(1) The document discusses two approaches to solving ordinary differential equations (ODEs) that model linear time-invariant (LTI) systems: (1) solve the ODE directly or (2) determine the zero-input response (ZIR) and zero-state response (ZSR). (2) It provides an example of using these approaches to analyze an RLC circuit with a voltage input of 10e-3tu(t). (3) The nature of the response of a second-order system depends on the damping ratio, and can be overdamped, critically damped, or underdamped.

1. CT-LTI System Analysis
Generalized I/O DE is
2
1 0 0
2
d y dy
a a y b x
dt dt
+ + =
Approach #I Steps:
(i) Derive the I/O DE
(ii) Solve the ODE using standard technique
 
 
( ) ( ) natural response/homogeneous solution of the ODE
( ) forced response/particular solution of the ODE
n
y t y t
y t

= +

2. Approach #II
Steps:
(i) Derive the I/O DE
(ii) Find ZIR from the homogeneous solution of the ODE using ICs
(iii)Find the impulse response(h(t)) ofthe system from the
homogeneous solution of the ODE using auxiliary ICs
(iii) Find ZSR by convolving h(t) with input x(t)
1
( ) ( ) ( )
j
n
t
j
j
y t c e x t h t

=
= + 

ZIR
ZSR

3. The technique discussed above is illustrated using
R= 3 W, L= 1 H, C=0.5 F and vs(t)=x(t)=10e-3tu(t).
The generalized ODE can be written in the operator notation as
2
( 3 2) 2
D D y x
+ + =
The characteristic equation is 2
( 3 2) 0
 
+ + =
The characteristic roots are 1 and -2
 = −
Natural response/
Homogeneous solution
of the ODE is
1 2
1 2
( ) t t
n
y t C e C e
− −
= +
Forced response/Particular solution of the ODE
Use method of undermined coefficient (page 140
Art. 2.5-1 BP Lathi

4. Here input x(t)=10e-3tu(t), the response will have form
Ae-3tu(t). Substituting the solution to the DE gives
3 3 3 3
3
3
9 9 2 2 10
since 0; 10.
( ) 10 ( )
t t t t
t
t
Ae Ae Ae e
e A
y t e u t

− − − −
−
−
− + = 
 =
=
2 3
1 2
0 0
( ) 10 ( )
: (0 ) (0 ) 5 (0)
(0 ) (0 ) 0 0
t t t
c c
c
t t
y t C e C e e u t
ICs v v V y
dv dy
i i C
dt dt
− − −
− +
− + = =
= + +
= = =
= = = = =

5. 1 2
0 1 2
1 2
2 3
(0) 10 5
2 30 0
: 20, 25
( ) (20 25 10 ) ( )
t
t t t
y C C
dy
C C
dt
Solving C C
y t e e e u t
=
− − −
= + + =
= − − − =
= = −
= − +
NaturalResponse Forced Response

6. Determination of ZIR
The characteristic roots are 1 and -2
 = −
Natural response/
Homogeneous solution
of the ODE is
1 2
1 2
( ) t t
n
y t C e C e
− −
= +
Approach #II
2
1 2
0 0
( )
: (0 ) (0 ) 5 (0)
(0 ) (0 ) 0 0
t t
c c
c
t t
y t C e C e
ICs v v V y
dv dy
i i C
dt dt
− −
− +
− + = =
= +
= = =
= = = = =

7. 1 2
0 1 2
1 2
2
(0) 5
2 0
: 10, 5
( ) (10 5 ) ( )
t
t t
ZSR
y C C
dy
C C
dt
Solving C C
y t e e u t
=
− −
= + =
= − − =
= = −
= −
Determination of Impulse Response
 
( ) ( ) ( ) ( ) ( )
( ) is the linear combination of the characteristics modes
of the system. Here n is the order of the system.
n n
n
h t b t P D y t u t
y t

= +

8. n 2
-t -2t
n 1 2
'
1 2 1 2
1 2
-t -2t
n
-t
-t -2t -t -2t
Here n=2, b =b =0, P(D)=2
y (t)=C e +C e
Using auxiliary ICs( y (0) 1, (0) 0)
2 1; 0
Solving: C =1, C =-1
y (t)=e -e
( ) ( ) ( )
( ) 0 ( ) 2(e -
2(e -e
e ) ( ) 2(e -e ) ( )
n n
ZSR
y
C C C C
y t
h t t u
t x
u
t
t
h
t

= =
− − = +
=
=
= 
=  +
=
-2t 3
2 3( ) 2 3
2 3
) ( ) 10 ( )
2( )10 10 20 10
( ) ( ) ( ) (20 25 10 ) ( )
t
t t t t t
o
t t t
ZIR ZSR
u t e u t
e e e d e e e
y t y t y t e e e u t
  

−
− − − − − − −
− − −

= − = − +
= + = − +


9. Nature of Response of a SO System
The nature of the response of a SO system depends on the
nature of the characteristic roots of the system.
2
1 0
2 2 2 2
1
1 2 1 0
0
( ) 0
1
/ ( 4 (
2 2
Where is the damping factor (neper/s) and
2
= is the undamped natural frequency(rad/s).
The ratio is called the damping ratio of the system.
Us
n
n
n
a a
a
a a
a
a
 
    





+ + =
= −  − = −  −
=
=
2 2
ing these notations, the characteristics equation of SOS
can be written as
2 0
n n
   
+ + =

10. 1 2
1 1
1 2
Case I: Overdamped response(when > )
and are real and negative. Natural response
y ( )
n
t t
n
t Ae A e
 
 
 
= +

11. 1 1
1 2
Case I: Critically damped response(when )
and are real, equal and negative. Natural response
y ( ) ( )
nI
t
n
t A A t e
 
 
=
= +

12. 1 1
2 2
1 2
2 2
Case III: underdamped response(when )
and are complex conjugate with negative real part.
/ =- j =- j
Where = is the damped natural frequency of the SOS
Natural response
y ( )
n
n d
d n
n
t
 
 
      
  

 − 
−
(- j ) (- j )
1 2 1 2
( cos sin )
d d
t t t
d d
Ae A e e B t B t
    
 
+ − −
= + = +

13. 1 1
2 2
1 2
(j ) ( j )
1 2 1 2
Case IV: undamped response(when 0)
and are complex conjugate with negative real part.
/ = j
Natural response
y ( ) ( cos sin )
d d
n
t t
n d d
t Ae A e B t B t
 

 
   
 
−
=
 −
= + = +

14. Symbolic Solution of ODE using Matlab
syms y(t)
Dy=diff(y)
vc=dsolve(diff(y,2)+3*diff(y)+2*y==20*exp(-3*t),y(0)==5,Dy(0)==0)
i=0.5*diff(vc)
vc =20*exp(-t) - 25*exp(-2*t) + 10*exp(-3*t)
i =25*exp(-2*t) - 10*exp(-t) - 15*exp(-3*t)
ezplot(vc, [0 10])

16. VC
i

17. VC
i
Undamped case

18. Finding Initial Values