Question 1 |

The characteristic equation of a linear time-invariant (LTI) system is given by

\Delta (s)=s^4+3s^3+3s^2+s+k=0

The system is BIBO stable if

\Delta (s)=s^4+3s^3+3s^2+s+k=0

The system is BIBO stable if

0 \lt k \lt \frac{12}{9} | |

k \gt 3 | |

0 \lt k \lt \frac{8}{9} | |

k \gt 6 |

Question 1 Explanation:

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}

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}

Question 2 |

The number of roots of the polynomial,

s^{7}+s^{6}+7s^{5}+14s^{4}+31s^{3}+73s^{2}+25s+200,

in the open left half of the complex plane is

s^{7}+s^{6}+7s^{5}+14s^{4}+31s^{3}+73s^{2}+25s+200,

in the open left half of the complex plane is

3 | |

4 | |

5 | |

6 |

Question 2 Explanation:

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.

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.

Question 3 |

The range of K for which all the roots of the equation s^{3}+3s^{2}+2s+K=0 are in the left half of the complex s-plane is

0 \lt K \lt 6 | |

0 \lt K \lt 16 | |

6 \lt K \lt 36 | |

6 \lt K \lt 16 |

Question 3 Explanation:

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

Using Routh's criterion, we get

K \lt 6 and K \gt 0 or 0 \lt K \lt 6

Question 4 |

A closed loop system has the characteristic equation given by s^{3}+Ks^{2}+(K+2)s+3=0. For this system to be stable, which one of the following conditions should be satisfied?

0 \lt K \lt 0.5 | |

0\lt K \lt 1 | |

0 \lt K \lt 1 | |

K \gt 1 |

Question 4 Explanation:

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.

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.

Question 5 |

The open loop transfer function of a unity feedback control system is given by

G(s)=\frac{K(s+1)}{s(1+Ts)(1+2s)}, K\gt 0,T \gt 0

The closed loop system will be stable if,

G(s)=\frac{K(s+1)}{s(1+Ts)(1+2s)}, K\gt 0,T \gt 0

The closed loop system will be stable if,

0\lt T \lt\frac{4(K+1)}{K-1} | |

0 \lt K \lt \frac{4(T+2)}{T-2} | |

0 \lt K \lt \frac{T+2}{T-2} | |

0 \lt T \lt \frac{8(K+1)}{K-1} |

Question 5 Explanation:

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}

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}

Question 6 |

Given the following polynomial equation s^{3}+5.5s^{2}+8.5s+3=0,

the number of roots of the polynomial, which have real parts strictly less than -1, is ________ .

the number of roots of the polynomial, which have real parts strictly less than -1, is ________ .

1 | |

2 | |

3 | |

4 |

Question 6 Explanation:

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.

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.

Question 7 |

The transfer function of a second order real system with a perfectly flat magnitude response of unity has a pole at (2-j3). List all the poles and zeroes.

Poles at (2\pmj3), no zeroes. | |

Poles at (\pm2-j3), one zero at origin. | |

Poles at (2-j3),(-2+j3), zeroes at (-2-j3),(2+j3). | |

Poles at (2\pmj3), zeroes at (-2\pmj3). |

Question 7 Explanation:

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

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

Question 8 |

A single-input single output feedback system has forward transfer function G(s)
and feedback transfer function H(s). It is given that |G(s)H(s)| \lt1. Which of
the following is true about the stability of the system ?

The system is always stable | |

The system is stable if all zeros of G(s)H(s) are in left half of the s-plane | |

The system is stable if all poles of G(s)H(s) are in left half of the s-plane | |

It is not possible to say whether or not the system is stable from the
information given |

Question 9 |

A system with the open loop transfer function

G(s)=\frac{K}{s(s+2)(s^{2}+2s+2)}

is connected in a negative feedback configuration with a feedback gain of unity. For the closed loop system to be marginally stable, the value of K is ______

G(s)=\frac{K}{s(s+2)(s^{2}+2s+2)}

is connected in a negative feedback configuration with a feedback gain of unity. For the closed loop system to be marginally stable, the value of K is ______

4 | |

5 | |

6 | |

7 |

Question 9 Explanation:

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 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

Question 10 |

For the given system, it is desired that the system be stable. The minimum value
of \alpha for this condition is ______.

0.22 | |

0.46 | |

0.86 | |

0.62 |

Question 10 Explanation:

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.

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.

There are 10 questions to complete.