This screencast will show how to solve equations in the Laplace domain or the s domain which
contain imaginary roots, and allow us to put them back into the time domain. To accomplish
this we'll use the equation shown on the screen here to show the fact that this is in fact
has imaginary roots all we have to do is recognize the s squared plus four s plus five part and
realizing that if using the quadratic equation if we look at the squared root term of the
quadratic equation if this term is less than zero that will tell us the fact that our roots
are in fact imaginary. So when we do this we have negative four. So therefore what this shows is
the fact the fact that we have two imaginary roots. Processes which have imaginary roots
are systems or processes which will oscillate in nature. From a mathematical standpoint
that represents the appearance of cosines and sines in our final solution. There are
some processes such as swings and pendulums which naturally have this, however most chemical
processes do not have imaginary roots. Where we sometimes see the presence of imaginary
roots in chemical process control is for some systems which are controlled using PID control
where it is possible to end up with imaginary roots. In fact, many tuning parameters take
advantage of this fact and are tuned to ensure the fact that the process obeys quarter decay,
which is an oscillatory response. Recall that systems with imaginary roots will have the
form of a plus or minus b i. But the question now becomes is when we're trying to convert
this from the Laplace domain to the time domain what functional form should we have? And the
best way to answer this questions is to look at the Laplace transforms that are available.
Looking at the transforms provided here what you'll see is the fact that there are no i
terms here, and the fact that everything is left in s squared plus omega squared, or s
plus b squared plus omega squared. So therefore we don't want to solve these using the value
i, which is equal to the square root of negative one. So our focus here will be to leave things
in s squared plus four s plus five terms. In other words our focus is going to be looking
at the last two equations that I have shown here in order to eventually take the inverse
Laplace transform. To do this what we're going to have to do is use partial fractions. So
what we'll do here is just rewrite what we have above, and then break it into its roots,
and what we'll have here is A over s, but we don't want to break the other root because
of the fact that we don't want to deal with i's because the i's provide us no help in
trying to use the Laplace transform table. But what this means is that the numerator
of this will not just be a constant, but it'll be of the functional form Bs plus C. To solve
for A, B, and C, our focus will shift to the numerator. What we'll now do is find a least
common factor, and multiply the A by the s squared plus four s plus five, and the B s
plus C term by s, of which this must equal the numerator we already have, three s squared,
plus five s, plus five. We'll provide that here. What you'll notice on the right hand
side is that the highest power of s that we have is s squared. So what we now want to
do is break this down into terms with varied powers of s, and we have three of them here,
we have s squared, we have s to the first, and we have s to the zero, in other words
the constant terms. So if we look at our right hand side we see that we have an A s squared
term, and a B s squared term. If we look at our s terms we have one for A, for A s, and
we have one for C, C times s. For constant the only one we have is five A. We then use
the left hand side to match what these coefficients have to equal. So if we look on the left hand
side we have three s squared, five s, and five. We can now solve for A, B, and C using
first, we'll start with the s to the zero equation, to find A. A equals one. We can
then substitute in A equals one to the s of the first term, and that will show that C
also equals one. And the s squared equation will allow us to solve for B, B equals two.
Now that we have solved for all of these constants we can substitute them back into the equation
shown above. When that substitution is done the equation is the one shown here. The inverse
Laplace transform for one over s is relatively straight forward, its just the inverse of
a step function which is just a constant, thats one. In order to take the inverse Laplace
transform of the second term we have to go back to our Laplace transform sheet. As noted
before our focus here will shift on the bottom two equations because we don't have a perfect
s squared plus omega squared in our denominator, but what we now need to do is find B and omega.
In order to do that we have to complete the square of the denominator. So in order to
complete the square s squared plus four s plus five, in order to complete that square
what you do is you take half of the value for the s term, which is four, and then take
the square of it. So therefore that would be s squared plus four s plus four. You then
add or subtract whatever the difference is with the original constant you have. So therefore
originally we had a constant value of five. We already have taken up four so therefore
what we're left with is one. The s squared plus four s plus four will be a perfect square.
In this case it will be half of the value of the constant, which was four, so therefore
this will become s plus two squared plus one is our new denominator. By mapping this to
the Laplace transform sheet we have shown the fact that b is two, and that omega is
equal to the square root of the constant term hanging out side, which in this case the value
is one, so the square root of that is also one. By looking at this we hit a little bit
of a problem. I we are to substitute omega and b in what we have is one, and for this
part here we have s plus two. The problem is is that our numerator for this is two s plus
one, so what we're going to have to do is find a way to turn two s plus one, our numerator,
into a linear combination of the one and the s plus two. There are a number of ways to
approach determining what the constant should be in front of the two terms found in Laplace
transform sheet. One will be provided here, and the methodology works very similarly to
the concept of partial fractions, where we're having a linear combination of multiple terms
such that the numerators are consistent. The easy this thing here is the fact that they
both have the same denominator so we don't have to worry about that step. So if we look
at what we have here we have that two s plus one is going to equal A plus B s plus two
B. By looking at this we see the fact that the only s term on the right hand side is
B s, so therefore that tells us that B equals two, and whats left of the constant terms
is that A plus two B equals one, which means the fact that, once we substitute in what
B is, that tells us that A equals negative three. So if we revisit our original equation
with the second term factored out to be more friendly for the Laplace transform table we
can see that when we take the inverse Laplace transform of our second term that just corresponds
to the sine term that we have here. So therefore this term becomes e to the negative b t, where
b has been shown to be two, multiplied by the sign of omega t, where here we showed
that omega equals one. And then we include the constant negative three. So for our third
term, it will correspond to the cosine term, which will be e to the negative b t, which
is still two, multiplied by cosine of omega t, which in this case is still one. We then
add the coefficient to our system, and when we put these all together we get our final
answer, which is that y in the time domain equals one minus three e to the negative two
t sinet, plus two e to the negative two t cos t. So an important part of dealing with
imaginary roots is that you are sure of the fact that when you are trying to solve the
problem you are gearing it towards the Laplace transform sheet that you have available. Different
transform sheets may show things in slightly different ways, so you're going to want to
gear it towards the Laplace transform sheet that is available to you.