written 5.8 years ago by | • modified 5.8 years ago |
Mumbai University > Electronics Engineering > Sem 4 > Linear Control Systems
Topic: Stability Analysis in Time Domain
Difficulty : High
Marks : 10M
written 5.8 years ago by | • modified 5.8 years ago |
Mumbai University > Electronics Engineering > Sem 4 > Linear Control Systems
Topic: Stability Analysis in Time Domain
Difficulty : High
Marks : 10M
written 5.8 years ago by | • modified 5.7 years ago |
${ s }^{ 8 }+5{ s }^{ 6 }+2{ s }^{ 4 }+3{ s }^{ 2 }+1=0$ |$s^8$|1|5 |2 |3 |1| |-|-|-|-|-|-| |$s^7$|0|0|0|0|0| The auxiliary equation is $A(s)= s^8 +5s^6 +2s^4 + 3s^2 + 1$ Differentiating $A(s)$ with respect to s, $\therefore \cfrac { dA(s) }{ ds } = 8s^7 + 30s^5 +8s^3 +6s$ |$s^8$|1| 5| 2| 3| 1| |---|---|----|---|---|---| |$s^7$|8|30|8| 6| 0| |$s^6$|1.25| 1|2.25| 1|0| |$s^5$|23.6|-6.4| -0.4| 0| 0| |$s^4$|1.33| 2.271|1|0| 0| |$s^3$|-46.6|-18|14| 0|0| 0| |$s^2$|1.75|1| 0| 0| 0| |$s^1$|8.48| 0| 0| 0| 0| |$s^0$|1|-|-|-|-|
Since there are two sign changes, two of the 8 closed loop poles lie on the RHS. Hence, the system is unstable.