# 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.]

Continues below

### 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.

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.

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.

eba

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 eb−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) :}

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)

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