Concepts Of Stability Practice Questions

Characteristic equation, $s^{7+}s^{6+}7s^{5+}14s^{4+}31s^{3+}73s^{2+}25s+200=0$ Auxillary equation, $A(s)=8s^{4+}48s^{2+}200$ $\frac{d}{ds}A(s)=32s^{3+}96s$ Total number of poles =7 Two sign change above auxillary equation=2 poles in RHS. Two sign changes below auxillary equation implies out of 4 symmetric roots about origin, two poles are in LHS and two poles are in RHS. Therefore 3 poles in LHS and 4 poles in RHS.

Characteristic equation is, $s^{3+}Ks^{2+}(K+2)s+3=0$ For this system to be stable, using Routh's criterion, we can write, $3 \lt K(K+2)$ or, $K^{2+}2K-3 \gt 0$ or, $(K+3)(K-1) \gt 0$ Here, the valid answer will be out of all the options given. $i.e K \gt 1.$

From the given equation, $s^{3+}3s^{2+}2s+K=0$ Using Routh's criterion, we get $K \lt 6$ and $K \gt 0$ or $0 \lt K \lt 6$

Open loop transfer function: $G(s)=\frac{K(s+1)}{s(1+Ts)(1+2s)};$ $\; K \gt 0 \;$ and $\; t \gt 0$ For closed loop system stability, characteristic equation is, $1+G(s)H(s)=0$ $1+\frac{K(s+1)}{s(1+Ts)(1+2s)}\cdot 1=0$ $s(1+Ts)(1+2s)+K(s+1)=0$ $2Ts^{3+}(2+T)s^{2+}(1+K)s+K=0$ Using Routh's criteria, For stability, $K \gt 0$ and $(2+T)(1+K)-2TK \gt 0$ $K(2+T-2T)+(2+T) \gt 0$ or $-(T-2)K+2(2+T) \gt 0$ $-K \gt -\frac{(2+T)}{(T-2)}$ or $K \lt \frac{T+2}{(T-2)}$ Hence for stability, $0 \lt K \lt \frac{T+2}{T-2}$

$s^{3+}5.5s^{2+}8.5s+3=0$ Putting, $s=s_1-1$ $(s_1-1)^{3+}5.5(s_1-1)^{2+}(8.5)(s_1-1)+3=0$ $s_1^{3+}2.5s_1^{2+}0.5s_1-1=1$ As there is one sign change, hence, two roots of given polynomial will lie to the left of $s=-1$ .

Response of transfer function is unit for all $\omega$ . $M=1; P_1=2-j3$ Second order system, hence number of poles =2 Therefore, second pole $P_2=2+j3$ Now for M=1, and due to x-axis symmetry of root locus of transfer function, position of zeroes must be $Z_1=-2-j3$ and $Z_2=-2+j3$

When all the element s of a row in Routh-Hurwitz array ends abruptly. i.e all elements of that row have a zero values then, the system will be either unstable or maginally stable. Marginally stable means it will have imaginary roots (two equal and opposite complex roots on imaginary axis of s-plane).

Given, $G(s)=\frac{(s+\alpha )}{s^{3+}(1+\alpha )s^{2+}(\alpha -1)s+(1-\alpha )}$ and $H(s)=1$ Characteristic equation of given control system is given by $1+G(s)H(s)=0$ or, 1+\left[ \frac{s+\alpha }{s^{3+}(1+\alpha )s^{2+}(\alpha -1)s+(1-\alpha )} \right]\=0$or,$s^{3+}(1+\alpha )s^{2+}(\alpha -1)s+(1-\alpha )+s+\alpha =0$or,$s^{3+}(1+\alpha )s^{2+}\alpha s+1=0$Routh's array is For the given system to be stable, there should not be any sign change \in the first column of Routh's array. Therefore,$1+\alpha \gt 0 ;or, ; \alpha \gt -1;; ...(i)$Also,$\frac{\alpha ^2 +\alpha -1}{1+\alpha } \gt 0$or,$\alpha ^2 +\alpha -1 \gt 0$or,$(\alpha +1.618)(\alpha -0.618) \gt 0$or,$\alpha \lt -1.618$or,$\alpha \gt 0.618$As$\alpha \gt -1$(using equation (i)) therefore,$\alpha \gt 0.618 ;;...(ii)$Combining condition (i) and (ii),$-1 \lt \alpha \lt 0.618$Thus, for the system to be stable minimum value of$\alpha =0.618.$

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Given, $G(s)=\frac{K}{s(s+2)(s^{2+}2s+2)}$ The characteristic equation of given unity-negative feedback control system is given by $1+G(s)H(s)=0$ or, $1+\frac{K}{s(s+2)(s^{3+}2s+2)}=0$ or, $s(s+2)(s^{3+}2s+2)+K=0$ or, $s^{4+}4s^{3+}6s^{2+}4s+K=0$ Forming Routh' array as shown below For stability of the system, $K \gt 0$ and $\frac{20-4K}{5} \gt 0$ or $K\gt 5$ $\therefore$ For stability, $0 \lt K \lt 5$ For given system to be marginally stable K=5

The characteristic equation, $\;\;1+G(s)H(s)=0$ $\Rightarrow s^{3+}as^{2+}(k+2)s+(k+1)=0$ Routh array of above equation, $as^{2+}(k+1)=0$ For $s=j\omega \; and \; \omega =2$ $-a\omega ^{2+}(k+1)=0$ $k=4a-1$ and $a=\frac{k+1}{k+2}$ $\Rightarrow \;\;\; a=0.75 \; and \; k=2$

$1+\frac{s-1}{(s+2)(s+3)}=0$ $s^{2+}6s+5=0$ There is no root in RHS of s-plane.

Routh-Array The third row vanishes. An auxillary equation is formed using elements pf 2nd row. Auxiliary equation $A(s)=4s^{2+}4=0\Rightarrow s=\pm j.$ The derivative of this auxiliary equation is taken w.r.t. s and the cofficient of the differntiated equation are taken as the elements of 3rd row. $\frac{dA(s)}{ds}=8s$ Routh-Array There is no root in RHS of s-plane. Two roots of $s=\pm j$ , so one root is in LHS of s-plane.

$G(s)=\frac{k}{s(s+3)(s+10)}$ and $H(s)=1$ Characteristic equation, $\Rightarrow 1+G(s)H(s)=0$ $\;\; 1+\frac{k}{s(s+3)(s+10)}=0$ $\Rightarrow s(s+3)(s+10)+k=0$ $\Rightarrow s^{3+}13s^{2+}30s+k=0$ Routh-Array According to Routh-Hurwitz criterian. For stable system, signs of first column do not change $k \gt 0$ and $\frac{13 \times 30-k}{13} \gt 0$ Therefore system to be stable $0 \lt k \lt 390.$

$(T/F)_1=\frac{\frac{s-1}{s+2}}{1+\frac{s-1}{s+2} \times \frac{1}{s-1}}$ $\;\;=\frac{s-1}{s+3}$ Pole is in LHS of s-plane, hence stable. $(T/F)_2=\frac{\frac{1}{s-1}}{1+\frac{1}{s-1} \times \frac{s-1}{s+2}}$ $\;\;=\frac{s+2}{(s-1)(s+3)}$ Hence, unstable as it ha pole at right hand side of s-plane.

Open loop transfer function, $G(s)H(s)=k\frac{s+3}{(s+8)^2}$ $\therefore$ Characteristic equation, $\;\;1+G(s)H(s)=0$ $\Rightarrow 1+k\frac{(s+3)}{(s+8)^2}=0$ $\Rightarrow s^{2+}16s+64+(s+3)=0$ $\Rightarrow s^{2+}(16+k)s+(64+3k)=0$ Routh-Array, For sustained oscillation, $\;\; 16+k=0$ $\Rightarrow k=-16$ Auxillary equation, $\;\;s^{2+}64+3k=0$ $\Rightarrow s^{2+}64+3x(-16)=0$ $\;\;s^{2+}16=0$ $\Rightarrow s=\pm j4$ So, frequency of oscillation in 4 rad/sec.

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Characteristic equation $\;\;1+G(s)H(s)=0$ $\Rightarrow 1+\frac{K(1-s)}{1+s}=0$ $\Rightarrow s+1+K(1-s)=0$ $\Rightarrow s(1-K)+1+K=0$ $\;\;s=\frac{K+1}{K-1}$ For a stable system pole lies in left hand side of s-plane, it means, must be negative for stable system. $s \lt 0$ $\frac{K+1}{K-1} \lt 0$ CASE-I: $K+1 \lt 0 \text{ and } K-1 \gt 0$ $K \lt -1 \text{ and } K \gt 1$ which is not possible CASE-II: $K+1 \gt 0 \text{ and } K-1 \lt 0$ $K \gt -1 \text{ and } K \lt 1$ $-1 \lt K \lt 1$ $|K| \lt 1$

$s^{3-}4s^{2+}s+6=0$ The routh-array is formed as follow: Number of sign changes in forst column of Routh array =2 According to Routh-Hurwitz criterion, the number of changes of sign in the first column gives the number of positive real part roots of the polynomial. So, number of roots in RHS of s-plane=2 Ttoal number of roots=3 Hence, no of roots in LHS of s-plane =3-2=1