Chapter 2
Systems Defined by
Differential or Difference
Equations
Linear I/O Differential Equations
with Constant Coefficients
• Consider the CT SISO system
x(t )
y (t )
System
described by
N 1
y
(N)
M
  ai y (t )  bi x (t )
(i )
(i )
i 0
i 0
i
ai  , bi 
(i )
y (t )
d y (t ) ( i )
x (t )
i
dt
i
d x(t )
i
dt
Initial Conditions
• In order to solve the previous equation for
t  0, we have to know the N initial
conditions
(1)
y(0), y (0),
,y
( N 1)
(0)
Initial Conditions – Cont’d
• If the M-th derivative of the input x(t)
contains an impulse k (t ) or a derivative of
an impulse, the N initial conditions must be

taken at time t  0 , i.e.,

(1)

y (0 ), y (0 ),
,y
( N 1)

(0 )
First-Order Case
• Consider the following differential equation:
dy (t )
 ay (t )  bx(t )
dt
• Its solution is
t
y (t )  y (0)e
 at
 e
 a ( t  )
bx ( )d , t  0
0
or
t
y (t )  y (0 )e

 at
 e
 a ( t  )
bx( )d , t  0
0
if the initial time is taken to be 0 
Generalization of the First-Order Case
• Consider the equation:
dy (t )
dx(t )
 ay (t )  b1
 b0 x(t )
dt
dt
• Define q (t )  y (t )  b1 x(t )
• Differentiating this equation, we obtain
dq (t ) dy (t )
dx(t )

 b1
dt
dt
dt
Generalization of the First-Order
Case – Cont’d
dy (t )
dx(t )
 ay (t )  b1
 b0 x(t )
dt
dt

dq (t ) dy (t )
dx(t )

 b1
dt
dt
dt

dq (t )
  ay (t )  b0 x(t )
dt
Generalization of the First-Order
Case – Cont’d
• Solving q (t )  y (t )  b1 x(t ) for y (t ) it is
y (t )  q (t )  b1 x(t )
dq (t )
which, plugged into
  ay (t )  b0 x(t ) ,
dt
yields
dq (t )
 a  q (t )  b1 x(t )   b0 x(t ) 
dt
 aq(t )  (b0  ab1 ) x(t )
Generalization of the First-Order
Case – Cont’d
dy (t )
If the solution of
 ay (t )  bx(t )
dt
t
is
y (t )  y (0)e
 at
 e
 a ( t  )
bx ( )d , t  0
0
dq(t )
then the solution of
 aq(t )  (b0  ab1 ) x(t )
dt
is
q (t )  q (0)e
t
 at
 e
0
 a ( t  )
(b0  ab1 ) x( )d , t  0
System Modeling – Electrical Circuits
resistor
v(t )  Ri (t )
t
dv(t ) 1
1
capacitor
 i (t ) or v(t )   i ( )d
dt
C
C 
di (t )
inductor v(t )  L
dt
t
1
or i (t )   v( )d
L 
Example: Bridged-T Circuit
Kirchhoff’s voltage law
loop (or mesh)
v1 (t ) 
equations

R1  i1 (t )  i2 (t )   x(t )
v1 (t )  v2 (t )  R2i2 (t )  0
 y (t )  x(t )  R i (t )
2 2

Mechanical Systems
• Newton’s second Law of Motion:
2
d y (t )
x(t )  M
2
dt
• Viscous friction:
dy (t )
x(t )  kd
dt
• Elastic force:
x(t )  k s y (t )
Example: Automobile Suspension
System
 d 2 q (t )
 dy (t ) dq (t ) 
 M 1 dt 2  kt [q (t )  x(t )]  k s [ y (t )  q (t )]  k d  dt  dt 




2
d
y (t )
 dy (t ) dq (t ) 
M
 k s [ y (t )  q (t )]  k d 

0
2
2


dt
dt 
 dt
Rotational Mechanical Systems
• Inertia torque:
d  (t )
x(t )  I
2
dt
2
• Damping torque:
d (t )
x(t )  kd
dt
• Spring torque:
x(t )  k s (t )
Linear I/O Difference Equation With
Constant Coefficients
• Consider the DT SISO system
x[n]
y[n]
System
described by
N
M
i 1
i 0
y[n]   ai y[n  i ]  bi x[n  i ]
ai  , bi 
N is the order or dimension of the system
Solution by Recursion
• Unlike linear I/O differential equations,
linear I/O difference equations can be
solved by direct numerical procedure (N-th
order recursion)
N
M
i 1
i 0
y[n]   ai y[n  i ]   bi x[n  i ]
(recursive DT system or recursive digital filter)
Solution by Recursion – Cont’d
• The solution by recursion for n  0 requires
the knowledge of the N initial conditions
y[ N ], y[ N  1],
, y[1]
and of the M initial input values
x[M ], x[M  1],
, x[1]
Analytical Solution
• Like the solution of a constant-coefficient
differential equation, the solution of
N
M
i 1
i 0
y[n]   ai y[n  i ]   bi x[n  i ]
can be obtained analytically in a closed form
and expressed as
y[n]  y zi [n]  y zs [n]
(total response = zero-input response + zero-state response)
• Solution method presented in ECE 464/564