Car suspension model
Mass – spring – viscous damper system
Free body diagram (no motion)
Force balance ➠
8
2
d
>
System ODE : < m x
dt2
(2nd order ordinary
FK
>
:
linear differential
Fdv2 x
Equation of motion:
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Mg
Mg
FK0
FK0
force due to spring
in equilibrium
force because spring
changes length
during motion
Model
equation)
Free body diagram
=) m
dt2
=
=
FK + Fv
=
=
-K(x - u)
-fv (ẋ - u̇)
Kx + Ku
d2 x
=) m 2 =
dt
force due to
viscous damping
Kx + Ku
2
fv ẋ + fv u̇
fv ẋ +=)
fv u̇ m d x + Kx + f ẋ = Ku + f u̇
v
v
dt2
d2 x
=) m 2 + Kx + fv ẋ = Ku + fv u̇
dt
2.004 System Dynamics and Control Spring 2013
1
General Linear Time-Invariant (LTI) system
dnx
d n −1 x
dnx
dx
dmy
d m −1 y
d1y
an n + an −1 n −1 + an − 2 n La1 + a0 x = bm m + bm −1 m −1 + Lb1 1 + b0u
dt
dt
dt
dt
dt
dt
dt
nth-order Linear Ordinary Differential Equation (ODE)
with constant coefficients (time-invariant)
general solution:
x(t) = xhomogeneous (t) + xforced (t)
➡ homogeneous solution: y=0 (no forcing term)
➡ forced solution: a “guess” solution for the system behavior when y(t)≠0
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Homogeneous and forced solutions
Homogeneous solution:
ω
Im(s)
x(t) = C0 + C1es1t + C2 es2 t + L + Cn esn t
where in general
si = σ i + jω (complex number)
s
σ
Re(s)
Forced solution: sometimes difficult to “guess”
but for specific forces of interest, quite easy.
For example, if y(t)=constant, then yforced=constant
as well (but a different constant!)
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2.004 System Dynamics and Control Spring 2013
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Commonly used input functions
Step function
(aka Heaviside)
step(t) =
⇢
0,
1,
Ramp
function
ramp(t) =
t < 0;
t > 0.
⇢
0,
t,
t < 0;
t > 0.
= t step(t)
Sinusoidal
function
⇢
0,
t < 0;
f (t) =
sin(!t), t 0.
= sin(!t) step(t)
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2.004 System Dynamics and Control Spring 2013
δ(t)
Impulse
(aka delta-function
or Dirac function)
t [sec]
4
1st order system
M v̇ + bv = f (t)
mass
viscous
damping
force
one time constant
τ
Impulse response:
equivalent to setting an initial condition v(t=0)
v(t = 0) = v0
v(t) = v0 e
t/⌧
0. 368 ⇡ e
t>0
,
time constant ⌧ =
1
M
b
Step response:
f(t) is the “step function” (or Heaviside function)
f (t) = F0 step(t) =
v(t = 0) = 0
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⇢
0,
F0 ,
t < 0;
t > 0.
F0 ⇣
v(t) =
1
b
e
t/⌧
⌘
,
t>0
2.004 System Dynamics and Control Spring 2013
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1st order system: step response
steady state
(final value)
0.632 ⇡ 1
e
1
Nise Figure 4.3
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Commons license. For more information, see http://ocw.mit.edu/help/faq-fair-use/.
2.004 Fall ’07
Lecture 06 – Monday, Sept. 17
How is this different than the car suspension system?
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2.004 System Dynamics and Control Spring 2013
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2nd order system: step response
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2.004 System Dynamics and Control Spring 2013
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Mechanical system components:
translation
Nise Table 2.4
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Commons license. For more information, see http://ocw.mit.edu/help/faq-fair-use/.
02/07/2013
2.004 System Dynamics and Control Spring 2013
9
MIT OpenCourseWare
http://ocw.mit.edu
2.04A Systems and Controls
Spring 2013
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