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PROFESSOR: Welcome back.

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So in this session, we're
going to look at unit step

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and impulse responses.

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So in this question, we ask
you to find the unit impulse

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response to these two equations,
x dot plus 2x equals f of t,

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and 2 x dot dot plus 27 x
dot-- oops, it should be a 7--

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plus 7 x dot plus
3x equals f of t.

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In the second part, you're asked
to find the unit step response

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for the first equation.

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So here, the key
points are really

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to remember what do we mean
by unit impulse and unit step

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response.

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Which initial condition
correspond to these responses?

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And what functions of f of t
do you choose in each case?

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So why don't you pause the video
and work through this problem?

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And I'll be right back.

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Welcome back.

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So let's look at the equation
A. So the unit impulse response

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is simply-- I'm going
to write this down,

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unit impulse response--
is simply the solution

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to the following problem, to
our differential equation,

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x dot plus 2x that we're
given, with the forcing

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in a delta function of magnitude
1 with rest initial conditions,

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which means in this case,
that x of 0 is equal to 0.

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x of 0 minus is equal to 0.

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So how do we go about
solving this equation?

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So here, first of all, let's
look at what this means.

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This equation could be
modeling, for example,

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the quantity of radioactive
chemical in a tank.

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And delta(t) here means that we
are giving a high disturbance

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on a very short
time on this system.

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So for example,
dumping a huge amount

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of chemical in that tank, and
then, just letting the system

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evolve after dumping all
that chemical in a tank.

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So really, what we could
see is that the delta

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can be identified to x
dot, which is the highest

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derivative in the system.

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And that would then
mean that we have

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a jump in the second
highest derivative, which

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would be in x.

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So delta t introduces
a jump in x.

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And what do I mean by
that is that we had rest

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initial conditions initially.

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Now, we have to go into a jump
where the quantity in the tank

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was raised to 1,
which is basically

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the magnitude on the
right-hand side and the fact

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that we have no
coefficients for the highest

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derivative in the system.

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So basically, we can
solve this equation

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by solving the equivalent
problem x dot plus 2x equals

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to 0, which is just
the homogeneous part,

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with our new initial conditions,
x of 0 minus equals to 1.

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And so here,
clearly from before,

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we saw that the solution would
be a decaying exponential.

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We would have a constant
of integration here, which

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would be just 1 given our
new initial condition.

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So this would be valid
for t larger than 0.

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And given that we started from
rest initial conditions prior

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to this perturbance
introduced by the delta,

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we could also write this
with a u of t step function

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if we were to consider
t in R. So if you

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want to look at how the solution
looks like, we would have

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basically rest
initial conditions,

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and then, a jump
introduced by our forcing,

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and after that, just a
decaying exponential,

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and a jump here to a new
initial condition of 1.

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So let's move to
the second problem.

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Now, we're dealing with a
second-order differential

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equation.

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2 x dot dot plus 7 x dot
plus 3x equals f of t.

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And we are asked again
to seek the unit impulse

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response to this problem.

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So the unit impulse response
is simply the solution of 2

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x dot dot plus 7 x dot
plus 3x equals-- Again,

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the unit impulse
means that we're just

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kicking the system with an
impulse or delta function.

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And we're starting from
rest initial conditions.

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So here, let's just look at
what could this be modeling.

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So this could be just
basically Newton's Second Law

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with acceleration and forces
applied to the system.

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For example, it could
be a mass hanged

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on a spring with certain
damping due to this term.

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And this delta here would
have basically the units

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corresponding to a force that
would be felt by the highest

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order of term in the integral
which could correspond

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to the acceleration multiplied,
for example, by a mass of 2.

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So really, the delta
can be identified

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to the highest-order
derivative, which

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means that we would have
a discontinuity from 0

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to 1 on the derivative one
degree lower, so in x dot.

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And so the delta introduces
the discontinuity

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on the second
highest derivative.

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And what I mean
by that is that we

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have a jump from our
original initial conditions

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of x dot 0 minus equals to
0 to new initial conditions,

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x dot of 0 minus
equals one half.

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Where did this one
half come from?

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It came from the fact
that I have a factor of 2

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in front of the
highest-order derivative.

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So if I'm identifying
this with delta,

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then integration of
this term would give me

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an x dot that corresponds to
the jump from 0 to 1 over 2,

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so one half.

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That's where it comes from.

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And each time that you would
have a coefficient in front

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of your highest order
derivative associated

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with the delta on the right,
you would have a jump of 1

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over that coefficient.

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So from this point,
what are we solving?

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What do we need to solve to
get the unit impulse response?

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It's equivalent to solving our
new system where we can then

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get rid of the delta with
our new initial conditions,

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dot equals one half.

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And of course, here,
I did mentioned

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by the initial condition on
x 0 minus 1, because we need

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two initial conditions for
the second order derivative,

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would still be 0, as the
discontinuity would not be

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felt by the function x itself.

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So we can go ahead and
solve this problem.

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So we use the characteristic
polynomial as we did before.

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This characteristic polynomial
would have a discriminant

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of 25, which gives us simple
roots that we can compute,

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so minus 7 plus or
minus the square root

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of the discriminant over 4.

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So we have two roots, minus
7 minus 5 over 4, minus 12

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over 4, which is minus
3; minus 7 plus 5 over 4,

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which just gives
us minus one half.

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So we can write down the
solution to this problem

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as c_1 exponential the
root that we just found,

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minus 3t, plus c_2
exponential the other root,

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minus one half of t.

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Now, to get c_1, c_2, we
need to take into account

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the initial conditions.

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So the first one tells
us that x at 0 is 0.

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So that will give us c_1
plus c_2 equals to 0,

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so basically, c_2
equals minus c_1.

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And the second initial
condition tells us

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that the derivative is 0.

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So that gives us minus 3c_1
minus one half c_2 equals to 0.

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Here, we have that c_2
is equal to minus c_1.

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So we can just factor
out everything.

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And we end up with 3 minus
one half-- whoops, sorry.

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Here it should be one half, our
new modified initial condition,

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equals to one half.

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And here, you would
have 5/2 equal to 1/2.

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So c_2 equals 1/5
equals minus c_1.

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And so plugging in c_1
and c_2 in this formula

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would give us the
general solution

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with c_2 exponential minus one
half t minus exponential minus

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3t.

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And all of this, remember, we're
solving for t larger than 0

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in our modified system with
our new initial conditions.

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So this is for t larger than 0.

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And again, here, if we wanted
to just write it for t in R,

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then we could just add
the step function u of t

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that would just signify that
we took rest initial conditions

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before x equals 0 minus.

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So that ends the first part.

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So now quickly, for
the second part,

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we're asked to
find the unit step

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response to the first system.

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So the unit step
response is just

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the solution to our original
ODE problem, x dot plus 2x.

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But now, f of t is the step
function, hence, the step

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here, with still rest
initial conditions.

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So really, the step
function is just

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0 everywhere before 0 minus.

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And it takes the
value of 1 after.

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And so basically, we
can just solve again

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for t positive, the modified
differential equation with u

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taking just the value of 1.

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And here, we just
get similar roots.

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So it would be a minus 2t.

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But then, we would have
a constant of integration

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to worry about.

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And a new particular
solution, a lucky guess

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would just be a constant.

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And that would give us the
one half from 1 over 2.

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So then, we just need to seek
the c_1 that would give us

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x of t equals to 0 at 0.

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Because in this
case, we don't need

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to modify the
initial conditions.

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They still remain the same,
rest initial conditions.

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And so we would get here just
c_1 equals minus one half.

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So we have 1 minus
exponential of minus 2t.

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So just here, something
I forgot to mention,

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what could this be modeling?

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So this could be
modeling, if I go back

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to my analogy of the radioactive
chemical in the tank,

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this would be telling me
that after a certain time,

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I start inputting
at a steady rate

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at a constant rate of u of
t, of one per unit of time,

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the amount in the tank
and then looking at how

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the system evolves to that.

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So there's no abrupt change
that introduces a discontinuity.

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So just to conclude, I just
want to sketch the solution

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just so that you see the
difference between the two.

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And here, what we have, again, I
could have introduced my u of t

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here if I want t
in R. So here, we

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have again a solution
that is 0 before.

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There is no discontinuity.

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It's still 0 at 0 minus.

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And then, we have
basically a solution

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that is growing until it
reaches an asymptote of one half

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when this exponential goes to 0.

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And so you can see that it's
a smooth transition because I

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just started
inputting the amount

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of the chemical in the
tank in a non-abrupt way.

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So this concludes this session.

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And the key here was
to really remember

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what do we mean by the
unit step response.

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What type of f of t
are we talking about?

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What initial
conditions do we need?

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And same thing for the
unit impulse response.