PROFESSOR: So in
this recitation,
we're going to look at step and
delta functions, integration
and generalized derivatives.
So the first part, you're
asked to compute the integral
from zero minus to infinity
delta(t) exponential t
squared dt.
The second one is from zero
minus to infinity delta(t-2)
exponential of t
squared sine t cos 2t.
The third one is zero plus to
infinity delta(t) exponential
of squared dt.
So know that it's the same
as a, except that the bounds
of integration changed.
The second part,
we're asked to define
the generalized derivatives
of these two functions,
where u here is just a step
function that you saw before.
So it's 3u(t) minus 2u(t-1).
And the second one is t
squared for t is negative
and exponential of
minus t for t positive.
So why don't you pause the video
and work through this example,
and I'll be right back.
Welcome back.
So let's compute the first one.
Zero minus, infinity, delta(t)
exponential t squared dt.
So just to remind you,
the delta function
is everywhere zero
except at the value zero.
And we represent
it with an arrow.
And the integral of the delta
would be 1 from minus infinity
to plus infinity.
In this integral, we're
integrating from zero
minus to infinity, which means
that the zero is included
in our interval from
zero minus to infinity.
Therefore, this
integral is basically
assigning the value to this
function exponential t squared.
And the value that
it's assigning
to it is the value
it would take at t
equals to zero where
the delta is non-zero.
So really clearly, this is
just exponential of zero.
And it gives us 1.
For the second integral, it goes
from zero minus to infinity,
delta t minus 2 exponential,
a more complicated function,
t square sine t cos 2t.
So now, let's represent
this delta function here.
So this is just our zero axis.
And the delta here
is zero everywhere
except at 2 where we
would represent it, again,
with an arrow at 2 amplitude 1.
So this delta is zero
everywhere except at 2
where it would assign
the value to the function
next to it at the
value t equals to 2.
So really this
integration gives us
just the value of this
function at t equal 2,
so 4 sine 2 cos 4.
And here, the key
was that again, this
interval of integration
from zero minus to infinity
and clearly it includes
the value at which delta
function is non-zero.
So for the last one, we return
to our first integral except
that now we are changing
the bounds of integration
to zero plus to infinity.
So now, if I do representation
of the delta function
that we're dealing with,
so delta centered at 1,
and the interval
of integration, we
have an open interval
now that does not
include the value at which
delta is non-equal to zero.
So everywhere this
function would just
be assigned its value at--
it'd just be multiplied
by the function that is zero.
And so basically, it's like
multiplying this function
by zero, and it
just gives us zero.
It's like the delta fell off
of our interval of integration,
so we're just left
with a zero function.
So let's move to
the second part.
The second part asked us to
find a generalized derivative
to f of t equals 3u of
t minus 2u of t minus 1.
So just to remind you
here of what u of t's are,
just want to sketch
a few things.
So first, u of t is
just the step function
that would be zero
everywhere and would
take the value 1 for
t larger than zero.
So this first part
here would just
be non-zero for t
larger than zero.
Instead of being
assigned value 1,
it's just the assigned
value 3 because we're
multiplying the u of t by 3.
So this first part
would look like this.
The second part here would
be u shifted by minus 1,
which means that u is zero
everywhere for t less than 1.
So we would have a
zero function here.
So let me just do dots, but
it should be on the same axis.
And 1 for t larger than 1.
But here, we're multiplying
it by factor minus 2.
So really, what we have
is another u function that
is shifted down to minus 2.
So the sum of these
two contributions
is zero for t negative,
takes the value
3 for t between zero and 1, and
the value 1, 3 minus 2, for t
larger than 1.
So clearly here, we have
discontinuities at t
equals to zero
and t equals to 1.
So let's just write
down the derivative.
The generalized derivative
here would lead us first
to compute the derivatives where
the function is continuous.
So for minus infinity to
zero, it's a constant,
derivative would be zero.
Between zero and
1, it's constant,
derivative would be zero.
And from 1 to infinity, it
would also give us zero.
So we would have a
zero contribution
from the continuous part of
the function, if you wish.
But we still need to account
for the discontinuities.
So at zero, we have a
jump from zero to 3.
And that we learned
can be written down
as a delta function of
magnitude 3 centered at zero.
After that, we
have another jump.
Now, it's from 3 to 1.
So it's a jump of minus 2
amplitude centered at 1.
So here, we can also
do that with the delta,
but we just need also
to shift it by minus 1
to show clearly that
the jump occurs at 1
and multiply this by minus 2 to
show the amplitude of the jump
down.
So if we were now to
represent this f prime,
basically, the
regular part is zero.
So there's nothing to write down
except just a zero function.
And these discontinuities
that I'm just
going to represent on the graph
would be the delta function
centered at zero magnitude 3
and delta function centered
around 1 of magnitude minus 2.
And the rest would just
be the zero function.
So that would be f prime.
f of t.
So for the second one
that we were given,
it's a function
f of t that takes
the value t squared for t
negative and exponential minus
t for t positive.
So a quick sketch here
tells that this function
looks like this, an exponential
minus t for t positive taking
a value here 1.
So clearly, there is a
jump here, discontinuity.
So how do we go about computing
this generalized derivative?
So let's look again at
the continuous parts.
So from minus infinity
to zero, we're
dealing with just the t squared.
So the derivative is just 2t.
For t between zero
and infinity, we're
just dealing with
exponential minus t.
So that's minus
exponential minus t.
But we need to account
for the discontinuity
the jump of
amplitude, 1 at zero.
So as we saw before,
this can just
be modeled with
a delta function.
And if we were to
represent this function,
then we would just need to add
the delta function of magnitude
1 here and then just sketch
2t, for example, and then
minus exponential of minus 2t.
That would give us
something like that.
And that ends the
problem for today.
And the key point here
were just to learn
how to manipulate
the step function
and how to use the
delta function when
you compute your integrals.
And be careful with the
bounds of integration.
