WEBVTT
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GILBERT STRANG: OK.
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So this is a video in which we
go for second-order equations,
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constant coefficients.
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We look for the
impulse response,
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the key function in
this whole business,
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and the step response, too.
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So those are the responses.
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So I'm going to call g--
that will be the impulse
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response, where the
right-hand side is
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a delta function, an impulse,
a sudden force at the moment t
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equals 0.
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So that's the equation.
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That's the impulse.
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And g is the response, and
we want a formula for it.
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Then the other possibility,
very interesting possibility,
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is when the right-hand
side is a step function.
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And then we want the
response to that function.
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I click a switch.
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The machine starts
working, and it
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approaches a steady response.
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The solution rises from 0.
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So it starts at r of 0
equals r prime of 0 equals 0.
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The step response
starts from rest.
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The action happens when I
click a switch at t equals 0,
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and then r of t will
rise to a constant.
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Very, very important solutions.
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But we'll focus
especially on this one.
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OK.
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So that's our equation with
the right-hand side delta.
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And of course, that
right-hand side
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is not totally familiar,
not as nice as e to the st.
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But there is something
that-- there is another way
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to approach it-- that's
a key idea here--
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that gives us this all-important
function from solving
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a null equation.
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How's that?
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I start with a null
equation, but now this
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had no initial condition.
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So this one started from g
of 0 and g prime of 0 both 0.
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And everything happened,
boom, from the delta function.
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This is the same function.
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Except when I look to
see what happens at t
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equals 0, what happens is g
prime immediately jumps to 1.
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So another way I can approach
g, the computation of g,
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is to think of it-- I'm just
looking for a null solution.
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I'm looking for the null
solution that starts from 0.
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But it starts with an initial
derivative, slope equal 1.
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So I know that g
is a combination.
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So I know how to solve equations
like that, null equation.
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You remember s1 and s2?
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I look at s squared-- I've
made this coefficient 1-- so s
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squared plus Bs plus C equals 0.
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That gives us s1 and s2.
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And now I'll tell
you what the g is.
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So that gives us the s1 and
the s2 in the null solution,
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and we're looking
for a null solution.
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So our g of t is
some combination of e
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to the s1t and e to the s2t.
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OK?
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It's some combination of those.
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And we want it to be 0.
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So no surprise, if I subtract
those-- I'm starting at t
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equals 0.
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I'm starting--
this is 1 minus 1.
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It's 0.
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And now I just have to
fix the initial slope,
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the first derivative, to be 1.
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Well, what's the
derivative of this?
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This brings down an s1.
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This brings down an s2.
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At t equals 0, I'm
getting an s1 minus s2.
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So I'll just divide
by that, s1 minus s2.
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There you go.
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That's the impulse
response-- a null solution
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that satisfies these
special initial conditions.
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So that's the function
in mathematics
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that's sometimes called
the fundamental solution.
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It's a solution from which
you can create all solutions.
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It's really the
mother of solutions
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to this second-order
differential equation.
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Because if I have
another forcing function,
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this tells me that growth rate.
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It's just like e to the at
for a first-order equation.
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Remember the growth
rate e to the
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at for the simple first-order
equation with interest
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rate coefficient a?
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Now we have two.
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Instead of a, we
have an s1 and an s2,
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and that's the special function.
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OK.
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We need to get more insight
on that for particular cases.
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So let me show you
the same function when
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I have no damping.
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Start with that case,
always the easiest case.
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When B is 0-- B was the
damping coefficient,
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the first derivative in
our differential equation.
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Can I just bring down the
differential equation?
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When B is 0 here,
that's no damping.
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I just have the second
derivative and the function.
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That's when things
oscillate forever.
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So that's what will happen.
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With B equals 0, I
have pure oscillation.
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The s1 and s2 are cosines
and sines that oscillate.
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Or it's neater to stay
with exponentials, the i
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omega and minus i
omega, where that's
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omega n, the natural frequency.
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Now, if I just plug in that
s1 and s2-- the plus is s1
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and the minus is s2--
I plug it in there,
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I get a nice formula for g of t.
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So that's what g of t
looks like for no damping.
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It just oscillates.
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OK.
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The next case is underdamping.
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It's good to see all
these cases each time.
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So that's a small value
of B. Underdamping
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means there is some
damping, but it's
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small enough so there's
now a real part,
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but there's still
an imaginary part.
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So this is, in a way, the trick
is the case when s is complex.
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If I go higher with the
dampening, increase B further,
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then I'll hit a
point where there
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are two real solutions equal.
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And if I push B beyond
that, I've got overdamping,
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and those two real
solutions separate.
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They're different,
but they're real.
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And then my formula,
in that case,
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overdamping, that would
be the best formula.
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But with underdamping, I
can see the oscillation.
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If I just plug in those two
solutions for s1 and s2,
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you'll see that I have the
e to the minus B over 2t
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appears throughout.
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But then I have the sine
of omega over omega,
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same as I had before, except
now the damping frequency
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is a bit slower than
the natural frequency.
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Damping slows that
frequency down.
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And in a different
video, we had a formula
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for omega damping,
omega damping.
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And then increase the damping
some more, then this part--
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this omega damping goes to 0.
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We don't see any imaginary
part in the solution.
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We see two equal real values.
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They're simple.
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They have to be just minus B/2.
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So that's a case of two
s's coming together.
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And when two things
come together,
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we're used to seeing
a factor t appear.
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So I have that they came
together at minus B/2.
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So I have the
exponential of that.
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But I have a factor t
from the merge of the two.
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And then if B increases beyond
that, that's my formula.
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The two s's are real.
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I don't think one
memorizes all this.
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I had to look them up and
write them on the board
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before starting this video.
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But I hope you see that
they're extremely nice.
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The no-damping case
with [? pure ?]
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frequencies and the underdamping
case with a real-- a decay.
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The critical damping when
you increase B further,
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you just have that
and no oscillation.
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And then beyond
that is overdamping.
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OK.
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So we're good for
the impulse response.
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And now I just have to say,
what's the step response?
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So can I end this video by
going back to my equation?
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I have to bring the board
down to show it to you.
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So now I'm going to deal
with the step response.
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So the equation is the same.
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I'm calling the
solution r for response.
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And the point is, the
right-hand side is now
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a step instead of a delta.
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So we'd like to solve that
equation starting from rest.
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So a switch went on, and I
want a formula for r of t.
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That's all that remains.
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And actually,
that's just-- well,
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you can see what the
particular solution is.
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We look at a
particular solution.
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Well, this right-hand side is 1.
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This right-hand side
is 1 beyond t equals 0.
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So I'm looking for a way
to get 1 out of that.
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Well, or it can be
just a constant.
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The particular solution
is the steady state
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that we're approaching.
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And there's one little
cool thing to do.
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Sometimes people who have the
dimensions and units of things
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clearly in their mind
will put a C in there.
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This is really the good thing
to do is to have that C in there
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because now the units
for r are the same as-- r
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is going to go to the 1.
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The steady state will
be 1 now because I
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have Cr equals C times 1.
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And out in infinity, the
simple solution is r equal 1.
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If when r is 1, its
derivative is 0.
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Second derivative's 0.
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r equal 1 is a solution.
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It's a particular solution.
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It's the steady-state solution.
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Good.
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But that r of t equal 1
does not start correctly.
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We want to start at
0, with a slope of 0.
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So I have to subtract off one
of these particular solutions
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with e to the s1.
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And now I have to
get it so that I
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have to subtract it off so
that this thing starts from 0.
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Let me see if I can do it.
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I think maybe if I
do an s2 e to the s1t
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and subtract off
an s1 e to the s2t.
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Do you see what that
one has achieved?
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At t equals 0, I have--
well, at least at t equals 0,
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I've made the derivative
0 because the derivative
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will bring down an s1 there.
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That derivative will
bring down an s2.
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And when I put t equals 0,
I get s1, s2 minus s1, s2.
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Good, good, good.
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OK.
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Now I think that together
they're all correct.
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I need to divide by s1 minus s2.
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Let me just say,
I think that's it.
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I think that's it.
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It wouldn't be a bad
idea if I just checked.
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And having checked, I've
learned that that's a plus sign.
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OK.
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So the graph of r.
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This is a graph of r of t.
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It starts from 0,
and it rises to 1.
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Asymptotically it's 1.
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This is a graph of r of t.
00:13:42.520 --> 00:13:45.480
And in practice, that's
a very important number.
00:13:45.480 --> 00:13:47.350
What is the rise time?
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How far do you have to go in
time before it rises up to,
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let's say, 95% of 1?
00:13:53.800 --> 00:13:56.930
All these questions are
extremely practical questions
00:13:56.930 --> 00:13:59.076
for an engineer.
00:13:59.076 --> 00:14:01.110
What's the rise time?
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And you're playing
with this formula.
00:14:03.290 --> 00:14:07.610
So let me just make
another comment
00:14:07.610 --> 00:14:11.740
on this r of t step response.
00:14:11.740 --> 00:14:15.160
My other comment is
I've emphasized g of t,
00:14:15.160 --> 00:14:19.240
the impulse response is like
responsible for everything.
00:14:19.240 --> 00:14:21.170
It's always with us.
00:14:21.170 --> 00:14:24.690
And how are those connected?
00:14:24.690 --> 00:14:27.450
That's my final
question in this video.
00:14:27.450 --> 00:14:32.270
How is r of t
connected to g of t?
00:14:32.270 --> 00:14:35.420
Well, let me ask about
the right-hand sides.
00:14:35.420 --> 00:14:38.370
How is the step
function connected
00:14:38.370 --> 00:14:40.720
to the delta function?
00:14:40.720 --> 00:14:46.220
Answer, the step function is the
integral of the delta function.
00:14:46.220 --> 00:14:47.780
The integral of
the delta function
00:14:47.780 --> 00:14:50.230
is 0 as long as
you're integrating off
00:14:50.230 --> 00:14:51.850
where the delta function is 0.
00:14:51.850 --> 00:14:55.910
But as soon as you
pass the big spike,
00:14:55.910 --> 00:15:00.900
then the integral jumps to 1,
and you have a step function.
00:15:00.900 --> 00:15:04.850
So the step function is the
integral of the delta function.
00:15:04.850 --> 00:15:10.650
So the step response is the
integral of the delta response.
00:15:10.650 --> 00:15:13.390
It's the integral of
the impulse response.
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r is the integral of g.
00:15:15.910 --> 00:15:21.850
r is the integral of g with
the correct initial conditions
00:15:21.850 --> 00:15:26.980
that gave us this and eventually
[INAUDIBLE] approach 1.
00:15:26.980 --> 00:15:34.490
So that is the two key
solutions, you could say.
00:15:34.490 --> 00:15:39.780
The impulse response important
in theory and in practice.
00:15:39.780 --> 00:15:42.820
The step response extremely
important in practice
00:15:42.820 --> 00:15:47.725
because turning on
a switch is so basic
00:15:47.725 --> 00:15:53.930
an operation in engineering.
00:15:53.930 --> 00:15:55.220
Good.
00:15:55.220 --> 00:15:57.120
Thank you.