# 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

mathcal(L){a · f(t)}=a · mathcal(L){f(t)}

### Example 1

mathcal(L){7\ sin\ t}=7\ mathcal(L){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

mathcal(L){a · f(t) + b · g(t)} = a · mathcal(L){f(t)} + b · mathcal(L){g(t)}

### Example 2

mathcal(L){3t + 6t^2} =3 · mathcal(L){t} + 6 · mathcal(L){t^2}

## Property 3. Change of Scale Property

If mathcal(L){f(t)}=F(s) then ccL{f(at)}=1/aF(s/a)

### Example 3

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

## Property 4. Shifting Property (Shift Theorem)

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

### Example 4

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

## Property 5.

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

See below for a demonstration of Property 5.

### 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

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

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

#### Demonstration of Property 5

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

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

Now F(s)=ccL{f(t)} =ccL{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:

ccL{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

#### Demonstration of Property 4: Shifting Property

For Example 6 we could have used:

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

Let g(t) = sin\ 3t

G(s)=ccL{g(t)}

=ccL{sin\ 3t}

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

=3/(s^2+9)

So

ccL{e^(2t)sin\ 3t} =G(s-a) =3/((s-2)^2+9)

This is the same result we obtained before for example (d).

## Exercises

Find Laplace Transforms of the following.

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

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

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

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

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

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