Signals and Systems 1
Lecture 8
Dr. Ali. A. Jalali
September 6, 2002
Signals and Systems 1
Lecture # 8
Differential Equation Model
EE 327 fall 2002
Differential Equation Model
1. Many physical systems are described by linear
2.
3.
4.
5.
6.
differential equations.
Reducing differential equations to algebraic
equations
Homogeneous solution, exponential solution
and natural frequencies.
Particular solution, system function and poleszeros
Total solution, initial condition and steadystate
Conclusion
The Nth-order Differential Equation Model
1.
2.
One important way of modeling LTI systems is by means of
linear constant-coefficient equations.
Example for DE model: Analog Filter
2nd-order ED
2
1
1
vt  
vt   2
v t   2
et .
RC1
R C1C2
R C1C2
The Nth-order Differential Equation Model
1. Example for DE model:
Instrument servo
3rd-order ED
kk p A
kk p A
RJ

BL
RB



t   
t  
 t  
t  
r t  .
 
 LJ 
LJ
LJ
LJ
The Nth-order Differential Equation Model
1. Example for DE model: Electric Network
2nd-order ED
The Nth-order Differential Equation Model
1. Example for DE model: Electric Network
2nd-order ED
The Nth-order Differential Equation Model
1. The general linear constant-coefficient Nth-Order
DE for SISO systems are:
dyt 
d N y t 
a0 yt   a1
a N
 b0 xt 
N
dt
dt
dxt 
d L x t 
b1
a L
.
L
dt
dt
OR:
d y t 
d x t 
ak
  bk
.

k
k
dt
dt
k 0
k 0
N
k
L
k
X(t) system input,
Y(t) system output and our practical restriction order
NL
Initial Condition Solution of Differential Equation
1. The character equation of DE of system is found by
substituting a trial solution
N

into the
k
d y(t )
ak
0

k
dt
k 0
homogeneous DE:
The result is:

y(t )  Cest




a0 s 0 Ce st  a1s1 Ce st    a N s N Ce st  0 .
Since the Ce st cannot be zero it can be factored out. The remaining term
must satisfy the algebraic equation.
a0 s  a1s    a N s  0
0
1
This is Characteristic Equation.
N
Initial Condition Solution of Differential Equation
This is Characteristic Equation
Can be written as:
Or as factored form:
a0 s 0  a1s1    a N s N  0
N
k
a
S
 k  0.
k 0
N
k
a
s
 k  aN s  r1 s  r2 s  rN   0
k 0
Characteristic roots are:
s1  r1 , s2  r2 ,, sN  rN
Where r1 , r2 , , rN may be real or complex (conjugate pairs).
Initial Condition Solution of Differential Equation
The solution of the homogeneous DE of:
k
N
d y(t )
ak
0

k
dt
k 0
For a given set of initial conditions:

yt  t 0 , dyt  dt  t 0 ,, d
N 1
yt  dt
N 1
Is called the initial condition (IC) solution and for simple (non
repeating) roots is of the form:
y IC t   C1e
r1t
 C2 e  C N e
r2 t
N
or:
y IC t    ck e rk t
k 1
rN t
.

t 0
Initial Condition Solution of Differential Equation
N
y IC t    ck e rk t
In IC solution:
k 1
k  1, 2,, N
are coefficients that must be determined in
order to satisfy the given set of initial conditions and
k  1, 2, , N
rk
are the characteristic roots.
Multiple roots:
If the CE contains multiple roots indicated by factor
terms of the form
 s  sk 
q
C1t q 1e sk t  C2 t q 2 e sk t Cq e sk t
will appear in IC solution.
System Stability in term of IC response
A casual system is stable if the IC response decays to zero
as:
t .
This happens if and only if all the rk ‘s are negative real, or
if there are any complex roots, their real parts must be negative.
Img
2
r1  1, r2  2, r3, 4  2  2 j
Real
-3
-2
-1
-2
Natural Frequency
The roots of the characteristic polynomial
are called natural frequencies.
These are frequencies which there is an
output in the absence of an input.
Also the roots of the characteristic
polynomial are called eigenvalues.
Example:
The flexible shaft subjected to an applied torque (t ) with the shaft angle
 (t ) described by the DE:

t   3 t   2t   1 3t 
1- Find CE?
2- Catachrestic roots?
3- Is this system stable?
4- Give the algebraic form of the initial condition response?
5- Find the constants in the IC solution if  (t )  1 and [d (t ) / dt ]
t 0
 0.
Example Solution
1- Find CE?
The homogeneous differential equation is:
s 2  3s  2  0.
Yielding the CE as:
2- CR?
Using the quadratic formula, CR are:
t   3t   2 t   0
r1, 2  1,  2.
3- Is this system stable?
Both roots are negative real so the system is stable.
4- Algebraic form of IC response?
 IC (t )  C1e t  C2 e 2t
5- Constant IC solution?  IC (0)  1  C1  C2
and [d IC (t ) / dt ] t 0  0  C1  2C2 .
Solving gives: C1  2 and C2  1. Thus  (t )  2e  t
IC
2 t
 e , t  0.
MATLAB Comment
Function roots in MATLAB:
If the CE is 5th order as:
5s5  4s 4  3s 3  2s 2  s  0
>>p=[5 4 3 2 1 0];
>>r=roots(p)
The results are:
>> p=[5 4 3 2 1 0];roots(p)
ans =
0
0.1378 + 0.6782i
0.1378 - 0.6782i
-0.5378 + 0.3583i
-0.5378 - 0.3583i
r1  0
r2 ,3  01378
.
 j0.6782
r4 ,5  05378
.
 j0.3583
The Unit Impulse Response Model
1.
The unit impale function is defined implicitly by
its sifting property:



 (t  t0 ) f (t )dt  f (t0 )
where f(t) is assumed to be continuous at t  t0 .
pt  .
Approximation to impulse function: t  t 0   lim
0
EE 327 fall 2002
Signals and Systems 1
The Unit Impulse Response Model
1.
Using the sifting property lead the
following product function.
Approximation of the sifting property


The value of integral is: f t 0   .   f t 0 
EE 327 fall 2002
Signals and Systems 1
Unit Impulse Response
The response of an LTI system to an input
of unit impulse function is called the unit
impulse response.
x(t)=(t)
LTI
y(t)=h(t)
Important: When determining the unit impulse
response h(t) of an LTI system, it is
necessary to make all initial conditions
zero. (output due to input not energy stored in system)
EE 327 fall 2002
Signals and Systems 1
Convolution
If the unit impulse response h(t) of a linear
continuous system is known, the system
output y(t) can be found for any input x(t).
Approximation by pulse

Sifting property of pulse
EE 327 fall 2002
x(t )   x( ) f (t   )d

Signals and Systems 1
Convolution Integral
1. The convolution integral is one of the most important
results used in the study of the response of linear
systems.
2. If we know the unit impulse response h(t) for a linear
system, by using the convolution integral we can
compute the system output for any known input x(t).
3. In the following integration integral h(t) is the
system’s unit impulse response.

x(t )   x( )h(t   )d

EE 327 fall 2002
Signals and Systems 1
Convolution Evaluation
1. The convolution integral can be evaluated in three
distinct ways.
a) Analytical method,
b) Graphical method,
c) Numerical convolution
We will discuss about these and about convolution
properties in class. (see class notes)
EE 327 fall 2002
Signals and Systems 1