Mathematical Foundations - Control of Color Imaging Systems: Analysis and Design - page 99

3.2.1 S OLUTION OF C ONSTANT -C OEFFICIENTS L INEAR D IFFERENTIAL E QUATIONS
We

first consider zero-input response that is the response of the system de ned by

Equation 3.1 when the input signal u ( t ) is zero. Consider an Nth-order constant-

, y 0 (

coef cients linear DE with zero input driven by initial conditions y ( 0 ) 0 ) ,...,and

y ( N 1 ) ( 0 ) :

N y ( t )
d t N

N 1

d

þ a N 1 d

y ( t )
d t N 1 þþ a 0 y ( t ) ¼ 0

( 3 : 2 )

Assuming a solution of the form y ( t ) ¼ Ce Dt , we would have

e Dt

CD N þ a N 1 D N 1 þ a N 2 D N 2

þþ a 0 ¼ 0 ( 3 : 3 )

Since Ce Dt is nonzero, then

D N þ a N 1 D N 1 þ a N 2 D N 2

þþ a 0 ¼ 0 ( 3 : 4 )

This polynomial is called characteristic equation of the DE. It is a polynomial of
degree N that has N roots D 1 , D 2 ,..., andD N . These roots are generally complex
and are called characteristic roots of the DE. The zero-input solution is given by

y ( t ) ¼ C 1 e D 1 t þ C 2 e D 2 t þþ C N e D N t

( 3 : 5 )

The constants C 1 , C 2 ,..., andC N are found by applying the N initial conditions.
The stability of the system is de

ned in terms of the zero-input response. The
discrete LTI system described by DE given in Equation 3.1 is stable if the zero-input
response decays to zero as t !1 . This implies that the characteristic roots must

have real parts less than zero, that is,

Re( D i ) <

for

i ¼

... ( : )

0 1, 2, , N 3 6

Example 3.1

Solve the following second-order DE with initial conditions y(0) ¼ 1 and

dy(t)
dt

j t¼0 ¼ 4:

d 2 y(t)
dt 2

3 dy(t) þ

dt þ

2y(t) ¼ 0

S OLUTION
The characteristic equation is D 2 þ 3D þ 2 ¼ 0, which can be written as

D 2 þ 3D þ 2 ¼ (D þ 1)(D þ 2) ¼ 0

Therefore, the roots are D 1 ¼ 1 and D 2 ¼ 2. The zero-input response is

¼ C 1 e D 1 t þ C 2 e D 2 t ¼ C 1 e t þ C 2 e 2t

y(t)