I Find the Laplace transforms of the following functions, using the information in Table 3.1.

(However, some of the individual terms in these functions may not have Laplace transforms.)

          (a) f(t)=5+e−3t +te−4t

          (b) f(t)=sin(4t)+t−3+e −(t−3) +5/t

          (c) f(t)=t cos(4t)+t/5

          (d) f(t)=(t−1) cos(4(t−1))+t2 

An a. f ( t )=5+ e−3t +te−4 t

s Using rules 2, 5, and 7 from the Table

5 1 1
F ( s) = + +
s s+ 3 ¿¿

5
− ( t −3 )
b. f ( t )=sin ( 4 t )+ t−3+e +
t
Transform using Table 3.1 from the book
4 e−3 s e−3 s 1
F (s) ¿ 2
s +16 s
+ 2 +
s+1
+5 L
t ()
1
¿ No Laplace Transform for
t

t
c. f ( t )=tcos ( 4 t ) +
5
Transform using Table 3.1 from the book; for functions multiplied with “t”, solve for the
Laplace of the other function then find its negative derivative with respect to the same
variable.
1
∗1
−d s 5
F ( s) =
(
ds s 2 +16
2
)
+ 2
s
−−s +16 1
F ( s) = 2 2
+ 2
( s +16 ) 5 s

d. f ( t )=( t−1 ) cos ( 4 ( t−1 ) ) +t 2

f ( t )=tcos ( 4 t−4 )−cos ( 4 t−4 ) +t 2
For the first term of the equation, use the same principle with that on letter C, and for the
rest, solve using Table 3.4 from the book.
−d cos ( 4 s ) +4 sin ( 4 ) cos ( 4 s )+ 4 sin ( 4 ) 2
F ( s) =
ds ( s 2+16
−
) s2 +16
+ 3
s
 −−cos ( 4 s 2 )−8 sin ( 4 s ) +16 cos ( 4 ) cos ( 4 s )+ 4 sin ( 4 ) 2
F ( s) = 2
− 2
+ 3
2
( s +16 ) s +16 s

II The dynamic model for a process is given by

d2y/ dt2 + 6 dy /dt +8y = 3u(t)

where u(t) is the input function and y(0) and dy/dt (0) are both zero.

What are the functions of the time(e.g.,e−t/τ) in the solution for each of the following cases?

          (a) u(t)=be−2t  

          (b) u(t)=ct

b and c are constants. Note:You do not have to find y(t) in these cases. Just determine the
functions of time that will appear in y(t). 
An
s
III Find the complete time domain solutions for the following differential equations using Laplace
transforms:

          a. d3x/dt3 + 4x = et with x(0) = 0 ,  dx(0)/dt = 0, d2x(0)/dt2 = 0

          b. dx/dt - 12x = sin 3t with x(0) = 0

         c. d2x/dt2 + 6dx/dt + 25x = e-t   with x(0) = 0 , dx(0)/dt = 0

An a.
s

Taking the Laplace Transform of the equation:

b.

Taking the Laplace Transform of the equation:

C.

d 2 x 6 dx
L ( dt 2
+
dt
+25 x=e−t )
25 1
s2 X ( s )+ 6 sX ( s ) + =
s 2
s +1

1 25
X ( s ) ( s 2 +6 s )= − 2
s+1 s

s2−25 s−25
X ( s )=
( s2 ) ( s+1 ) ( s 2+ 6 s )

s2−25 s−25 A B C D E
X ( s )= = 3+ 2+ + +
( s ) ( s+1 ) ( s +6 ) s s s s+1 s+6
3

A s 2 +7 As+ 6 A
B s 3+7 B s2 +6 Bs

C s 4 +7 C s 3 +6 C s2

D s 4 +6 D s 3

E s 4 + E s3

@ s0

6 A=−25

−25
A=
6

@ s1
−25
7( ) 6
+6 B=−25
175
6 B=−25+
6
25 1 25
B= ()
6 6 36
=

@ s2
−25 25
6
+7 ( )
36
+6 C=1
25
6 C=1−
36
36−25 11
6 C= =
36 36
11
C=
216
3
@s
25 11
36
+7 ( )
216
+6 D+ E=0

@ s4
11
+ D+ E=0
216
25 11 11
36
+7 ( )
216
+6 D+ E=
216
+ D+ E
−150−66
5 D=
216
−1
D=
5
 11 1
− + E=0
216 5
1 11
E= −
5 216
216−55 161
E= =
1080 1080
−25 25 11 1 161
L−1 X ( s )= 3 + + − +
6 s 36 s 216 s 5(s+1) 1080( s+6)
2

−25 t 2 25t 11 e−t 161e−6 t

f ( t )= + + − +
12 36 216 5 1080

IV Expand each of the following s-domain functions into partial fractions:

          a. Y(s) = 6(s+1)/s2(s+1)

          b. Y(s) = 12(s+2)/s(s2 + 9)

         c. Y(s) = (s+2)(s+3)/(s+4)(s+5)(s+6)

         d. Y(s) = 1/(s+1)2(s+2)

An 6 (s+1) 6 α 1 α 2
s a. Y(s) = = = +
s 2 (s +1) s 2 s s2

6
α 2=s2
s2 |
s=0
=6 α 1=0

6
Y ( s )=
s2

12(s +2) α 1 α 2 s +α 3
b. Y(s) = 2
= + 2
s (s + 9) s s +9

Multiply both sides by s( s2 +9)

12 ( s+ 2 )=α 1 ( s2 + 9 ) +(α 2 s+ α 3 )( s)
12 s+ 24=( α 1+ α ) s 2 +α 3 s+9 α 1

Equating coefficients of like powers of s,

s2 :α 1 +α 2=0
 s1 :α 3 =12
s0 :9 α 1 =24

Solving simultaneousy,

8 −8
α 1 = , α 2= , α 3=12
2 3

−8
s +12)
(
81 3
Y ( s )= +
3s s 2+ 9

(s+2)(s+3) α α α
c. Y(s) = = 1 + 2 + 3
(s+ 4)(s+5)(s+6) s+ 4 s +5 s+5

( s+2)(s +3)
α 1=
(s +5)(s +6) s=−4
=1 |
( s+2 ) ( s +3 )
α 2=
( s+ 4 )( s+6 ) |
s=−5
=−6

( s +2 ) ( s +3 )
α 3=
( s+ 4 )( s+5 ) |
s=−6
=6

1 6 6
Y ( s )= − +
s +4 s+ 5 s+6

d.
End of Assignment No 4