(1) Find steady state error.

(2) If it is desired to have steady state error to be 0.2 find corresponding value of ‘k’

(3) Find steadystate error if input is changed to 2+6t and value of ‘k’ to 10.

G(S) =$\frac{K}{(S(1+S)(1+0.4S)}$

H(S) =1, unity feedback.

(1) For Ramp input

r (t)=4t,

k=2

Steady state error,

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

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

=$\lim_(s\to 0)$ $S\frac{k}{(S(1+S)(1+0.4S)}$

=$\lim_(s\to 0)$ $S\frac{2}{(S(1+S)(1+0.4S)}$ =2/(1)(1)

$k_v $ = 2

$e_ss$=$\frac{A}{k_v}$ ,

As A=2 for Ramp input

$e_ss$=2/2 e_ss=1

(2) $e_ss$=0.2, k=?

same input Ramp of magnitude

A=2

$k_v$=$\lim_(s\to 0)$ S. G(S) H(S) =$\lim_(s\to 0)$ $S\frac{k}{(S(1+S)(1+0.4S)}$

$k_v$=k/(1)(1)

$k_v$=k

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

=$\frac{A}{k}$

0.2=2/k

k=2/0.2

k=10

(3) Steady state error for input step of magnitude A_1=2 and Ramp of magnitude A_2=6.

k=10.

For step input, position error coefficient

$k_p$=$\lim_(s\to 0)$ G(S) H(S)

=$\lim_(s\to 0)$ $\frac{k}{(S(1+S)(1+0.4S)}$ . 1

=$\lim_(s\to 0)$ $\frac{10}{(S(1+S)(1+0.4S)}$ =$\lim_(s\to 0)$ $\frac{10}{(0(1)(1)}$ =10/0 $k_p$=∞

For Ramp input velocity error coefficient,

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

=$\lim_(s\to 0)$ S$\frac{k}{(S(1+S)(1+0.4S)}$ . 1

=$\lim_(s\to 0)$ S$\frac{10}{(S(1+S)(1+0.4S)}$ .1

=$\lim_(s\to 0)$ $\frac{10}{(1)(1)}$

=10

$k_v$=10

$e_ss$=$\frac{A_1}{(1+k_p)}$+$\frac{A_2}{k_v}$

=$\frac{2}{∞}$+$\frac{6}{10}$

=0+0.6

$e_ss$=0.6