Question 1)The feedback control system has G(s)=K(s+6)/(s+4) and an ideal integrator with unity gain in feedback path.Determine the following with reference to root locus technique: (i)K=0 and K=infinity points on root loci (ii)number of asymptotes (iii)breakaway points,if any 2)if root locus of s(s+2)+K(s+4)=0 is a circle.What are the coordinates of the centre of this circle? 3)Also sktech the asymptotes for K>0 and K<0 for G(H)=K/s(s+2)(s+1+j)(s+1-j)

30QBND The Asker · Chemical Engineering

1)The feedback control system has
G(s)=K(s+6)/(s+4)
and an ideal integrator with unity gain in feedback path.Determine the following with reference to root locus technique:
(i)K=0 and K=infinity points on root loci
(ii)number of asymptotes
(iii)breakaway points,if any
2)if root locus of s(s+2)+K(s+4)=0 is a circle.What are the coordinates of the centre of this circle?

3)Also sktech the asymptotes for K>0 and K<0 for
G(H)=K/s(s+2)(s+1+j)(s+1-j)

More

Community Answer

Honor Code

Solved 1 Answer

Answer:Given: \rightarrow\begin{aligned}G(s) & =\frac{k(s+6)}{(s+4)} \\H(s) & =\frac{1}{s} \\G(s) H(s) & =\frac{k(s+6)}{s(s+4)}\end{aligned}(i) K=0, represents the location of open leop poles i-e s=-4 and s=0 in the given case.K=\infty, represents the open leop sero i.es=-6 \text { and } s=-\infty \text {. }(ii)\text { Number of asymptotes }=\text { no. of open loop poles }(P)-\text { no: of }\begin{aligned}& \text { open loop zeros (z) } \\= & (2-1)=1\end{aligned}(iii) Characteristic equation:\begin{array}{r}1+\frac{k(s+6)}{s(s+4)}=0 \\k=\frac{-s(s+4)}{(s+6)}\end{array}Fer breakaway points, \frac{d k}{d s}=0\begin{array}{c}\Rightarrow\left[\frac{(2 s+4)(s+6)-\left(s^{2}+4 s\right)}{(s+6)^{2}}\right]=0 \\2 s^{2}+16 s+24-s^{2}-4 s=0 \\s^{2}+12 s+24=0 \\s=-2.536,-9.464 \\\uparrow\end{array}Breakauay Break in Point point\begin{array}{c}s(s+2)+k(s+4)=0 \\1+\frac{k(s+4)}{s(s+2)}=0 \quad \text { (Dividing by } s(s+2) \\\therefore \quad G(s) H(s)=\frac{k(s+4)}{s(s+2)}=\frac{k(s+b)}{s(s+a)} \text { (Comparing) } \\\text { Centre }=(-b, 0)=(-4,0) .\end{array}Therefere, the co-ardinates of the centre of this circle is (-4,0)3)Given,G(H)=\frac{K}{s(S+2)(s+1+j)(s+1-j)}The center of asymptotes is\sigma_{c}=-\frac{(0+2+1+j+1-j)}{4}=-1Fer k>0, the angles of the asymptotes areQ=45^{\circ}, 135^{\circ}, 225^{\circ} \text {, and } 315^{\circ}Fer k<0, the angles of the asymptotes are\theta=0^{\circ}, 90^{\circ}, 180^{\circ} \text {, and } 270^{\circ}The plot's are shoun abore. ...