3. Some Properties of Laplace Transforms

We saw some of the following properties in the Table of Laplace Transforms.

Property 1. Constant Multiple

If a is a constant and f(t) is a function of t, then

`Lap{a · f(t)}=a · Lap{f(t)}`

Example 1

`Lap{7\ sin t}=7\ Lap{sin t}`

[This is not surprising, since the Laplace Transform is an integral and the same property applies for integrals.]

Property 2. Linearity Property

If a and b are constants while f(t) and g(t) are functions of t, then

`Lap{a · f(t) + b · g(t)}` `= a · Lap{f(t)} + b · Lap{g(t)}`

Example 2

`Lap{3t + 6t^2}` `=3 · Lap{t} + 6 · Lap{t^2}`

Property 3. Change of Scale Property

If `Lap{f(t)}=F(s)` then `Lap{f(at)}=1/aF(s/a)`

Example 3

`Lap{f(5t)}=1/5F(s/5)`

Property 4. Shifting Property (Shift Theorem)

`Lap {e^(at)f(t)} = F(s-a)`

Example 4

`Lap {e^(3t)f(t)} = F(s-3)`

Property 5.

`Lap{tf(t)}=-F^'(s)=-d/(ds)F(s)`

See below for a demonstration of Property 5.

Continues below

Example 5

Obtain the Laplace transforms of the following functions, using the Table of Laplace Transforms and the properties given above.

(We can, of course, use Scientific Notebook to find each of these. Sometimes it needs some more steps to get it in the same form as the Table).

(a) `f(t) = 4t^2`

Answer

We use:

`Lap{t^n}=(n!)/(s^(n+1))`

and the Constant Multiple property from above, to obtain:

`Lap{4t^2}=4xx(2!)/(s^(2+1))=8/s^3`

(b) `v(t) = 5\ sin 4t`

Answer

We use: `Lap{sin\ omega t}=omega/(s^2+omega^2)`

Clearly, `ω = 4`.

`Lap{5\ sin\ 4t}` `=5xx4/(s^2+4^2)=20/(s^2+16)`

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(c) `g(t) = t\ cos 7t`

Answer

We use `Lap{t\ cos\ omega t}=(s^2-omega^2)/((s^2+omega^2)^2)`

We substitute to obtain

`Lap{t\ cos\ 7 t}` `=(s^2-7^2)/((s^2+7^2)^2)=(s^2-49)/((s^2+7^2)^2)`

Demonstration of Property 5

Example (c) is of the form `Lap{tf(t)}`.

We could have also used Property 5, `Lap{tf(t)}` `=-F'(s)=-(d/(ds)F(s))`, with `f(t) = cos 7t`.

Now `F(s)=` `Lap{f(t)}=` `Lap{cos 7t}` `=s/(s^2+7^2)`

So

`d/(ds)F(s)=d/(ds)s/(s^2+7^2)`

`=-(s^2-7^2)/(s^2+7^2)^2`

`=-(s^2-49)/(s^2+49)^2`

Then we have:

`Lap{t\ cos 7t}=-(-(s^2-49)/(s^2+49)^2)`

`=(s^2-49)/(s^2+49)^2`

This is the same result that we obtained using the formula.

For a reminder on derivatives of a fraction, see Derivatives of Products and Quotients.

Example 6

Find the Laplace Transform of `f(t)=e^(2t)sin 3t`

Answer

We use

`Lap{e^(at)\ sin\ omega t}=omega/((s-a)^2+omega^2)`

and simple substitution yields:

`Lap{e^(2t)\ sin\ 3t}` `=3/((s-2)^2+3^2)` `=3/((s-2)^2+9)`

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Demonstration of Property 4: Shifting Property

For Example 6 we could have used:

`Lap {e^(at)g(t)} = G(s-a)`

Let `g(t) = sin 3t`

`G(s)=Lap{g(t)}`

`=Lap{sin 3t}`

`=3/{s^2+3^2)`

`=3/(s^2+9)`

So

`Lap{e^(2t)sin 3t} =G(s-a)`

`=3/((s-2)^2+9)`

This is the same result we obtained before for example 6.

Exercises

Find Laplace Transforms of the following.

1. `f(t)=t^4e^(-jt)`

Answer

From the Laplace table, we have:

`Lap{e^(at)t^n}=(n!)/((s-a)^(n+1))`

So

`Lap{t^4e^(-jt)}` `=(4!)/((s--j)^(4+1))` `=24/((s+j)^5)`

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2. `f(t) = te^(-t)\ cos 4t`

Answer

We will use:

`Lap{t*g(t)}` `=-G^'(s)` `=-d/(ds)G(s)`

Let `g(t) = e^(-t)\ cos\ 4t`

Then

`G(s)= Lap{e^(-t)cos\ 4t}`

`=(s+1)/((s+1)^2+16)`

`=(s+1)/(s^2+2s+17)`

Now

`d/(ds)(s+1)/((s+1)^2+16)` `=-(s^2+2s-15)/((s^2+2s+17)^2)`

So `Lap{t*e^(-t)*cos\ 4t}` `=(s^2+2s-15)/((s^2+2s+17)^2)`

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3. `f(t) = t^2sin 5t`

Answer

We will use `Lap{t^ng(t)}=(-1)^n(d^nG(s))/(ds^n)`, with `n=2`.

Now

`G(s)= Lap{sin 5t}`

`=5/(s^2+5^2)`

`=5/(s^2+25)`

The first derivative:

`d/(ds)5/(s^2+25)=(-10s)/((s^2+25)^2`

Now for the second derivative:

`d/(ds)(-10s)/((s^2+25)^2)=10(3s^2-25)/((s^2+25)^3)`

For the formula, we need:

`(-1)^2=1`

So

`Lap{t^2\ sin\ 5t}=10(3s^2-25)/((s^2+25)^3)`

4. `f(t) = t^3cos t = t^2(t\ cos t)`

Answer

So we are letting `g(t)=t\ cos\ t`.

This time we need the 2nd derivative of `G(s)`.

`G(s)=` `Lap{t\ cos\ t}=(s^2-1)/((s^2+1)^2)`

(This is from the Table of Laplace Transforms.)

First derivative:

`d/(ds)(s^2-1)/((s^2+1)^2)=-2s(s^2-3)/((s^2+1)^3)`

Second derivative:

`(d^2)/(ds^2)(s^2-1)/((s^2+1)^2)=6(s^4-6s^2+1)/((s^2+1)^4)`

Now `(-1)^2=1`.

So

`Lap{t^3\ cos\ t}=6(s^4-6s^2+1)/((s^2+1)^4)`

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5. `f(t)=cos^2 3t` given that `Lap{cos^2t}=(s^2+2)/(s(s^2+4))`

Answer

For this one, we need to apply the Scale Property:

`Lap{f(at)}=1/aF(s/a)`

Here, `a = 3` so

`Lap{cos^2 3t}=1/3((s/3)^2+2)/((s/3)((s/3)^2+4))`

`=(s^2/9+2)/(s(s^2/9+4))`

`=(s^2+18)/(s(s^2+36))`

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