Inverse Laplace Transform

Related Calculator: Inverse Laplace Transform Calculator

As was seen previously Laplace transform is `F(s)=L(f(t))` , in other words we are given function `f(t)` and we need to find `F(s)` . Inverse Laplace transform is operation of finding `f(t)` given `F(s)` .

Operation of Laplace transform was clear: we need to calculate imporper integral and we are done. Finding inverse Laplace transform is more tedious and complicated. Table of transforms is very useful for this.

Also, following property holds:

`L^-1(aF(s)+bG(s))=aL^-1(F(s))+bL^-1(G(s))=af(t)+bg(t)` , where a and b are some constants.

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

Proof.

From property of Laplace transform `aF(s)+bG(s)=L(af(t)+bf(t))` . Taking inverse Laplace transform of both sides yields

`L^-1(aF(s)+bG(s))=L^-1(L(af(t)+bg(t)))`

Or

`L^-1(aF(s)+bG(s))=af(t)+bg(t)` , since `L^-1(L(f(t)))=f(t)` .

This completes the proof.

Let's go through a couple of examples.

Example 1. Calculate `L^-1(5/s+2/(s-2)+5/(s^2+1)+1/(s+3))`

`L^-1(5/s+2/(s-2)+5/(s^2+1)+1/(s+3))=5L^-1(1/s)+2L^-1(1/(s-2))+5L^-1(1/(s^2+1)+1/(s-(-3)))=5*(1)+2e^(2t)+5sin(t)+e^(-3t)=5+2e^(2t)+5sin(t)+e^(-3t)`

Above inverse Laplace transforms can be seen in table of Laplace transforms.

Example 2. Calculate `L^-1((s+1)/(s^2+3)+1/(5s-7))`

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

On this stage look at the denominator of first two terms. Since `1/(s^2+3)=1/(s^2+(sqrt(3))^2)` then it seems that first term must be `cos(sqrt(3)t)` and seond term is `cos(sqrt(3)t)` . However there is not enough `sqrt(3)` in numerator for `sin(sqrt(3)t)` . So,

`L^-1(s/(s^2+3))+L^-1(1/(s^2+3))+L^-1(1/(5(s-7/5)))=cos(sqrt(3)t)+L^-1(sqrt(3)/sqrt(3)1/(s^2+3))+1/5L^-1(1/(s-7/5))=cos(sqrt(3)t)+1/sqrt(3)L^-1((sqrt(3))/(s^2+3))+1/5e^(7/5t)=cos(sqrt(3)t)+1/sqrt(3)sin(sqrt(3)t)+1/5e^(7/5t)`

Denominator is the key part in taking the inverse transform. Identify what function are you dealt with based on denominator. If there is more than one possibility, examine numerator. Then correct numerator to get the form you need.

Example 3. Calculate `L^-1((7s-5)/(s^2+2s+5))`

Again examine denominator: `s^2+2s+5=s^2+2s+1+4=(s+1)^2+2^2=(s-(-1))^2+2^2`

Seems like we are dealing with either `e^(-t)cos(2t)` or `e^(-t)cos(2t)` . Now take a look at the numerator. It is `7s-5` or `7(s-5/7)` . For `e^(-t)cos(2t)` we need `s+1` in numerator and for `e^(-t)sin(2t)` we need 2 in numerator, so rewrite numerator as `7(s+1-1-5/7)=7(s+1-12/7)=7(s+1)-12=7(s+1)-2*6` .

So,

`L^-1((7s-5)/(s^2+2s+5))=L^-1((7(s+1))/((s+1)^2+2^2))-L^-1((2*6)/((s+1)+2^2))=7L^-1((s+1)/((s+1)^2+2^2))-6L^-1(2/((s+1)+2^2))=7e^(-t)cos(2t)-6e^(-t)sin(2t)=e^(-t)(7cos(2t)-6sin(2t))`

Example 4. Calculate `L^-1(s/(2s^2-4s-9))`

Examine denominator: `2s^2-4s-9=2(s^2-2s-9/2)=2(s^2-2s+1-1-9/2)=2((s-1)^2-11/2)` .

So, we are dealing with `e^tsinh(sqrt(11/2)t)` and `e^tcos(sqrt(11/2)t)` . For term involving `cosh` we need `s-1` in numerator and for term involving `sinh` we need `sqrt(11/2)` , so rewrite numerator as `s=s-1+1=(s-1)+sqrt(2/11)*sqrt(11/2)` .

`L^-1(s/(2s^2-4s-9))=L^-1(((s-1)+sqrt(2/11)*sqrt(11/2))/(2((s-1)^2-(sqrt(11/2))^2)))=1/2L^-1((s-1)/((s-1)^2+(sqrt(11/2))^2))+sqrt(1/22)L^-1((sqrt(11/2))/((s-1)^2-(sqrt(11/2))^2))=1/2e^tcosh(sqrt(11/2)t)+sqrt(1/22)e^tsinh(sqrt(11/2)t)=e^t/22(sqrt(22)sinh(sqrt(11/2)t)+11cosh(sqrt(11/2)t))`

Next examples are examples that you will be dealt with most often. They require knowledge of partial fraction decomposition.

Example 5. Calculate `L^-1((s+2)/((s+3)(s-1)))`

Looking at the denominator we see that it has terms `s+3` and `s-1` and it looks like that inverse transform will contain `e^(-3t)` and `e^t` .

However, to do that we need to split "product" into sum.

To do this assume that `(s+2)/((s+3)(s-1))=A/(s+3)+B/(s-1)` where A and B are unknown constants.

Then

`A/(s+3)+B/(s-1)=(A(s-1)+B(s+3))/((s-1)(s+3))=(s(A+B)-A+3B)/((s-1)(s+3))`

And this must equal `(s+2)/((s+3)(s-1))` , so we obtain following system:

`{(A+B=1),(-A+3B=2):}`

Which has solution `A=1/4` `B=3/4` .

So, `(s+2)/((s+3)(s-1))=1/4 1/(s+3)+3/4 1/(s-1)`

This is called partial fraction decomposition.

Finally,

`L^-1((s+2)/((s+3)(s-1)))=1/4 L^-1(1/(s+3))+3/4 L^-1(1/(s-1))=1/4 e^(-3t)+3/4 e^t`

Note, that we could write denominator as `(s+3)(s-1)=s^2+2s-3=s^2+2s+1-4=(s+1)^2-4=(s+1)^2-2^2` and solve it like in previous example.

However, there are problems when this approach won't work.

In general if you have in denominator `(a_nx^n+a_(n-1)x^(n-1)+...+a_0)^k` then corresponding term in partial fraction decomposition is `(A_(n-1)x^(n-1)+A_(n-2)x^(n-2)+...+A_0)/(a_nx^n+a_(n-1)x^(n-1)+...+a_0)+(A_(n-1)x^(n-1)+A_(n-2)x^(n-2)+...+A_0)/(a_nx^n+a_(n-1)x^(n-1)+...+a_0)^2+...+(A_(n-1)x^(n-1)+A_(n-2)x^(n-2)+...+A_0)/(a_nx^n+a_(n-1)x^(n-1)+...+a_0)^k`

In particular,

  1. `ax+b->A/(ax+b)`
  2. `ax^2+bx+c->(Ax+B)/(ax^2+bx+c)`
  3. `(ax+b)^2->A/(ax+b)+B/(ax+b)^2`
  4. `(ax^2+bx+c)^2->(Ax+B)/(ax^2+bx+c)+(Cx+D)/(ax^2+bx+c)^2`

Example 6. Calculate `L^-1(1/(s^2(s+1)^2))`

Partial fraction decomposition is

`1/(s^2(s+1)^2)=A/s+B/s^2+C/(s+1)+D/(s+1)^2`

`A/s+B/s^2+C/(s+1)+D/(s+1)^2=(As(s+1)^2+B(s+1)^2+Cs^2(s+1)+Ds^2)/(s^2(s+1)^2)=(As^3+2As^2+As+Bs^2+2Bs+B+Cs^3+Cs^2+Ds^2)/(s^2(s+1)^2)=`

`=(s^3(A+C)+s^2(2A+B+C+D)+s(A+2B)+B)/(s^2(s+1))^2`

It should equal `1/(s^2(s+1)^2)` , so we have following system:

`{(A+C=0),(2A+B+C+D=0),(A+2B=0),(B=1):}`

From last equation `B=1` , from third equation `A=-2` , from first equation `C=2` and from second equation `D=1` , so

`1/(s^2(s+1)^2)=-2/s+1/s^2+2/(s+1)+1/(s+1)^2` .

Therefore,

`L^-1(1/(s^2(s+1)^2))=-2L^-1(1/s)+L^-1(1/s^2)+2L^-1(1/(s+1))+L^-1(1/(s+1)^2)=-2*1+t+2e^(-t)+L^-1(1/(s+1)^2)`

To find `L^-1(1/(s+1)^2)` note that `-d/(ds)1/(s+1)=1/(s+1)^2` . Now using property 3 of laplace transform and fact that `L(e^-t)=1/(s+1)` we obtain that

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

Taking inverse transform of both sides yields

`L^-1(1/(s+1)^2)=te^-t`

Finally,

`L^-1(1/(s^2)(s+1)^2)=-2+t+2e^-t+te^-t`

As can be seen to successfully take inverse Laplace transform you should identify denominator and master partial fraction decomposition. These are two essential parts and only a lot of practice can improve your skills.