In this screencast we will explore dead time in the Laplace domain and the usage of the
Real Translation Theorem. To motivate it we'll start with the problem provided below where
we're going to determine the inverse Laplace transform of the function provided. Looking
at this in the Laplace domain dead time is always represented by the presence of an
exponential in the s domain, so this represents dead time. The dead time, which is also known
as the transportation lag is a common occurrence in chemical processes. Why? the major reason
is piping, so in large chemical plants a fluid has to go from one point to another. However,
these distances are often times non-negligible, so therefore a fluid that is at point A, in
order for it to get to point B it could take a couple minutes in order for this to happen.
Therefore, the process is not acting in terms of changing the inputs effect on the output
is not occurring, so therefore that represents this lag time, which in a graphical standpoint
can be somewhat represented like this, where this here is the dead time. And again, because
chemical processes have significant piping, or other means where nothing exactly is happening,
where another example is some certain types of equipment, for example gas chromatographs,
there is a time it takes for the experiment to run before you can obtain a composition.
Because of the importance of dead time in chemical processes, there is a theorem provided
in the Laplace domain to handle dead time, and it is the Real Translation Theorem, which
is shown here. Where the left hand side of Laplace of a function multiplied by a step
function will equal the exponential times a function itself. Lets take a look at the
real translation theorem, and what information we can get from it. First lets focus on this
s term, the step function, and what this term tells us is that at times less than t zero,
the value of the step is going to be zero. So in other words the term is zero, the entire
term, where as for times greater than or equal to the t naught, s, the value of the step,
will be one, meaning that the value of this function will be the f t minus t naught part
shown here. We can look at the effects on the right hand side by looking at the graph
above, where the dead time represents the exponential, the e to the negative s t naught where
dead time equals t naught. The graph here would represent the function we would want
to take the Laplace of. So looking at this equation what does it mean in terms of how
we apply this mathematically. Well, our focus here will be how to take an inverse Laplace
transform. And the way we can do it is as such. Step one is to ignore the e to the negative
s t naught term. Step two is to take the inverse Laplace transform, and then step three is where
ever e to the negative s t naught is seen you do the following, andhere are two things
that you're going to do. The first is replace t with t minus t naught, and b, show that
the term is valid only when t is greater than or equal too t naught, and that can be done
in one of two ways, it can be written either as the step function shown above, or in a
more common mathematical terms like when t is greater than or equal too t naught. Either
way would be acceptable. With all this information provided about the real translation theorem
and its utility for processes with dead time now lets solve the original question given,
which has been re-provided below. The first step is to break it into parts. When we do
this the four over two s plus one term, we take an inverse Laplace transform of this,
this is not affected by dead time, so therefore this will just act as a regular function,
which it is, and therefore the inverse Laplace transform of this is the constant four multiplied
by the exponential of negative t over two. Now we will look at the term with the e to
the negative three s in it. If we follow the steps provided earlier, the first step is
to ignore the e to the negative s t naught term, in other words this one over two s plus
one. The next step is to take the inverse Laplace transform, similar to the one we just
did. So therefore that will be e to the negative t over two. Now we have to evaluate step three.
The first thing we do is replace t in this with t minus t naught, where t naught is the
dead time here, which is three, and the next step is to show this is valid only when t
is greater than or equal to t naught, so therefore here we'll just write this as the step function
of t minus three, which indicates that this function is zero when t is less than three, and its
itself the exponential of t minus three over two when t is greater than or equal too three.
We now bring back the plus term to show the sum of these, and we get our final answer,
which we'll represent in a few different ways. So the first way, which we written with the
step term in there for e to the negative t over two plus e to the negative t minus three
over two times s the t minus three, or that y of t equals four e to the negative t over
two for times in between zero, where we've implied changes occurred, to until the dead
time ends at time equals three, and then the entire function for all other times. An important
point to note here is that these two functions are continuous because at t equals three this
term will equal zero. The second answer shows that this is a continuous function, which
is what we expect it to be.