Properties of the Laplace Transform
Related Calculators: Laplace Transform Calculator , Inverse Laplace Transform Calculator
The Laplace transform has a number of interesting properties.
Property 1. Linearity of the Laplace transform: `L(af(t)+bg(t))=aL(f(t))+L(g(t))`, where `a` and `b` are some constants.
The proof is straightforward through the 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 the property will still hold.
Example 1 . Calculate `L(5e^(3t)-2sin(7t)+cos(t))`.
From the 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 the 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 a positive constant.
The proof is straightforward:
`L(f(at))=int_0^oo e^(-st)f(at)dt`.
Using the change of variable technique, we make the substituion `u=at`; then, `du=adt`, which yields `dt=(du)/a`. Since `t` is changing from `0` to `oo`, we have that `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 the 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 2. Calculate `L(tsin(t))`.
Since `L(sin(t))=1/(s^2+1)`, 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 the 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 the 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`, we have that `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 3. Calculate `L((sin(t))/t)`.
Since `L(sin(t))=1/(s^2+1)`, using property 4, we obtain that
`L((sin(t))/t)=int_s^oo 1/(p^2+1)dp=lim_(m->oo)int_s^m 1/(p^2+1)dp=lim_(m->oo)atan(p)|_s^m=`
`=lim_(m->oo)atan(m)-atan(s)=pi/2-atan(s)=acot(s)`.