Compute the following Laplace transform. f(t)=sin(t−3)theta(t) or f(t)=sin(t−3)theta(t−3), where theta(t) is the Heaviside function?

Maribel Mcintyre

Maribel Mcintyre

Answered question

2022-10-20

Compute the following Laplace transform.
f ( t ) = sin ( t 3 ) θ ( t ) or f ( t ) = sin ( t 3 ) θ ( t 3 ) ,
where θ ( t ) is the Heaviside function?

Answer & Explanation

megagoalai

megagoalai

Beginner2022-10-21Added 22 answers

We have to remember the Laplace's delay property:
L [ f ( t T ) ] = e T s F ( s )
So, in the case
f ( t ) = s i n ( t 3 ) H ( t 3 ) becomes:
F ( s ) = e 3 s s 2 + 1
Using: L [ s i n ( a t ) ] = a s 2 + a 2
The first case f ( t ) = s i n ( t 3 ) H ( t ) is more complex. The function is a sinusoid with a different phase (because it not starts from 3, but 0)
It's equal to: c o s ( 3 ) s i n ( t ) s i n ( 3 ) c o s ( t )
So, the laplace transform is:
c o s ( 3 ) s 2 + 1 s s i n ( 3 ) s 2 + 1
Using: L [ c o s ( a t ) ] = s s 2 + a 2
Amiya Melendez

Amiya Melendez

Beginner2022-10-22Added 2 answers

0 f ( t ) θ ( t t ) d t = 0 t f ( t ) θ ( t t ) d t + t f ( t ) θ ( t t ) d t
for t ( 0 , t ) , θ ( t t ) = 0 and for t t , θ ( t t ) = 1, so above integral becomes
0 f ( t ) θ ( t t ) d t = t f ( t ) d t
That would give
L ( sin ( t 3 ) θ ( t ) ) = 0 e s t sin ( t 3 ) d t
and,
L ( sin ( t 3 ) θ ( t 3 ) ) = 3 e s t sin ( t 3 ) d t

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