The complete solution to systems with inputs
Eytan Modiano
Eytan Modiano
Slide 1
Learning Objectives
•
Analyze linear time-invariant systems with inputs
–
Solve for the homogeneous response of the system
Natural response without inputs
–
Solve for the particular solution
Identify forced response for different input functions
–
Obtain the complete solution using initial conditions
Complete solution = homogeneous solution + particular solution
–
Eytan Modiano
Slide 2
Derive the transfer function
Systems with input
•
In general, systems have inputs
–
–
Applied force in mechanical systems
Voltage and current sources in circuits
E.g., battery, power-supply, antenna, scope probe, etc.
•
Systems also have outputs
–
•
Displays, speakers, voltmeters, etc.
We need to be able to analyze the system response to inputs
–
Two methods:
Solution to linear constant-coefficients differential equations
Transfer function methods
Eytan Modiano
Slide 3
Linear constant coefficient differential equations
dx(t)
+ 2x(t)=!u(t)
dt
•
E.g.,
•
Where x is the state variable and u is the input
•
The complete solution is of the form:
x(t)=!x p (t) + xh (t)
where!x p !is!the!particular!solution!(when!input spefified)
and x h !is!the!homogeneous!solution!to!the!DE!when!u(t) = 0
dx(t)
!!!!!!!!!!!!!!!!!!!!!!!!i.e.,!
+ 2x(t)= 0
dt
•
Eytan Modiano
Slide 4
Thus far we have only considered homogeneous systems
The particular solution
"0
dx(t)
+ 2x(t)= u(t),!!!!u(t) = # 3t
dt
$e
•
t<0
t!0
A common method for solving for the particular solution is to try a
solution of the same form as the input
–
This is called the “forced response”
x p (t) = ae3t !!
•
So try,
•
To solve for the constant a, we plug the solution to the original
equation
Eytan Modiano
Slide 5
dx(t)
+ 2x(t)= e3t ! 3ae3t + 2ae3t = e3t ! a = 1 / 5
dt
e3t
Particular!solution :!!x p (t)=
,!!t > 0
5
The complete solution
homogeneous!solution:!!!xh (t) = Best !!
Bsest + 2Best = 0!!!!Bs + 2B = 0 ! B(s + 2) = 0 ! s = "2
! xh (t) = Be"2t !
e3t
x(t)!=! + Be"2t !!!,!!t > 0
5
In order to solve for B, must know initial conditions.
e0
1
0
E.g., x(0)!=!0 ! + Be = 0 ! B = "
5
5
Eytan Modiano
Slide 6
1 3t
x(t) = #$ e " e"2t %& ,!!!t > 0
5
Key points
•
Solution consists of homogeneous and particular solution
– Homogeneous solution is also called the “natural response”
It is the response to zero input
– The particular solution often takes on the form of the input
It is therefore referred to as the “forced response”
•
The complete solution requires specification of initial conditions
–
An nth order system would have n initial condition
–
Apply initial conditions to the complete solution in order to obtain the
constants
The initial conditions are on the complete solution, not just the
homogeneous part
Eytan Modiano
Slide 7
Example: RC circuit with inputs
e1
= !i1
R
dv1 i1
dv1 !e1
= "
=
u(t) = e
dt C
dt
RC
e1 (t) = v1 (t) + u(t)
dv1 !v1 (t) u(t)
"
=
!
dt
RC
RC
dv1
!C=1, !R=1 "
= !v1 (t) ! u(t)
dt
!5t
- Vc
+
-
C
+
i1
R
Y(t)
C = 1 F, R = 1 ohm
Homogeneous!Solution:!!u(t) = 0
Guess!v1 (t) = aest ! asest = "aest ! as = "a ! s = "1
Eytan Modiano
Slide 8
! vH = ae"t
The complete solution
Forced!Response: vF (t) = Be!5t ,! v!F = !5Be!5t
dv1
= !v1 (t) ! u(t) " !5Be!5t = !Be!5t ! e!5t
dt
e!5t
" !5B = !B ! 1 " B = 1 / 4 " vF (t) =
4
!5t
e
v1 (t) = vH (t) + vF (t) = ae!t +
4
0
e
1
1
0
Initial!conditions:!v1 (0) = 0v " ae + " a + = 0 " a = !
4
4
4
!5t
!t
!e!t + e!5t
e
!
e
v1 (t) =
,!!y(t) = v1 (t) + u(t) = e!5t +
4
4
Eytan Modiano
Slide 9
Example: RLC circuit with inputs
R
u(t) = 1V
+
-
i1
C
+ +
v1 v2
- -
i2
L
Y(t)
Initial!conditions:!!Vc (0) = !2V,! iL (0) = 2A
Output = Y(t) = Voltage across inductor = v2 (t)
Eytan Modiano
Slide 10
v1 (t) ! u(t)
v1 u
Node!equation!at!v1 :
+ i1 + i2 = 0 " i1 = + ! i2
R
R R
dv1 i1
dv1
v1
u
i2
= "
=
+
!
dt C
dt CR CR C
di2 1
v1
= v2 =
dt L
L
The homogeneous solution
(aka: the natural response)
d ! v1 $ ! '1 / RC '1 / C $ ! v1 $ !1 / RC $
=#
+#
u(t)
#
&
&
#
&
&
0 % " i2 % " 0 %
dt " i2 % " 1 / L
homogeneous solution: take u(t)=0
d ! v1 $ ! '1 / RC '1 / C $ ! v1 $
=#
#
&
0 &% #" i2 &%
dt " i2 % " 1 / L
!##
#"###
$
A
! -2 -1$
! s+2 1$
Let!C=1 F,!R=1/2 ",!L=2 H ( A= #
( SI ' A = #
&
&
1/2
0
-1/2
s
"
%
"
%
characteristic equation: s 2 + 2s + 1 / 2 = 0
1
1
( s1 = '1 +
, s1 = '1 '
2
2
Eytan Modiano
Slide 11
Natural!response:!!e('1+1/
2 )t
,!!e('1'1/
2 )t
The homogeneous solution, continued
Finding the eigen-vectors:
# ! 2 &
# ! 2 &
1
1
%
(
s1
s1 = !1 +
" E = % 2 + 1 ( , !!s 2 = !1 !
" E s2 = %% 2 ! 1 ((
2
2
%$ 1 ('
%$ 1 ('
! 2 (!1+
v1n = a(
)e
2 +1
1
2
)t
! 2 (!1!
+ b(
)e
2 !1
1
2
)t
,!!i2n = ae
Complete!solutions:!!v1 = v1n + v1 f ,!!!i2 = i2n + i2 f
Eytan Modiano
Slide 12
(!1+
1
2
)t
+ be
(!1!
1
2
)t
The particular solution
(aka: the forced response)
u(t) = 1v
The forced response would be a constant. I.e., v1 f = A,!i2 f = B
!v1 f
u
i2 !A
1
B
=0=
+
! =
+
!
dt
RC RC C RC RC C
dv1 f
C = 1F, R = 1 / 2", L = 2H # 0 = 2A + 2 ! B # 2A + B = 2
di2 f
v1 A
= = = 0 # A = 0 # v1 f = 0V
dt
L L
2A + B = 2 # B!= 2 # i2 f = 2A
Eytan Modiano
Slide 13
Does this solution make sense?
The complete solution
Initial!conditions:v1 (0) = 2,!i2 (0) = 2
Complete!solutions:!!v1 = v1n + v1 f ,!!!i2 = i2n + i2 f
! 2
v1 = a(
)e
2 +1
i2 = ae
(!1+
1
2
)t
(!1+
+ be
1
2
! 2
+ b(
)e
2 !1
)t
(!1!
1
2
)t
(!1!
1
2
)t
+0
+2
#
! 2
! 2
1
!1
v1 (0) = 2 " a(
) + b(
) = 2%
,!!b =
$"a=
2 +1
2 !1
2
2
%
i2 (0) = 2 " a + b = 2 " a = !b
&
v1 (t) =
Eytan Modiano
Slide 14
!1
e
2 +1
1
i2 (t) =
e
2
(!1+
(!1+
1
2
)t
1
2
)t
1
+
e
2 !1
1
!
e
2
(!1!
1
2
)t
(!1!
+2
1
2
)t
Forced response, v1f = 0
Forced response, i2f = 2