State Space Approach to Solving RLC circuits

[Pages:14]State Space Approach to Solving RLC circuits

Eytan Modiano

Eytan Modiano Slide 1

Learning Objectives

? Analysis of basic circuit with capacitors and inductors, no inputs, using

state-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 using

state-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 Modiano Slide 2

Second order RC circuits

e1

R1

+ i1

v1

C1

-

e3

R3

R2

e2

i2 +

C2

v2

-

R1 = R2= R3 = 1 C1 = C2= 1F

Eytan Modiano Slide 3

!!!!!!!i1

=

C1

dv1 dt

!!!!!!i2

=

C2

dv2 dt

Node equations :

e1 :!!!i1 + (e1 ! e3 ) / R1 = 0 e2 :!!!!(e2 ! e3 ) / R2 +!i2 = 0 e3 :!!!(e3 ! e1) / R1 + (e3 ! e2 ) / R2 + e3 / R3 = 0

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 Modiano Slide 4

State equations for RLC circuits

? We want to obtain state equations of the form: x"!(t) = f (x"(t))

? Where f is a linear function of the states

? In our example,

x(t

)

=

!"#vv12

(t ) $ (t )%&

,!!and!we!need!to!find,

d dt

v1(t) =

d f1(v1(t), v2 (t))!!and !! dt

v2 (t) =

f2 (v1(t ), v2 (t ))

Eytan Modiano Slide 5

Obtaining the state equations

We!have,

d dt

v1(t )

=

i1(t)!!and !! d

C1

dt

v2 (t )

=

i2 (t ) C2

? 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

e1 :!!!i1 + (e1 ! e3 ) / R1 = 0 e2 :!!!!(e2 ! e3 ) / R2 +!i2 = 0 e3 :!!!(e3 ! e1) / R1 + (e3 ! e2 ) / R2 + e3 / R3 = 0

?

Note that e2, e3

the

equations

at

e1

and

e2

give

us

i1

and

i2

directly

in

terms

of

e1,

? Also note that v1 = e1 and v2 = e2

e = e +3 e ? Equation at e3 gives e3 in terms of e1 and e2

12 3

Eytan Modiano Slide 6

Obtaining the state equations...

? We now have,

dv1 dt

=

i1

=

!2 3 v1

+

1 3

v2

;!!

dv2 dt

=

i2

=

1 3 v1

!

2 3 v2

"#$vv!!12

% &'

=

"!2 / 3 #$ 1 / 3

1/ !2

3% / 3&'

" #$

v1 v2

% &'

? Guessing an exponential solution to the above ODE's we get,

Eytan Modiano Slide 7

v1(t ) = E1est ,!v2 (t ) = E2est

E1sest

+

2 E1e st 3

!

E2est 3

= 0 " E1(s + 2 / 3) ! E2

/3= 0

E2 sest

!

E1est 3

+

2 E2 e st 3

= 0 " !E1

/ 3 + E2 (s + 2 / 3) = 0

The non-trivial solution

"s + 2 / 3 #$ !1 / 3

!1 / s+2

3% / 3&'

" #$

E1 E2

% &'

=

0

? The above equations have a non-trivial (non-zero) solution if

equations are linearly dependent. From linear algebra we know

this implies:

"s + 2 / 3 det #$ !1 / 3

!1 / s+2

3% / 3&'

=

0

(

(s

+

2

/

3)2

!

1

/

9

=

0

s2

+

4 3

s

+1/

3

=

0

(

s1

=

!1,! s2

=

!1 /

3

Eytan Modiano Slide 8

s1

=

!1

(

v1(t )

=

E s1 1

e!

t

,!!

v2

(t

)

=

E s1 2

e!

t

s2

=

!1 /

3(

v1(t )

=

E s2 1

e!

t

/

3

,!!

v2

(t

)

=

E e s2 !t /3 2

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