Step response using Laplace transform
First order systems
Problem:
1 dy
+ y = u(t)
(1)
a dt
Solve for y(t) if all initial conditions are zero.
Show that y(∞) = 1. Use the Laplace transform.
Solution. Use Table A and Table B. If all initial conditions are zero, applying Laplace transform, we have
Y (s) =
1
1
a
= −
s(s + a)
s s+a
So
y(t) = 1 − e−at
Use Final value theorem (Item 11 Table B)
s
y(∞) = lim sY (s) = lim (1 −
)=1
s→0
s→0
s+a
1
General second order systems
Problem: If all initial conditions are zero. Find
the general function form of y(t) for the general second order system
1 d2y
dy
2
(
+
2ζω
+
ω
y) = u(t)
n
n
2
2
ωn dt
dt
where ωn is called the natural frequency. ζ is
the damping ratio. Solve the problem in each
of the following cases:
1. ζ > 1 (overdamped)
2. ζ = 1 (critically damped)
3. 0 < ζ < 1 (underdamped)
4. ζ = 0 (undamped)
respectively.
2
Solution: Use Table A and Table B. If all initial conditions are zero, applying Laplace transform, we have Y (s)
2
ωn
Y (s) =
2)
s(s2 + 2ζωns + ωn
(i) ζ > 1 (overdamped)
2 = (s + σ )(s + σ )
s2 + 2ζωn s + ωn
1
2
where
σ1 = ζωn +
q
σ2 = ζωn −
q
ζ 2 − 1ωn
ζ 2 − 1ωn
are real and distinct (Case 1).
Y (s) =
K1
K2
K3
+
+
s
s + σ1
s + σ2
The general form of y(t) is
y(t) = K1 + K2e−σ1t + K3e−σ2t
3
(ii) ζ = 1 (critically damped)
2
s2 + 2ζωn s + ωn
= (s + σ)2
where
σ = ωn
are real and repeated (Case 2).
K2
K3
K1
+
+
Y (s) =
2
s
(s + σ)
s+σ
The general form of y(t) is
y(t) = K1 + K2te−σt + K3e−σt
4
(iii) 0 < ζ < 1 (Undamped)
2
s2 + 2ζωn s + ωn
= (s + σ1)(s + σ2)
where
q
σ1 = ζωn + j 1 − ζ 2ωn
q
σ2 = ζωn − j 1 − ζ 2ωn
are complex (Case 3).
Partial fraction expansion of Y (s)
2
K1
K2s + K3
ωn
=
+
2)
2
s(s2 + 2ζωn s + ωn
s
s2 + 2ζωn s + ωn
Find K1, K2, K3
2
ωn
K1 = 2
|
=1
2 s→0
s + 2ζωn s + ωn
5
2
2
ωn
= (s2 + 2ζωn s + ωn
) + s(K2s + K3)
Balancing the coefficients for s1 and s2,
s1 :
s2 :
0 = 2ζωn + K3
0 = 1 + K2
K2 = −1, K3 = −2ζωn . Then we have
1
s + 2ζωn
Y (s) = −
2(1 − ζ 2 )
s (s + ζωn )2 + ωn
1
= −
s
q
s + ζωn + (ωn 1 − ζ 2) √ ζ
1−ζ 2
2 (1 − ζ 2 )
(s + ζωn )2 + ωn
The general form of y(t) is
y(t) = 1 − e
−ζωn t
cos(ωn
p
ζ
2
1 − ζ t) +
1
p
1 − ζ2
sin(ωn
p
1 − ζ 2t)
q
=1−q
e−ζωn t cos(ωn 1 − ζ 2t − φ)
1 − ζ2
where φ = arctan( √ ζ
1−ζ 2
).
6
!
(iv) ζ = 0 (Undamped)
2
2
s2 + 2ζωn s + ωn
= s2 + ωn
where two complex roots are +jωn and −jωn,
still Case 3.
Partial fraction expansion of Y (s)
2
ωn
K1
K2s + K3
=
+
2)
2
s(s2 + ωn
s
s2 + ωn
Find K1 = 1 as usual. Find K2, K3
2
2
ωn
= (s2 + ωn
) + s(K2s + K3)
Balancing the coefficients for s1 and s2,
s1 :
0 = K3
s2 :
0 = 1 + K2
K2 = −1, K3 = 0. Then we have
Y (s) =
s
1
−
2
s s + ωn
y(t) = 1 − cos ωnt
7
Figure above: Second order system response
as a function of damping ratio.
8
Underdamped second order systems
Figure below: Second order underdamped response y(t) for damping ratio values.
y(t) = 1 − q
1
1 − ζ2
q
e−ζωn t cos(ωn 1 − ζ 2t − φ)
where φ = arctan( √ ζ
1−ζ 2
).
Peak time Tp is
found by differentiating y(t) and then the first
zero crossing after t = 0.
2
d
ωn
L[ y(t)] = sY (s) = 2
2)
dt
(s + 2ζωn s + ωn
=
√ ωn
1−ζ
q
2
ω
1
−
ζ
n
2
2 (1 − ζ 2 )
(s + ζωn )2 + ωn
9
d
ωn
q
e−ζωn t sin(ωn 1 − ζ 2t) = 0
y(t) =
dt
1 − ζ2
q
q
ωn 1 − ζ 2t = π
Tp =
π
q
ωn 1 − ζ 2
The maximum ymax is
√
−ζπ/
y(Tp) = 1 − e
1−ζ 2
The final value y(∞) = 1. Thus
√
ymax − y(∞)
−ζπ/ 1−ζ 2
=e
× 100
%OS =
y(∞)
We can similarly found that the settling time
(defined as the time required to settle with
±2% of the final value) can be approximated
4 .
by Ts = ζω
n
10
Figure above:
underdamped
part; (b) with
with constant
Step responses of second order
systems. (a) with constant real
constant imaginary part and (c)
damping ratio.
11