Question

List five specifications for a second order underdamped system

Answer 1

Lecture Notes on Control Systems/D. Ghose/2012 57
10.7 Performance of Second-Order System (Unit Step
Response)
Consider the second order system
a2y¨ + a1y˙ + a0y = b0r
So,
Y (s)
R(s) = b0
a2s2 + a1s + a0
= K ω2
n
s2 + 2ζωns + ω2
n
where,
ω2
n = a0
a2
, K = b0
a0
, ζ = a1
2
√a0a2
Apply a unit step input r(t) = u(t) = 1
s
Then,
Y (s) = K ω2
n
s (s2 + 2ζωns + ω2
n)
This can be written as,
Y (s) = k1
s
+
k2s + k3
s2 + 2ζωns + ω2
n
Let us find the values of k1, k2, and k3 by equating the coefficients of the numerator
polynomials.
Y (s) = K ω2
n
s(s2 + 2ζωns + ω2
n) = (k1 + k2)s2 + (2ζωnk1 + k3)s + ω2
nk1
s(s2 + 2ζωns + ω2
n)
Comparing the coefficients,
k1 + k2 = 0
2ζωnk1 + k3 = 0
ω2
nk1 = Kω2
n
which can be solved to yield
k1 = K
k2 = −K
k3 = −2ζωnK