In this screencast we will show the usage of the initial value theorem and the final
value theorem, and their utility in analyzing systems. We'll accomplish this by using example
of solving a differential equation provided on the top line to solve in the Laplace domain.
Part (a) asks to convert this equation into the Laplace domain to find y of s, so to do
that we're going to have to take Laplace transforms of our equation. So to take the Laplace transform
of this, Laplace transform of dy dt via the real differentiation theorem will equal s
times the Laplace variable y of s, minus y of zero, where y of zero here is equal to
one. The Laplace transform of little y is big y, and the Laplace transform of a number
is just that number over s. So when we put this all together we have three times s multiplied
by the Laplace variable minus y of zero, which is one, plus two times the Laplace variable,
equals one over s. To solve for y we just need to do some rearranging and some algebra.
This is our answer in the Laplace domain. Part (b) is asking us to use the Initial Value
Theorem to find y of zero. So first lets state the Initial Value Theorem which is at the
limit as t approaches zero of a function y of t, which we have yet to find, equals the
limit as s approaches infinity of s multiplied by y of s. One of the major utilities of the
initial Value Theorem is that it allows us to do a quick check to make sure that what
we put in Laplace domain is correct. it is not a fail-safe, but if we know the fact that
the Initial Value Theorem doesn't hold in the Laplace domain that means we made an error in our
analysis. So on the left hand side as t goes to zero we know what y of zero is, its one,
it was given as the initial condition in the problem statement. So now what we're going
to do is evaluate the limit as s approaches infinity of s multiplied by our answer y.
The s's will cancel out, so were left with the limit of three s plus one over three s
plus two. Here we would have a limit of infinity over infinity, but if we use L'Hospital's
Rule we can easily see the fact that this limit will approach three over three which
equals one. So therefore this is a check on our development of our Laplace variable for
y. Part (c) of the question asks us to use the Final Value Theorem to find the steady
state value of y. In other words the value of y as it approaches infinity. The nice thing
about using the Final Value Theorem is the fact that we can determine the steady state
value of this system without having to do any potentially complicated partial fractions
or mathematical analysis, all we need is this variable in the y domain. So the Final Value
Theorem states the fact that the limit of y of t as t goes to infinity equals the limit
as s approaches zero of s y of s. So again, the convenient thing about this is we can
determine our steady state limit as t goes to infinity without actually calculating y
of t, which we have yet to do. So therefore out limit here as t goes to infinity of y
of t would equal the limit as s approaches zero of s, scroll back just a little bit we
can show what our y of s was, that was three s plus one divided by s times three s plus
two, the s's again cancel, and if we substitute zero in here we end up with one half. This
would show the fact that we would expect this differential equation to converge to one half.
For part (d) we're asked to find y of t and to validate the answer to part (c) to show
the fact that the function converges to one half at steady state. So to do this we start
with our function. The roots have already been factored so therefore we can rewrite this
as A over s plus B over three s plus two, and we can solve this either using partial
fractions of the Heaviside Method. Here we'll solve using the Heaviside Method. The Heaviside
Method equation has been recreated here. We can use this equation to determine A and B, because
both of them are non repeating roots. We'll start by using this to solve A. So A will
be the limit as s approaches the root here, which would be zero because of the fact that
its just s. So the limit as s approaches zero of s multiplied by y. We see cancellation,
then we evaluate this at s equals zero, we get the fact that A is one half. To solve
for B we use the same methodology, here the root is negative two thirds so this will be
s plus two thirds multiplied by three s plus one divided by s times three s plus two. So
in order to get the s plus two thirds to cancel out here which we need to avoid a zero over
zero what we do is factor out the three in the three s plus two term. So they will cancel
and that leaves us with three s plus one over three s, which will equal a half. But there
is a very important point to note here is what we have done here is we have not found
the B that corresponds to three s plus two, we have found the b that corresponds to s
plus two thirds based on the methodology which the Heaviside Method uses. So this one half
corresponds to a denominator of s plus two thirds. Therefore we can restate y as y equals
one half over s plus one half over s plus two thirds. A point to note here is that if
you we're to solve this using partial fractions you would have ended up with y equals one
half over s plus three halves over three s plus two, which when we go through the math
of the inverse Laplace transforms will be identical because the second terms are in
fact identical, which is the fact that in the equation here the numerator and denominator
have both been multiplied by three. So by using the Laplace transform table we can easily
take the inverse Laplace of both of these terms.The one half over s, the inverse Laplace
of this will be the unit step function multiplied by the constant one half. For our second term
we'll have the one half constant and the inverse Laplace transform of one over s plus B as
e to the B t, we end with e to the negative two thirds t, and we have actually taken the
inverse Laplace transform so the inverse Laplace transform of big y is going to be little y in
the time domain, so we have now found y of t. Now that we have found y of t lets check
what happens at time equals zero and time equals infinity. Y of zero would equal one
half plus one half e to the negative two thirds time zero, which means that the exponential
would be one, so would be one half plus one half times one, which is one half, which equals
one, and this was our initial condition. As y goes to infinity this would equal one half
plus one half e to the negative two thirds times infinity, and in that case the exponential
will go to zero, and this will equal a half, which is consistent with what was determined
from the Final Value Theorem. So this screencast shows the utility of using Initial Value Theorem
to determine whether or not your transformed function is potentially correct. The nice
thing here is it shows the utility of the Final Value Theorem which gives us a, the
value of the steady state without having to do the math. Here we used a relatively simple
example to show the matching between the Final Value Theorem and actually taking the inverse
Laplace transform, however in many situations these mathematics may not be as simple making,
the Final Value Theorem an invaluable resource to evaluate steady state characteristics.