# 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

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

Easy to understand math videos:
MathTutorDVD.com

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

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)

Easy to understand math videos:
MathTutorDVD.com

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

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

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)

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

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)

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

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)

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)

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

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))