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}

There are 5 questions to complete.