general response of second order system -...

Eytan ModianoSlide 1

General Response of Second Order System

Eytan Modiano


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


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


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


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!!!


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


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


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


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


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


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)


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


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

#

$%

&

'(


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%&'

('


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


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


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

general response of second order system -...

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