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

!

!