Properties of the Laplace Transform

Related Calculators: Laplace Transform Calculator , Inverse Laplace Transform Calculator

Laplace transform has some interesting properties.

Property 1. Linearity of Laplace transform: `L(af(t)+bg(t))=aL(f(t))+L(g(t))`, where a and b are some constants.

Proof is straightforward through definition:

`L(a(f(t)+bg(t)))=int_0^ooe^(-st)(af(t)+bg(t))dt=int_0^oo(af(t)e^(-st)+bg(t)e^(-st))dt`.

Now using linearity we can write that `int_0^oo(af(t)e^(-st)+bg(t)e^(-st))dt=a int_0^ooe^(-st)f(t)dt+b int e^(-st)g(t)dt=aL(f(t))+bL(g(t))`

This completes the proof.

Note, that there can be more than 2 terms and property will still hold.

Example. Calculate `L(5e^(3t)-2sin(7t)+cos(t))`

From table of laplace transforms it is known that `L(e^(3t))=1/(s-3)` , `L(sin(7t))=7/(s^2+7^2)` and `L(cos(t))=s/(s^2+1^2)`, so using Property 1 we obtain following:

`L(5e^(3t)-2sin(7t)+cos(t))=5L(e^(3t))-2L(sin(7t))+L(cos(t))=`

`=5/(s-3)-2 7/(s^2+49)+s/(s^2+1)=5/(s-3)-14/(s^2+49)+s/(s^2+1)`

Property 2. Scaling by constant: `L(f(at))=1/sF(s/a)`, where a is positive constant.

Proof is straightforward:

`L(f(at))=int_0^oo e^(-st)f(at)dt`

Using change of variable technique we make substituion `u=at` then `du=adt`, which yields `dt=(du)/a` . Since t is changing from 0 to `oo` then u is changing from `a*0=0` to `a*oo=oo` .

So,

`int_0^oo e^(-st)f(at)dt=int_0^oo e^(-su/a)f(u)(du)/a=1/a int_0^oo e^(-s/au)f(u)du=1/aF(s/a)`

This completes the proof.

Property 3. `L(tf(t))=-d/(ds)F(s)`, where `L(f(t))=F(s)`

Proof.

From the definition it follows that

`F(s)=int_0^oo e^(-st)f(t)dt`

Differentiating this equality with respect to s gives (note that limits of integration are constant):

`d/(ds)F(s)=int_0^oo d/(ds)e^(-st)f(t)dt`

Or

`d/(ds)F(s)=int_0^oo -te^(-st)f(t)dt`

Which finally gives

`-d/(ds)F(s)=int_0^oo te^(-st)f(t)dt=L(tf(t))`

This completes the proof.

Example. Calculate `L(tsin(t))`

Since `L(sin(t))=1/(s^2+1)` then using property 3 we obtain that

`L(tsin(t))=-d/(ds)1/(s^2+1)=1/(s^2+1)^2*(d(s^2+1))/(ds)=(2s)/(s^2+1)^2`

Property 4. `L(f(t)/t)=int_s^ooF(u)du` where `F(s)=L(f(t))`

Proof.

This property follows from property 3.

Let `g(t)=f(t)/t` and `L(g(t))=G(s)` . We need to prove that `G(s)=int_s^oo F(u)du`

So, `f(t)=tg(t)`.

Taking Laplace transform of both sides gives:

`L(f(t))=L(tg(t))`

Or

`F(s)=-d/(ds)G(s)` (using property 3)

Integrating this equality with respect to s and taking limits `s` and `oo` yields

`int_s^oo -d/(ds)G(s)ds=int_s^ooF(u)du`.

`-(lim_(a->oo)G(a)-G(s))=int_s^oo F(u)du`

Since `G(s)=int_0^oo e^(-st)g(t)dt` then `lim_(s->oo)G(s)=0` .

So,

`-(0-G(s))=int_s^oo F(u)du`

`G(s)=int_s^oo F(u)du`

This completes the proof.

Example. Calculate `L((sin(t))/t)`

Since `L(sin(t))=1/(s^2+1)` then using property 4 we obtain that

`L((sin(t))/t)=int_s^oo 1/(p^2+1)dp=arctan(p)|_s^oo=arctan(oo)-arctan(s)=pi/2-arctan(s)=a r c c o t(s)`