Begin2.DVI

Example 12-14. First shift property

If F(s) = L{f(t)}, show that L

{

eatf(t)

}

=F(s−a)or eatf(t) = L−^1 {F(s−a)}

Solution

By definition

F(s) = L{ f(t)}=

∫∞

0

f(t)e−st dt

so that if one replaces sby s−athere results

F(s−a) =

∫∞

0

f(t)e−(s−a)tdt =

∫∞

0

eatf(t)e−stdt =L

{

eatf(t)

}

or

L−^1 {F(s−a)}=eatf(t)

Example 12-15. (Second shift property)

If F(s) = L{f(t)}, show L{ f(t−α)H(t−α)}=e−αsF(s)or

f(t−α)H(t−α) = L−^1

{

e−αsF(s)

}

where H(t)is the Heaviside step function defined by

H(t) =

{ 0 , t < 0

1 , t > 0
Solution

By definition

F(s) =

∫∞

0

f(t)e−st dt

so that replacing tby uand then multiplying by e−αs one finds

e−αsF(s) =

∫ ∞

0

f(u)e−su e−αs du =

∫∞

0

f(u)e−s(u+α)du

Make the change of variables t=u+αwith dt =du and then calculate the appropriate

limits of integration to show

e−αsF(s) =

∫∞

t=α

f(t−α)e−stdt =

∫a

0

(0) ·f(t−α)e−st dt +

∫∞

α

(1) ·f(t−α)dt

This last integral has the more compact form

e−αsF(s) =

∫∞

0

f(t−α)H(t−α)e−st dt =L{f(t−α)H(t−α)}

or

L−^1

{

e−αsF(s)

}

=f(t−α)H(t−α)

As an integration exercise one can verify the various transform properties listed

on the next page.