Solution to ECE Test #2 Su06
1.
2s ( s + 5)
can be expressed in the form
s2 + 9
x ( t ) = Aδ ( t ) + B  cos ( at ) + C sin ( bt )  u ( t ) .
The inverse Laplace transform of X ( s ) =
Find numerical values for A, B, C, a and b.
A = 2 , B = 10 , C = −0.6 , a = 3 , b = 3 .
)
2
s + 9 2 s + 10 s
2
2
2 s 2 + 18
10 s − 18
⇒ X(s) = 2 +
10 s − 18
1.8 3 
 s
= 2 + 10  2
−

2
s +9
s +9
3 s2 + 9 
x ( t ) = 2δ ( t ) + 10  cos ( 3t ) − 0.6 sin ( 3t )  u ( t )
2.
The inverse Laplace transform of X ( s ) =
3s − 2
can be expressed in the form
( s + 4 )2
x ( t ) = A (1 + Bt ) eat u ( t ) .
Find numerical values for A, B and a.
A = 3 , B = −14 / 3 , a = −4 .
X(s) =
−14
(s + 4)
2
+
(
)
3
⇒ x ( t ) = 3e−4 t − 14 te−4 t u ( t ) = ( 3 − 14 t ) e−4 t u ( t )
s+4
x ( t ) = 3(1 − (14 / 3) t ) e−4 t u ( t )
Alternate Solution:
Start with
3 14 3s − 14
L
3 u ( t ) − 14 t u ( t ) ←
→ − 2 =
s s
s2
Let s → s + 4 . Then, using the complex-frequency shifting property,
L
x ( t ) = e−4 t  3 u ( t ) − 14 t u ( t )  = 3(1 − (14 / 3) t ) e− t u ( t ) ←
→
3s − 2
( s + 4 )2
= X(s)
Second Alternate Solution:
X(s) =
(
3s − 2
(s + 4)
2
=
3s
(s + 4)
2
−
2
( s + 4 )2
)
d
3s
2
L
3te−4 t u ( t ) +  3te−4 t u ( t ) t = 0− − 2te−4 t u ( t ) ←
→
2 −
dt
( s + 4 ) ( s + 4 )2
(
)
3s
2
L
3 te−4 tδ ( t ) + u ( t ) −4 te−4 t + e−4 t  − 2te−4 t u ( t ) ←
→
2 −
( s + 4 ) ( s + 4 )2
L
3(1 − (14 / 3) t ) u ( t ) e−4 t ←
→
3s
(s + 4)
2
−
2
( s + 4 )2
3.
Given the Laplace-transform pair
3s 2 + 9 s − 14
x ( t ) ←→
s s 2 + 2 s + 10
L
(
)
and the fact that x ( t ) is continuous at t = 0 ,
(a)
Find the Laplace transform of
d
( x ( t − 2 )) .
dt
d
3s 2 + 9 s − 14
L
− x 0−
x ( t )) ←→ s
(
2
dt
s s + 2 s + 10
(
( ) ( )
x 0− = x 0+
( )
)
(from continuity at t = 0 )
From the initial-value theorem
( )
x 0 + = lim s
s →∞
3s 2 + 9 s − 14
=3
s s 2 + 2 s + 10
(
)
d
3s 2 + 9 s − 14
L
( x (t )) ←→ s 2 + 2 s + 10 − 3
dt

 3s 2 + 9 s − 14
d
L
2
←

→
− 3 e−2 s
x
t
−
(
)
(
)

2
dt
 s + 2 s + 10

(b)
Find the numerical value of lim x ( t ) if it exists. If it does not exist explain
t →∞
how you know that.
Poles of s X ( s ) are at −1 ± j 3 in the open left half-plane. Therefore the
final-value theorem applies.
L
lim x ( t ) ←
→ lim s
t →∞
s→ 0
3s 2 + 9 s − 14
= −1.4
s s 2 + 2 s + 10
(
)
4.
In the standard feedback configuration (illustrated below), H1 ( s ) =
K
and
s −1
4
. What numerical range of values of K makes the overall feedback
s+5
system stable?
H2 ( s) =
X(s)
E(s)
H1(s)
Y(s)
H2(s)
K
K ( s + 5)
s −1
H (s) =
=
4K
( s − 1) ( s + 5 ) + 4 K
1+
( s − 1) ( s + 5 )
For stability the poles must be in the open left half-plane. Poles are at
( s − 1) ( s + 5 ) + 4 K = 0
s2 + 4 s + 4 K − 5 = 0 ⇒ s =
−4 ± 16 − 16 K + 20 −4 ± 36 − 16 K
=
2
2
If K > 9 / 4 the poles are complex, the real parts are both -2, therefore in the open
left half-plane and the system is stable.
If K ≤ 9 / 4 the poles are real.
If 36 − 16 K ≥ 16 then one real pole will be not be in the open left halfplane. This translates into 16 K ≤ 20 ⇒ K ≤ 5 / 4 .
So, overall, for stability, K > 5 / 4 .