Concepts of Stability

Routh array is

For BIBO stability,
\frac{\left ( \frac{8}{3}-3k \right )}{\left ( \frac{8}{3} \right )} \gt 0
\Rightarrow k \lt \frac{8}{9}
and k \gt 0
\therefore \;\; 0 \lt k \lt \frac{8}{9}

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.

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

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.

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

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

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.