Automatic Control Systems
Assessment 3 Answers
Year 2025
0027 63 985 5033
,
, QUESTION 1
(a) 𝑠 3 + 25𝑠 2 + 10𝑠 + 450 = 0
Routh’s Tabulation
𝑠3 1 10
𝑠2 25 450
250 − 450
𝑠1 = −8 0
25
𝑠0 450
There are two sign changes in the first column therefore the characteristic
equation has two roots in the right-half of the s-plane and zero roots on the 𝑗𝜔
axis.
Using MATLAB to find the roots:
The system is unstable.
Note: A system is considered stable if all the roots of its chacteristic equation
have negative real parts.
Assessment 3 Answers
Year 2025
0027 63 985 5033
,
, QUESTION 1
(a) 𝑠 3 + 25𝑠 2 + 10𝑠 + 450 = 0
Routh’s Tabulation
𝑠3 1 10
𝑠2 25 450
250 − 450
𝑠1 = −8 0
25
𝑠0 450
There are two sign changes in the first column therefore the characteristic
equation has two roots in the right-half of the s-plane and zero roots on the 𝑗𝜔
axis.
Using MATLAB to find the roots:
The system is unstable.
Note: A system is considered stable if all the roots of its chacteristic equation
have negative real parts.
