# 4. Laplace Transforms of the Unit Step Function

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

Recall `u(t)` is the unit-step function.

1. ℒ`{u(t)}=1/s`

2. ℒ`{u(t-a)}=e^(-as)/s`

3. Time Displacement Theorem:

If `F(s)=` ℒ`{f(t)}` then ℒ`{u(t-a)*g(t-a)}=e^(-as)G(s)`

[You can see what the left hand side of this expression means in the section Products Involving Unit Step Functions.]

### Examples

Sketch the following functions and obtain their Laplace transforms:

(a) `f(t)={ {: (0,t < a), (A, a < t < b), (0, t > b) :}`

Assume the constants *a*, *b*, and *A* are positive, with *a* < *b*.

Answer

The function has value *A* between *t *=* a* and *t *=* b * only.

Graph of `f(t)=A*[u(t-a)-u(t-b)]`.

We write the function using the rectangular pulse formula.

`f(t)=A*[u(t-a)-u(t-b)]`

We use `Lap{u(t-a)}=(e^(-as))/s`

We also use the linearity property since there are 2 items in our function.

`Lap{f(t)}=A[(e^(-as))/s-(e^(-bs))/s]`

(b) `f(t)={ {: (0,t < a), (e^(t-a), a < t < b), (0, t > b) :}`

Assume the constants *a* and *b *are positive, with *a* < *b*.

Answer

Our function is `f(t)=e^(t-a)`. This is an exponential function (see Graphs of Exponential Functions).

When `t = a`, the graph has value `e^(a-a)= e^0= 1`.

^{b−a}

Graph of `f(t)=e^(t-a)*{u(t-a)-u(t-b)}`.

The function has the form:

`f(t)=e^(t-a)*{u(t-a)-u(t-b)}`

We will use the Time Displacement Theorem:

`Lap{u(t-a)*g(t-a)}=e^(-as)G(s)`

Now, in this example, `G(s)=` `Lap{e^t}=1/(s-1)`

`Lap{e^(t-a)*[u(t-a)-u(t-b)]}``=` `Lap{e^(t-a)*u(t-a)-e^(t-a)*u(t-b)}`

We now make use of a trick, by noting `(t-a) = (b-a ) + (t-b)` and re-writing `e^(t-a)` as `e^(b-a)e^(t-b)`:

`= Lap{e^(t-a)*u(t-a)` `{:-e^(b-a)e^(t-b)*u(t-b)}`

[We have introduced *e ^{b−a}*, a constant, for convenience.]

`=` `Lap{e^(t-a)*u(t-a)}-` `e^(b-a)Lap{e^(t-b)*u(t-b)}`

[Each part is now in the form `u(t − c) · g(t − c)`, so we can apply the Time Displacement Theorem.]

`=e^(-as)xx1/(s-1)` `-e^(b-a)xxe^(-bs)xx1/(s-1)`

`=(e^(-as))/(s-1)-(e^(b-a-bs))/(s-1)`

`=(e^(-as)-e^(b-a-bs))/(s-1)`

(c) `f(t)={ {: (0,t < 0), (sin\ t, 0 < t < pi), (0, t > pi) :}`

Answer

Here is the graph of our function.

Graph of `f(t) = sin t * [u(t) − u(t − π)]`.

The function can be described using Unit Step Functions, since the signal is turned on at `t = 0` and turned off at `t=pi`, as follows:

`f(t) = sin t * [u(t) − u(t − π)]`

Now for the Laplace Transform:

`Lap{sin\ t * [u(t)-u(t-pi)]}` `=` `Lap{sin\ t * u(t)}- ` `Lap{sin\ t * u(t - pi)}`

Now, we need to express the second term all in terms of `(t - pi)`.

From trigonometry, we have:

`sin(t − π) = -sin\ t`

So we can write:

`Lap{sin\ t * u(t)}- ` `Lap{sin\ t * u(t - pi)}`

`= ` `Lap{sin\ t * u(t)}+ ` `Lap{sin(t - pi)* u(t - pi)}`

`=1/(s^2+1)+(e^(-pis))1/(s^2+1)`

`=(1+e^(-pis))/(s^2+1)`

# Problem Solver

Need help solving a different Calculus problem? Try the Problem Solver.

**Disclaimer:** IntMath.com does not guarantee the accuracy of results. Problem Solver provided by Mathway.

### Search IntMath

### Online Calculus Solver

This calculus solver can solve a wide range of math problems.

Go to: Online calculus solver