state space approach to solving rlc...
TRANSCRIPT
Eytan ModianoSlide 1
State Space Approach to Solving RLC circuits
Eytan Modiano
Eytan ModianoSlide 2
Learning Objectives
• Analysis of basic circuit with capacitors and inductors, no inputs, usingstate-space methods
– Identify the states of the system– Model the system using state vector representation– Obtain the state equations
• Solve a system of first order homogeneous differential equations usingstate-space method
– Identify the exponential solution– Obtain the characteristic equation of the system– Obtain the natural response of the system using eigen-values and vectors– Solve for the complete solution using initial conditions
Eytan ModianoSlide 3
Second order RC circuits
ii22 +v2-
ii11+v1-
R1 R2
R3C1 C2
ee11 ee22ee33
R1 = R2= R3 = 1Ω
C1 = C2= 1F
!!!!!!!i1= C
1
dv1
dt!!!!!!i
2= C
2
dv2
dt
Node equations :
e1
:!!!i1+ (e
1! e
3) / R
1= 0
e2
:!!!!(e2! e
3) / R
2+!i
2= 0
e3
:!!!(e3! e
1) / R
1+ (e
3! e
2) / R
2+ e
3/ R
3= 0
Eytan ModianoSlide 4
State of RLC circuits
• Voltages across capacitors ~ v(t)• Currents through the inductors ~ i(t)
• Capacitors and inductors store energy– Memory in stored energy– State at time t depends on the state of the system prior to time t– Need initial conditions to solve for the system state at future times
E.g, given state at time 0, can obtain the system state at timest > 0
State at time 0 ~ v1(0), v2(0), etc.
Eytan ModianoSlide 5
State equations for RLC circuits
• We want to obtain state equations of the form:
– Where f is a linear function of the states
• In our example,
!"x(t) = f (
"x(t))
x(t) =v1(t)
v2(t)
!
"#
$
%&,!!and!we!need!to!find,
d
dtv1(t) = f
1(v1(t),v
2(t))!!and !!
d
dtv2(t) = f
2(v1(t),v
2(t))
Eytan ModianoSlide 6
Obtaining the state equations
• So we need to find i1(t) and i2(t) in terms of v1(t) and v2(t)– Solve RLC circuit for i1(t) and i2(t) using the node or loop method
• We will use node method in our examples
• Note that the equations at e1 and e2 give us i1 and i2 directly in terms of e1,e2, e3– Also note that v1 = e1 and v2 = e2– Equation at e3 gives e3 in terms of e1 and e2
We!have, d
dtv
1(t) =
i1(t)
C1
!!and !!d
dtv
2(t) =
i2(t)
C2
e1:!!!i
1+ (e
1! e
3) / R
1= 0
e2:!!!!(e
2! e
3) / R
2+!i
2= 0
e3:!!!(e
3! e
1) / R
1+ (e
3! e
2) / R
2+ e
3/ R
3= 0
e3=e1+ e
2
3
Eytan ModianoSlide 7
Obtaining the state equations…
• We now have,
• Guessing an exponential solution to the above ODE’s we get,
dv1
dt= i
1=!2
3v1+1
3v2;!!dv
2
dt= i
2=1
3v1!2
3v2
!v1
!v2
"
#$
%
&' =
!2 / 3 1 / 3
1 / 3 !2 / 3
"
#$
%
&'v1
v2
"
#$
%
&'
v1(t) = E
1est,!v
2(t) = E
2est
E1se
st+2E
1est
3!E2est
3= 0" E
1(s + 2 / 3) ! E
2/ 3 = 0
E2se
st!E1est
3+2E
2est
3= 0"!E
1/ 3+ E
2(s + 2 / 3) = 0
Eytan ModianoSlide 8
The non-trivial solution
• The above equations have a non-trivial (non-zero) solution ifequations are linearly dependent. From linear algebra we knowthis implies:
s + 2 / 3 !1 / 3
!1 / 3 s + 2 / 3
"
#$
%
&'E1
E2
"
#$
%
&' = 0
dets + 2 / 3 !1 / 3
!1 / 3 s + 2 / 3
"
#$
%
&' = 0( (s + 2 / 3)
2 !1 / 9 = 0
s2+4
3s +1 / 3 = 0( s
1= !1,!s
2= !1 / 3
s1= !1( v
1(t) = E
1
s1e! t,!!v
2(t) = E
2
s1e! t
s2= !1 / 3( v
1(t) = E
1
s2 e! t /3,!!v
2(t) = E
2
s2 e! t /3
Eytan ModianoSlide 9
Obtaining the complete solution
• We must solve for the values of E1 and E2
• Any multiple of the above is also a valid choice for E1,E2– Notice that with this choice for si, the equations are linearly
dependant, which implies solution is not unique
• With this choice for Ei’s and si’s we get the following solutions:
s1= !1"
!1 / 3 !1 / 3
!1 / 3 !1 / 3
#
$%
&
'(E1
E2
#
$%
&
'( = 0" E
1= !E
2,!!chooseE
s1 =!1
1
#
$%
&
'(
s2= !1 / 3"
1 / 3 !1 / 3
!1 / 3 1 / 3
#
$%
&
'(E1
E2
#
$%
&
'( = 0" E
1= E
2,!!chooseE
s2 =1
1
#
$%&
'(
s1= !1" v
1(t) = !1e
! t,!!v
2(t) = 1e
! t
s2= !1 / 3" v
1(t) = 1e
! t /3,!!v
2(t) = 1e
! t /3
Eytan ModianoSlide 10
The complete solution
• Since the system is linear (and homogeneous - I.e., no inputs) anylinear combination of the above solution is also a solution. I.e.,
• In the above, A and B are constants– Can solve for A and B using initial conditions V1(0) and V2(0)
v1(t) = !Ae
! t+!Be
! t /3
v2(t) = Ae
! t+ Be
! t /3
Eytan ModianoSlide 11
Solving the state equations using Eigen-valuesand Eigen-Vectors
• Any linear homogeneous system can be expressed in the form:
• Guess a solution of the form:
!x = Ax,!!where!A!isannxnmatrix
!v1
!v2
!
"#
$
%& =
'2 / 3 1 / 3
1 / 3 '2 / 3
!
"#
$
%&v1
v2
!
"#
$
%& ( A =
'2 / 3 1 / 3
1 / 3 '2 / 3
!
"#
$
%&
!x =!Ve
st
!"x ="Vse
st
!"x = Ax"x ="Ve
st
!
"#
$#%"Vse
st= A"Ve
st
s"V = A
"V % (sI & A)
"V = 0; (I :nxn identitymatrix)
Eytan ModianoSlide 12
The characteristic equation
• A non-trivial solution to exists only if:
• Values of s satisfying the characteristic equation are Eigen-values of the matrix A
– Corresponding solution for V is an eigen-vector
• General solution to is:
(sI ! A)
!V = 0
det[sI ! A] = 0 (characterstic equation for A)
!x = Ax
x(t) = ai
i
!!Vie"#
it
#i~ areeigen-values!of!A!Vi~ corresponding!eigen-vector
ai~ constant!that depends!on initial!conditions
Eytan ModianoSlide 13
Example
!v1
!v2
!
"#
$
%& =
'2 / 3 1 / 3
1 / 3 '2 / 3
!
"#
$
%&v1
v2
!
"#
$
%& ( A =
'2 / 3 1 / 3
1 / 3 '2 / 3
!
"#
$
%&
sI ! A =s + 2 / 3 !1 / 3
!1 / 3 s + 2 / 3
"
#$
%
&'
dets + 2 / 3 !1 / 3
!1 / 3 s + 2 / 3
"
#$
%
&' = 0( (s + 2 / 3)
2 !1 / 9 = 0
s2+4
3s +1 / 3 = 0( s
1= !1,!s
2= !1 / 3!!"eigen ! values"
eigen ! vectors :!!V1=
!1
1
"
#$
%
&',V2 =
1
1
"
#$%
&'
Eytan ModianoSlide 14
Example, cont.
!v(t) = a1
!V1e
! t+ a2
!V2e
! t /3
v1(t) = !a1e! t+ a2e
! t /3
v2 (t) = a1e! t+ a2e
! t /3
Initial !conditions : v1(0) = 10,!v1(0) = 0"
!a1 + a2 = 10
a1 + a2 = 0
#$%" a1 = !5,!a2 = 5
Solution : v1(t) = 5e! t+ 5e
! t /3,!v2 (t) = !5e! t + 5e! t /3
!
!