We're asked to find the inverse
Laplace Transform of F[s] which will give us
the function f[t].
For this example we're asked to give f[t]
as an exponential function as well as
a hyperbolic function.
And actually in our homework system
they want f[t] as an exponential function
but again we will show both.
The first step is to begin with F[s]
and find the inverse Laplace Transform
of both sides of the equation.
So we'd have the inverse Laplace Transform of F[s]
equals the inverse Laplace Transform
of one divided by the quantity s squared minus 16.
And looking at the table of Laplace transforms,
because we're given F[s] and we're trying
to determine f[t] we're using the table
in this direction here.
So let's see if we can recognize the formula
we'll use to find this inverse Laplace transform.
Let's focus on the denominator which is s squared - 16.
Notice how these two have a denominator
of s squared minus a squared, but here we have
a numerator of s and here we have a numerator
of a which is a constant which means we'll use
this row to determine the inverse Laplace transform
involving hyperbolic sign.
So we'll first find f[t] using a hyperbolic function
then we'll find it using an exponential function.
So on the left side the inverse Laplace transform
of F[s] is equal to f[t] equals,
and before we find this inverse Laplace transform
notice how looking at our formula
our denominator is s squared - 16 which is in
the form of s squared minus a squared,
where a squared equals 16 and a equals 4.
So notice how here we must have a numerator of four
in order to apply this formula.
So let's write this is as the inverse Laplace transform of,
let's write one as four over four
or 4/4 and we still have a denominator
of the quantity s squared - 16.
And now if we want the denominator to be four,
not four over four, we can factor out 1/4
and write this as f[t] = 1/4 * the inverse
Laplace transform of four divided
by the quantity s squared minus.
To make it better fit this formula,
let's write 16 as four squared.
So now we can easily see a = 4 which means
the inverse Laplace transform is equal
to hyperbolic sign 4t, which means f[t]
equals 1/4 hyperbolic sign 4t.
So here is f[t] expressed as a hyperbolic function
and now I'll show how we can express
this using an exponential function.
And we'll show this two ways.
We'll actually use the definition of hyperbolic sign
to write this as an exponential function
and then we'll also come back and perform
partial fraction decomposition on F[s]
and then use this formula to find the inverse
Laplace transform of F[s].
Let's first write f[t] as an exponential function
using the definition of hyperbolic sign x.
Hyperbolic sign x = 1/2 * the quantity e
to the x minus e raised to the power of -x.
Which means you can write f[t] as f[t]
equals 1/4 * 1/2 * the quantity,
now the input is 4t rather than x
so we'd have e raised to the power of 4t
minus e raised to the power of -4t.
Which means you can write f[t] as f[t]
equals 1/8 e to the 4t minus 1/8 e to the -4t.
So this would be one way to express f[t]
as an exponential function.
Another way would be to go back and start with
F[s] where F[s] is equal to one divided
by the quantity s squared minus 16
and perform partial fraction decomposition
on to one divided by the quantity s squared - 16.
And factored to form this would be one divided
by the quantity [s-4] times the quantity [s+4].
Performing partial fraction decomposition
we would have the fraction in factored form
equals a divided by the liner factor of [s-4]
plus b divided by linear factor of [s+4].
From here we'd multiply both sides of this equation
by the quantity [s-4] * [s+4] which would give us
the equation 1 = a * [s+4] + b * the quantity [s-4].
And then I'll find the values of a and b,
we'll select convenient values of s,
for example if s = -4 notice how this factor
would be zero, this would give us the equation
one equals, and this would be a * 0 plus
b * -8, so we have 1 = -8b, so B = -1/8.
And now to find a let's select s = 4.
So if s is 4 this factor would be zero,
so we'd have 1 = 8A + 0,
so notice that A is equal to positive 1/8.
Which means you can write F[s] as A
divided by the quantity [s-4] plus B
divided by the quantity [s+4], so we can write F[s]
as again A is positive 1/8, so we'd have 1/8
divided by the quantity [s-4] plus B is -1/8
so let's write this as -1/8 divided
by the quantity [s+4].
So now we'll take this form of F[s]
and take the inverse Laplace transform
of both sides of the equation.
So we have the inverse Laplace transform
of F[s] equals the inverse Laplace transform
of 1/8 divided by the quantity [s-4]
and then we have minus inverse Laplace transform
of 1/8 divided by the quantity [s+4].
Well the inverse Laplace transform of F[s]
is f[t] which is what we're looking for equals,
let's go ahead and factor out the 1/8 and write
this as 1/8 * the inverse Laplace transform
of now we have one divided by the quantity [s-4].
Notice how this form fits the formula perfectly
where a = 4, let's factor out 1/8 here.
So you have -1/8 * the inverse Laplace transform
of we'd have a numerator of one,
and now let's change the form of [s+4] to [s - -4].
Now this also fits the form probably where a = -4.
So now have f[t] = 1/8, this inverse Laplace transform
is equal to e raised to the power of positive 4t
because A is 4, so we have 1/8 e to the
positive 4t, and here we have -1/8 times,
for this inverse Laplace transform A is equal to -4,
so we'd have e raised to the power of -4t,
so we have -1/8 e to the -4t.
And of course we get the same result.
Again here we have f[t] expressed
as an exponential function.
Here we have it expressed as the same
exponential function using the definition
of hyperbolic sign, and then here we have
f[t] expressed equivalently as the hyperbolic sign function.
Again we're asked to enter the exponential function
which again would be 1/8 e to the 4t
minus 1/8 e to the negative 4t.
I hope you found this helpful.