Analysis of RC circuits
Eytan Modiano
Eytan Modiano
Slide 1
Learning Objectives
•
Analysis of basic circuit with capacitors, no inputs
–
•
Solve a system of first order homogeneous differential equations
using classical method
–
–
–
–
Eytan Modiano
Slide 2
Derive the differential equations for the voltage across the capacitors
Identify the exponential solution
Obtain the characteristic equation of the system
Obtain the natural response of the circuit
Solve for the complete solution using initial conditions
Second order RC circuits
e1
+
v1
-
R1
e3
e2
R2
i1
i2
C1
R3
C2
+
v2
-
R1 = R2= R3 = 1Ω
C1 = C2= 1F
dv1
dv2
!!!!!!!i1 = C1
!!!!!!i2 = C2
dt
dt
Eytan Modiano
Slide 3
Node equations :
e1 :!!!i1 + (e1 ! e3 ) / R1 = 0
e2 :!!!!(e2 ! e3 ) / R2 +!i2 = 0
e3 :!!!(e3 ! e1 ) / R1 + (e3 ! e2 ) / R2 + e3 / R3 = 0
The differential equations
!!!!!!!!!!!!!!!!Note :!!!v1!= e1 !;v2 != e2
Differential equations:
de1
C1
+ (e1 ! e3 ) / R1 = 0
dt
de2
C2
+ (e2 ! e3 ) / R2 = 0
dt
(e3 ! e1 ) / R1 + (e3 ! e2 ) / R2 + e3 / R3 = 0
Plugging!in!values! for!resistors!capacitors :
(1) e!1 + e1 ! e3 = 0
(2) e!2 + e2 ! e3 = 0
(3)!3e3 ! e2 ! e1 = 0
Eytan Modiano
Slide 4
Guessing a solution
We “guess” that the solution is an exponential of the
form:
Guess!ei = Ei est !! e!i = Ei sest
E1sest + E1est ! E3est = 0 "
E1s + E1 ! E3 = 0
$
st
st
st
E2 se + E2 e ! E3e = 0 # & E2 s + E2 ! E3 = 0
3E3 ! E2 ! E1 = 0
3E3est ! E2 est ! E1est = 0 $%
E1 (1 + s)
!! ! E3 = 0
E2 (1 + s) ! E3 = 0
!E1 !!!!!!!E2 !!!!!!!!!!!+3E3 = 0
Eytan Modiano
Slide 5
Solution to homogeneous equations
•
If equations are linearly independent we have one unique solution:
–
–
•
In order to obtain non-trivial solution, the equations cannot be linearly
independent
–
•
E1=E2=E3=0
“trivial” solution
For what values of s are the equations not linearly independent?
Rewrite equations in the form: AE = 0
0
!1% " E1 %
"(1 + s)
$ 0
' $E ' = 0
1
+
s
!1
$
'$ 2'
$# !1
!1
3 '& $# E3 '&
•
Rows of A are linearly dependent if determinant of A = 0
–
Eytan Modiano
Slide 6
Non-trivial solution
The characteristic equation
det A = 0 (The characteristic equation)
see!linear!algebra!primer!on taking!determinants
A = 3(1 + s)(1 + s) ! (1 + s) ! (1 + s) = 0
!4 ± 16 ! 12 !4 ± 2
3s + 4s + 1 = 0 " s =
=
6
6
" s1 = !1,!s2 = !1 / 3
2
Two!solutions :
s = !1 " ei = Ei e!t
Eytan Modiano
Slide 7
s = !1 / 3 " ei = Ei e!t / 3
Non-trivial solution
•
Now we must solve for the values of Ei’s
•
Recall
•
E1 (1 + s)
!! ! E3 = 0
E2 (1 + s) ! E3 = 0
!E1 !!!!!!!E2 !!!!!!!!!!!+3E3 = 0
For s = -1 we have,
E1 (0)
!! ! E3 = 0 " E3 = 0
E2 (0) ! E3 = 0 " E3 = 0
!E1 !!!!!!!E2 !!!!!!!!!!!+3E3 = 0 " E1 =!!E2 !
Solution!not !unique,!choose :!!E2 = 1,!E1 = !1
" e1 = !1e!t ,!!e2 = 1e!t ,!e3 = 0
Eytan Modiano
Slide 8
Non-trivial solution, continued
•
Similarly, for s = -1/3
!! ! E3 = 0 " E1 = 3 / 2E3
!!E2 (2 / 3) ! E3 = 0 " E2 = 3 / 2E3
!E1 !!!!!!!E2 !!!!!!!!!!!+3E3 = 0
E1 (2 / 3)
Solution!not !unique,!choose :!!E3 = 1,!E1 = E2 = 3 / 2
" e1 = 3 / 2e!t / 3 ,!!e2 = 3 / 2e!t / 3 ,!e3 = e!t / 3
•
Eytan Modiano
Slide 9
Note that in general we can solve for the E’s using row reduction, etc. (see
linear algebra notes)
The complete solution
•
We now have two possible solutions:
(1) :!!!e1 = 3 / 2e!t / 3 ,!!e2 = 3 / 2e!t / 3 ,!e3 = e!t / 3
(2) :!!e1 = !1e!t ,!!e2 = 1e!t ,!e3 = 0
•
Since the system is linear and homogeneous, any linear combination of
these solution is also a solution. Hence,
" !1%
"3 / 2%
!
e(t) = a $$ 1 '' e!t + b $$ 3 / 2 '' e!t / 3
$# 0 '&
$# 1 '&
•
Where a, and b are constants that depend on initial conditions
–
Eytan Modiano
Slide 10
Given initial values for e1 = v1 and e2 = v2 we can solve for a and b
The complete solution, continued
•
Suppose we are given the initial voltage values across the
capacitors: e1(0) = v1(0), e2(0) = v2(0)
•
Plug these into the solution
to obtain:
Eytan Modiano
Slide 11
e1 (0) = !ae0 + 3 / 2be0 = !a + 3 / 2b "$
#&
0
0
e2 (0) = ae + 3 / 2be =!!a + 3 / 2b $%
e2 (0) " e1 (0)
e1 (0) + e2 (0)
!a=
,!!b =
2
3
Finally,
e1 (0) " e2 (0) "t e1 (0) + e2 (0) "t / 3
e1 (t) =
e +
e
2
2
e2 (0) " e1 (0) "t e1 (0) + e2 (0) "t / 3
e2 (t) =
e +
e
2
2
e1 (0) + e2 (0) "t / 3
e3 (t) =
e
3