In this screen cast we will analyze the stability
and oscillatory nature of a second order differential equation.
Provided in the problem statement we have
a differential equation that shows the relationship between
a manipulated variable (x) and a controlled variable (y).
We know that y at time equals zero and the derivative of y at time equals zero are zero.
The question is asking for what range of K will the system have an oscillatory response and
for what range of K will the system have an overdamped response.
The first thing we can look at here is that we have a second order differential equation.
So we have a second order differential equation
What that means is the fact that we have a second order transfer function.
With second order transfer functions we can analyze
stability and oscillations via zeta.
Where zeta come from the standard form of a second order differential equation
which is k over tau squared s squared plus
two tau zeta s plus one.
Where zeta represents the damping coefficient,
and the damping coefficient provides some insight into if a process does oscillate
how quickly the oscillations will subside.
So our first step will be to take the Laplace transform of our differential equation and get it into
the standard form shown above.
So when we do that the Laplace transform of a second derivative where the
y of zero and y prime of zero is zero is just s squared multiplied by y.
The Laplace transform of the first derivative where y of zero is zero
is just s times y and the coefficient K maintains
and then the Laplace transform of for y will simply be four multiplied by the Laplace variable y.
and the Laplace transform of the variable x will just be x.
If we factor this out
we can factor the y out of the left hand side, we end up with y multiplied by s squared
plus Ks plus four equals x
giving us a transfer function of y over x equals 1 over s squared plus Ks plus four.
But the important point here is if we look at the denominator of the standard form
we see the fact that the coefficient at the end has to be one
So in order to modify this we are going to have to multiply both
our numerator and our denominator by one quarter.
And when we do this we end up with our transfer function of
y over x equals one quarter over one quarter s squared plus one quarter K times s plus one.
Now that we are in standard form
we can now match our denominator with the standard form denominator
so in other words what the means is
is that one fourth is tau squared and that one fourth K is going to be two times tau times zeta
So from this we can determine both tau
and the more important part in terms of determining stability and oscillatory nature of our process
here zeta
So we will start with tau. We see that fact that tau squared equals a quarter
So we take the square root of both sides
we show the fact that tau equals one half.
For the sake of our standard form here it is assumed the fact that tau is always a positive value.
So to find zeta we know the fact that two tau zeta
equals K over four
If we substitute in the one half for tau
we can easily see the fact that zeta will be K over four.
So now that we have a relationship for zeta we have to revisit
what ranges of zeta lead to what particular responses,
and that table is provided here.
So from this table of zeta we can
see what values of zeta correspond to either monotonic or oscillatory responses,
and what values of zeta correspond to stable or unstable processes.
So we can use this table to find the ranges of K for our two parts of the question.
So part (a) asks for when the response will be oscillatory,
and we can see for that, that that will occur in the range shown here.
So in other words, the range of when the absolute value of zeta is less than one,
or in other words, when negative one is less than zeta which is less than one.
We can substitute zeta in to get negative one is less than K over four which is less than one,
and if we multiply everything by four, we get our acceptable range of K.
which is the fact that negative four is less than K, which is less than four.
Part (b) asks for what range of K will get an overdamped response
and that can be found by looking at our table
and seeing that an overdamped response corresponds to a response that is both monotonic and stable.
So what we are going to be looking for are values of zeta
that are greater than or equal to one.
So again we substitute in the fact that zeta for our process here is K over four
and we can determine the fact that our process will be overdamped
when K is greater than or equal to four.
So what this shows is that over a particular range of a constant,
it doesn't have to be the one in front of the dy/dt term,
it can be in front of any of them,
it shows that this very range of K can allow for different types of responses
Therefore think about it from my control scheme standpoint
is by adjusting the parameters for Kc, the controller gains
tau(i), the integral time, and tau(t), the reset time
we can change the values of these constants in our transfer function
which can change our type of response
making it going from overdamped to underdamped to undamped,
or in the opposite order depending on what value of K is being used.