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`.
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) :}`
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)`
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