Welcome to a lesson
on the inverse Laplace Transform.
From the previous lesson,
we know the Laplace transform will allow us
to convert a differential equation
into an algebraic equation.
Once we solve the algebraic equation
in the frequency domain in terms of s,
we will want to get back to the time domain t,
as that is what we are solving for.
Given a function big F of s,
we wish to find a function little f of t
such that the Laplace transform of little f of t
equals big F of s.
From the uniqueness theorem,
we know little f of t is unique.
As a result, the definition follows.
Suppose big F of s
is equal to the Laplace transform
of little f of t for some function little f of t.
We define the inverse Laplace transform
as the inverse Laplace transform of big F of s
equals little f of t.
There is an integral formula
for the inverse Laplace transform
but it is not as simple as the Laplace transform itself.
It requires complex numbers and path integrals.
For us it will suffice to compute
the inverse Laplace transform using table 6.1 below.
As an example, we are given big F of s
equals one divided by the quantity s plus one
and we're asked to find the inverse Laplace transform.
Looking at our table, we use the formula
at the bottom of the first column,
which I've enlarged here on the left.
The Laplace transform of e to the power of negative at
equals one divided by the quantity s plus a.
In our case, notice a is equal to one
and therefore, the inverse Laplace transform
of one divided by the quantity s plus one
equals e to the power of negative one 1t
or just e to the negative t.
As the Laplace transform is linear,
the inverse Laplace transform is also linear.
That is the inverse Laplace transform of A times big F of s
plus B times big G of s equals A
times the inverse Laplace transform of big F of s
plus B times the inverse plus transform of big G of s.
Of course, we also have the inverse Laplace transform
of A times big F of s equals A
times the inverse Laplace transform a big F of s.
Let us demonstrate how linearity can be used.
For the next example, we are given big F of s
equals s squared plus s plus one,
all divided by the quantity s cubed plus s.
And we're asked to find the inverse Laplace transform.
We first use the method of partial fraction decomposition
to write big F in a form where we can use table 6.1
to find the inverse Laplace transform.
This indicates we begin by factoring the denominator
as s times the quantity s squared plus one.
Notice how we have a linear factor and a quadratic factor.
So now we write the given function big F of s
as A divided by s plus the quantity Bs plus C
divided by the quantity s squared plus one.
Recover, when we have a linear factor,
the numerator is a constant.
When we have a quadratic factor, the numerator is linear.
And now, for the next step,
which I've shown here in blue on the left,
we multiply both sides of the equation by the LCD,
which results in the equation s squared plus s
plus one equals A times the quantity s squared plus one
plus the quantity Bx plus C times s.
And now we equate the coefficients,
which I've done in the next line.
Because s squared must equal A times s squared,
we know A is equal to one.
We can also tell us by looking at the constant term one.
And then we also know that Bx plus C
must equal the coefficient of s, which is one.
This indicates that B is zero and C is one.
And now we know the partial fraction decomposition
is one over s plus one over the quantity s squared plus one.
And now we take the inverse Laplace transform
of both sides of the equation
and use the property of linearity.
And now we find the inverse Laplace transform of one over s.
And then we add the inverse Laplace transform
of one divided by the quantity s squared plus one.
To find the inverse Laplace transform of one divided by s,
we use the first formula in the first column
where the inverse Laplace transform of a constant
is equal to the constant divided by s.
Notice in our case, the constant is one
and therefore, the inverse Laplace transform
of one divided by s is equal to one.
And therefore, the inverse Laplace transform
of one divided by the quantity s squared plus one.
We use the first formula in the second column
where the Laplace transform of sine omega t
is equal to omega divided by the quantity s squared
plus omega squared.
Notice in our case, because the denominator
is s squared plus one, omega squared and omega equal one,
and therefore the inverse Laplace transform is sine t.
And therefore the inverse Laplace transform
of the original function, big F of s, is one plus sine t.
Another useful property is a so-called shifting property
or first shifting property,
which states the Laplace transform of e
to the power of negative at
times little f of t equals big F of the quantity s plus a
where big F of s is the Laplace transform of little f of t.
The shifting property can be used, for example,
when the denominator is a more complicated quadratic
that may come up in the method of partial fractions.
We complete the square and write such quadratics
as the square of s plus a plus b
and then use the shifting property.
As an example, let's find the inverse Laplace transform
of one divided by the quantity s squared plus 4s plus eight.
To begin, notice the denominator doesn't factor
and therefore, we begin by completing the square
on s squared plus 4s,
which I've shown here on the right.
To complete the square on s squared plus 4s,
we add half of the square of the coefficient of s,
which is four.
Half of four is two, two squared is four,
and therefore we add four.
But so that we don't change the expression,
we now have to subtract four.
And now s squared plus 4s plus four
is a perfect square trinomial.
It's equal to two factors of s plus two
and eight minus four gives s plus four.
So now we know, look at our formula,
we have big F of s plus two is equal to one divided
by the sum of the square of s plus two and four.
We also know a is equal to two and big F of s,
which is big F without the shift
is one divided by s squared plus four.
And now in the formula above, to find little f of t,
we need to find the inverse Laplace transform
of big F of s, which, again,
is one divided by the square of s plus four.
Looking at our table of formulas,
more specifically the first formula in the second column,
we have the Laplace transform of sine omega t
is equal to omega divided by the quantity s squared
plus omega squared.
Notice in our case, omega squared is equal to four
and therefore, omega is equal to two,
or more specifically, on the far left in blue,
we now know the inverse Laplace transform of two
divided by the quantity s squared plus four equals sine 2t.
But because our numerator is one, not two,
the inverse Laplace transform of one divided
by the square s plus four is equal to 1/2 sine 2t,
which gives us little f of t.
So now putting all the pieces together,
where we know little f of t is equal to 1/2 sine 2t,
we also know big F of s plus two
is equal to one divided by the sum of the square of s
plus two and four, as well as a is equal to two.
We have the inverse Laplace transform
of one divided by the quantity s squared plus 4s plus eight
equals the inverse Laplace transform
of one divided by the square of s plus two plus four,
which, using our formula,
is equal to e to the power of negative at
times little f of t, where in our case,
a is equal to two and the little f of t
is equal to 1/2 sine 2t,
giving us an inverse Laplace transform of 1/2
e to the power of negative 2t sine 2t.
In general, we want to apply the inverse Laplace transform
to rational functions,
meaning in the form of big F of s divided by big G of s,
where big F of s and big G of s are polynomials.
Since normally, for the functions that we are considering,
the Laplace transform goes to zero as s approaches infinity,
it's not hard to see the degree of big F of s
must be smaller than that of big G of s.
Such rational functions are called proper rational functions
and we can always apply the method of partial fractions.
Of course, this means we need to be able
to factor the denominator into linear and quadratic factors,
which involves finding the roots of the denominators.
But we also now know if the denominator
is a quadratic that doesn't factor,
we can try completing the square
and using the shifting property.
I hope you found this helpful.