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Laplace Transform 1

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    I'll now introduce you
    to the concept
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    of the Laplace Transform.
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    And this is truly one of the
    most useful concepts that
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    you'll learn, not just in
    differential equations, but
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    really in mathematics.
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    And especially if you're going
    to go into engineering, you'll
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    find that the Laplace Transform,
    besides helping you
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    solve differential equations,
    also helps you transform
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    functions or waveforms from
    the time domain to the
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    frequency domain, and study
    and understand a
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    whole set of phenomena.
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    But I won't get into
    all of that yet.
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    Now I'll just teach
    you what it is.
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    Laplace Transform.
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    I'll teach you what it is, make
    you comfortable with the
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    mathematics of it and then in
    a couple of videos from now,
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    I'll actually show you how it
    is useful to use it to solve
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    differential equations.
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    We'll actually solve some of the
    differential equations we
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    did before, using the
    previous methods.
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    But we'll keep doing it, and
    we'll solve more and more
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    difficult problems.
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    So what is the Laplace
    Transform?
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    Well, the Laplace Transform,
    the notation is the L like
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    Laverne from Laverne
    and Shirley.
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    That might be before many
    of your times, but
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    I grew up on that.
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    Actually, I think it was even
    reruns when I was a kid.
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    So Laplace Transform
    of some function.
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    And here, the convention,
    instead of saying f of x,
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    people say f of t.
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    And the reason is because in
    a lot of the differential
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    equations or a lot of
    engineering you actually are
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    converting from a function
    of time to
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    a function of frequency.
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    And don't worry about
    that right now.
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    If it confuses you.
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    But the Laplace Transform
    of a function of t.
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    It transforms that function into
    some other function of s.
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    and And does it do that?
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    Well actually, let me just do
    some mathematical notation
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    that probably won't
    mean much to you.
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    So what does it transform?
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    Well, the way I think of
    it is it's kind of
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    a function of functions.
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    A function will take you from
    one set of-- well, in what
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    we've been dealing with-- one
    set of numbers to another set
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    of numbers.
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    A transform will take you from
    one set of functions to
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    another set of functions.
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    So let me just define this.
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    The Laplace Transform for our
    purposes is defined as the
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    improper integral.
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    I know I haven't actually done
    improper integrals just yet,
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    but I'll explain them
    in a few seconds.
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    The improper integral from 0 to
    infinity of e to the minus
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    st times f of t-- so whatever's
    between the Laplace
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    Transform brackets-- dt.
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    Now that might seem very
    daunting to you and very
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    confusing, but I'll now do
    a couple of examples.
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    So what is the Laplace
    Transform?
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    Well let's say that f
    of t is equal to 1.
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    So what is the Laplace
    Transform of 1?
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    So if f of t is equal to 1--
    it's just a constant function
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    of time-- well actually, let me
    just substitute exactly the
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    way I wrote it here.
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    So that's the improper integral
    from 0 to infinity of
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    e to the minus st
    times 1 here.
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    I don't have to rewrite it here,
    but there's a times 1dt.
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    And I know that infinity is
    probably bugging you right
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    now, but we'll deal
    with that shortly.
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    Actually, let's deal with
    that right now.
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    This is the same thing
    as the limit.
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    And let's say as A approaches
    infinity of the integral from
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    0 to Ae to the minus st. dt.
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    Just so you feel a little bit
    more comfortable with it, you
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    might have guessed that this
    is the same thing.
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    Because obviously you can't
    evaluate infinity, but you
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    could take the limit as
    something approaches infinity.
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    So anyway, let's take the
    anti-derivative and evaluate
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    this improper definite
    integral, or
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    this improper integral.
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    So what's anti-derivative
    of e to the minus st
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    with respect to dt?
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    Well it's equal to minus 1/s
    e to the minus st, right?
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    If you don't believe me, take
    the derivative of this.
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    You'd take minus s times that.
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    That would all cancel out, and
    you'd just be left with e to
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    the minus st. Fair enough.
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    Let me delete this here,
    this equal sign.
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    Because I could actually use
    some of that real estate.
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    We are going to take the limit
    as A approaches infinity.
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    You don't always have to do
    this, but this is the first
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    time we're dealing with
    improper intergrals.
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    So I figured I might as
    well remind you that
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    we're taking a limit.
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    Now we took the anti-derivative.
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    Now we have to evaluate it at A
    minus the anti-derivative
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    evaluate it at 0,
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    and then take the limit of
    whatever that ends up being as
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    A approaches infinity.
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    So this is equal to the limit
    as A approaches infinity.
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    OK.
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    If we substitute A in here
    first, we get minus 1/s.
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    Remember we're, dealing
    with t.
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    We took the integral
    with respect to t.
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    e to the minus sA, right?
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    That's what happens when
    I put A in here.
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    Minus -
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    Now what happens when I put
    t equals 0 in here?
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    So when t equals 0, it becomes
    e to the minus s times 0.
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    This whole thing becomes 1.
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    And I'm just left
    with minus 1/s.
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    Fair enough.
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    And then let me scroll
    down a little bit.
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    I wrote a little bit bigger
    than I wanted
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    to, but that's OK.
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    So this is going to be the limit
    as A approaches infinity
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    of minus 1/s e to the
    minus sA minus minus 1/s.
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    So plus 1/s.
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    So what's the limit as A
    approaches infinity?
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    Well what's this term
    going to do?
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    As A approaches infinity, if
    we assume that s is greater
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    than 0-- and we'll make that
    assumption for now.
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    Actually, let me write
    that down explicitly.
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    Let's assume that s
    is greater than 0.
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    So if we assume that s is
    greater than 0, then as A
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    approaches infinity, what's
    going to happen?
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    Well this term is going to go to
    0, right? e to the minus--
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    a googol is a very,
    very small number.
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    And an e to the minus googolplex
    is an even smaller number.
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    So then this e to the minus
    infinity approaches 0, so this
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    term approaches 0.
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    This term isn't affected because
    it has no A in it, so
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    we're just left with 1/s.
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    So there you go.
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    This is a significant
    moment in your life.
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    You have just been exposed to
    your first Laplace Transform.
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    I'll show you in a few videos,
    there are whole tables of
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    Laplace Transforms, and
    eventually we'll
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    prove all of them.
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    But for now, we'll just
    work through some of
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    the more basic ones.
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    But this can be our
    first entry in our
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    Laplace Transform table.
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    The Laplace Transform of
    f of t is equal to
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    1 is equal to 1/s.
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    Notice we went from a function
    of t-- although obviously this
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    one wasn't really dependent
    on t-- to a function of s.
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    I have about 3 minutes left,
    but I don't think that's
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    enough time to do another
    Laplace Transform.
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    So I will save that for
    the next video.
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    See you soon.
Title:
Laplace Transform 1
Description:

Introduction to the Laplace Transform

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Video Language:
English
Duration:
08:01

English subtitles

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