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DAVID SHIROKOFF: Hi, everyone.

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Welcome back.

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So today I'd like to take a
look at gain and phase lag.

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And we're going to consider
this simple problem.

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So first off, find a periodic
solution to x dot dot plus 8x

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dot plus 7x equals F_0
times cosine omega*t.

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So basically, it's just
forcing a differential equation

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with some frequency
cosine omega t.

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And the problem we're
interested in today

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is to give the gain and phase
lag to this periodic solution.

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So I'll let you take a look at
this problem and I'll be back

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in a minute.

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Hi, everyone.

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Welcome back.

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OK, so we're interested in
finding this periodic solution

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to the differential equation.

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And we see here that we're
forcing it with right-hand side

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of cosine omega*t.

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So the standard procedure
is to first complexify.

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So we're going to consider the
differential equation x dot dot

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plus 8x dot plus 7x equals
F_0 e to the i*omega*t.

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And then at the end,
we're going to take

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the real part of our solution
to this differential equation.

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And we do this because cosine
omega*t is the real part of e

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to the i omega t.

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Now, for this equation,
we see that we're

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forcing it with an exponential.

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So we can just use the
exponential response formula.

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And this gives us a
particular solution.

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So the exponential response
formula is 1 over p of i*omega

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times the right-hand side,
which is F_0 e to the i*omega*t.

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And in this case, the
characteristic polynomial,

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p of s, is s squared
plus 8s plus 7.

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So that p of i*omega is going
to be 7 minus omega squared--

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so the omega squared comes
from s squared-- plus 8i*omega.

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Now, I'd just like
to take a moment

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and step back for a second.

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If we take a look at the
differential equation,

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the input on the right-hand
side is F_0 e to the i*omega*t.

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Now notice how this
characteristic-- sorry,

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the exponential response formula
gives us a particular solution,

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which is 1 over p of i*omega
F_0 e to the i*omega*t,

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which is a periodic solution.

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But moreover, the output
of this formula shows that

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it's actually the input forcing
multiplied by this factor 1

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over p of i*omega.

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So this factor right
here, 1 over p of i omega

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relates the input forcing
to the output forcing.

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Or sorry, to the
output response.

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And this factor is sometimes
called the complex gain.

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And the reason it's the complex
gain is because if we take

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a look at p of i*omega, it has
a real part and an imaginary

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part.

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So specifically, p of
i*omega contains two pieces

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of information.

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Because it's a
complex number, we

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can think of it as being
an amplitude and a phase.

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So the amplitude of p of
i*omega is It's sometimes called

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the gain.

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And the phase of p of
i*omega is the phase lag.

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So p of i*omega contains
two pieces of information,

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which relate the input forcing
signal to the output response

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of the differential equation.

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Now, I'd like to go ahead and
write x of t, decomplexify.

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So we're going to take the real
part of 1 over p of i*omega

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times F_0 e to the i*omega*t.

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So we have 7 minus omega
squared plus 8i*omega.

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Upstairs is F_0 e
to the i*omega*t.

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And I'm going to just put
this in amplitude-phase form.

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And when I take the real part,
I'll do it in two steps first.

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8 omega squared.

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Square rooted.

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F_0 e to i*omega*t
divided by e to the i*phi.

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Where here phi is the
phase lag and tangent phi

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is going to be the imaginary
part over the real part.

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And just quickly note that as
omega goes from 0 to infinity,

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tangent phi will
go from 0 to pi.

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So this is the range of phi
that we're interested in.

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And now, when we
take the real part,

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we end up with F_0 1 over 7
minus omega squared quantity

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squared plus 8 omega
squared, all squared rooted,

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cosine omega*t minus phi.

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And now we'd like to look at
what the amplitude and phase

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are as a function of omega.

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So for each fixed
omega, the output

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is going to be a
sinusoid which oscillates

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at the same frequency
as the input.

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However, the only difference
is that it's going

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to have a rescaled amplitude.

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And it will have a
shifted phase as well.

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So for the amplitude response,
we're interested in plotting 1

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over the amplitude
of p of i*omega.

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So here's the amplitude.

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And this function is just going
to decrease and asymptotically

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approach infinity.

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So this is the
amplitude response.

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This is 1 over 7 minus
omega squared square rooted

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plus 8omega squared,
quantity square rooted.

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This is omega.

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And then in addition to
the amplitude response,

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we also have the phase.

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And the phase, if I go
back up here for a second,

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we can write it
explicitly as phi

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is equal to tan inverse 8omega
divided by 7 minus omega

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squared.

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So I'm going to plot omega on
this axis and phi on this axis.

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And, OK, so to plot this curve,
we note that when omega is 0,

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tan inverse of 0 is 0.

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So we're going to start at 0.

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Typically, what I usually
like to do in this case

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is look at the denominator
in the arctangent.

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We see that when omega is
equal to the square root of 7,

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this argument blows up.

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It goes to infinity.

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And the arctangent of
infinity is pi over 2.

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So I can draw a curve in
here, which is pi over 2.

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So we know that phi is pi over
2 when omega is equal to root 7.

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And note how this is the
natural frequency if there

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was no damping in the system.

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And then lastly, there's
another curve here at pi.

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We see that as omega
approaches infinity, again,

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this argument approaches 0.

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And if you were to take the
derivative of this whole

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quantity, we would see that
it's-- of tan inverse of this

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argument, we would see that the
function's increasing for all

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omega.

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So it actually looks
something like this.

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It looks like an s-shaped curve
that asymptotically approaches

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pi as omega goes to infinity.

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So just to quickly recap.

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When we look at the
differential equation,

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or a particular solution to
a differential equation which

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is being forced by a
periodic sinusoidal function,

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the output is always
going to be rescaled

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and phase shifted,
but still oscillating

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at the same frequency.

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And specifically,
depending on which

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frequency the differential
equation's forced at,

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the amplitude will take
on different values.

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The amplitude response will
take on different values.

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And the phase shift will also
have different values depending

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on the frequency of forcing.

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So I'd just like
to conclude here,

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and I'll see you next time.