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Laplace Transformation

This document provides the Laplace transforms of 4 functions: 1) f(t)=5+e−3t +te−4t has a Laplace transform of F(s)=5/(s+3)+1/s+1/(s+4) 2) f(t)=sin(4t)+t−3+e−(t−3)+5/t has a Laplace transform of F(s)=4/(s2+16)+2/s+e−3s/ (s+1)+5/L(1/t) 3) f(t)=tcos(4t)+t/5 has a Laplace transform of F(s)=-d/

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252 views2 pages

Laplace Transformation

This document provides the Laplace transforms of 4 functions: 1) f(t)=5+e−3t +te−4t has a Laplace transform of F(s)=5/(s+3)+1/s+1/(s+4) 2) f(t)=sin(4t)+t−3+e−(t−3)+5/t has a Laplace transform of F(s)=4/(s2+16)+2/s+e−3s/ (s+1)+5/L(1/t) 3) f(t)=tcos(4t)+t/5 has a Laplace transform of F(s)=-d/

Uploaded by

Seanne Cruz
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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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

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