Welcome back, this is Dr. Ferri.
Today's lesson will be on second order differential equations.
In previous lessons we've looked at solutions to first order differential
equations and we've applied that to the analysis of RC and RL circuits.
So looking at our module on reactive circuits here.
that covers, that first order analysis covers up through RC and RL circuits.
So now we want to introduce second-order differential equations, so that we can
examine more complicated circuits, particularly ones that involve RLC
circuits. They include both inductors and
capacitors, as well as resistors. Our lesson objectives are to examine
second-order differential equations with a constant input.
We want to be able to determine the steady-state solution and determine the
type of transient response that we can get out of it.
And then we want to see what the actual response looks like if we plot it and
recognize the characteristics of that. So as we talked about before with first
order differential equations, And ordinary Differential equation is just an
equation with functions and variables of their derivatives.
Looking at this equation right here, we've got a second derivative, so that
makes it a second order differential equation and were just assuming that
we've got constant coefficients here. And that our forcing term here is just a
constant. Many physical systems are governed by
differential equations like this second order differential equations.
Where the forcing term is the input to our system.
And then the output is the solution to our differential equation.
Vibratory system and RLC circuit are examples of this.
In a vibratory system our forcing term, our actual physical force that we apply
at the system is the input. And then the solution to the differential
equation is a position. In our RLC circuit, our input is a
voltage, or current source, and our output is a voltage or current across a
particular element. So let's look at a generic second-order
differential equation. And I say generic so that we can apply it
to Rlc circuits, vibratory systems or anything that's governed by second order
differential equations. And in, in the system that were looking
at, were assuming that these are constant but unknown parameters, a1 and a2 and k
is also constant. We assume that we have known values for
initial conditions, so y of 0 is known, That's time equals zero and the first
derivative at time equals zero is also known.
Our solution is made up of a steady state part plus a transient.
Plus steady state part is pretty easy to solve for and it's exactly the same way
we solve for the steady state in a first order differential equation.
We just take this value K I redivided by this last coefficient, a sub 2.
And that's what happens to y of t as it becomes very large.
It approaches this value. The transient, well the transient's a
little bit more difficult than it was in the first order differential equation.
Because we have 3 possible forms, depending on the roots of this
polynomial. Okay.
Where do I get that from? I can look up at the differential
equation, and wherever I see a second derivative, I'm going to replace it with
an s squared. That's a sub 1.
My first derivative, I'm going to replace with an s.
Plus a 2, equals 0. Those of you who have studied
differential equations before might recognize this as a, as a characteristic
equation. Those of you that have seen Laplace
transform before may recognize it as a Laplace transform.
In our case, we're just going to use. This define the solution.
I'm not going to assume that you had any other background in differential
equations than what I'm covering here. So the transient response again is
governed by that polynomial and we've got three possible cases.
If I've got real and distinct roots. Say, R1 and R2.
Then I've got this case right here, and my solution, kind of my generic form of
my transient response looks like this. So remember, I've got a steady state plus
this part, here's my solution to the differential equation.
Now, if R1 And r two are both less than zero.
Then my transient suppose it starts up here with initial condition up here.
It's going to go like this. It's going to look exponential versus
time. And that's because one of these roots.
It usually is, is much it decays much faster than the other one.
So if this decays much faster compared to this one, then my response will look
exponential. In the critically damped case, I have two
roots that are equal, r and r. So my solution looks like this.
So I get this, this sort of strange term in here.
T times e to the r t. Well it turns out r is less than zero
then the exponential goes to zero faster than t gets large.
So my response for that case. Also looks exponential, and that's if the
case where r is less than 0. We oftentimes call these stable systems,
where the transient decays to 0. Now the under-damped case, that's when
I've got complex roots, and I'll say the complex roots are equal to A plus or
minus j v. Where j is the square root of minus one.
By imaginary term. So in that case I get this sinusoidal.
Where the real part, the real part goes as the exponential.
And the imaginary part shows up as a frequency.
In this case, if a is less than 0, then I've got e to that a t with a negative
exponent. That's going to decay.
And let me show you what this looks like. Given some initial condition, the
transient Will look sort of like that. Where if I were to look at this
amplitude, the amplitude would be exponential because it follows this.
This looks like an amplitude of this sine way.
We often call this a damped sign wave. So this is a damp sign wave and it's
damped by this real part of my root. So my overdamped case looks exponential,
my underdamped case looks oscillatory. In the cases we're going to be examining,
it's, we're not going to be really worried about the critically damped case,
because critically damped, we have to have your, your system perfectly match.
So that you end up with real and equal roots.
And the chances of that happening in a physical system aren't very good.
Unless you really are tuning your system so with resistors that have tolerances to
them. It's unlikely you actually come up with a
critically damped system. So we're just going to worry about
overdamped and underdamped. Lets look at some sample examples.
Sample problems. Given this problem, the steady state
value is one over four. And if I see, what happens is y gets very
large as it approaches 0.25. So that a steady-state value.
The transient is found by solving this differential equation, s squared plus 5 s
plus 4 equals 0. If I find the roots of that, I get roots
equal to minus 1 and minus 4. That means my transient is some constant,
k 1, times e to the minus 1 times t, plus some constant k 2.
E to the minus 4 times t. Well in this case you see that e to the
minus 4 to k is much faster than e to minus t.
So this system kind of dominates by this exponential here.
And that's what we see right here. We see something that looks almost
exponential. The, the transient has an exponential
decay until it reaches a steady-state value.
And, in fact, if I look at a value of 1, this, this looks to be, having pretty
close to a time constant of 1, because that's when the transient decays to about
gets to about two-thirds of the way to the steady-state value.
So it's pretty close to a time constant of 1.
So this case, real and distinct roots, it's overdamped.
And it looks sort of exponential. Now if I look at this, the next case, the
steady state. Just like it was in the previous case,
it's equal to one over four. And we see that it does approach one over
four here. Transient I have to solve s square plus
.8 s. Plus four equals zero.
Well the roots to this one are complex. It's equal to minus .4s .4 plus or minus
1.96 j. It's complex so that means it's
underdamped. And I serve the general form of the
transient, is k, you know some constant case of one.
E to the minus .4 t. Sin of 1.96.
T Plus some angle. This is my sine wave, and you see that if
I were to look at the amplitude here, the amplitude looks like this, it's a
decaying exponential. With that decay rate.
One thing I want to point out about this is, if I look at a case of a low value
here, compared to the previous example, the lower the value here, it turns out
the more oscillatory the system is. And there's a parameter here in this
coefficient that we're going to call Damping factor or damping ratio.
And so as the damping ratio goes down, as this coefficient goes down, gets
smaller,then it becomes more oscillatory and it peaks up a little bit more So this
is, this coefficient is proportional for the damping ratio and as damping ratio
goes down then it becomes more oscillatory.
[SOUND] And again it peaks out more. So in this case we've got complex roots.
It's underdamped and the lower that value, the more it peaks up.
In summary, we've examined in generic second order differential equations.
And, we've talked about how it these apply to vibratory systems and RLC
circuits. We've showed how to find the steady state
solution. We've showed generic transient solutions
to underdamped and overdamped responses. And then, we looked at characteristics of
plots of underdamped and overdamped responses.
Constant input applied at time equal to zero.
In our next lesson, we will demonstrate RLC circuit equations, And their
responses and show how we use these solutions to second order differential
equations to analyze those sorts of circuits.
Now please visit the forms And I'll see you online.
Thank you.