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PROFESSOR: So in
this recitation,

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we're going to look at step and
delta functions, integration

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and generalized derivatives.

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So the first part, you're
asked to compute the integral

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from zero minus to infinity
delta(t) exponential t

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squared dt.

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The second one is from zero
minus to infinity delta(t-2)

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exponential of t
squared sine t cos 2t.

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The third one is zero plus to
infinity delta(t) exponential

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of squared dt.

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So know that it's the same
as a, except that the bounds

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of integration changed.

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The second part,
we're asked to define

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the generalized derivatives
of these two functions,

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where u here is just a step
function that you saw before.

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So it's 3u(t) minus 2u(t-1).

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And the second one is t
squared for t is negative

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and exponential of
minus t for t positive.

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So why don't you pause the video
and work through this example,

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and I'll be right back.

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Welcome back.

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So let's compute the first one.

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Zero minus, infinity, delta(t)
exponential t squared dt.

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So just to remind you,
the delta function

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is everywhere zero
except at the value zero.

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And we represent
it with an arrow.

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And the integral of the delta
would be 1 from minus infinity

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to plus infinity.

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In this integral, we're
integrating from zero

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minus to infinity, which means
that the zero is included

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in our interval from
zero minus to infinity.

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Therefore, this
integral is basically

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assigning the value to this
function exponential t squared.

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And the value that
it's assigning

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to it is the value
it would take at t

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equals to zero where
the delta is non-zero.

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So really clearly, this is
just exponential of zero.

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And it gives us 1.

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For the second integral, it goes
from zero minus to infinity,

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delta t minus 2 exponential,
a more complicated function,

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t square sine t cos 2t.

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So now, let's represent
this delta function here.

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So this is just our zero axis.

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And the delta here
is zero everywhere

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except at 2 where we
would represent it, again,

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with an arrow at 2 amplitude 1.

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So this delta is zero
everywhere except at 2

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where it would assign
the value to the function

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next to it at the
value t equals to 2.

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So really this
integration gives us

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just the value of this
function at t equal 2,

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so 4 sine 2 cos 4.

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And here, the key
was that again, this

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interval of integration
from zero minus to infinity

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and clearly it includes
the value at which delta

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function is non-zero.

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So for the last one, we return
to our first integral except

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that now we are changing
the bounds of integration

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to zero plus to infinity.

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So now, if I do representation
of the delta function

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that we're dealing with,
so delta centered at 1,

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and the interval
of integration, we

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have an open interval
now that does not

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include the value at which
delta is non-equal to zero.

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So everywhere this
function would just

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be assigned its value at--
it'd just be multiplied

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by the function that is zero.

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And so basically, it's like
multiplying this function

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by zero, and it
just gives us zero.

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It's like the delta fell off
of our interval of integration,

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so we're just left
with a zero function.

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So let's move to
the second part.

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The second part asked us to
find a generalized derivative

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to f of t equals 3u of
t minus 2u of t minus 1.

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So just to remind you
here of what u of t's are,

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just want to sketch
a few things.

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So first, u of t is
just the step function

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that would be zero
everywhere and would

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take the value 1 for
t larger than zero.

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So this first part
here would just

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be non-zero for t
larger than zero.

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Instead of being
assigned value 1,

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it's just the assigned
value 3 because we're

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multiplying the u of t by 3.

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So this first part
would look like this.

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The second part here would
be u shifted by minus 1,

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which means that u is zero
everywhere for t less than 1.

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So we would have a
zero function here.

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So let me just do dots, but
it should be on the same axis.

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And 1 for t larger than 1.

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But here, we're multiplying
it by factor minus 2.

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So really, what we have
is another u function that

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is shifted down to minus 2.

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So the sum of these
two contributions

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is zero for t negative,
takes the value

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3 for t between zero and 1, and
the value 1, 3 minus 2, for t

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larger than 1.

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So clearly here, we have
discontinuities at t

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equals to zero
and t equals to 1.

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So let's just write
down the derivative.

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The generalized derivative
here would lead us first

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to compute the derivatives where
the function is continuous.

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So for minus infinity to
zero, it's a constant,

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derivative would be zero.

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Between zero and
1, it's constant,

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derivative would be zero.

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And from 1 to infinity, it
would also give us zero.

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So we would have a
zero contribution

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from the continuous part of
the function, if you wish.

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But we still need to account
for the discontinuities.

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So at zero, we have a
jump from zero to 3.

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And that we learned
can be written down

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as a delta function of
magnitude 3 centered at zero.

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After that, we
have another jump.

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Now, it's from 3 to 1.

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So it's a jump of minus 2
amplitude centered at 1.

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So here, we can also
do that with the delta,

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but we just need also
to shift it by minus 1

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to show clearly that
the jump occurs at 1

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and multiply this by minus 2 to
show the amplitude of the jump

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down.

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So if we were now to
represent this f prime,

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basically, the
regular part is zero.

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So there's nothing to write down
except just a zero function.

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And these discontinuities
that I'm just

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going to represent on the graph
would be the delta function

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centered at zero magnitude 3
and delta function centered

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around 1 of magnitude minus 2.

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And the rest would just
be the zero function.

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So that would be f prime.

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f of t.

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So for the second one
that we were given,

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it's a function
f of t that takes

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the value t squared for t
negative and exponential minus

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t for t positive.

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So a quick sketch here
tells that this function

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looks like this, an exponential
minus t for t positive taking

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a value here 1.

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So clearly, there is a
jump here, discontinuity.

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So how do we go about computing
this generalized derivative?

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So let's look again at
the continuous parts.

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So from minus infinity
to zero, we're

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dealing with just the t squared.

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So the derivative is just 2t.

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For t between zero
and infinity, we're

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just dealing with
exponential minus t.

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So that's minus
exponential minus t.

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But we need to account
for the discontinuity

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the jump of
amplitude, 1 at zero.

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So as we saw before,
this can just

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be modeled with
a delta function.

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And if we were to
represent this function,

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then we would just need to add
the delta function of magnitude

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1 here and then just sketch
2t, for example, and then

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minus exponential of minus 2t.

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That would give us
something like that.

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And that ends the
problem for today.

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And the key point here
were just to learn

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how to manipulate
the step function

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and how to use the
delta function when

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you compute your integrals.

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And be careful with the
bounds of integration.