WEBVTT
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GILBERT STRANG: OK.
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So this is the next step for
a first-order differential
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equation.
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We take-- instead
of an exponential,
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now we have an oscillating.
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Exponentials, the previous
lecture, grew or decayed,
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now we have an oscillate.
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We have AC, alternating
current in this problem,
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instead of real exponentials,
we have oscillation, vibration,
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all the applications that
involve circular motion,
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going around and around instead
of going off exponentially.
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OK.
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So here's the point.
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Again I'm looking for
a particular solution.
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The particular solution--
it would be nice
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if we could say the
particular solution
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was just some multiple
of the cosine.
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But that won't work.
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So that makes this
problem one step harder
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than the exponential.
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We need to allow
the signs in there.
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Because, if I look for a cosine,
if I tried only this part,
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I could match that.
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I'd have a times the cosine,
that would be cosine.
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But the derivative of a
cosine is a sine function.
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So signs are going
to get in there
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and we have to allow
them into the solution.
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OK.
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So that's the right
thing to assume.
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Actually there will
be, you'll see,
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three different ways to write
the answer to this problem.
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And this is the first sort
of most straightforward,
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but not the best
in the long run.
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OK.
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Straightforward one, I'm
going to substitute that
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into the equation and find
M and N. That's my job.
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Find these numbers.
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So put that into the equation.
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On the left side I
want the derivative,
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so that we'll be omega--
well derivative of the cosine
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is minus omega m sine omega t.
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The derivative brought
out this factor omega.
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The derivative of
cosine was sine.
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Now the derivative
of this brings out
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a factor of omega--
omega N cosine omega t.
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And that should equal
a times y-- there's y,
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so I just multiply by
a-- a M cosine omega t,
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and a N sine omega t.
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That's the ay part.
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And now I have the source
term plus cosine omega t.
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And that has to be
true for all time.
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And now I need the equation.
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What do I do with this?
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I'm looking for two
things, M and N.
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I'm looking for two equations.
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So I match the cosine terms.
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I match that term, that cosine
term, and the source term.
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So they all multiply
cosine omega t.
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So I want omega
n-- so I bring this
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over on the other side--
minus a m plus omega n
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equals-- here I just have
one cosine-- equals 1.
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Minus a M plus omega N equals 1.
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And now I'll match
the sine terms.
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So in the sine terms, I have
a minus omega M, sine omega t.
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And I have to bring
this on the other side,
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so that'll be a minus
a N sine omega t.
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And there's no sine
omega t in the source.
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There's my two equations.
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Those are my two
equations for M and N.
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So I just solve
those two equations
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and I've got the particular
solution that I look for.
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So, it's two equations,
two unknowns.
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It's the basic problem
of linear algebra.
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I'm inclined to just write
down the answer, which
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I prepared in advance.
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And it turns out
to be minus a over
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omega squared plus a squared.
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And N turns out to have
that same omega squared
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plus a squared, and
above it goes omega.
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If you check this equation,
for example, omega times the M
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will give me a a omega
with a minus with a minus.
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And then a times N will
also have an a omega.
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And the same omega
squared plus a
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squared, they cancel to give 0.
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And this equation
is also solved.
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So one more important
problem solved.
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Well, we found the
particular solution.
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I haven't added in-- I haven't
match the initial condition.
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Now in many, many cases,
it's this particular solution
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that's of interest.
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This here-- let me put a
box around our solution--
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and we substituted that in
the differential equation.
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We discovered M. We
discovered N. We've
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got this particular solution.
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And that's the oscillation
that keeps going.
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That if we're listening
to radio or if we
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have alternating current,
this is what we see,
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the null solution.
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The thing that's coming
with no source term.
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Usually a is negative
and that disappears.
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That's called the
transient term.
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So the null solution would be
have an ae to the at as always.
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But I'm not so interested in
that because it disappears.
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You don't hear it
after a minute.
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And this is the solution that
you're-- this is what your ear
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is hearing.
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OK.
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So we've got one
form of the answer.
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Now, that's a pretty nice
form, but it's not perfect.
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I can't see exactly--
this can be simplified
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in a really nice way.
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So when we work with
sines and cosines,
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it's this next step
that's important.
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I believe that that
same yp of t can
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be written in a different
way as what also--
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another form, a different--
well I should say,
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another form for
the same y of t.
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Another form will
be the same y of t.
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You see what I don't like is
having a cosine and a sine
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because those are out of
phase, and they're combining it
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into something, and
I want to find out
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what their combining into.
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And it's really nice.
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They're combining into a single
cosine, but not just omega t,
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there's a lag, a phase shift.
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The angle involved is
often called the phase.
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So the two, sine
and cosine, combine
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to give a phase shift
with some amplitude,
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maybe I'll call it G, the gain.
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Or often it would be
called capital R just
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for-- because it's-- sort of
what you're seeing here is
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polar coordinates.
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So I want to match
this, which has the G
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and the alpha-- polar
coordinates is really
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the right way to think of this.
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G and an alpha, a
magnitude and an angle.
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I want to match that with
the form I already had.
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So I'll use a little
trigonometry here
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to remember that
this is equal to-- I
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have a G. Do you
remember a formula
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for the cosine of a minus b?
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The cosine of a difference
is cosine of omega t,
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cosine of alpha,
plus-- it's a plus
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here because it's
a minus there--
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sine omega t sine alpha.
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So I have just written this
out in the two-term form,
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and I did that so that I could
match the two-term form I
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already had.
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So can I just do that matching?
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The cosine omega t, the
M must be G cosine alpha.
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And they N must be G sine alpha.
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So I now have two equations.
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The M and the N, I still
remember what those are.
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I figured those out.
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But now I want to convert the
M N form to the G alpha form,
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and this is what I have to do.
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And it's the usual thing
with polar coordinates.
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How can I get-- how do I
discover what G is there,
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and what alpha is?
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The trick is-- the one
fundamental identity
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when you see cosines and sines--
is to remember that cosine
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squared plus sine squared is 1.
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I'm going to use
that, have to use it.
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So I'll square both sides.
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I'll have M squared,
and I'll add.
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So I'll have M squared
plus N squared is
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G squared cosine squared alpha.
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G squared times
cosine squared alpha--
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when I square that one--
and sine squared alpha
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when I square that one.
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And again, the
point is that's one.
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So that's just G squared.
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So what do I learned?
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G is the square root of this.
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G is the square root of
M squared plus N squared.
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And I'm always freedom plug-in
the M and the N that I found.
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OK that's-- ah,
what about alpha?
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That's the angle.
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So I have to-- again,
I'm thinking trig here.
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How am I going to
get alpha here?
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I want to get G
out of this formula
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now, and just
focus on the alpha.
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Previously I got alpha
out of it and got the G.
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Now the way to do
is take the ratio.
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If I take the ratio
of that to that,
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divide one by the other,
the G's will cancel.
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So I'll take the
ratio of that to that
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to get G sine alpha divided by
g cosine alpha is then N over M.
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And the G's cancel as I wanted.
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So now I have an
equation for alpha.
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Or more exactly, I have an
equation for tangent of alpha.
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Sine over cosine is tangent
of alpha is N over M.
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So this is called
the-- you could call it
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the sinusoidal identity.
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What is that word sinusoid?
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Sinusoid is a word for any
mixture of sines and cosines,
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any mixture of sines and
cosines of the same omega t.
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So the sinusoidal
identity says that I
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can rewrite that solution
into this solution.
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And I really see the key
number in the whole thing
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is the gain, the magnitude.
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It's how loud the
station comes through
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if we're tuning a radio.
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So and again, this
is the response
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that's keeps going because
the cosine oscillates forever.
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There will be also
something coming
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from the initial condition
that we expect to die out.
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So I mentioned at the
very start that there
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were three forms of the
answer to this cosine input,
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and I've given you two.
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I've given you the M and N form.
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You could say rectangular
coordinates-- cosines
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and sines.
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I've given you the
polar form, which
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is a gain, a
magnitude, and a phase.
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And the third one
involves complex numbers.
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I have to make that a separate
lecture, maybe even two.
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So complex numbers,
where do they come in?
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It's a totally real equation.
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If I think about all
this that I've done,
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it was all totally
real, but there's
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a link-- the key fact about
complex numbers, Euler's
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great formula will give me a
connection between cosine omega
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t and sine omega t with
e to the I omega t.
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So at the price of introducing
that complex number,
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imaginary number i, or j
for electrical engineers,
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we're back to exponentials.
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We're back to exponentials.
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So that'll come in
the next lecture.
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This is one more example
of a nice source function.
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Maybe I could just say, what
are the nicest source functions?
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So this is the source
function here and it was nice.
00:15:28.040 --> 00:15:30.080
Exponential was even nicer.
00:15:30.080 --> 00:15:32.170
Constant was best of all.
00:15:32.170 --> 00:15:35.970
And I want to-- another
one I want to introduce
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is a delta function.
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So that's-- a delta
function is an impulse,
00:15:41.310 --> 00:15:43.390
something that
happens in an instant.
00:15:43.390 --> 00:15:46.870
And that's an interesting, very
interesting and very important
00:15:46.870 --> 00:15:47.740
possibility.
00:15:47.740 --> 00:15:48.240
OK.
00:15:48.240 --> 00:15:50.070
Thank you.