(i) Gain

(ii) Damping ratio

(iii) Natural Frequency

(iv) Transfer function

From the given figure,

$M_p$=3.45-2

=1.45

$t_p$=0.55 sec

$t_s$=6 sec

$M_p$=e$^\frac{-πξ_2}{{\sqrt{(1-ξ_2 2))}}}$

1.45=e$^\frac{-πξ_2}{\sqrt{(1-ξ_2 2)}}$

Log 1.45=$\frac{-πξ}{\sqrt{(1-ξ2)}}$ log e

0.161=$\frac{-πξ}{\sqrt(1-ξ2)}$ x 0.4342

0.3707=$\frac{-πξ}{\sqrt{(1-ξ2)}}$

0.118=$\frac{ξ}{\sqrt{(1-ξ2}}$

Squaring both the sides

0.0139=$\frac{ξ^2}{(1-ξ^2 )}$

0.0139-0.0139$ξ^2$=$ξ^2$

1.0139$ξ^2$=0.0139 ξ= 0.117

Peak time,$t_p$=0.55 sec=$\frac{π}{w_d}$

$w_d$ =5.70 rad/sec

$w_d$=$w_n \sqrt{(1-ξ2)}$

$w_n$=$\frac{5.70}{\sqrt{(1-5.70^2 )}}$

$w_n$=5.789 rad/sec

$w_n$->Natural frequency

T.F=$\frac{(w_n^2)}{(S^2+2ξw_n S+w_n^2 )}$ …