Lecture 17
Second-Order Systems
Second-Order System Response (pp.118-125)
• For step-by-step derivatives of all second-order system equations, see: http://www.nd.edu/ pdunn/www.text/derivations.pdf
• The general expression for a second-order ODE is
a2ÿ + a1ẏ + a0y = F (t).
(1)
• The is a linear second-order ODE that can be rearranged as
1
2ζ
ÿ
+
ẏ + y = KF (t),
ωn2
ωn
(2)
r
where ωn = a0/a2 denotes the natural frequency and ζ =
√
a1/2 a0a2 damping ratio of the system.
• Note that when 2ζ >> 1/ωn, the second derivative term
in Equation 2 becomes negligible with respect to the other
terms, and the system behavior approaches that of a firstorder system with a system time constant equal to 2ζ/ωn.
r
r
• ωn = 1/LC for a RLC circuit and k/m for a springmass-damper system.
r
√
• ζ = R/ 4L/C for a RLC circuit and γ/ 4km for a springmass-damper system.
1
• The governing equation for a spring-mass-damper system is
m d2y γ dy
1
+
+
y
=
F t.
k dt2 k dt
k
• The governing equation for a RLC circuit is
dI
d2 I
dEi(t)
LC 2 + RC + I = C
.
dt
dt
dt
(3)
(4)
EXAMPLE PROBLEM: Determine the input-output
expression for the RLC circuit shown in Figure 1.
• Kirchhoff’s voltage law (KVL)can be used to determine the
expression that relates the circuit’s output voltage , Eo, to
its input voltage, Ei, and the values of R, C, and L.
Figure 1: RLC circuit diagram.
2
• Consider the loop from the ground of Ei through R, L, and
C back to ground. KVL gives:
• Now recall that I = dQ/dt. Substitution gives:
which is a linear, second-order ODE.
• Now examine another loop from the C’s ground through to
Eo’s ground. Application of KVL yields:
• So, to find Eo, we must first find Q. Thus, we must integrate
the derived second-order ODE. Choose a forcing function
of Ei = A sin(ωt). Integration (see the system equation
derivations) results in:
• Now use the equation involving Eo from the second loop.
Substitution gives the final equation:
3
• Finally, examine what happens when L is removed from the
circuit. Letting L = 0 in the above solution results in:
which is the same as Eqn. 4.38 for the response of a firstorder system to sinusoidal-input forcing.
Solutions for Step-Input Forcing
• For step-input forcing, there are three specific solutions of
the governing ODE. This is because there are three different
roots in the characteristic equation
1 2 2ζ
r + r + 1 = 0.
ωn2
ωn
• Application of the quadratic formula simplifies to
r
r1,2 = −ζωn ± ωn ζ 2 − 1.
(5)
(6)
• Two initial conditions are required to obtain a solution: ẏ =
0 and y(0) = 0.
• The exact solution is of the form y(t) = yh + yp where
yp = KF (t).
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• The form of the homogeneous solution, yp, is governed by
√
the the value of the discriminant ζ 2 − 1.
• When ζ 2 − 1 > 0: the roots are real, negative, and distinct.
The general form of the solution is
yh(t) = c1er1t + c2er2t.
(7)
• When ζ 2 − 1 = 0: the roots are real, negative, and equal to
−ωn. The general form of the solution is
yh(t) = c1ert + c2tert.
(8)
• When ζ 2 − 1 < 0: the roots are complex and distinct. The
general form of the solution is
yh(t) = c1er1t + c2er2t = eλt(c1 cos µt + c2 sin µt),
(9)
using Euler’s formula eit = cos t + i sin t and noting that
r1,2 = λ ± iµ,
√
with λ = −ζωn and µ = ωn 1 − ζ 2.
5
(10)
2
1.8
ζ=0
1.6
0.2
1.4
0.4
M(t)
1.2
0.6
0.8
1
1.0
1.7
0.8
2.5
0.6
0.4
0.2
0
0
2
4
6
8
10
12
14
ωnt
Figure 2: Second-order system response to step-input forcing.
• The exact solutions when F (t) = KA are given by Equations (4.57 through (4.60) in the text.
• All of the solutions when presented in a nondimensional
form can be plotted, as shown in Figure 2.
SUGGESTED PROBLEMS: RP4.3, RP4.4, RP 4.9,
RP4.20, HP4.8, HP4.10, HP4.14
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