general response of second order system -...
TRANSCRIPT
Eytan ModianoSlide 1
General Response of Second Order System
Eytan Modiano
Eytan ModianoSlide 2
Learning Objectives
• Learn to analyze a general second order system and to obtain thegeneral solution
– Identify the over-damped, under-damped, and critically dampedsolutions
– Convert complex solution to real solution
– Suspended “mass-spring-damper” equivalent system
Eytan ModianoSlide 3
Second order RC circuit
• System with 2 state variables– Described by two coupled first-order differential equations
• States– Voltage across the capacitor - V1– Current through the inductor - iL
• What to obtain state equations of the form: x’ = Ax– Need to obtain expression for dv1/dt in terms of V1 and iL– Need to obtain expression for dil/dt in terms of V1 and iL
iicc+v1-
RC
vv11
LiiLL
Eytan ModianoSlide 4
State Differential Equations
• For the capacitor and inductor we have,
• Above immediately gives us an expression for dil/dt in terms of V1• Need to obtain expression for dv1/dt in terms of V1 and iL
• Write node equation at v1 to obtain:
• Hence the state equations are given by:
dv1
dt=ic
C,di
L
dt=vL
L=v1
L
ic+ i
L+ v
1/ R = 0
(1)!!!dv
1
dt=!i
L
C!v1
RC
(2)!!!di
L
dt=v1
L
Eytan ModianoSlide 5
Solving the state differential equations
d
dt
v1
iL
!
"#
$
%& =
'1 / RC '1 /C1 / L 0
!
"#
$
%&
A! "### $###
v1
iL
!
"#
$
%&!!!
! sI ' A[ ] =s +1 / RC 1 /C
'1 / L s
!
"#
$
%&
((s) =!characteristic!eqn!=!det(sI ' A) = 0
det(sI ' A) = s2 +s
RC+1
LC= 0
solve!quadratic!equation!to!obtain :
!!!!!!!!!!s1=
'1
RC+
1
RC
)*+
,-.2
'4
LC
2,!!!s
2=
'1
RC'
1
RC
)*+
,-.2
'4
LC
2!!!
Eytan ModianoSlide 6
The natural response of RLC circuits
• Three cases
– Over-damped response: Characteristic equation has two (negative) real roots Response is a decaying exponential No oscillation (hence the name over-damped, because the resistor damps out the
frequency of oscillation)
– Under-damped response: Characteristic equation has two distinct complex roots Response is a decaying exponential that oscillates
– Critically-damped response: Characteristic equation has two read, distinct roots Solution no longer a pure exponential Response is on the verge of oscillation
• Analogy to oscillating suspended spring-mass-damper system; whereenergy is stored in the spring and mass
Eytan ModianoSlide 7
Over-damped response
• Characteristic equation has two distinct real roots; s1, s2– Both s1 > 0, s2 > 0 (why?)
• Solution of the form:
• Decaying exponential response
1
(RC)2>4
LC
v1= aE
1
s1e
s1t+ bE
1
s2 e
s2t
il= aE
2
s1e
s1t+ bE
2
s2 e
s2t
Over-damped response
Eytan ModianoSlide 8
Critically-damped response
• Characteristic equation has two real repeated roots; s1, s2– Both s1 = s2 = -1/2RC
• Solution no longer a pure exponential– “defective eigen-values” ⇒ only one independent eigen-vector
Cannot solve for (two) initial conditions on inductor and capacity
• However, solution can still be found and is of the form:
• See chapter 5 of Edwards and Penny; or 18.03 notes
1
(RC)2=4
LC
v1= a
1est+ b
1te
st
il= a
2es1t+ b
2te
s2t
Eytan ModianoSlide 9
Critically-damped response, cont.
• Response is on the verge of oscillation• Known as “critically damped”
critically-damped response
v1= ae
! t /2RC+ bte
! t /2RC
Eytan ModianoSlide 10
Under-damped response
• Characteristic equation has two distinct complex roots; s1, s2
• Two distinct eigenvectors, V1 and V1*
– Complex eigenvalues and eigenvectors
• Solution of the form:
1
(RC)2<4
LC
v1= a
1es1t+ a
1
*es2t
il= a
2es1t+ a
2
*es2t
s1= !" + j#,!!s
2= s
1
*= !" ! j#
" = 1 / 2RC, # =1 / (RC)
2 ! 4 / LC
2
Eytan ModianoSlide 11
Under-damped response
v1= a
1e(!" + j# )t
+ a1
*e(!" ! j# )t
Under-damped response, cont.
Euler 's formula :!!e! + j"
= e!(cos(") + j sin("))
# v1= a
1e$! t(cos("t) + j sin("t)) + a
1
*e$! t(cos($"t) + j sin($"t))
cos($") = cos("),!!sin($") = $ sin(")
# v1= 2e
$! tRe[a
1]cos("t) $ Im[a
1]sin("t)[ ]
Decaying oscillation between -Re(a1) and + Re(a1) or- Im(a1) and + Im(a1)
Eytan ModianoSlide 12
Example of over-damped response(complex eigen-values and eigen-vectors)
• Consider the same circuit with the following values– C=1/2 F, L=1/5 H, R=1– Initial conditions, V1(0)=1v, ic(0)=1A
d
dt
v1
iL
!
"#
$
%& =
'1 / RC '1 /C
1 / L 0
!
"#
$
%&
A! "### $###
v1
iL
!
"#
$
%&!!!=
'2 '2
5 0
!
"#
$
%&v1
iL
!
"#
$
%&!!
! sI ' A[ ] =s + 2 2
'5 s
!
"#
$
%&
s + 2 2
'5 s
!
"#
$
%& = s
2+ 2s +10 = 0
s1= '1+ 3 j,!!!s
2= '1' 3 j
Eytan ModianoSlide 13
Example, continued: Eigen-vectors
• Now find the eigen-vectors
s1 = !1+ 3 j" sI ! A =1+ 3 j 2
!5 !1+ 3 j
#
$%
&
'(V1
s1
V2s1
#
$%
&
'( = 0
"!5V1s1+ (!1+ 3 j)V2
s1= 0
"V1s1= !1 / 5 + 3 / 5 j[ ]V2
s1
"Vs1=V1
s1
V2s1
#
$%
&
'( =
!1 / 5 + 3 / 5 j
1
#
$%
&
'(
s2 = s1*= !1! 3 j
Vs2=V1
s1
V2s1
#
$%
&
'(
*
=!1 / 5 + 3 / 5 j
1
#
$%
&
'(
*
=!1 / 5 ! 3 / 5 j
1
#
$%
&
'(
Eytan ModianoSlide 14
The total solution
V1(t) = a1(!1 / 5 + 3 / 5 j)e(!1+3 j )t
+ a2 (!1 / 513 / 5 j)e(!1!3 j )t
ic (t) = a1(1)e(!1+3 j )t
+ a2 (1)e(!1!3 j )t
Use!V1(0) = 1,!!ic (0) = 1!!to!obtain :!!a1 = 1 / 2 ! j,!!a2 = 1 / 2 + j
V1(t) = (1 / 2 +1 / 2 j)e(!1+3 j )t
+ (1 / 2 !1 / 2 j)e(!1!3 j )t
ic (t) = (1 / 2 ! j)e(!1+3 j )t + (1 / 2 + j)e(!1!3 j )t
to!express!as!real!values!use,
Euler's!formula:!!e" +j# = e" (cos(#) + j sin(#))
some!algebra$V1(t) = cos(3t) ! sin(3t)[ ]e! t
ic (t) = cos(3t) + 2sin(3t)[ ]e! t%&'
('
Eytan ModianoSlide 15
2nd order mechanical systemsmass-spring-damper
• Force exerted by spring is proportional to the displacement (x) of the massfrom its equilibrium position and acts in the opposite direction of thedisplacement
– Fs = -kx– Fs < 0 if x > 0 (I.e., spring stretched)– Fs > 0 if x < 0 (spring compressed)
• Damper force is proportional to velocity (v = x’) and acts in oppositedirection to motion
– Fr = -bv = -bx’
M
frictionless support
Damper (dashpot)massspring
X=0 X>0
Eytan ModianoSlide 16
Spring-mass-damper system
• If Fr and Fs are the only forces acting on the mass m, then
– F = ma where a = dv/dt = x’’– F = m x’’ = Fs + Fr = -kx - bx’
• So the differential equation for the mass position, x, is given by:
mx’’ + bx’ + kx = 0
– This is a second order system
– Guessing x = Aest for the solution we get,
mAs2est + bAsest + kAest = 0
– The characteristic equation is thus
s2 + (b/m)s + k/m = 0
s1=(!b / m) + (b / m)
2! 4(k / m)
2,!!!s
2=(!b / m) ! (b / m)
2! 4(k / m)
2
Eytan ModianoSlide 17
Response of Spring-mass-damper system
• Note that for this system the state can be described by– Position, x(t), Velocity, x’(t)– Hence, the initial conditions would be x(0) and x’(0)
• Note similarity to RLC circuit response:
• Notice relationship between 1/R in RLC circuit and damping factor(b) in spring-mass-damper system
– B ~ 0 ⇒ un-damped system ⇒ oscillation– This is the basis for the terminology, over-damped, under-damped, etc.– Over-damped system ⇒ damping factor is large and system does not
oscillate (just exponential decay)
x(t) = A1es1t + A2e
s2 t
where!A1 !and!A2 !are!chosen!to!satisfy!initial!conditions
!!!!s1=
!1RC
+ 1RC( )
2
! 4LC
2,!!!s
2=
!1RC
! 1RC( )
2
! 4LC
2