Show that $\frac{(Tk_1-1)}{(Tk_2-1)}$=43.33, where $k_1$ & $k_2$ are values of ‘k’ for 0.6 & 0.2 overshoot respectively

T.F=$\frac{(C(S))}{(R(S))}$

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

=$\frac{(\frac{k}{(S(1+TS})}{(1+\frac{k}{(S(1+TS)}.1)}$

=$\frac{(\frac{k}/{T})}{(S^2+\frac{S}{T}+\frac{k}{T})}$

Comparing with standard form

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

$w_n^2$ =$\frac{k}{T}$

Therefore, $w_n$=$\sqrt{\frac{k}{T}}$

$2ξw_n$=$\frac{1}{T}$

Therefore, ξ=$\frac{1}{2}\sqrt{kT}$

Now for $M_P$=0.6,

Let ξ=$ξ_1$

Therefore, 0.6=$e^\frac{-πξ_1}{\sqrt{(1-ξ_1 2)}}$

Therefore,$ξ_1$=0.1602

For $M_P$=0.2,

Let ξ=$ξ_2$

Therefore, 0.2=$e^\frac{-πξ_2}{\sqrt{(1-ξ_2 2)}}$

Therefore, $ξ_2$=0.4559

$ξ_1$=0.1602, k=$k_1$

&$ξ_2$=0.4559, k=$k_2$

$ξ_1$=$\frac{1}{2}{\sqrt{(k_1 T)}}$

Therefore, 0.1602=$\frac{1}{2}{\sqrt{(k_1 T)}}$…………………(1)

$ξ_2$=$\frac{1}{2}{\sqrt{(k_2 T)}}$

Therefore, 0.4559=$\frac{1}{2}{\sqrt{(k_2 T)}}$)…………………(2)

For equation(1),

${\sqrt{(k_2 T)}}$=$\frac{1}{2}$ x 0.1602 =3.1210

$k_1 $T=9.74

$k_1$ T-1=8.7412……………….(3)

From equation (2),

${\sqrt{(k_2 T)}}$=$\frac{1}{2}$ x 0.4559)

=1.09673

$k_2$ T=1.2028

$k_2$ T-1=0.2028…………………..(4)

From equation (3) & (4),

$\frac{(k_1 T-1)}{(k_2 T-1)}$=$\frac{8.7412}{0.2028}$

=43.33