Chapter 9 - The Laplace Transform
Selected Solutions
1. Sketch the pole-zero plot and region of convergence (if it exists) for these signals.
ω
ROC
(a)
x( t) = e
(b)
x( t) = e 3 t cos(20πt) u(− t)
−8 t
[s]
u( t)
σ
s = −8
ω
(c)
x( t) = e u(− t) − e
2t
−5 t
u( t)
[s]
ROC
s = −5 s = 2
2. Starting with the definition of the Laplace transform,
Solutions 9-1
σ
L ( g ( t )) = G ( s ) =
∞
∫ g (t ) e
− st
dt ,
0−
find the Laplace transforms of these signals.
(a)
x( t) = e t u( t)
(b)
x( t) = e 2 t cos(200πt) u( t)
(c)
x( t) = ramp( t)
X( s) =
∞
∫ x(t)e
∞
− st
dt =
0−
∫ ramp(t)e
∞
− st
dt =
0−
∫ te
− st
dt
0−
Using
e ax
∫ xe dx = a2 (ax − 1)
∞
 e − st

1
X( s) = 
= 2 , Re( s) = σ > 0
2 (− st − 1) 
 (− s)
0− s
ax
(d)
x( t) = te t u( t)
3. Using the time-shifting property, find the Laplace transform of these signals.
(a)
x( t) = u( t) − u( t − 1)
(b)
x( t) = 3e −3( t − 2) u( t − 2)
(c)
x( t) = 3e −3 t u( t − 2)
L
3 u( t) ←
→
3
s
Using the time shifting property,
L
3 u( t − 2) ←
→
3e
Alternate solution:
−3 t
3e −2 s
s
3e −2( s+ 3)
u( t − 2) ←→
s+3
L
x( t) = 3e −6e −3( t − 2) u( t − 2)
Using the time shifting property,
Solutions 9-2
X( s) =
(d)
3e −6e −2 s 3e −2 s− 6
=
s+3
s+3
x( t) = 5 sin(π ( t − 1)) u( t − 1)
4. Using the complex-frequency-shifting property, find and sketch the inverse Laplace
transform of
1
1
+
.
X( s) =
(s + j 4) + 3 (s − j 4) + 3
5. Using the time-scaling property, find the Laplace transforms of these signals.
(a)
x( t) = δ ( 4 t)
(b)
x( t) = u( 4 t)
L
u ( t ) ←
→
L
u ( 4 t ) ←
→
1
, Re ( s ) > 0
s
1 1 1
= , Re ( s ) > 0
4 s s
4
6. Using the time-differentiation property, find the Laplace transforms of these signals.
(a)
x( t) =
d
(u(t))
dt
( )
d
L
→ s G ( s ) − g 0−
g ( t )) ←
(
dt
L
u ( t ) ←
→
1
, Re ( s ) > 0
s
( )
d
1
L
→ s − u 0 − = 1 , All s
u ( t )) ←
(
dt
s
=0
(b)
x( t) =
d −10 t
(e u(t))
dt
(c)
x( t) =
d
(4 sin(10πt) u(t))
dt
(d)
x( t) =
d
(10 cos(15πt) u(t))
dt
Solutions 9-3
7. Using multiplication-convolution duality, find the Laplace transforms of these signals
and sketch the signals versus time.
(a)
x( t) = e − t u( t) ∗ u( t)
(b)
x( t) = e − t sin(20πt) u( t) ∗ u( t)
L
e− t sin ( 20π t ) u ( t ) ∗ u ( t ) ←
→
( s + 1)
20π
2
+ ( 20π )
2
1
s
20π
1
20π
1
As + B
=
+
2
2
2
(s + 1) + (20π ) s 1 + (20π ) s (s + 1) 2 + (20π ) 2
Multiply through by s, let s approach infinity and solve for A. After finding A, let
s = 1 and solve for B,
X( s) =
1

20π
1
20π
s +1
2  −
2
2 −
2
2
20π ( s + 1) + (20π ) 
1 + (20π )  s ( s + 1) + (20π )
x( t) =
sin(20πt) 
20π 
−t 
 u( t)
2 1 − e cos(20πt) +
20π 
1 + (20π ) 

(c)
 πt 
x( t) = 8 cos  u( t) ∗ [u( t) − u( t − 1)]
 2
(d)
x( t) = 8 cos(2πt) u( t) ∗ [u( t) − u( t − 1)]
After time, t = 1, the solution is zero.
8. Using the initial and final value theorems, find the initial and final values (if possible) of
the signals whose Laplace transforms are these functions.
Initial Value Theorem
g(0 + ) = lim s G( s)
s→∞
Final Value Theorem
limg( t) = lim s G( s)
t →∞
(a)
X( s) =
10
s+8
s→ 0
, if limg( t) exists
t →∞
, One pole in open LHP
x(0 + ) = lim s
s→∞
10
= 10
s+8
Solutions 9-4
lim x( t) = lim s
t →∞
s→ 0
10
=0
s+8
and the limit exists because the only pole of X( s) =
(b)
X( s) =
s+3
, Poles at −3 ± j 2
(s + 3) 2 + 4
(c)
X( s) =
s
, Poles at ± j2
s +4
10
is in the open LHP
s+8
2
x(0 + ) = lim s
s→∞
s
=1
s +4
2
Final-value theorem does not apply because there are two poles on the ω axis
(d)
X( s) =
(e)
X( s) =
(f)
10 s
, Poles at −5 ± j16.583
s + 10 s + 300
2
8
, Poles at 0 and -20
s( s + 20)
8
X( s) = 2
, Double pole at zero.
s ( s + 20)
9. Find the inverse Laplace transforms of these functions.
(a)
X( s) =
24
s( s + 8)
X(s) =
3
3
−
s s+8
x( t) = 3(1 − e −8 t ) u( t)
(b)
X( s) =
20
s + 4s + 3
(c)
X( s) =
5
s + 6 s + 73
(d)
X( s) =
10
s( s + 6 s + 73)
X( s) =
2
2
2
 10 1

10 1
3
8
s+6
s+3
−
 −
=
 −

2
2
2
73  s ( s + 3) + 64  73  s ( s + 3) + 64 8 ( s + 3) + 64 
Solutions 9-5
X( s) =
(e)
4
s ( s + 6 s + 73)
2
2
X( s) =
2
Using the “cover up” method, A =
4
≅ 0.0548 . Using
73
d m−k 
1
s − pq
( m − k )! ds m − k 
,
K qk =
B=
4
A B
Cs + D
= 2+ + 2
s s + 6 s + 73
s ( s + 6 s + 73) s
2
(

1 d 
4
 2

1! ds  ( s + 6 s + 73) 
)
m
H ( s )
 s → pq
[
= −4 ( s2 + 6 s + 73)
s→ 0
−2
(2s
2
k = 1, 2,
]
+ 6)
s→ 0
=−
,m
24
≅ −0.0045
(73) 2
24
4
4
Cs + D
(73) 2
73
X( s) = 2 2
= 2 −
+ 2
s
s + 6 s + 73
s ( s + 6 s + 73) s
Multiply through by s and let s approach infinity,
0=−
24
24
≅ 0.0045
2 +C ⇒C =
(73)
(73) 2
24
24
4
2
2 s+ D
4
73
73
(
)
(
)
X( s) = 2 2
= 73 −
+
s
s ( s + 6 s + 73) s2
(s + 3) 2 + 64
Then let s = 1.
24
+D
4
4
24
21464
148
(73) 2
⇒ D= 4−
=
−
= −0.0278
2 +
2 = −
80 73 ( 73)
80
(73)
(73) 2
24
24
148
4
37 

2
2 s−
2
s−


1
292
24
(
)
(
)
(
)
73
73
73
6
=
X( s) = 73
+ 2
+ 24
2  2 −
2
2 −
s
s
s + 6 s + 73
s
(73)  s
(s + 3) + 64 


X( s) =


s+3
1  292 24
55
8
+ 24 
−
2  2 −
2
2

s
(73)  s
 ( s + 3) + 64 48 ( s + 3) + 64  
Solutions 9-6
x( t) =
1 
55

−3 t 
 cos(8 t) − sin(8 t)  u( t)
2 292 t − 24 + 24 e


48
(73) 
(f)
X( s) =
2s
s + 2 s + 13
(g)
X( s) =
s
s+3
(h)
X( s) =
s
s + 4s + 4
(i)
X( s) =
s2
s2 − 4 s + 4
(j)
X( s) =
10 s
s + 4 s2 + 4
2
2
4
j5 2
4
X( s) =
s+ j 2
(
j5 2
4
−
s− j 2
) (
2
)
2
⇒ x( t) =
(
j5 2 − j
te
4
2t
− te j
2t
) u(t)
10. Using a table of Laplace transforms, find the CTFT’s of these signals.
(a)
x( t) = 10e −100 t u( t)
(b)
x( t) = 3e −50 t cos(100πt) u( t)
11. Using the Laplace transform, solve these differential equations for t ≥ 0.
(a)
x′ ( t) + 10 x( t) = u( t)
,
x(0 − ) = 1
s X( s) − x(0 − ) + 10 X( s) =
1
s
1
+1
s +1
s
X( s) =
=
s + 10 s( s + 10)
1
9
1 + 9e −10 t
X( s) = 10 + 10 ⇒ x( t) =
u( t)
10
s s + 10
Solutions 9-7
Checking initial conditions,
x(0 + ) = 1
which agrees with the initial condition, x(0 − ) = 1. For this system and this excitation the
response cannot change instantaneously.
(b)
x′′ ( t) − 2 x′ ( t) + 4 x( t) = u( t) ,
d

x(0 − ) = 0 ,  x( t) 
=4
 dt
t = 0−
(c)
x′ ( t) + 2 x( t) = sin(2πt) u( t)
x(0 − ) = −4
,
12. Using the Laplace transform, find and sketch the time-domain response, y( t) , of the
systems with these transfer functions to the sinusoidal excitation, x( t) = A cos(10πt) u( t) .
(a)
H( s) =
1
s +1
2
 1
 1

10π
A
s + (10π ) 
A
s
10
Y( s) =
−
+
−
+
+
π
=



2
2
2
2
2
1 + (10π )  s + 1 s2 + (10π )  1 + (10π )  s + 1 s2 + (10π )
s2 + (10π ) 
(b)
H( s) =
s− 2
(s − 2) 2 + 16
s− 2
As
s2 − 2 s
Y( s) =
=A 2
2
2
2
2
(s − 2) 2 + 16 s2 + (10π ) 2
s ( s − 2) + 16 s2 + (10π ) ( s − 2) + 16(10π )
Y( s) =
(
s2 − 2 s
)
s4 − 4 s3 + s2 20 + (10π ) − 4 (10π ) s + 20(10π )
2
2
s2 − 2 s
Y( s) = 4
s − 4 s3 + 1006.97 s2 − 3947.84 s + 19739.2
Using MATLAB,
»X
Transfer function:
s
--------s^2 + 987
»H
Solutions 9-8
2
Transfer function:
s - 2
-------------s^2 - 4 s + 20
»Y
Transfer function:
s^2 - 2 s
-----------------------------------------s^4 - 4 s^3 + 1007 s^2 - 3948 s + 1.974e04
»[z,p,k] = zpkdata(Y,'v') ;
»z
z =
0
2
»p
p =
-0.00000000000000
-0.00000000000000
2.00000000000000
2.00000000000000
+31.41592653589795i
-31.41592653589795i
+ 4.00000000000000i
- 4.00000000000000i
r =
-0.00105906221326
-0.00105906221326
0.00105906221326
0.00105906221326
+
+
-
0.01610704734047i
0.01610704734047i
0.00203398509622i
0.00203398509622i
 −0.00106 − j 0.01611 −0.00106 + j 0.01611 
+


s − j10π
s + j10π


Y( s) = A
 0.00106 + j 0.00203 0.00106 − j 0.00203 
+
+

s − 2 − j4
s − 2 + j4


 −0.00212 s + 1.0122 0.00212 s − 0.02048 
Y( s) = A 
+

2
s2 + (10π )
(s − 2) 2 + 16 

 s − 9.66
s + 477.45 
Y( s) = 0.00212 A 
− 2
2
2
 ( s − 2) + 16 s + (10π ) 


s− 2
s
7.66
4
477.45
10π
Y( s) = 0.00212 A 
−
− 2
2
2
2 −
2
2
4 ( s − 2) + 16 s + (10π )
10π s + (10π ) 
 ( s − 2) + 16
{
}
y( t) = 0.00212 A e 2 t [cos( 4 t) − 1.915 sin( 4 t)] − cos(10πt) − 15.2 sin(10πt) u( t)
Solutions 9-9
y(t)
5
-1
5
t
-10
13. Write the differential equations describing these systems and find and sketch the
indicated responses.
(a)
x( t) = u( t)
,
y( t) is the response, y(0 − ) = 0
∫
x(t)
y(t)
4
The solution is continuous at t = 0 because, if it were not the discontinuity would
cause an impulse on the left-hand side of the equation which could not be equated to the step
excitation on the right-hand side.
(b)
v(0 − ) = 10 , v( t) is the response
+
C = 1 µF v(t)
R = 1 kΩ
C v′ ( t) +
[
v( t)
=0
R
]
C s V( s) − v(0 − ) +
V( s) =
10C
1
sC +
R
V( s)
=0
R
= 10
1
s+
Solutions 9-10
1
RC
v( t) = 10e
−
t
RC
u( t) = 10e −1000 t u( t) , t > 0
v(0 + ) = 10 = v(0 − ) . Check.
The solution is continuous at t = 0 because the capacitor voltage cannot change
instantaneously.
x(t)
10
0.004
14.
t
Find the three parts, x ac ( t), x 0 ( t) and x c ( t) , of the following signals.
(a)
x( t) = e −10 t u( t) − e 2 t u(− t)
x ac ( t) = −e 2 t u(− t) , x 0 ( t) = 0 , x c ( t) = e −10 t u( t)
(b)
x( t) = K
(c)
x( t) = u( t)
(d)
x( t) =
d
(u(t))
dt
15. Find the bilateral Laplace transforms of these signals.
(a)
x( t) = 3e −7 t u( t) − 12e 4 t u(− t)
x c ( t) = 3e −7 t u( t) ⇒ X c ( s) =
3
, Re( s) > −7
s+7
x 0 ( t) = 0 ⇒ X 0 ( s) = 0
x ac ( t) = −12e 4 t u(− t) ⇒ x ac (− t) = −12e −4 t u( t) ⇒ X ac (− s) = −
X ac ( s) =
X( s) =
12
, Re( s) > −4
s+4
12
, Re( s) < 4
s− 4
5 s + 24
3
12
+
=3 2
, − 7 < Re( s) < 4
s+ 7 s− 4
s + 3s − 28
Solutions 9-11
(b)
x( t) = 50e −10 t
16. Find the responses, y( t) , of these systems to these excitations.
(a)
h( t) = e −5 t u( t)
,
x( t) = 3e −7 t u( t) − 12e 4 t u(− t)
(b)
h( t) = tri( t)
,
x( t) = e − t u( t)
Using
s
− 
 2s
2
e −e 
L
tri ( t ) ←
→


s


s
− 
 2s
2
e
−
e

H( s) = 

s 


Therefore
2
, All s
2
, All s
and
X( s) =
1
, Re( s) > −1
s +1
2
s
− 
 2s
2
1 e − e 
es − 2 + e−s
=
Y( s) =
, Re( s) > −1
s + 1
s 
s2 ( s + 1)


1 
1 1
Y( s) = (e s − 2 + e − s ) 2 − +
 , Re( s) > −1
s
s s + 1

ramp( t + 1) − 2 ramp( t) + ramp( t − 1)


y( t) = − u( t + 1) + 2 u( t) − u( t − 1)

+ e −( t +1) u( t + 1) − 2e − t u( t) + e −( t −1) u( t − 1) 


[
]
 ramp( t + 1) − 1 + e −( t +1) u( t + 1) 


y( t) = −2 ramp( t) − 1 + e − t u( t)



−( t −1)
u( t − 1)
+ ramp( t − 1) − 1 + e
[
(c)
h( t) = e −10 t u( t)
[
]
,
]
x( t) = 50e −10 t
17. Sketch the pole-zero plot and region of convergence (if it exists) for these signals.
(a)
x( t) = e − t u(− t) − e −4 t u( t)
(b)
x( t) = e −2 t u(− t) − e t u( t)
Solutions 9-12
18. Using the integral definition find the the unilateral Laplace transform of these time
functions.
(a)
g( t) = e − at u( t)
(b)
g( t) = e − a ( t −τ ) u( t − τ ) , τ > 0
(c)
g( t) = e − a ( t +τ ) u( t + τ ) , τ > 0
G( s) = e
(d)
g( t) = sin(ω 0 t) u( t)
(e)
g( t) = rect ( t)
(f)
 1
g( t) = rect  t − 
 2
∞
− aτ
e − aτ
 1 −( s + a ) t 
− s + a e
 − = s + a , τ > 0
0
19. Using MATLAB (or any other appropriate computer mathematics tool) do the inversion
integral of
1
G( s ) =
s + 10
numerically. That is, approximate the inversion integral with a summation of the form
g ( t ) L 1 ( G ( s )) =
1
j 2π
e sn t
1
∑ s + 10 ∆sn = j 2π
n=− N n
N
e(σ + jn∆ω )t
j ∆ω , σ > 0 .
∑
n = − N σ + jn ∆ω + 10
N
Choose the combination of large N and small ∆ω so that the summation will range over a
contour from well below to well above the real axis. Plot g(t) versus t by computing the
value of g(t) at every value of t from the above summation approximation to the inversion
integral. Compare to the analytical result. Try at least three different values of σ to see the
effect on the result. (Ideally there is no effect of changing σ as long as it is greater than
-10, but actually, in this numerical approximation, there will be some small effects.)
%
Program to demonstrate the inverse Laplace transform numerically.
close all ;
tau = 0.1 ; dw = 1 ; p = -1/tau ;
w = dw*[-2000:2000]' ; ds = j*dw*ones(length(w),1) ;
allint = [] ;
for sigma = 0:5,
s = sigma + j*w ;
int = [] ; tv = [] ;
for t = -tau*2:tau/20:tau*4,
Solutions 9-13
f = exp(s.*t)./(s - p) ; tv = [tv;t] ; int =
[int;sum(f.*ds)/(j*2*pi)] ;
end
int = real(int) ; allint = [allint,int] ;
end
subplot(3,2,1) ; h = plot(tv, allint(:,1), 'k') ; set(h, 'LineWidth',2)
;
xlabel('Time, t (s)') ; ylabel('h(t)') ; title('sigma = 0') ;
grid ; axis([-0.2, 0.4, -0.1, 1]) ;
subplot(3,2,2) ; h = plot(tv, allint(:,2), 'k') ; set(h, 'LineWidth',2)
;
xlabel('Time, t (s)') ; ylabel('h(t)') ; title('sigma = 1') ;
grid ; axis([-0.2, 0.4, -0.1, 1]) ;
subplot(3,2,3) ; h = plot(tv, allint(:,3), 'k') ; set(h, 'LineWidth',2)
;
xlabel('Time, t (s)') ; ylabel('h(t)') ; title('sigma = 2') ;
grid ; axis([-0.2, 0.4, -0.1, 1]) ;
subplot(3,2,4) ; h = plot(tv, allint(:,4), 'k') ; set(h, 'LineWidth',2)
;
xlabel('Time, t (s)') ; ylabel('h(t)') ; title('sigma = 3') ;
grid ; axis([-0.2, 0.4, -0.1, 1]) ;
subplot(3,2,5) ; h = plot(tv, allint(:,5), 'k') ; set(h, 'LineWidth',2)
;
xlabel('Time, t (s)') ; ylabel('h(t)') ; title('sigma = 4') ;
grid ; axis([-0.2, 0.4, -0.1, 1]) ;
subplot(3,2,6) ; h = plot(tv, allint(:,6), 'k') ; set(h, 'LineWidth',2)
;
xlabel('Time, t (s)') ; ylabel('h(t)') ; title('sigma = 5') ;
grid ; axis([-0.2, 0.4, -0.1, 1]) ;
20. Using a table of unilateral Laplace transforms and the properties find the unilateral
Laplace transforms of the following functions.
(a)
g( t) = 5 sin(2π ( t − 1)) u( t − 1)
(b)
g( t) = 5 sin(2πt) u( t − 1)
sin(2πt) = sin(2π ( t − 1))
Therefore
L
5 sin ( 2π t ) u ( t − 1) ←
→
(c)
10π e− s
s 2 + ( 2π )
2
g( t) = 2 cos(10πt) cos(100πt) u( t)
Use the trigonometric identity,
cos(10πt) cos(100πt) =
1
[cos(10πt − 100πt) + cos(10πt + 100πt)] ,
2
then complete the solution as usual.
Solutions 9-14
(d)
(e)
g( t) =
g (t ) =
d
(u(t − 2))
dt
t
∫ u (τ ) dτ
0−
(f)
g( t) =

d  − t −2τ
u( t − τ ) , τ > 0
 5e
dt 

Use these properties:
Frequency Shifting
Time Shifting
Linearity
Time Differentiation Once
(g)
g( t) = 2e −5 t cos(10πt) u( t)
(h)
π

x( t) = 5 sinπt −  u( t)

8
π 
π 
π cos  − sin  s
 8
 8
X( s) = 5
2
2
s +π
21. Given
L
g ( t ) ←
→
s +1
s(s + 4)
find the Laplace transforms of
(a)
g(2t)
(b)
d
(g(t))
dt
Time Differentiation Once
s +1
d
g( t)) ←
L→ s
− g(0 − )
(
s( s + 4 )
dt
d
s +1
L
→
− g 0−
g ( t )) ←
(
dt
s+4
( )
g(0 + ) = lim s G( s) = 1
Initial Value Theorem
s→∞
d
s +1
L
→
−1
g ( t )) ←
(
dt
s+4
d
3
L
→−
g ( t )) ←
(
dt
s+4
Solutions 9-15
(This is correct if g(0 − ) = g(0 + ) . That is, if g is continuous at time, t = 0.)
(c)
g( t − 4 )
g( t) ∗ g( t)
(d)
22. Find the time-domain functions which are the inverse Laplace transforms of these
functions. Then, using the initial and final value theorems verify that they agree with the
time-domain functions.
(a)
G( s) =
4s
(s + 3)(s + 8)
32
 12

g( t) =  − e −3 t + e −8 t  u( t)
 5

5
 12
32


limg( t) = lim − e −3 t + e −8 t  u( t)  = 0
t →∞
t →∞ 

5
 5

lim+ s G( s) = lim+
s→ 0
(b)
(d)
s→ 0
4 s2
= 0 , Check.
(s + 3)(s + 8)
 12
32


lim+ g( t) = lim+  − e −3 t + e −8 t  u( t)  = 4

t→0
t → 0 
5
5

2
4s
= 4 , Check.
lim s G( s) = lim
s→∞
s→∞ ( s + 3)( s + 8)
s
4
(c)
G( s) =
G( s) = 2
s + 2s + 2
(s + 3)(s + 8)
e −2 s
G( s) = 2
s + 2s + 2
23. Given
L
e−4 t u ( t ) ←
→ G (s)
find the inverse Laplace transforms of
(a)
 s
G 
 3
 s
L
3e−12 t u ( 3t ) ←
→G 
 3
Frequency Scaling
3e −12 t u( 3t) = 3e −12 t u( t)
(b)
G( s − 2) + G( s + 2)
(c)
G( s)
s
24. The CTFT of
Solutions 9-16
x( t) = e − t
exists but the (unilateral) Laplace transform does not. Why?
25. Compare the CTFT and the Laplace transform of a unit step. Why can the CTFT not be
found from the Laplace transform?
F
u ( t ) ←
→
1
+ πδ (ω )
jω
and
L
u ( t ) ←
→
1
s
26. Show that the common Laplace transform pairs
L
u ( t ) ←
→
1
1
1
L
L
, e−α t u ( t ) ←
→
, te−α t u ( t ) ←
→
s
s +α
( s + α )2
L
sin (ω 0t ) u ( t ) ←
→
L
e−α t sin (ω 0t ) u ( t ) ←
→
ω0
s
L
, cos (ω 0t ) u ( t ) ←
→ 2
2
s + ω0
s + ω 02
2
s +α
ω0
L
, e−α t cos (ω 0t ) u ( t ) ←
→
2
2
(s + α ) + ω0
( s + α )2 + ω 02
L
can be derived from only the impulse transformation, δ ( t ) ←
→ 1 , and the properties of the
Laplace transform.
L
u ( t ) ←
→
1
:
s
t
∫ g (τ ) dτ ←→
Integration
L
0−
t
G (s)
s
1
∫ δ ( λ ) dλ ←→ s
L
0−
L
u ( t ) ←
→
L
e−α t u ( t ) ←
→
1
s
1
:
s +α
Frequency Shifting
L
te−α t u ( t ) ←
→
1
( s + α )2
:
Solutions 9-17
L
e−α t u ( t ) ←
→
1
s +α
ω0
:
s + ω 02
s
L
cos (ω 0t ) u ( t ) ←
→ 2
:
s + ω 02
L
sin (ω 0t ) u ( t ) ←
→
2
L
e−α t sin (ω 0t ) u ( t ) ←
→
ω0
:
( s + α )2 + ω 02
L
e−α t cos (ω 0t ) u ( t ) ←
→
s +α
:
( s + α )2 + ω 02
27. Given an LTI system transfer function, H( s) , find the time-domain response, y( t) to the
excitation, x( t) .
1
s +1
(a)
x( t) = sin(2πt) u( t) , H( s) =
(b)
x( t) = u( t) , H( s) =
3
s+2
(c)
x( t) = u( t) , H( s) =
3s
s+2
(d)
x( t) = u( t) , H( s) =
5s
s + 2s + 2
(e)
x( t) = sin(2πt) u( t) , H( s) =
2
5s
s + 2s + 2
2
X( s) =
Y( s) =
2π
2
s + (2π )
2
s
2π
5s
= 10π 2
2 2
2
s + (2π ) s + 2 s + 2
s + (2π ) ( s2 + 2 s + 2)
[
2
Y( s) = 10π
[s
s
2
+ (2π )
2
](s
2
+ 2 s + 2)
=
]
As + B
Cs + D
2 + 2
s + 2s + 2
s + (2π )
2
Letting s be zero,
0=
B
D
2 +
(2π ) 2
Multiplying through by s and letting s approach infinity,
0= A+C
Solutions 9-18
Letting s = 1,
A+B
C+D
2π
.
2 =
2 +
5
1 + (2π )
1 + (2π )
Letting s = −1,
−A + B
−10π
2 =
2 −C + D .
1 + (2π )
1 + (2π )
Arranging the equations in matrix form,

0


1

1

 1 + (2π ) 2

1
−
2
 1 + (2π )
Solving,
1
(2π ) 2
0
1
2
1 + (2π )
1
2
1 + (2π )
1
0


2  A  


0

1 0  B  
2π

1 1   = 
2
+
(
)
π
1
2




5 5 C



D − 10π 

2
−1 1 
1 + (2π ) 


0
 A  −0.7535 
 B   1.5875 

 = 
C   0.7535 

  
D −0.0804 
where
Y( s) =
−0.7535 s + 1.5875 0.7535 s − 0.0804
+
2
s2 + 2 s + 2
s2 + (2π )


 s +1

s
2.107
2π
1
+
Y( s) = −0.7535  2
.
.
0
7535
−
1
1067



2 −
2
2
2
2π s2 + (2π ) 
(s + 1) + 1
 s + (2π )
 ( s + 1) + 1
{
}
y( t) = −0.7535[cos(2πt) − 0.3353 sin(2πt)] + 0.7535e − t [cos( t) − 1.1067 sin( t)]
28. Write the differential equations describing these systems and find and sketch the
indicated responses.
(a)
x( t) = u( t) , y( t) is the response ,
y(0 − ) = −5
Solutions 9-19
d

,  ( y( t)) 
= 10
 dt
t = 0−
∫
x(t)
∫
y(t)
2
10
Y( s) =

s +1
1 1
49
3
− 51
+
2
2

10  s
(s + 1) + 9 3 (s + 1) + 9
Both the response and its first derivative must be continuous in response to a step
excitation because of the double integration between excitation and response.
(b)
is ( t) = u( t) , v( t) is the response , No initial energy storage
i(t)
i s(t)
C 1 = 3 µF
R 1 = 2 kΩ
+
C 2 = 1 µF v(t)
R 2 = 1 kΩ
-

v( t) 
= C1 v{
is ( t) − C2 v′ ( t) +
′s ( t)
R2 
144244
voltage
3
across
i( t )
current
source

v( t) 
R
v s ( t) = v( t) + C2 v′ ( t) +
R2  1
144
42444
3
voltage across R1
Combining equations,
R1R2C1C2 v′′ ( t) + ( R2C2 + ( R2 + R1 )C1 ) v′ ( t) + v( t) = R2 is ( t)
R1R2C1C2 s2 V( s) + ( R2C2 + ( R2 + R1 )C1 ) s V( s) + V( s) =
R2
s
1.667 × 10 8
1000
73.52
1073.52
V( s) =
=
+
−
s( s + 1560)( s + 106.8)
s
s + 1560 s + 106.8
Solutions 9-20
(c)
is ( t) = cos(2000πt) u( t) , v( t) is the response , No initial energy storage
i(t)
C 1 = 3 µF
i s(t)
R 1 = 2 kΩ
+
C 2 = 1 µF v(t)
R 2 = 1 kΩ
-
From part (b)
R1R2C1C2 v′′ ( t) + ( R2C2 + ( R2 + R1 )C1 ) v′ ( t) + v( t) = R2 is ( t)
R1R2C1C2 s2 V( s) + ( R2C2 + ( R2 + R1 )C1 ) s V( s) + V( s) = R2
V( s) = −3.96
29.
s
2
s + (2000π )
2
s
6626
2000π
4.269
0.3104
−
2 +
2 +
2
2000π s + (2000π )
s + 1560 s + 106.84
s + (2000π )
2
Find the three parts, x ac ( t), x 0 ( t) and x c ( t) , of the following signals.
(a)
x( t) = t
(b)
x( t) = sin(ωt)
(c)
x( t) =
d
(sgn(t))
dt
x( t) = 2δ ( t)
x ac ( t) = 0 , x 0 ( t) = 2δ ( t) , x c ( t) = 0
30. Find the bilateral Laplace transforms of these signals.
(a)
x( t) = rect ( t)
1
2
x c ( t) = rect ( t) u( t)
1
−
s
 e − st  2 e 2 − 1 1 − e
X c ( s) = ∫ e − st dt = 
 = −s = s
s
−
+

0+
0
x 0 ( t) = 0 ⇒ X 0 ( s) = 0
Solutions 9-21
−
s
2
, Any s
x ac ( t) = rect ( t) u(− t) ⇒ x ac (− t) = rect (− t) u( t)
X ac (− s) =
1
2
∫e
− st
0+
X( s) =
1− e
s
s
2
+
s
2
s
2
−
s
2
e 
e − 1 1− e
dt = 
=
=

s
−s
 − s 0−
X ac ( s) =
−
1
2
− st
e −1 e − e
=
s
s
−
s
2
s
2
−
s
2
, Any s
s
2
1− e
e −1
=
, Any s
−s
s
=
e
−j
js
2
−e
s
j
js
2
 js
sin − 
 2
 js 
=
= sinc −  , Any s
s
 2π 
−j
2
ω
Notice that if we make the change of variable, s = jω , we get X( jω ) = sinc 
 2π 
which is the CTFT of x( t) = rect ( t) and this is allowed because the region of
convergence is the entire s plane.
(b)
x( t) = rect ( t) sin(20πt)
(c)
x( t) = e −2 t u( t) − e 2 t u(− t) sin(2πt)
[
]
Solutions 9-22