WEBVTT
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Today we are going to do a last
serious topic on the Laplace
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transform, the last topic for
which I don't have to make
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frequent and profuse apologies.
One of the things the Laplace
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transform does very well and one
of the reasons why people like
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it, engineers like it,
is that it handles functions
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with jump discontinuities very
nicely.
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Now, the OR function with a
jump discontinuity is-- Purple.
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Is a function called the unit
step function.
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I will draw a graph of it.
Even the graph is
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controversial,
but everyone is agreed that it
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zero here and one there.
What people are not agreed upon
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is its value at zero.
And some people make it zero,
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some people make it one,
and some equivocate like me.
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I will leave it undefined.
It is u of t.
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It is called the unit step.
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Because that is what it is.
And let's say we will leave u
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of zero undefined.
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If that makes you unhappy,
get over it.
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Of course, we don't always want
the jump to be at zero.
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Sometimes we will want to jump
in another place.
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If I want the function to jump,
let's say at the point a
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instead of jumping at zero,
I am going to start doing what
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everybody does.
You put in the vertical lines,
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even though I have no meaning,
whatever, but it makes the
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graph look more connected and a
little easier to read.
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So that function I will call u
sub a is the function
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which jumps at the point a.
How shall I give its
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definition?
Well, you can see it is just
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the translation by a of the unit
step function.
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So that is the way to write it,
u of t minus a.
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Now I am not done.
There is a unit box function,
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which we will draw in general
terms like this.
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It gets to a,
then it jumps up to one,
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falls down again at b and
continues onto zero.
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This happens between a and b.
And the value to which it
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arises is one.
I will call this the unit box.
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It is a function of t,
a very simple one but an
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important one.
And what would be the formula
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for the unit box function?
Well, in general,
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almost all of these functions,
as you will see when you use
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jump discontinuity,
the idea is to write them all
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cleverly using nothing but u of.
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Because it is that will have
the Laplace transformer.
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The way to write this is (u)ab.
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And, if you like,
you can treat this as the
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definition of it.
Let's make it a definition.
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Okay, three lines.
Or, better yet,
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a colon and two lines.
I am defining this to be,
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what would it be?
Make the unit step at a step up
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at a, but then I would continue
at one all the time.
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I should, therefore,
step down at b.
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Now, the way you step down is
just by taking the negative of
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the unit step function.
I step down at b by subtracting
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u sub b of t .
In other words,
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it is u of t minus a minus u of
t minus b.
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And now I have expressed it
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entirely in terms of the unit
step function.
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That will be convenient when I
want take the Laplace transform.
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What is so good about these
things?
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Well, these functions,
when you use them in
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multiplications,
they transform other functions
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in a nice way.
Not transform.
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That is not the right word.
They operate on them.
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They turn them into other
strange creatures,
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and it might be these strange
creatures that you are
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interested in.
Let me just draw you a picture.
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That will be good enough.
Suppose we have some function
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like that, f of t,
what would the function u sub
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ab,
I will put in the variable,
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t times f of t,
what function would that be?
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I am just going to draw its
graph.
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What would its graph be?
Well, in between a and b this
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function (u)ab of t has the
value one.
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All I am doing is multiplying f
of t by one.
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In short, I am not doing
anything to it at all.
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Outside of that interval,
(u)ab has the value zero,
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so that zero times f of t makes
zero.
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And, therefore,
outside of this it is zero.
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The effect of multiplying an
arbitrary function by this unit
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box function is,
you wipe away all of its graph
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except the part between a and b.
Now, that is a very useful
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thing to be able to do.
Well, that is enough of that.
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Now, let's get into the main
topic.
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That is just preliminary.
I will be using these functions
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all during the period,
but the real topic is the
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following.
Let's calculate the Laplace
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transform of the unit step
function.
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Well, this is no very big deal.
It is the integral from zero to
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infinity e to the minus s t
times y of t, dt.
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But, look, when t is bigger
than zero,
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this has the value one.
So it is the same of the
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Laplace transform of one.
In other words,
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it is one over s for
positive values of s.
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Or, to make it very clear,
the Laplace transform of one is
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exactly the same thing.
As you see, the Laplace
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transform really is not
interested in what happens when
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t is less than zero
because that is not part of the
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domain of integration,
the interval of integration.
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That is fine.
They both have Laplace
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transform of one over s.
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What is the big deal?
The big deal is,
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what is the inverse Laplace
transform of one over s?
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Will the real function please
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stand up?
Which of these two should I
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pick?
Up to now in the course we have
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been picking one just because I
never made a fuss over it and
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one was good enough.
For today one is no longer
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going to be good enough.
And we have to first
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investigate the thing in a
slightly more theoretical way
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because this problem,
I have illustrated it on the
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inverse Laplace transform of one
over x,
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but it occurs for any inverse
Laplace transform.
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Suppose I have,
in other words,
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that a function f of t has as
its Laplace transform capital F
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of s?
And now, I ask what the inverse
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Laplace transform of capital F
of s is.
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Well, of course you want to
write f of t.
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But the same thing happens.
I will draw you a picture.
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Suppose, in other words,
that here is our function f of
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t.
Well, one answer certainly is f
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of t.
That is okay.
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That is the answer we have been
using up until now.
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But, you see,
I can complete this function in
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many other ways.
Suppose I haven't told you what
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it was for s less than zero.
Any of these possibilities all
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will produce the same Laplace
transform.
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In fact, I can even make it
this.
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That is okay.
Each of these,
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f of t with any one of
these tails, all have the same
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Laplace transform.
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Because the Laplace transform,
remember the definition,
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integral zero to infinity,
e to the negative s t,
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f of t, dt
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because the Laplace transform
does not care what the function
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was doing for negative values of
t.
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Now, if we have to have a
unique answer --
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And most of the time you don't
because, in general,
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the Laplace transform is only
used for problems for future
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time.
That is the way the engineers
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and physicists and other people
who use it habitually think of
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it.
If your problem is starting now
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and going on into the future and
you don't have to know anything
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about the past,
that is a Laplace transform
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problem.
If you also have to know about
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the past, then it is a Fourier
transform problem.
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That is beyond the scope of
this course, you will never hear
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that word again,
but that is the difference.
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We are starting at time zero
and going forward.
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All right.
It does not care what f of t
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was doing for negative values of
t.
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And that gives us a problem
when we try to make the Laplace
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transform unique.
Now, how will I make it unique?
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Well, there is a simple way of
doing it.
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Let's agree that wherever it
makes a difference,
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and most of the time it
doesn't, but today it will,
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whenever it makes a difference
we will declare,
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we will by brute force make our
function zero for negative
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values of t.
That makes it unique.
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I am going to say that to make
it unique, now,
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how do I make f of t zero
for negative values
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of t?
The answer is multiply it by
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the unit step function.
That leaves it what it was.
00:11:32.000 --> 00:11:38.000
It multiplies it by one for
positive values but multiplies
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it by zero for negative values.
The answer is going to be u of
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t times f of t.
That will be the function that
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will look just that way that I
drew, but I will draw it once
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more.
It is the function that looks
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like this.
And when I do this,
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it makes the inverse Laplace
transform unique.
00:12:04.000 --> 00:12:10.000
Out of all the possible tails I
might have put on f of t,
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it picks the least
interesting one,
00:12:11.000 --> 00:12:17.000
the tail zero.
00:12:23.000 --> 00:12:29.000
That is a start.
But what we have to do now is
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--
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What I want is a formula.
What we are going to need is,
00:12:38.000 --> 00:12:44.000
as you see right even in the
beginning, if for example,
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if I want to calculate the
Laplace transform of this,
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what I would like to have is a
nice Laplace transform for the
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translate.
If you translate a function,
00:12:52.000 --> 00:12:58.000
how does that effect this
Laplace transform?
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In other words,
the formula I am looking for is
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--
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I want to express the Laplace
transform of f of t minus a.
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In other words,
00:13:15.000 --> 00:13:21.000
the function translated,
let's say a is positive,
00:13:19.000 --> 00:13:25.000
so I translate it to the right
along the t axis by the distance
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a.
I want a formula for this in
00:13:27.000 --> 00:13:33.000
terms of the Laplace transform
of the function I started with.
00:13:34.000 --> 00:13:40.000
Now, my first task is to
convince you that,
00:13:37.000 --> 00:13:43.000
though this would be very
useful and interesting,
00:13:40.000 --> 00:13:46.000
there cannot possibly be such a
formula.
00:13:43.000 --> 00:13:49.000
There is no such formula.
00:13:56.000 --> 00:14:02.000
Why not?
Well, I think I will explain it
00:13:59.000 --> 00:14:05.000
over there since there is a
little piece of board I did not
00:14:05.000 --> 00:14:11.000
use.
Waste not want not.
00:14:07.000 --> 00:14:13.000
Why can't there be such a
formula?
00:14:10.000 --> 00:14:16.000
What is it we are looking for?
Let's take a nice average
00:14:15.000 --> 00:14:21.000
function f of t.
It has a Laplace transform.
00:14:20.000 --> 00:14:26.000
And now I am going to translate
it.
00:14:23.000 --> 00:14:29.000
Let's say this is the point
negative a.
00:14:29.000 --> 00:14:35.000
And so the corresponding point
positive a will be around here.
00:14:34.000 --> 00:14:40.000
I am going to translate it to
the right by a.
00:14:38.000 --> 00:14:44.000
What is it going to look like?
Well, then it is going to start
00:14:43.000 --> 00:14:49.000
here and is going to look like
this dashy thing.
00:14:48.000 --> 00:14:54.000
That is f of t minus a.
00:14:51.000 --> 00:14:57.000
That is not too bad a picture.
It will do.
00:14:54.000 --> 00:15:00.000
I just took that curve and
shoved it to the right by a.
00:15:01.000 --> 00:15:07.000
Now, why is it impossible to
express the Laplace transform of
00:15:09.000 --> 00:15:15.000
the dashed line in terms of the
Laplace transform of the solid
00:15:18.000 --> 00:15:24.000
line?
The answer is this piece.
00:15:23.000 --> 00:15:29.000
I will write it this way.
The trouble is,
00:15:28.000 --> 00:15:34.000
this piece is not used for the
Laplace transform of f of t.
00:15:37.000 --> 00:15:43.000
Why isn't it used?
00:15:41.000 --> 00:15:47.000
Well, because it occurs to the
left of the vertical axis.
00:15:44.000 --> 00:15:50.000
It occurs for negative values
of t.
00:15:47.000 --> 00:15:53.000
And the Laplace transform of f
of t simply does not
00:15:51.000 --> 00:15:57.000
care what f of t was
doing to the left of that line,
00:15:55.000 --> 00:16:01.000
for negative values of t.
It does not enter into the
00:15:58.000 --> 00:16:04.000
integral.
It was not used when I
00:16:01.000 --> 00:16:07.000
calculated this piece of the
curve.
00:16:04.000 --> 00:16:10.000
It was not used when I
calculated the Laplace transform
00:16:07.000 --> 00:16:13.000
of f of t.
On the other hand,
00:16:10.000 --> 00:16:16.000
it is going to be needed.
It occurs here,
00:16:13.000 --> 00:16:19.000
after I shift it to the right.
It is going to be needed for
00:16:17.000 --> 00:16:23.000
the Laplace transform of f of t
minus a,
00:16:21.000 --> 00:16:27.000
because I will have to start
the integration here,
00:16:25.000 --> 00:16:31.000
and I will have to know what
that is.
00:16:34.000 --> 00:16:40.000
In other words,
when I took the Laplace
00:16:37.000 --> 00:16:43.000
transform, I automatically lost
all information about the
00:16:41.000 --> 00:16:47.000
function for negative values of
t.
00:16:44.000 --> 00:16:50.000
If I am later going to want
some of that information for
00:16:48.000 --> 00:16:54.000
calculating this,
I won't have it and,
00:16:51.000 --> 00:16:57.000
therefore, there cannot be a
formula expressing one in terms
00:16:56.000 --> 00:17:02.000
of the other.
Now, of course,
00:16:59.000 --> 00:17:05.000
that cannot be the answer,
otherwise I would not have
00:17:02.000 --> 00:17:08.000
raised your expectations merely
to dash them.
00:17:05.000 --> 00:17:11.000
I don't want to do that,
of course.
00:17:07.000 --> 00:17:13.000
There is a formula,
of course.
00:17:09.000 --> 00:17:15.000
It is just, I want to emphasize
that you must write it my way
00:17:13.000 --> 00:17:19.000
because, if you write it any
other way, you are going to get
00:17:17.000 --> 00:17:23.000
into the deepest trouble.
The formula is-- the good
00:17:21.000 --> 00:17:27.000
formula, the right formula --
00:17:30.000 --> 00:17:36.000
-- accepts the given.
It says look,
00:17:32.000 --> 00:17:38.000
we have lost that pink part of
it.
00:17:35.000 --> 00:17:41.000
Therefore, I can never recover
that.
00:17:37.000 --> 00:17:43.000
Therefore, I won't ask for it.
The translation formula I will
00:17:42.000 --> 00:17:48.000
ask for is not one for the
Laplace transform of f of t
00:17:46.000 --> 00:17:52.000
minus a,
but rather for the Laplace
00:17:50.000 --> 00:17:56.000
transform of this thing where I
have wiped away that pink part
00:17:54.000 --> 00:18:00.000
from the translated function.
In other words,
00:17:59.000 --> 00:18:05.000
the function I am talking about
now is the formula for,
00:18:04.000 --> 00:18:10.000
I will put it over here to show
you the function what we are
00:18:10.000 --> 00:18:16.000
talking about.
It is the function f of,
00:18:13.000 --> 00:18:19.000
well, in terms of the pink
function it is,
00:18:17.000 --> 00:18:23.000
I will have to reproduce some
of that picture.
00:18:21.000 --> 00:18:27.000
There is f of t.
f of t minus a,
00:18:26.000 --> 00:18:32.000
then, looked like this.
And so the function I am
00:18:32.000 --> 00:18:38.000
looking for is,
this is the thing translated,
00:18:36.000 --> 00:18:42.000
but when I get down to the
corresponding,
00:18:39.000 --> 00:18:45.000
this is the point that
corresponds to that one,
00:18:43.000 --> 00:18:49.000
I wipe it away and just go with
zero after that.
00:18:48.000 --> 00:18:54.000
So this is u of t minus a times
f of t minus a times f of t
00:18:53.000 --> 00:18:59.000
minus a.
What is this Laplace transform?
00:19:00.000 --> 00:19:06.000
Now that does have a simple
answer.
00:19:04.000 --> 00:19:10.000
The answer is it is e to the
minus as,
00:19:10.000 --> 00:19:16.000
a funny exponential,
times the Laplace transform of
00:19:16.000 --> 00:19:22.000
the original function.
Now, this formula occurs in two
00:19:23.000 --> 00:19:29.000
forms.
This one is not too bad
00:19:26.000 --> 00:19:32.000
looking.
The trouble is,
00:19:29.000 --> 00:19:35.000
when you want to solve
differential equations you are
00:19:33.000 --> 00:19:39.000
going to be extremely puzzled
because the function that you
00:19:37.000 --> 00:19:43.000
will have to take to do the
calculation on will not be given
00:19:41.000 --> 00:19:47.000
to you in the form f of t minus
a.
00:19:44.000 --> 00:19:50.000
It will look sine t
or t squared or some
00:19:48.000 --> 00:19:54.000
polynomial in t.
It will not be written as t
00:19:50.000 --> 00:19:56.000
minus a.
What do you do?
00:19:53.000 --> 00:19:59.000
If your function does not look
like that but instead,
00:19:56.000 --> 00:20:02.000
in terms of symbols looks like
this, you can still use the
00:20:00.000 --> 00:20:06.000
formula.
Just a trivial change of
00:20:03.000 --> 00:20:09.000
variable means that you can
write it instead.
00:20:06.000 --> 00:20:12.000
Now, this is one place,
there is no way of writing the
00:20:10.000 --> 00:20:16.000
answer in terms of capital F of
s.
00:20:13.000 --> 00:20:19.000
This is one of those cases
where this notation is just no
00:20:17.000 --> 00:20:23.000
good anymore.
I am going to have to write it
00:20:20.000 --> 00:20:26.000
using the L notation.
The Laplace transform of f of,
00:20:24.000 --> 00:20:30.000
and wherever you see a t,
you should write t plus a.
00:20:28.000 --> 00:20:34.000
Basically, this is the same
00:20:31.000 --> 00:20:37.000
formula as that one.
But I will have to stand on my
00:20:35.000 --> 00:20:41.000
head for one minute to try to
convince you of it.
00:20:38.000 --> 00:20:44.000
I won't do that now.
I would like you just to take a
00:20:42.000 --> 00:20:48.000
look at the formula.
You should know what it is
00:20:45.000 --> 00:20:51.000
called.
There are a certain number of
00:20:48.000 --> 00:20:54.000
idiots who call this the
exponential shift formula
00:20:51.000 --> 00:20:57.000
because on the right side you
multiply by an exponential,
00:20:55.000 --> 00:21:01.000
and that corresponds to
shifting the function.
00:21:00.000 --> 00:21:06.000
Unfortunately,
we have preempted that.
00:21:02.000 --> 00:21:08.000
We are not going to call it
this.
00:21:05.000 --> 00:21:11.000
I will call it what your book
calls it.
00:21:08.000 --> 00:21:14.000
The difficulty is there is no
universal designation for this
00:21:13.000 --> 00:21:19.000
formula, important as it is.
However, your book calls this
00:21:17.000 --> 00:21:23.000
t-axis translation formula.
Translation because I am
00:21:21.000 --> 00:21:27.000
translating on the t-axis.
And that is what I do to the
00:21:25.000 --> 00:21:31.000
function, essentially.
And this tells me what its new
00:21:30.000 --> 00:21:36.000
Laplace transform is.
00:21:38.000 --> 00:21:44.000
The other formula,
remember it?
00:21:40.000 --> 00:21:46.000
The exponential shift formula,
the shift or the translation
00:21:44.000 --> 00:21:50.000
occurs on the s-axis.
In other words,
00:21:47.000 --> 00:21:53.000
the formula said that F of s
minus a,
00:21:51.000 --> 00:21:57.000
you do the translation in s
variable corresponded to
00:21:55.000 --> 00:22:01.000
multiplying this by e
to the a t.
00:22:00.000 --> 00:22:06.000
In other words,
the formulas are sort of dual
00:22:02.000 --> 00:22:08.000
to each other.
This guy translates on the left
00:22:05.000 --> 00:22:11.000
side and multiplies by the
exponential on the right.
00:22:08.000 --> 00:22:14.000
The formula that you know
translates on the right and
00:22:12.000 --> 00:22:18.000
multiplies by the exponential on
the left.
00:22:14.000 --> 00:22:20.000
What are we going to calculate?
I am trying to calculate,
00:22:18.000 --> 00:22:24.000
so I am trying to prove this
first formula.
00:22:20.000 --> 00:22:26.000
The second one will be an easy
consequence.
00:22:23.000 --> 00:22:29.000
I am trying to calculate the
Laplace transform of that thing.
00:22:27.000 --> 00:22:33.000
What is it?
Well, it is the integral from
00:22:31.000 --> 00:22:37.000
zero to infinity of e to the
minus st times u of t minus a,
00:22:35.000 --> 00:22:41.000
f of t minus a times dt.
00:22:42.000 --> 00:22:48.000
That is the formula for it.
But I am trying to express it
00:22:46.000 --> 00:22:52.000
in terms of the Laplace
transform of f itself.
00:22:49.000 --> 00:22:55.000
Now, it is trying to be the
Laplace transform of f.
00:22:54.000 --> 00:23:00.000
The problem is that here,
a (t minus a) occurs,
00:22:58.000 --> 00:23:04.000
which I don't like.
I would like that to be just a
00:23:03.000 --> 00:23:09.000
t.
Now, in order not to confuse
00:23:05.000 --> 00:23:11.000
you, and this is what confused
everybody, I will set t1 equal
00:23:09.000 --> 00:23:15.000
to t minus a.
I will change the variable.
00:23:13.000 --> 00:23:19.000
This is called changing the
variable in a definite integral.
00:23:17.000 --> 00:23:23.000
How do you change the variable
in a definite integral?
00:23:21.000 --> 00:23:27.000
You do it.
Well, let's leave the limits
00:23:24.000 --> 00:23:30.000
for the moment.
e to the minus s
00:23:27.000 --> 00:23:33.000
times --
Now, t, remember you can change
00:23:31.000 --> 00:23:37.000
the variable forwards,
direct substitution,
00:23:33.000 --> 00:23:39.000
but now I have to use the
inverse substitution.
00:23:37.000 --> 00:23:43.000
It's trivial,
but t is equal to t1 plus a.
00:23:40.000 --> 00:23:46.000
To change this I must
00:23:42.000 --> 00:23:48.000
substitute backwards and make
that t1 plus a.
00:23:45.000 --> 00:23:51.000
How about the rest of it?
Well, this becomes u of t1.
00:23:49.000 --> 00:23:55.000
This is f of t1.
00:23:51.000 --> 00:23:57.000
I have to change the dt,
too, but that's no problem.
00:23:55.000 --> 00:24:01.000
dt1 equals dt
because a is a constant.
00:24:00.000 --> 00:24:06.000
That is dt1.
And the last step is to put in
00:24:03.000 --> 00:24:09.000
the limits.
Now, when t is equal to zero,
00:24:06.000 --> 00:24:12.000
t1 has the value
negative a.
00:24:10.000 --> 00:24:16.000
So this has to be negative a
when t is infinity.
00:24:13.000 --> 00:24:19.000
Infinity minus a is still
infinity, so that is still
00:24:17.000 --> 00:24:23.000
infinity.
In other words,
00:24:19.000 --> 00:24:25.000
this changes to that.
These two things,
00:24:22.000 --> 00:24:28.000
whatever they are,
they have the same value.
00:24:25.000 --> 00:24:31.000
All I have done is changed the
variable.
00:24:30.000 --> 00:24:36.000
Make it change a variable.
But now, of course,
00:24:32.000 --> 00:24:38.000
I want to make this look
better.
00:24:34.000 --> 00:24:40.000
How am I going to do that?
Well, first multiply out the
00:24:37.000 --> 00:24:43.000
exponential and then you get a
factor e to the minus s(t1).
00:24:40.000 --> 00:24:46.000
That is good.
00:24:42.000 --> 00:24:48.000
That goes with this guy.
Now I get a factor e to the
00:24:44.000 --> 00:24:50.000
minus s times a from
the exponential law.
00:24:47.000 --> 00:24:53.000
But that does not have anything
to do with the integral.
00:24:51.000 --> 00:24:57.000
It is a constant as far as the
integral is concerned because it
00:24:54.000 --> 00:25:00.000
doesn't involve t1.
And, therefore,
00:24:56.000 --> 00:25:02.000
I can pull it outside of the
integral sign.
00:25:00.000 --> 00:25:06.000
And write that e to the minus s
times a.
00:25:05.000 --> 00:25:11.000
Let's write it the other way.
Times the integral of what?
00:25:10.000 --> 00:25:16.000
Well, e to the negative st1.
00:25:14.000 --> 00:25:20.000
Now, u of t1,
f of t1 times dt1.
00:25:18.000 --> 00:25:24.000
Still integrated from minus a
00:25:22.000 --> 00:25:28.000
to infinity.
And now the final step.
00:25:29.000 --> 00:25:35.000
This u of t1 is zero
for negative values of t.
00:25:34.000 --> 00:25:40.000
And, therefore,
it is equal to one for positive
00:25:38.000 --> 00:25:44.000
values of t.
It is equal to zero for
00:25:42.000 --> 00:25:48.000
negative values of t,
which means I can forget about
00:25:47.000 --> 00:25:53.000
the part of the integral that
goes from negative a to zero.
00:25:52.000 --> 00:25:58.000
I better rewrite this.
00:25:56.000 --> 00:26:02.000
Okay, leave that.
In other words,
00:26:00.000 --> 00:26:06.000
this is equal to e to the minus
as times the
00:26:04.000 --> 00:26:10.000
integral from zero to infinity
of e to the minus s --
00:26:08.000 --> 00:26:14.000
Let me do the shifty part now.
00:26:17.000 --> 00:26:23.000
And this is since u of t1 is
equal to zero for
00:26:23.000 --> 00:26:29.000
t1 less than zero.
That is why I can replace this
00:26:29.000 --> 00:26:35.000
with zero.
Because from negative a to
00:26:33.000 --> 00:26:39.000
zero, nothing is happening.
The integrand is zero.
00:26:37.000 --> 00:26:43.000
And why can I get rid of it
after that?
00:26:40.000 --> 00:26:46.000
Well, because it is one after
that.
00:26:43.000 --> 00:26:49.000
And what is this thing?
This is the Laplace transform.
00:26:47.000 --> 00:26:53.000
No, it is not the Laplace
transform they said.
00:26:51.000 --> 00:26:57.000
Because you had t1 there,
not t.
00:26:53.000 --> 00:26:59.000
It is the Laplace transform
because this is a dummy
00:26:58.000 --> 00:27:04.000
variable.
The t1 is integrated out.
00:27:01.000 --> 00:27:07.000
It is a dummy variable.
It doesn't matter what you call
00:27:05.000 --> 00:27:11.000
it.
It is still the Laplace
00:27:07.000 --> 00:27:13.000
transform if I make that wiggly
t or t star or tau or u.
00:27:12.000 --> 00:27:18.000
I can call it anything I want
and it is still the Laplace
00:27:16.000 --> 00:27:22.000
transform of f of t.
00:27:24.000 --> 00:27:30.000
What is the answer?
That is e to the negative as
00:27:27.000 --> 00:27:33.000
times the Laplace transform of
the function f.
00:27:33.000 --> 00:27:39.000
That is what I promised you in
that formula.
00:27:35.000 --> 00:27:41.000
Now, how about the other
formula?
00:27:37.000 --> 00:27:43.000
Well, let's look at that
quickly.
00:27:39.000 --> 00:27:45.000
That is, as I say,
just sleight of hand.
00:27:42.000 --> 00:27:48.000
But since that is the formula
you will be using at least half
00:27:46.000 --> 00:27:52.000
the time you better learn it.
This little sleight of hand is
00:27:50.000 --> 00:27:56.000
also reproduced in one page of
notes that I give you,
00:27:53.000 --> 00:27:59.000
but maybe you will find it easy
to understand if I talk it out
00:27:57.000 --> 00:28:03.000
loud.
The problem now is for the
00:28:00.000 --> 00:28:06.000
second formula.
I am going to have to recopy
00:28:03.000 --> 00:28:09.000
out the first one in order to
make the argument in a form in
00:28:07.000 --> 00:28:13.000
which you will understand it,
I hope.
00:28:09.000 --> 00:28:15.000
This goes to e to the minus as
F s,
00:28:13.000 --> 00:28:19.000
except I am now going to write
that not in F of s;
00:28:16.000 --> 00:28:22.000
since I will not be able to
write the second formula using F
00:28:20.000 --> 00:28:26.000
of s, I am not going to write
the first formula that way
00:28:24.000 --> 00:28:30.000
either.
I will write it as the Laplace
00:28:26.000 --> 00:28:32.000
transform with f of t.
00:28:30.000 --> 00:28:36.000
Now, formally if somebody says,
okay, how do I calculate the
00:28:34.000 --> 00:28:40.000
Laplace transform of this thing?
I say put down this.
00:28:39.000 --> 00:28:45.000
Well, that has no t in it.
It doesn't have the f in it
00:28:43.000 --> 00:28:49.000
either.
Then write this.
00:28:45.000 --> 00:28:51.000
What formula did I do?
I looked at that and changed t
00:28:50.000 --> 00:28:56.000
minus a to t. Now,
how did I change t minus a?
00:28:55.000 --> 00:29:01.000
The way to say it is
00:28:58.000 --> 00:29:04.000
I changed t.
Because the t is always there.
00:29:02.000 --> 00:29:08.000
t to t plus a.
You get this by replace t by t
00:29:08.000 --> 00:29:14.000
plus a to get the right-hand
side.
00:29:16.000 --> 00:29:22.000
I replace this t by t plus a,
and that turns this
00:29:20.000 --> 00:29:26.000
into f of t.
And that is the f of t
00:29:22.000 --> 00:29:28.000
that went in there.
That is the universal rule for
00:29:26.000 --> 00:29:32.000
doing it.
Now I am going to use that same
00:29:28.000 --> 00:29:34.000
rule for transforming u of t
minus a times f of t.
00:29:32.000 --> 00:29:38.000
See, the problem is now I have
00:29:36.000 --> 00:29:42.000
a function like t squared
or sine t,
00:29:39.000 --> 00:29:45.000
which is not written in terms
of t minus a.
00:29:42.000 --> 00:29:48.000
And I don't know what to do
with it.
00:29:44.000 --> 00:29:50.000
The answer is,
by brute force,
00:29:46.000 --> 00:29:52.000
write it in terms of t minus a.
00:29:49.000 --> 00:29:55.000
What is brute force?
Brute force is the following.
00:29:52.000 --> 00:29:58.000
I am going to put a t minus a
there if it kills me.
00:29:56.000 --> 00:30:02.000
t minus a plus a.
No harm in that,
00:30:01.000 --> 00:30:07.000
is there?
Now there is a t minus a there,
00:30:04.000 --> 00:30:10.000
just the way there was up
there.
00:30:06.000 --> 00:30:12.000
And now what is the rule?
I am just going to follow my
00:30:11.000 --> 00:30:17.000
nose.
What's sauce for the goose is
00:30:13.000 --> 00:30:19.000
sauce for the gander.
Minus as, Laplace transform of
00:30:17.000 --> 00:30:23.000
f of, now what am I going to
write here?
00:30:20.000 --> 00:30:26.000
Wherever I see a t,
I am going to change it from t
00:30:24.000 --> 00:30:30.000
plus a.
Here I see a t.
00:30:29.000 --> 00:30:35.000
I will change that
to t plus a.
00:30:33.000 --> 00:30:39.000
What do I have?
t plus a minus a plus a,
00:30:37.000 --> 00:30:43.000
well,
if you can keep count,
00:30:41.000 --> 00:30:47.000
what does that make?
It makes t plus a in
00:30:46.000 --> 00:30:52.000
the end.
00:31:02.000 --> 00:31:08.000
The peace that passeth
understanding.
00:31:04.000 --> 00:31:10.000
Let's do some examples and
suddenly you will breathe a sigh
00:31:08.000 --> 00:31:14.000
of relief that this all is
doable anyway.
00:31:11.000 --> 00:31:17.000
Let's calculate something.
I hope I am not covering up any
00:31:16.000 --> 00:31:22.000
crucial, yes I am.
I am covering up the u of t's,
00:31:19.000 --> 00:31:25.000
but you know that by now.
Let's see.
00:31:22.000 --> 00:31:28.000
What should we calculate first?
What I just covered up.
00:31:27.000 --> 00:31:33.000
Let's calculate the Laplace
transform of (u)ab of t.
00:31:31.000 --> 00:31:37.000
What is that going to be?
00:31:34.000 --> 00:31:40.000
Well, first of all,
write out what it is in terms
00:31:38.000 --> 00:31:44.000
of the unit step function.
00:31:45.000 --> 00:31:51.000
Remember that formula?
There.
00:31:47.000 --> 00:31:53.000
Now you see it.
Now you don't.
00:31:50.000 --> 00:31:56.000
Its Laplace transform is going
to be what?
00:31:54.000 --> 00:32:00.000
Well, the Laplace transform of
t minus a,
00:31:59.000 --> 00:32:05.000
that is a special case here
where this function is one.
00:32:05.000 --> 00:32:11.000
Well, that one.
Either one.
00:32:07.000 --> 00:32:13.000
It makes no difference.
It is simply going to be the
00:32:11.000 --> 00:32:17.000
Laplace transform of what f of t
would have been,
00:32:15.000 --> 00:32:21.000
which is --
00:32:26.000 --> 00:32:32.000
See, the Laplace transform of u
of t is what?
00:32:30.000 --> 00:32:36.000
That's one over s,
right?
00:32:31.000 --> 00:32:37.000
Because this is the function
one, and we don't care the fact
00:32:36.000 --> 00:32:42.000
that it is zero or negative
values of t.
00:32:39.000 --> 00:32:45.000
That is my f of s.
And so I multiply it by e to
00:32:43.000 --> 00:32:49.000
the minus as times one over s.
00:32:46.000 --> 00:32:52.000
I am using this formula,
e to the minus as times the
00:32:50.000 --> 00:32:56.000
Laplace transform of the unit
step function,
00:32:53.000 --> 00:32:59.000
which is one over s.
How about the translation?
00:32:59.000 --> 00:33:05.000
That was taken care of by the
exponential factor.
00:33:03.000 --> 00:33:09.000
And it's minus because this is
minus.
00:33:06.000 --> 00:33:12.000
The same thing with the b.
This is the Laplace transform
00:33:10.000 --> 00:33:16.000
of the unit box function.
It looks a little hairy.
00:33:14.000 --> 00:33:20.000
You will learn to work with it,
don't worry about it.
00:33:19.000 --> 00:33:25.000
How about the Laplace transform
of --
00:33:23.000 --> 00:33:29.000
Okay.
Let's use the other formula.
00:33:25.000 --> 00:33:31.000
What would be the Laplace
transform of u of t minus one
00:33:29.000 --> 00:33:35.000
times t squared, for example?
00:33:33.000 --> 00:33:39.000
See, if I gave this to you and
you only had the first formula,
00:33:38.000 --> 00:33:44.000
you would say,
hey, but there is no t minus
00:33:41.000 --> 00:33:47.000
one in there.
There is only t squared.
00:33:44.000 --> 00:33:50.000
What am I supposed to do?
00:33:46.000 --> 00:33:52.000
Well, some of you might dig way
back into high school and say
00:33:51.000 --> 00:33:57.000
every polynomial can be written
in powers of t minus one,
00:33:56.000 --> 00:34:02.000
that is what I will do.
That would give the right
00:34:02.000 --> 00:34:08.000
answer.
But in case you had forgotten
00:34:05.000 --> 00:34:11.000
how to do that,
you don't have to know because
00:34:08.000 --> 00:34:14.000
you could use the other formula
instead, which,
00:34:12.000 --> 00:34:18.000
by the way, is the way you do
it.
00:34:15.000 --> 00:34:21.000
What are we going to do?
It goes into e to the minus s.
00:34:19.000 --> 00:34:25.000
The a is one in this case,
00:34:22.000 --> 00:34:28.000
plus one.
e to the minus s times the
00:34:25.000 --> 00:34:31.000
Laplace transform of what
function?
00:34:30.000 --> 00:34:36.000
Change t to t plus one.
00:34:33.000 --> 00:34:39.000
The Laplace transform of t plus
one squared.
00:34:38.000 --> 00:34:44.000
What is that?
That is e to the minus s times
00:34:41.000 --> 00:34:47.000
the Laplace transform of t
squared plus 2t plus one.
00:34:46.000 --> 00:34:52.000
What's that?
00:34:49.000 --> 00:34:55.000
Well, by the formulas which I
am not bothering to write on the
00:34:54.000 --> 00:35:00.000
board anymore because you know
them, it is e to the minus s
00:34:59.000 --> 00:35:05.000
times --
Laplace transform of t squared
00:35:04.000 --> 00:35:10.000
is two factorial over s cubed.
00:35:07.000 --> 00:35:13.000
Remember you always have to
00:35:09.000 --> 00:35:15.000
raise the exponent by one.
This is two factorial,
00:35:12.000 --> 00:35:18.000
but that is the same as two.
Plus two.
00:35:15.000 --> 00:35:21.000
This two comes from there.
The Laplace transform of t is
00:35:19.000 --> 00:35:25.000
one over s squared.
00:35:21.000 --> 00:35:27.000
And, finally,
the Laplace transform of one is
00:35:24.000 --> 00:35:30.000
one over s.
You mean all that mess from
00:35:30.000 --> 00:35:36.000
this simple-looking function?
This function is not so simple.
00:35:34.000 --> 00:35:40.000
What is its graph?
What is it we are calculating
00:35:38.000 --> 00:35:44.000
the Laplace transform of?
Well, it is the function t
00:35:42.000 --> 00:35:48.000
squared.
But multiplying it by that
00:35:45.000 --> 00:35:51.000
factor u of t minus one
means that the only part of
00:35:50.000 --> 00:35:56.000
it I am using is this part,
because u of t minus one is one
00:35:55.000 --> 00:36:01.000
when t is
bigger than one.
00:36:00.000 --> 00:36:06.000
But when t is less than one it
is zero.
00:36:02.000 --> 00:36:08.000
That function doesn't look all
that simple to me.
00:36:05.000 --> 00:36:11.000
And that is why its Laplace
transform has three terms in it
00:36:09.000 --> 00:36:15.000
with this exponential factor.
Well, it is a discontinuous
00:36:13.000 --> 00:36:19.000
function.
And it gets discontinuous at a
00:36:16.000 --> 00:36:22.000
very peculiar spot.
You have to expect that.
00:36:19.000 --> 00:36:25.000
Where in this does it tell you
it becomes discontinuous at one?
00:36:23.000 --> 00:36:29.000
It is because this is e to the
minus one times s.
00:36:29.000 --> 00:36:35.000
This tells you where the
discontinuity occurs.
00:36:32.000 --> 00:36:38.000
The rest of it is just stuff
you have to take because it is
00:36:37.000 --> 00:36:43.000
the function t squared.
It's what it is.
00:36:41.000 --> 00:36:47.000
All right.
I think most of you are going
00:36:44.000 --> 00:36:50.000
to encounter the worst troubles
when you try to calculate
00:36:49.000 --> 00:36:55.000
inverse Laplace transforms,
so let me try to explain how
00:36:53.000 --> 00:36:59.000
that is done.
I will give you a simple
00:36:56.000 --> 00:37:02.000
example first.
And then I will try to give you
00:37:01.000 --> 00:37:07.000
a slightly more complicated one.
But even the simple one won't
00:37:06.000 --> 00:37:12.000
make your head ache.
We are going to calculate the
00:37:11.000 --> 00:37:17.000
inverse Laplace transform of
this guy, one plus e to the
00:37:16.000 --> 00:37:22.000
negative pi s --
00:37:24.000 --> 00:37:30.000
-- divided by s squared plus
one.
00:37:35.000 --> 00:37:41.000
All right.
Now, the first thing you must
00:37:37.000 --> 00:37:43.000
do is as soon as you see
exponential factors in there
00:37:40.000 --> 00:37:46.000
like that you know that these
functions, the answer is going
00:37:44.000 --> 00:37:50.000
to be a discontinuous function.
And you have got to separate
00:37:48.000 --> 00:37:54.000
out the different pieces of it
that go with the different
00:37:52.000 --> 00:37:58.000
exponentials.
Because the way the formula
00:37:54.000 --> 00:38:00.000
works, it has to be used
differently for each value of a.
00:37:59.000 --> 00:38:05.000
Now, in this case,
there is only one value of a
00:38:02.000 --> 00:38:08.000
that occurs.
Negative pi.
00:38:03.000 --> 00:38:09.000
But it does mean that we are
going to have to begin by
00:38:07.000 --> 00:38:13.000
separating out the thing into
one over s squared plus one
00:38:11.000 --> 00:38:17.000
and this other
factor e to the negative pi s
00:38:15.000 --> 00:38:21.000
divided by s squared plus one.
00:38:19.000 --> 00:38:25.000
Now all I have to do is take
the inverse Laplace transform of
00:38:23.000 --> 00:38:29.000
each piece.
The inverse Laplace transform
00:38:26.000 --> 00:38:32.000
of one over s squared plus one
is --
00:38:32.000 --> 00:38:38.000
Well, up to now we have been
saying its sine t,
00:38:36.000 --> 00:38:42.000
right?
If you say it is sine t you are
00:38:38.000 --> 00:38:44.000
going to get into trouble.
How come?
00:38:41.000 --> 00:38:47.000
We didn't get into trouble
before.
00:38:43.000 --> 00:38:49.000
Yes, but that was because there
were no exponentials in the
00:38:47.000 --> 00:38:53.000
expression.
When there are exponentials you
00:38:50.000 --> 00:38:56.000
have to be more careful.
Make the inverse transform
00:38:54.000 --> 00:39:00.000
unique.
Make it not sine t,
00:38:56.000 --> 00:39:02.000
but u of t sine t.
00:39:00.000 --> 00:39:06.000
You will see why in just a
moment.
00:39:02.000 --> 00:39:08.000
If this weren't there then sine
t would be perfectly okay.
00:39:07.000 --> 00:39:13.000
With that factor there,
you have got to put in the u of
00:39:11.000 --> 00:39:17.000
t, otherwise you won't
be able to get the formula to
00:39:16.000 --> 00:39:22.000
work right.
In other words,
00:39:18.000 --> 00:39:24.000
I must use this particular one
that I picked out to make it
00:39:23.000 --> 00:39:29.000
unique at the beginning of the
period.
00:39:26.000 --> 00:39:32.000
Otherwise, it just won't work.
Now, I know that is fine.
00:39:31.000 --> 00:39:37.000
But now what is the inverse
Laplace transform of e to the
00:39:35.000 --> 00:39:41.000
minus pi?
In other words,
00:39:38.000 --> 00:39:44.000
it is the same function,
except I am now multiplying it
00:39:41.000 --> 00:39:47.000
by e to the negative pi s.
00:39:44.000 --> 00:39:50.000
Well, now I will use that
formula.
00:39:46.000 --> 00:39:52.000
My f of s is one over s squared
plus one,
00:39:50.000 --> 00:39:56.000
and that corresponds it to
sine t.
00:39:54.000 --> 00:40:00.000
If I multiply it by e to the
minus pi s,
00:39:57.000 --> 00:40:03.000
just copy it down.
It now corresponds,
00:40:02.000 --> 00:40:08.000
the inverse Laplace transform,
to what the left side says it
00:40:08.000 --> 00:40:14.000
does.
u of t minus pi
00:40:12.000 --> 00:40:18.000
times, in other words,
this corresponds to that.
00:40:17.000 --> 00:40:23.000
Then if I multiply it by e to
the minus pi s,
00:40:23.000 --> 00:40:29.000
it corresponds to change the t
to t minus pi.
00:40:38.000 --> 00:40:44.000
What is the answer?
The answer is you sum these two
00:40:43.000 --> 00:40:49.000
pieces.
The first piece is u of time
00:40:47.000 --> 00:40:53.000
sine t.
The second piece is u of t
00:40:53.000 --> 00:40:59.000
minus pi sine t of minus pi.
00:41:00.000 --> 00:41:06.000
Now, if you leave the answer in
that form it is technically
00:41:05.000 --> 00:41:11.000
correct, but you are going to
lose a lot of credit.
00:41:09.000 --> 00:41:15.000
You have to transform it to
make it look good.
00:41:13.000 --> 00:41:19.000
You have to make it
intelligible.
00:41:16.000 --> 00:41:22.000
You are not allowed to leave it
in that form.
00:41:20.000 --> 00:41:26.000
What could we do to it?
Well, you see,
00:41:24.000 --> 00:41:30.000
this part of it is interesting
whenever t is positive.
00:41:30.000 --> 00:41:36.000
This part of it is only
interesting when t is greater
00:41:34.000 --> 00:41:40.000
than or equal to pi because
this is zero.
00:41:37.000 --> 00:41:43.000
Before that this is zero.
What you have to do is make
00:41:41.000 --> 00:41:47.000
cases.
Let's call the answer f of t.
00:41:44.000 --> 00:41:50.000
The function has to be
00:41:47.000 --> 00:41:53.000
presented in what is called the
cases format.
00:41:50.000 --> 00:41:56.000
That is what it is called when
you type in tech,
00:41:54.000 --> 00:42:00.000
which I think a certain number
of you can do anyway.
00:42:00.000 --> 00:42:06.000
You have to make cases.
The first case is what happened
00:42:04.000 --> 00:42:10.000
between zero and pi?
Well, between zero and pi,
00:42:08.000 --> 00:42:14.000
only this term is operational.
The other one is zero because
00:42:12.000 --> 00:42:18.000
of that factor.
Therefore, between zero and pi
00:42:16.000 --> 00:42:22.000
the function looks like,
now, I don't have to put in the
00:42:21.000 --> 00:42:27.000
u of t because that is
equal to one.
00:42:24.000 --> 00:42:30.000
It is equal to sine t
between zero and pi.
00:42:30.000 --> 00:42:36.000
What is it equal to bigger than
pi?
00:42:32.000 --> 00:42:38.000
Well, the first factor,
the first term still obtains,
00:42:36.000 --> 00:42:42.000
so I have to include that.
But now I have to add the
00:42:40.000 --> 00:42:46.000
second one.
Well, what is the second term?
00:42:43.000 --> 00:42:49.000
I don't include the t minus
pi, u of t minus pi
00:42:48.000 --> 00:42:54.000
anymore because
that is now one.
00:42:51.000 --> 00:42:57.000
That has the value one.
It is sine of t minus pi.
00:42:55.000 --> 00:43:01.000
But what is sine of t minus pi?
00:43:00.000 --> 00:43:06.000
You take the sine curve and you
translate it to the right by pi.
00:43:07.000 --> 00:43:13.000
So what happens to it?
It turns into this curve.
00:43:13.000 --> 00:43:19.000
In other words,
it turns into the curve,
00:43:18.000 --> 00:43:24.000
what curve is that?
Minus sine t.
00:43:30.000 --> 00:43:36.000
The other factor,
this factor is one and this
00:43:33.000 --> 00:43:39.000
becomes negative sine t.
00:43:42.000 --> 00:43:48.000
And so the final answer is f of
t is equal to sine t
00:43:50.000 --> 00:43:56.000
between zero and pi and
zero for t greater than or equal
00:43:58.000 --> 00:44:04.000
to pi.
That is the right form of the
00:44:03.000 --> 00:44:09.000
answer.