Question

The transfer function of a position servo system is given as
$G\left(s\right)=\frac{1}{s(s+1)}.$

A first order compensator is designed in a unity feedback configuration so that poles of the compensated system are placed at $-1\pm j1$ and -4. The transfer function of the compensated system is

• A. $\frac{s+3}{2(s+3)}$
• B. $\frac{2s+3}{(s+5)}$
• C. $\frac{5(s+1.6)}{s+5}$
• D. $\frac{3(2s+3)}{s+4}$

Answer 1

The correct option is C $\frac{5(s+1.6)}{s+5}$

Assume compensator T.F.
$G_c\left(s\right)=\frac{k(s+z_1)}{(s+p_1)}$


T.F. of closed loop,
$\frac{C\left(s\right)}{R\left(s\right)}=\frac{k(s+z_1)}{s(s+p_1)(s+1)+k(s+z_1)}$

$=\frac{k(s+z_1)}{s^3+p_1{s}^2+s^2+p_{1}s+ks+k{z}_1}$

$\frac{C\left(s\right)}{R\left(s\right)}=\frac{k(s+z_1)}{s^3+(p_1+1)s^2+(p_1+k)s+k{z}_1}$

$\therefore$ characteristic equation,

$s^3+(p_1+1)s^2+(p_1+k)s+k{z}_1=0$ ...(i)

and form given data,

$\left[\right(s+1{)}^2+1\left]\right(s+4)=0$

$(s^2+2s+2)(s+4)=0$

$s^3+2s^2+2s+4s^2+8s+8=0$

$\therefore$ $s^3+6s^2+10s+8=0$ ...(ii)

comparing equation (i) and (ii),

$P_1+1=6\Rightarrow P_1=5$

$P_1+k=10\Rightarrow k=10-5$

$\Rightarrow k=5$

$k{z}_1=8\Rightarrow z_1=8/5=1.6$

$\therefore G_c\left(s\right)=\frac{5(s+1.6)}{(s+5)}$