left(s\right)\left[1-\frac{Y\left(s\right)}{R\left(s\right)}\right]$

E(s) = SR(s) [1 - CLTF]

For steady state error

S → 0

$e_{ss}=\mathrm{l}\mathrm{i}\mathrm{m}s\times \frac{1}{s}\left[1-\frac{1}{s^3+\left(4+k_3\right)s^2+\left(3+k_2+k_3\right)s+k_1}\right]$

= 0

= 1 – K₁ = 0

Concept:

Error = input – output

e(t) = r(t) – c(t)

taking replace

$e\left(s\right)=\underset{s\to 0}{\mathrm{l}\mathrm{t}}s\left[R\left(s\right)-C\left(s\right)\right]$ 

$E\left(s\right)=\underset{s\to 0}{\mathrm{l}\mathrm{t}}sR\left(s\right)\left[1-\frac{C\left(s\right)}{R\left(s\right)}\right]$ 

$E\left(s\right)=\underset{s\to 0}{\mathrm{l}\mathrm{t}}sR\left(s\right)\left[1-CLTF\right]$ 

Application:

$E\left(s\right)=\underset{s\to 0}{\mathrm{l}\mathrm{t}}s\times \frac{1}{s}\left[1-\frac{\frac{4}{s+2}}{1+\left(\frac{4}{s+2}\right)H\left(s\right)}\right]=0$ 

$=\underset{s\to 0}{\mathrm{l}\mathrm{t}}\left[1-\frac{\frac{4}{s+2}}{\frac{s+2+4H\left(s\right)}{s+2}}\right]=0$ 

$=\underset{s\to 0}{\mathrm{l}\mathrm{t}}\frac{4}{s+2+4H\left(s\right)}=0$ 

$4=2+4\underset{s\to 0}{\mathrm{l}\mathrm{t}}H\left(s\right)$ 

$\frac{2}{4}=\underset{s\to 0}{\mathrm{l}\mathrm{t}}H\left(s\right)$ 

$\underset{s\to 0}{\mathrm{l}\mathrm{t}}H\left(s\right)=\frac{1}{2}$ 

Concept:

The Steady-state error for different types of system is as shown:

Where ‘Kp, Kv, and Ka are the system gain.

Calculation:

$G\left(s\right)H\left(s\right)=\frac{k}{\left(1+s\right)\left(1+4s\right)}$

Since this is a type-zero system, the steady-state error will be:

$e_{ss}=\frac{1}{1+K_p}$

${K}_p=\underset{s\to \mathrm{\infty }}{lim}G\left(s\right)$

$K_p=\underset{s\to \mathrm{\infty }}{lim}\frac{k}{\left(1+s\right)\left(1+4s\right)}=k$

${e}_{ss}=\frac{1}{1+K_p}=0.2$

$\frac{1}{1+K}=0.2$

K = 4

Closed loop transfer function,

 $\mathrm{T}(\mathrm{s})=\frac{\mathrm{G}(\mathrm{s})\mathrm{H}(\mathrm{s})}{1+\mathrm{G}(\mathrm{s})\mathrm{H}(\mathrm{s})}=\mathrm{C}{\left[\mathrm{s}\mathrm{I}-\mathrm{A}\right]}^{-1}\mathrm{B}$

Now, we have

$\left(\mathrm{s}\mathrm{I}-\mathrm{A}\right)=\left[\begin{array}{cc}\mathrm{s}& -2\\ 4& \mathrm{s}+8\end{array}\right]$

${\left(\mathrm{s}\mathrm{I}-\mathrm{A}\right)}^{-1}=\frac{1}{\left({\mathrm{s}}^2+8\mathrm{s}+8\right)}\left[\begin{array}{cc}\mathrm{s}+8& 2\\ -4& \mathrm{s}\end{array}\right]$

${\left(\mathrm{s}\mathrm{I}-\mathrm{A}\right)}^{-1}\mathrm{B}=\frac{1}{\left({\mathrm{s}}^2+8\mathrm{s}+8\right)}\left[\begin{array}{cc}\mathrm{s}+8& 2\\ -4& \mathrm{s}\end{array}\right]\left[\begin{array}{c}0\\ 2\end{array}\right]=\frac{1}{({\mathrm{s}}^2+8\mathrm{s}+8)}\left[\begin{array}{c}4\\ 2\mathrm{s}\end{array}\right]$

$\mathrm{C}\left[\mathrm{s}\mathrm{I}-{\mathrm{A}}^{-1}\right]\mathrm{B}=\frac{1}{\left({\mathrm{s}}^2+8\mathrm{s}+8\right)}\left[\begin{array}{cc}2& 2\end{array}\right]\left[\begin{array}{c}4\\ 2\mathrm{s}\end{array}\right]=\frac{4\left(\mathrm{s}+2\right)}{\left({\mathrm{s}}^2+8\mathrm{s}+8\right)}$

Open loop transfer function $=\mathrm{G}\left(\mathrm{s}\right)\mathrm{H}\left(\mathrm{s}\right)=\frac{\mathrm{T}(\mathrm{s})}{1-\mathrm{T}(\mathrm{s})}=\frac{4\left(\mathrm{s}+2\right)}{\left({\mathrm{s}}^2+8\mathrm{s}+8\right)-4\left(\mathrm{s}+2\right)}$

$=\frac{4\left(\mathrm{s}+2\right)}{{\mathrm{s}}^2+4\mathrm{s}}$

$=\frac{4\left(\mathrm{s}+2\right)}{\mathrm{s}\left(\mathrm{s}+4\right)}$

∴ Position error constant ${\mathrm{K}}_{\mathrm{p}}=\underset{\mathrm{s}\to 0}{lim}\mathrm{G}\left(\mathrm{s}\right)\mathrm{H}\left(\mathrm{s}\right)=\mathrm{\infty }$

∴ Steady state error $={\mathrm{e}}_{\mathrm{s}\mathrm{s}}=\frac{1}{1+{\mathrm{K}}_{\mathrm{p}}}=0$

Note:

For step input, steady state error ${\mathrm{e}}_{\mathrm{s}\mathrm{s}}=\left[1+\mathrm{C}\left({\mathrm{A}}^{-1}\right)\mathrm{B}\right]$

For ramp input, steady rate error ${\mathrm{e}}_{\mathrm{s}\mathrm{s}}=\underset{\mathrm{t}\to \mathrm{\infty }}{lim}\left[\left[1+\mathrm{C}\left({\mathrm{A}}^{-1}\right)\mathrm{B}\right]\mathrm{t}+\mathrm{C}{\left({\mathrm{A}}^{-1}\right)}^2\mathrm{B}\right]$

When reference input is zero i.e. $\mathrm{R}\left(\mathrm{s}\right)=0$, the transfer function between $\mathrm{C}(\mathrm{s})\mathrm{a}\mathrm{n}\mathrm{d}{\mathrm{T}}_{\mathrm{L}}(\mathrm{s})$ is given by

$\begin{array}{l}\frac{C\left(s\right)}{T_L\left(s\right)}=\frac{\frac{-1}{0.15s^2+s}}{1+\frac{5}{0.15s^2+s}}=\frac{-1}{0.15s^2+s+5}\\ \therefore \mathrm{C}\left(\mathrm{s}\right)=\frac{-1}{0.15{\mathrm{s}}^2+\mathrm{s}+5}\left(\frac{1}{\mathrm{s}}\right)\end{array}$

We know error $\mathrm{E}\left(\mathrm{s}\right)=\mathrm{R}\left(\mathrm{s}\right)-\mathrm{C}\left(\mathrm{s}\right)$

$\therefore \mathrm{E}\left(\mathrm{s}\right)=0-\mathrm{C}\left(\mathrm{s}\right)=\frac{1}{0.15{\mathrm{s}}^2+\mathrm{s}+5}\ast \frac{1}{\mathrm{s}}$

The steady state error due to ${\mathrm{T}}_{\mathrm{L}}$ is given by

$\begin{array}{l}{e}_{ss}=\begin{array}{c}1t\\ s\to 0\end{array}s.E\left(s\right)\\ =\begin{array}{c}1t\\ s\to 0\end{array}\mathrm{s}\frac{1}{0.15{\mathrm{s}}^2+\mathrm{s}+5}\ast \frac{1}{\mathrm{s}}\\ =\frac{1}{0+0+5}=0.2\\ \therefore {\mathrm{e}}_{\mathrm{s}\mathrm{s}}=0.2\end{array}$