WEBVTT

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PROFESSOR: Convolution
is a tricky concept

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and hard to understand.

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But it gives so much
insight that it's worthwhile

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coming to grips with it.

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The Convolution:
Accumulation mathlet

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provides one perspective on
the convolution integral.

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I'll explain the
applet using a story.

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There's a lake in Iowa with
a stream running into it

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and a stream running out of it.

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Farm runoff puts
phosphate into the lake.

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The rate of deposition of the
phosphate varies over the year.

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Most in the summer, none
in the dead of winter.

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Suppose the farm is new
and that t equals 0,

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there is no phosphate
in the lake.

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This is called rest
initial conditions.

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The question is, how
much phosphate is there

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in the lake at some
later time, say t.

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This will be an accumulation
of contributions from earlier

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times, but the
earlier contributions

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will have decayed
somewhat by time t.

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We can visualize this process
using the Convolution:

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Accumulation mathlet.

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I'll set it up to model
what we're looking at.

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The signal f of t is the
rate of input of phosphate

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into the lake.

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And I will model it by the menu
item f of t is 1 plus cosine

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b*t.

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You can see it drawn in the
lower graphing window here.

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A certain proportion of
the phosphate in the lake

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leaves it during
each unit of time.

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This results in an exponential
decay of each contribution.

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This decay profile is called the
weight function, denoted here

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by g of t.

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And I will select
the weight function,

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giving exponential decay
e to the minus a*t.

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Now let's see what
happens if I let

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a small amount of time elapse.

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How small?

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Well I can select that
by picking the step size.

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And I will select a
step size of one eighth.

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So now let's let a little
bit of time elapse.

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Say, amount of
time of one eighth.

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If I click on the time slider at
one eighth, I get this picture.

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t equals 0 happens to
occur at mid-summer,

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so that the signal
has a value of 2

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in kilograms per unit time.

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This phosphate decays away
as water comes into the lake

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and carries it off according
to the weight function.

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And this is indicated by the
exponential decay function

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here in the bottom window.

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The weight function is a rate.

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It needs to get multiplied
by the step size

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before it gives an actual
amount of phosphate in the lake.

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The step size is one
eighth, and the product

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is the signal, the
value of the signal at t

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equals 0, that's 2, times
the weight function, that's

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the exponential decay, times the
step size, which is one eighth.

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So that gives all
together one quarter

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times that exponential decay.

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And you see that drawn in
the top graphing window here,

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the value near t
equals 0 is a quarter

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and it decays away
exponentially.

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Now let's move on to
the next time interval.

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I'll click here, and
you see what happens.

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The signal is now
a little bit less

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and the time is a
little bit later.

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So the weight function is
scaled and shifted a little bit

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differently in this
bottom graphing window.

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It gets multiplied by the
step size one eighth again,

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and laid down on top of
the graph of the phosphate

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arising from the
first time interval.

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So the sum of those
two is a record

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of the phosphate in the lake
arising from these first two

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time intervals as it
decays away in later time.

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This process continues.

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You can see the next few
steps by clicking them up.

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Here's the effect of
the third time interval

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as it decays away, but laid
down on top of the preceding

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contributions to
the lake, and so on.

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Or I can animate
the entire process

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and watch the thing
evolve as time increases.

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So you can see, the
effect is a steady state

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is reached after a while, you
get a sinusoidal total amount

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of phosphate in the lake.

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It's delayed a little
bit from the peak.

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The maximum amount of
phosphate in the lake

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occurs a little
after mid-summer.

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Now imagine shrinking
the step size.

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In the limit, as the
step size goes to 0,

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this process is described
by the convolution integral.

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At time t, all the contributions
from time 0 to time t

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are present.

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And so we're going to take an
integral from u equals 0 to u

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equals t.

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The contribution from time u
decays by the factor of g of t

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minus u by time t.

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So the contribution from between
time u and time u plus du

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is given by the product f of u,
the signal of time u, times g

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of t minus u, the weight
function evaluated

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at t minus u, times du.

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And the integral of
this differential,

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from u equals 0
to u equals t, is

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the amount of phosphate
in the lake at time t.

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And it's given by the
convolution integral.

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This system is modeled by
a differential equation,

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namely dx/dt plus
a*x equals f of t.

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The weight function of the
operator dx/dt plus a*x is e

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to the minus a*t.

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And the solution to the
differential equation

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with rest initial
conditions is given

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by the convolution integral
of the input signal

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with the weight function.

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This statement is a general
fact for LTI operators,

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not just first-order operators.