Notes
Example 1:
Deriving equations for this system:
>System is Frictionless
>Define a variable (with respect to the origin).
>Displacement (), equilibrium (),
>Coordinate with equil. Position and displacement of the mass
>Rate of change of , d/dt = (whe
Notes
Characterization of Transients
( i ) T p=
A Swiftness of response
( Exact for 2 nd order , underdamped )
n 1 2
( i ) PO=100 e 1
PO
( 100 )
=
PO
+ ln (
100 )
2
2
C Characterization of SteadyState
2.16 + .6
( for .3 .8 )
n
ln
B Closeness of Respon
Notes
Electric Motors-
T m ( t )=K 1 K f i (t ) i a ( t )
1)Field (current) Controller-
T m ( t )=K m , f i f ( t )
InputsV f ( t ) , d ( t )
Output
s
J + bs T d
K m ,f
1
( s )=
V f ( s)
( J s +bs )( Lf s + Rf )
2
s
V f ( s)
[
]
1
Lf ( s ) [ K m , f
Notes
Example 1: Positioning the reading head of a disk drive.
Plant:
( Js +b ) ( La s + Rac ) + K m K e
s
K
G p ( s )= m
2
Values: J =1 Nm
s
rad
b=20
Nms
rad
R=1 ohm
La=1 mH
K m=
5 Nm
A
Control of the reading head of a disk drive:
Open Loop Control
Km
Rq
Notes
L cfw_ output
=transfer function at zero initial conditions Example (cont.)
L cfw_input
M 1 x 1+ ( b1 +b2 ) x 1b1 x 2=r ( 1 )
M 2 2+ b1 x 2 +k 2 x 2b1 x 1=0 ( 2 )
x
Output :could be x 1 , x 1x 2 , x 2
Input :r ( t )
Transfer Function from r(t) to
Notes
Laplace Transform Objectives: Conclude how a linear system evolves without explicitly stating the solution of the
ODE that describes the system.
Ingredients :
Some complex variable s= +i
Function f ( t ) with f ( t )=0 for t< 0.
The LaplaceTransfo
Notes
Model-
m ( t ) + c ( t ) +kx ( t ) =F0 cos ( t )
x
x
X=
F0
( k m ) +( c )
2 2
sin ( )=
2
cw
2 2
( k m ) + ( cw )
2
cos ( ) =
k m 2
( km ) +( cw )
2 2
2
+ 2 n x + 2 x =m + c +kx
x
x
n
In nondimensional form:
X
1
=
, where st =the static deflectio
Notes
+ 25 n + 2 x=0
x
x n
n=natural frequency
n=
=damping ratio
In this case,
k
c
; = ; c cr =2 ( mk )
m
c cr
Solution types:
1) Underdamped, <1
[
x ( t )=e t C 1 ei 1
I)
Charged values : S 1,2=( i 1 2 ) n
II)
2
+C2 e i 1
2
]
d= 1 2 n=the frequency
Notes
Example 1
Balancing equation:
F=mx -> -kx -cx + F(t) = mx
m + c +kx=F ( t )
x x
The general solution is the sum of the homogenous solution
(Xh(t) capturing the component of the motion when F(t) is not
applied, plus the particular solution Xp(t) cap
Notes
+2 x=0
x n
x ( t )=C 1 e i t +C 2 e i t
x ( t )= Acos ( n t )
n
n
x ( t )= A1 cos ( n t ) + A 2 sin ( n t )
A= A2 + A 2 ; =tan 1
1
2
A2
A1
( )
A 1=C 1+ C2 . A 2=i ( C 1C 1 )
x ( t )= A o sin ( n t+ o )
A o =A ; 0=tan1
A1
A2
Example:
A1 =1 ; A2=1
=t
Notes
Y ( s )=
G (s )
R (s )
1+G ( s ) H ( s )
The poles of the closed loop transfer function
above are the roots of the denominator, or the
characteristic polynomial.
Question: How do the closed loop poles move in
the complex plane as a parameter changes