Using first principle

i. Determine the position error & steady state error for a unit step input.

ii. Obtain the equation for close loop response in time domain C(t) due to step input.

iii. What is steady state error of close loop response for ramp and parabolic inputs?

G(S)=$\frac{1}{(S^2+2S+3)}$

Position error $k_p$= $\lim_(S\to 0)$ G(S) H(S), H(S)=1

=$\lim_(S\to 0)$ $\frac{1}{(S^2+2S+3)}$ x 1

$k_p$=$\frac{1}{3}$

For unit step input, magnitude A=1

$e_ss$=$\frac{A}{(1+k_p )}$=$\frac{1}{(1+1/3)}$

$e_ss$=0.75

Comparing the given equation with second order standard equation

$\frac{(w_n^2)}{(S^2+2ξw_n S+w_n^2 )}$

$w_n^2$=3

Therefore,$w_n$=1.732

$2ξw_n$=2

Therefore, ξ=0.577 underdamped system

Θ=〖tan^(-1)$\sqrt{\frac{(1-ξ2)}{ξ}]}$

Θ=tan^(-1)$\sqrt{\frac{(1-0.577)}{0.577}}]$

=54.76

Θ=0.9557 radians $w_d$=$w_n$ $\sqrt{(1-ξ2)}$

=1.732$\sqrt{(1-0.577)}$

=1.414 rad/sec

We know that for underdamped system, the output(response) in time domain is,

C(t)=1-e$^\frac{(-ξw_n t)}{\sqrt{(1-ξ2)sin(w_d+θ)}}$

=1-e$^\frac{(-0.999t)}{0.8167sin(1.414+0.9557)}$

C(t)=1-1.2244e$^(-0.999t) sin(1.414+0.9557)$

Ramp input, magnitude$ A_2$

$k_v$ =$\lim_(S\to 0)$ S G(S) H(S)

=$\lim_(S\to 0)$ $\frac{S}{(S^2+2S+3)}$ x 1

=0/3

$k_v$=0

$e_ss$=$\frac{A_2}{k_v}$

=$A_2$/0

$e_ss$=∞

Let $A_3$ is the magnitude for parabolic input

$k_a$=$\lim_(s\to 0)$ $S^2$ G(S) H(S)

=$k_v$=$\lim_(s\to 0) S^2$ $\frac{1}{(S^2+2S+3)}$ x 1

=0/3

$k_a$ =0

$e_ss$=$\frac{A_2}{k_a}$

=$A_3$/0

$e_ss$=∞