Eq
Fall EL611
Differential 2008
1. Find the System functions (transfer functions) of the systems described by the
following differential equations. Also, assuming the systems are causal, state whether
or not they are BIBO stable.
a) 2 y (t ) 8 y(t ) 6 y(t ) x(t ) x(t )
Taking laplace transform of both sides using the derivative theorem yields
2 s 3Y (s) 8s 2Y (s) 6 sY (s) s 2 X (s) X (s)
2 s3 8s 2 6 sY (s) s 2 1X (s)
So,
H ( s)
Y ( s)
s2 1
( s 1)(s 1)
( s 1)
X ( s) 2 s 3 8s 2 6 s 2s( s 1)(s 3) 2s( s 3)
The system is not BIBO stable because there is a pole ( s 0 ) on the j - axis.
b) y (t ) y(t ) 2 y(t ) x(t ) x(t )
H ( s)
s 1
s2 s 2
s 1
1
( s 1)(s 2) ( s 2)
The system is BIBO stable because all poles are in the left half-plane, Re{s} 0 .
2. Find the differential equation of a system with the transfer function below.
H ( s)
3s 3 2s 2 5s
( s 1)( s 0.5) 3
Expanding out the denominator gives
H ( s)
3s 3 2s 2 5s
s 4 1 / 2 s 3 3 / 4 s 2 5 / 8 s 1 / 8
.
The differential equation is, then
y (4) (t ) 1 / 2 y (t ) 3 / 4 y (t ) 5 / 8 y (t ) 1 / 8 y(t ) 3 x(t ) 2 x(t ) 5 x(t )
3. Find the differential equation of systems with the impulse responses below.
a) ) h(t t e 2 t U (t )
H ( s)
1
( s 2) 2
1
s2 4 s 4
.
y (t ) 4 y (t ) 4 y(t ) x(t )
b) h(t ) 2 e 3 t U (t ) 5 et U (t )
H ( s)
3 s 17
3 s 17
2
5
s 3 s 1 ( s 1)(s 3) s 2 2 s 3
y(t ) 2 y (t ) 3 y(t ) 3 x(t ) 17 x(t )
4. A system is defined by the differential equation
y (t ) 3 y (t ) 2 y(t ) x(t ) 2 x(t ) .
a) Find the output, y (t ) , if the input is x(t ) t 2 U (t ) .
H ( s)
X (s)
s2 2
s2 3s 2
2
s3
s2 2
( s 1)(s 2)
2 s 2 2
2s 2 4
.
Y ( s)
s 3 s 2 3s 2 s 3 ( s 1)(s 2)
Expanding in partial fractions yields
Y ( s)
2 s2 4
s 3 ( s 1)(s 2)
9/ 2 3
2
6
3/ 2
.
s
s 2 s3 s 1 s 2
Inverse transforming yields
9 / 2U (t ) 3 t U (t ) t 2 U (t ) 6 e t U (t ) 3 / 2 e 2 t U (t )
b) Find the output if the input is x(t ) e j t U (t ) . Show that the steady-state
response is equal to y ss (t ) H ( j ) e j t .
Now X ( s)
Y ( s)
1
.
s j
c0
c
c
s2 2
1 2,
( s j )(s 1)(s 2) s j s 1 s 2
Where
c0
( j )2 2
H ( j) ,
( j 1)( j 2)
c1
(1) 2 2
3
(1 j )(1 2) (1 j )
c2
(2) 2 2
6
.
(2 j )(2 1) (2 j )
y(t ) c0 e j t U (t ) c1 e t U (t ) c2 e 2 t U (t ) .
y ss (t ) c0 e j t H ( j ) e j t .