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PROFESSOR: Hello and welcome.
So today is mostly distinguished
by what happens tomorrow.
Not surprisingly, but of
course, you all know that.
So tomorrow we have
our first quiz.
Tomorrow evening 7:30
to 9:30, no recitation.
We've been through
this several times.
I won't spend any time on it
other than to ask if there are
any questions, so if I
don't hear any questions,
the idea is going to be at
the end of this lecture,
the next time we'll see
you is in office hours,
and, or tomorrow, Wednesday,
7:30 on the third floor,
building 26.
Questions or comments
about the exam?
Yep.
AUDIENCE: [INAUDIBLE]
PROFESSOR: So that one
page of notes, 8 and 1/2
by 11, front and back.
You can write as
small as you like.
In fact, later in
today's lecture,
I'll show you how
a microscope works
and you're welcome to use a
microscope because they're
completely non-electronic.
Oh, as long as you use
an optical microscope.
Other questions about the exam?
OK, then for today, so
far, since the beginning
of the term, we
thought about a number
of different kinds
of representations
for both DT systems--
discrete time-- and CT systems--
continuous time systems--
and we saw that
we were interested
in that large number
of representations,
because each of them had
some particular aspect that
made it particularly
convenient sometimes.
So, for example, in both CT
and DT we looked at verbal,
but not so much.
That was mostly in the homework.
We looked at difference
in differential equations,
mostly because they were
so compact, so concise, so
precise, they told you
exactly what the system does.
No fluff, this is it.
So that was nice.
Block diagrams, by
contrast, are less concise,
but they tell you the
way a signal propagates
through the system on its way
from the input to the output,
and that can be very
helpful for understanding
why certain behaviors occur,
especially when we talk
about things like feedback.
We looked at operator
representations.
They were nice because
we could transform
the way we think about
systems into the way
we think about polynomials,
so we reduced a college level
thing to a high
school level thing.
That's always nice, and then
we looked at transforms.
The distinguishing feature
of transform representation
was that we took an
entire function of time,
and turned it into an
algebraic expression.
So we turned a
differential system
that described a system
of differential equations
that describes a system into a
system of algebraic equations
that describes a system.
All of those were useful
for different ways
and what I wanted to talk
about today is yet another way
to represent a system
and that is to represent
a system by a single signal.
So in some sense
we're going backwards,
because we're taking
what we would normally
think of as an entire
system and reducing it,
getting rid of the
system altogether,
so all we're going
to have is signals.
That turns out to be a
particular powerful way
to do some sorts of
operations, and is actually
the first instance,
the first step,
in that we will take
in a major field
thought of signal processing.
When you reduce the entire
behavior of a system
to a signal, we then regard
the whole processing task
as a signal processing task.
Signal processing
task, not system.
So far, we have focused
on the responses
to the most elementary
kinds of signals--
unit sample signal,
unit impulse signal--
but generally speaking,
we're interested in much more
complicated signals.
As you've already
seen, I've already
asked you to calculate things
like unit step responses.
So generally, we're going
to be interested in much
more complicated signals--
the responses of systems to
much more complicated signals.
That's not hard.
The reason we
skipped over it was
that you can always
figure out the response
to a more complicated
signal at least
by falling back on some
of our more primitive ways
of thinking about systems.
So for example, if we think
about a difference equation
or a block diagram,
we can think about how
a more complicated signal
excites a response by simply
thinking about the system
operating on a sample
by sample basis.
So you, of course,
all know that,
and to prove that
you all know that,
answer the following question.
Here is a system.
Here is a signal that is more
complicated than just a unit
sample or unit sample signal.
Figure out what is
the third response
of this system to that signal?
You should be absolutely quiet.
That was sarcasm,
just so you know.
Lots of self-satisfied looks
so I assume everybody's done.
So what's the answer?
Raise the number of
fingers that corresponds
to the answer y of 3.
Raise your hand so
that I can see them.
Drop the ones with
the wrong answers.
I don't want to see those.
OK, about 50%, so maybe
take another 10 seconds.
Notice that I'm
asking for y of 3.
So what's the answer y of
3, and raise your hand.
Everybody has the right
answer, raise your hand.
Much better.
OK, so now the overwhelming
majority says the answer is 2.
That's not very hard.
If we think about the block
diagram representation,
and if we think
about propagating
the more complicated signal
represented over here
through that signal,
through that system,
we start with the
system at rest.
That means there's 0's
coming out of all the delays,
at time n equals
minus 1, x is 0,
combined with the initial
0's, we get the answer is 0.
And then, at time n equals
0, the input becomes 1,
so now the output goes to
1, at time 1, it goes to 2.
At time 2, it goes to 3.
Times 3, it goes to 2.
This is obvious, right?
Et cetera.
So the answer is 2.
y of 3 is 2, and the point
is that it's very trivial
to think about by thinking about
the system in a sort of sample
by sample way.
Not surprisingly,
the point of today
is to not think of the
system sample by sample,
but to elevate the conversation
from samples to signals.
The first step in thinking
about it as signals
is to realize that you can
think about the response
of the system by decomposing
the input into additive parts.
I can think about x which
is this 3 sample signal.
I can decompose it
into single samples,
and then think about the
response to each of those,
and it may not surprise you
that if the system were linear,
then the response
to the sum would
be the sum of the responses.
So you can sort
of see that there
would be a way of
adding together
these rectangular pulses
to get a triangle pulse.
So that works simply because
the system is linear.
The system has the property
that the output for a sum
is the sum of the outputs for
the individual components,
and we can write that this
way, so a system is linear.
We can define linear in a more
rigorous mathematical sense
by saying that a
system is linear
if the response to a
weighted sum of inputs
is the similarly
weighted sum of outputs.
So imagine that I
have a system whose
output when the
input is x1 is y1
and whose output when
the input is x2 is y2.
We'll say the
system is linear if
and only if the weighted sum
of inputs alpha x1 plus beta x2
gives the same wakings
alpha y1 plus beta y2
for all possible values
of alpha and beta.
If that's true, then we'll
say the system is linear.
So then if it's
linear, then we're
allowed to do this
decomposition,
because all we're
doing is decomposing
the input into a sum of inputs.
We'll always be able to do that
operation of the decomposition
if the system is linear
according to the definition I
already showed.
If the system also
has the property
that we will call
time invariance,
then the response
to the parts will be
particularly easy to calculate.
Time invariance has a
similar formal definition.
We will call a
system time invariant
if given that the input--
the response to x of n is y.
Given that, we'll say the
system is time invariant
if a shifted
version of the input
simply shifts the response.
That seems like a kind of
gobbledygook sort of thing.
It's a very simple minded
notion that you should all
have from common experience.
All it says is that if
I do something today,
and I get a response, if I
do the same thing tomorrow,
I should get the same response
just delayed by a day.
That's all it's saying.
So basically the system
behaves sort of the same
now as it did previously,
and as it will in the future.
That's what time
invariance means.
So if the response
is time invariant,
then we can compute the response
to a shifted unit sample.
Notice that this is the
unit sample response.
This is a shifted unit
sample so if the system
is time invariant, then the
response to a shifted unit
sample is the shifted
unit sample response which
you can see from the picture.
So the idea then is
that superposition
is a very easy way to think
about the response of a system.
If the system is linear
and time invariant,
linearity let's us break it up,
and think about the response
to each part.
Shift invariance,
or time invariance,
allows us to shift the
input, and know automatically
what's the response
going to look like.
And there's a formal way we
can think about the way you
do that operation.
How do you implement
this superposition thing?
You think about a system as
having a unit sample response.
This is the unit sample signal.
We'll call the unit
sample response h of n,
then a shifted unit sample
will give a shifted unit
sample response.
That's time invariance.
Then a weighted shifted impulse
gives the same weighted shifted
impulse response.
Then the sum of such things
gives the sum of such things.
That's just a formal
derivation of a process
that we will call convolution.
So the response to an
arbitrary DT signal
that excites a linear
time invariant system
can be described by the
convolution of the input
with the unit sample response.
We'll call that formula--
we'll call that
operation convolution.
Convolution is completely
straightforward.
For that reason, we try to make
it a little bit more confusing
by using terribly
confusing notation.
That too is sarcastic, I mean.
So the only thing
that's at all confusing
about convolution-- convolution
is completely trivial.
Here's the way we would
write it. x convolves with h.
The signal x convolves with the
signal h to give a new signal.
Being a signal, I can ask
what's the nth sample look like,
and what that symbol
means is this sum.
The confusing thing is that
most people in the field
write it this way.
The signal x of n convolves
with the signal h of n.
The reason that's confusing,
and the thing you will never do,
because you are here.
The thing that you
will never do is
confuse the meaning
of that statement
with what looks like an
operation on samples.
Had I said multiply,
you would have said,
if this were a multiply operator
instead of a convolution
multiplier operator--
if that were multiply
instead of convolve,
you would have
said, oh, that means
the oneth sample multiplied
by the oneth sample
is the product of
the oneth samples.
This is not true.
This is not generally true.
The convolution operation
means take the whole signal x.
That's why we think about it
as an operation on signals,
not an operation on samples.
Convolution means take
the whole signal x,
and convolve it with
the whole signal h
to get a brand new
signal x convolved h,
and then take the nth sample.
So the only thing that's at
all confusing about convolution
is remembering the convolution
is an operation that is applied
to signals, not samples.
So structure of convolution.
So I just showed you a
mathematical formula.
What I'd like you to
have is a little bit
of an intuition for what happens
when you convolve two signals.
So let's think about the
structure of this operation.
What are we doing?
Imagine that we're going back
to that original problem.
What happens when you take x of
n and convolve it with h of n?
All we need to do is
this formula, right?
That's all we need to do.
What's that formula say?
Well let's think about how you
would compute the 0-th output.
According to that formula all
I did was substitute n equal 0
every place there
was an n, and what
I see is I have to multiply
x of k times h of minus k,
but I've got x of n in h of n.
So the first thing I do is I
flip the axises and I make them
k's.
That's not hard.
Then the x looks OK,
but the h doesn't.
I need h of minus k,
so I have to flip it.
So I'm flipping about
the n equals 0 axis.
So that positive
n becomes minus n,
positive k becomes
minus k, because I want
this to be minus k up here.
Then generally, I have
some shift thing here.
This 0, because I was
looking for the 0-th sample.
In general, that
might be different.
That might be 7 if I
wanted to find y of 7.
Then I'd have a 7 over here, and
that number represents a shift.
In the case of 0,
it's a 0 shift.
Then I have to
multiply these two,
so I just place this thing
over here so I can multiply.
I multiply down.
You can see that there's only
one sample that in the two
is both non-zero.
Therefore, I get a
single non-zero answer,
and then according
to the formula,
I have to sum so the 0-th answer
is flip, shift, multiply, sum,
and you just repeat that for
all the different answers.
So at time equals 0,
the answer at time 0
is flip, shift by
0, multiply, sum.
The answer is 1.
If I want to find the one answer
now the shift is shift by 1.
So now instead of
having flip which
would have put the 3
samples here, I shift by 1
so now they're over there.
So now when I do the multiply,
I pick up 2 non-zero answers
and the answer is 2.
If I wanted y equals
2, I do the same thing
but now I shift by 2.
Flip, shift, multiply,
sum, that's all I do.
It's completely trivial.
If I continue, the
shift becomes larger
and now it's
falling off the end.
Continue, continue,
and I get in general
that's the prescription.
So what I've tried
to show is two ways
of thinking about this
convolution thing.
The first was by
superposition, where I just
think about breaking the
input into a bunch of samples,
thinking about the response
to each of those samples
and adding.
That's an input centric way
of thinking about things,
because I think of the
input being broken up
by a bunch of samples.
This convolution formula
is an output centric way
of thinking about
things I tell you.
I'd like to know the
output at time p,
and to compute the
output of time p,
you say, well that's easy.
Flip, shift by p, multiply, sum.
So input centric, that's
the superposition way
of thinking about things.
Output centric, that's
the convolution way
of thinking about things.
So now that you know
about convolution,
find which plot below 1, 2,
3, 4, or none of the above
shows the result of convolving
the two functions shown above.
You're so quiet.
I assume you're practicing
for the exam tomorrow.
You're allowed to talk.
So which one's right?
1, 2, 3, 4, or 5?
See if I know.
It looks good.
About I only see one wrong,
two wrong, so 95% or so.
So how do I think about this?
What's the way that I should
think about convolving those?
Easiest, most straightforward
way, go back to the formula.
That will always work.
Can somebody tell me a more
intuitive, insightful way
of thinking about what will
be the result of convolving
those top two functions?
Tell me a property of
the result of convolving.
Yes, yes, flip,
shift, multiply, sum.
What's the answer at n equals 1?
1.
So I've got two things that
look kind of like geometrics.
Imagine for the moment--
that was intended to be a hint--
imagine for the moment that
the sequence looks like 1,
2/3, 4/9, 8/27,
blah, blah, blah.
Imagine that it's a
geometric sequence
with the base of about 2/3.
How would I compute the answer
when I convolve that sequence
with itself at zero?
Flip, shift, multiply,
divide, so I started out
with two things that were
both starting at zero.
You flip one of them.
How much overlap is there?
Just the 1.
Just the n equals 0, what's
the answer at n equals 0?
1, right?
So if I imagine that this is
the sequence after I flip it.
There's a 1 under the 1.
The 0's here kill
the terms down here.
The 0's here kill
the terms up there.
The only thing that
lives is 1 times 1 is 1,
so the answer at
y equals 0 is 1.
What's the answer at y equals 1?
Flip, shift, multiply, sum.
So I flip, shift.
So when I shift, the new answer
looks like 1, 2/3, 4/9, 8/27,
et cetera.
Multiply and sum,
what's the answer?
AUDIENCE: [INAUDIBLE]
PROFESSOR: 4/3.
So the only non-zero answers are
1 times 2/3 plus 2/3 times 1--
4/3.
If I want to compute y of
0 1 2, I shift it further.
So I do 1, 2/3, 4/9,
8/27, blah, blah, blah.
So flip, shift, I
shift 1 more, multiply,
sum, multiply 1 times
4/9, and I get 4/9.
Multiply 2/3 times
2/3, I get 4/9.
Multiply 4/9 times 1, I get 4/9.
The answer to the
sum of those is 4/3.
So I get one 4/3, 4/3.
So you can see it's tracing out.
This wave form so far,
this is the only one that
has up and then flat, and
if I continue that process
it will start to fall off.
If you're exclusively
mathematically minded,
you can also just
do it with math.
All you do is think about
a mathematical description
of the left signal.
Say 2/3 of the nu of n and
the right signal, and now all
I need to do is think
about that formula.
So do a sum, taking
this, the function of n
and turning it into
a function of k.
The second one, I want to
make a function of n minus k.
I have to shift both.
I have to change the
exponent as well as the index
into the unit sample signal,
same thing over here.
Now when I think about
multiplying them, u of k
kills all the terms
for k less than 0,
so I can start at 0
instead of minus infinity.
This u kills everything for
which n minus k is less than 0.
That means k less than n--
less than or equal to n.
So I end up with this.
This product is
particularly easy
because it's the the
k into the minus k,
so the answer is to the n,
and now I'm summing over k,
but there are no k's, so
that's summing over 1.
And so my answer is
just n plus 1 and 2/3
to the nu of n which is the
same thing by thinking about it
intuitively.
So the point is that the
operation is friendly,
and so the idea then the
big picture was convolution
is a different way to
represent a system.
Using convolution, we
represent an entire system
by a single signal.
That signal, the
unit sample response,
is sufficient to characterize
the output of the system
for any possible input.
We just saw how the operation
is called convolution,
so that enables us
the big picture.
We've represented an entire
system by a single signal,
in this case h of n.
That's what convolution is.
It's a new representation.
You can do exactly the
same thing for a CT system,
and the reason we use delta
to represent the unit sample
signal and the unit
impulse response.
The unit impulse
signal is clear,
because the representation
of an arbitrary signal,
in terms of delta functions,
looks much the same
in CT and in DT.
You can get there by thinking
about the limiting argument
for how to interpret
the unit impulse.
The unit impulse
function was a function
that is easiest to
think about in a limit.
Imagine that I
have a signal that
is a square pulse whose area,
regardless of width, is 1.
That's what a unit
impulse function is.
Imagine how you would construct
an arbitrary signal x of t
by having such a signal.
You could take a signal and it's
shifted version, and come up
with a weighted sum
of impulse functions
or rectangular approximations
to impulse functions
to represent an
arbitrary signal.
If you did that, you would get
an approximation to the signal
x which could be
written as a limit.
So if you think about each of
these being of width capital
delta, then the height
has to be 1 over delta.
So the area remains one.
Then if I want to build
an arbitrary function x
out of such signals,
I need a sum of them,
and each one of these p's has
to be multiplied by delta.
So that when I multiply
by the value of x at one
point k delta.
I get the right height
independent of what is delta.
So for a given delta, I get
a sum that looks like that,
and then in keeping with
the idea of thinking
about a unit impulse
function as a limit,
I take the limit of that.
The result is a
function that looks
very much like the decomposition
of a signal in terms
of the unit sample.
In the previous case, we sum
together a weighted version
of a unit sample signal.
Here the sum is
replaced by an integral,
and is weighted just
like it was before.
The point is that the
mathematics for CT and DT
look very similar.
I decompose in the case of the
CT, and arbitrary signal x of t
into an entire row of weighted
unit impulse functions.
Once I have it in that
form, the argument's
precisely the same for
CT as it was in DT.
Imagine that I have a linear
time invariant system.
Linear means that I can
compute the response to a sum
as the sum of the responses.
Time invariant means
that shifting the input
merely shifts the output.
Doing the experiment
tomorrow is the same
as doing the experiment today,
except it's now a day later.
So if the response
of a system is
h of t when the
input is delta of t,
if the system is shift
invariant, shifting this by tau
is the same as
shifting that by tau.
A weighted sum of such things is
a weighted sum of such things,
and a sum of such things
is the sum of such things,
so I get an
expression which we'll
think of as convolution for
CT that looks just the same.
So in DT, we thought about
if you convolve x with h,
you take the first
index, x of n,
and turn it into x of
k, a dummy variable.
You take the second
one, and do n minus k.
Here we do the same
thing. t goes to tau,
and the second one
goes to t minus tau.
The sum up here turns
into an integral.
Otherwise, it's
exactly the same thing.
So to show your mastery of
such things, what signal would
result if you convolved e
to the minus tu of t with e
to the minus tu of t?
1, 2, 3, 4, or none?
Well, the place is
quiet so I assume
that means you stopped
talking, so that
means you've all agreed, yes?
So which wave form
best represents
the convolution of the top
two signals, 1, 2, 3, or 4?
Almost 100% correct.
Most people say 4.
How do you get 4?
Yeah?
AUDIENCE: Same
reason [INAUDIBLE]
PROFESSOR: So what would I do?
What would be my first step?
So I imagine that
I want to think
of flip so this gets multiplied
by the flip of the other one.
So at time t equals
0, the answer is--
AUDIENCE: 0
PROFESSOR: --0, because
there's no overlap.
AUDIENCE: [INAUDIBLE]
PROFESSOR: OK so
that it starts at 0,
so that means that this is out.
OK, OK, OK, fine.
Now shift.
What do I shift?
Which one do I shift which way?
AUDIENCE: Why is there a
[INAUDIBLE] flipping is there
a value that equals 0?
PROFESSOR: That's
a valid question.
So the question is
if I'm integrating
over a function that goes from
minus infinity to 0, and from 0
to infinity.
Let's say the answer
right at 0 is 1.
You could say that
there is a single point
whose value is non-zero.
What would happen if I
integrated a function that is 0
everywhere except at a point?
So it's 0 everywhere up to here,
then a 0 everywhere after that,
and at zero it's not zero.
What's the integral of
a function that differs
from 0 at a single point?
AUDIENCE: [INAUDIBLE]
PROFESSOR: 0, it's a little
bit of a trick question,
because we will later have some
functions for which that's not
true.
What kind of a function
would that not be true for?
AUDIENCE: [INAUDIBLE]
PROFESSOR: Delta.
If I were convolving
delta with delta,
then you can integrate over
an infinitesimal area region,
and get something that's not 0.
So a little bit of
a caveat, so as long
as the function
that I'm convolving
doesn't have an
impulse in it or worse.
We will talk later in the
course about things worse
than impulses.
If there's nothing as high
as an impulse or worse,
so that has a step in it.
We would think of a step
as a singularity that
is better, less ill
behaved than an impulse.
As long as the function does
not have an impulse or a worse,
when you flip it, you'll get
zero contribution at zero.
That all makes sense?
So this is zero at zero, but
the reasoning is a little bit
complicated.
So now what do I get when t
gets a little bit bigger than 0?
What's the result of
convolving when the time is
slightly bigger than time 0?
All flip, shift,
multiply, integrate.
So I have to shift
one of those, so now
instead of having this one, I
might have shifted a little bit
to the right.
So it might look like that.
So now what happens
when I multiply?
Well you don't get 0 anymore.
So as for small t for t
on the order of epsilon,
how does the
integral grow with t?
So if I want to make a
plot of the convolution,
so if I want to think
about e to the minus
tu of t convolved with e to
the minus tu of t versus t,
I already know that that's
like that for t small.
How will the function grow?
Linear so if this
is very small then
the deviations from the height
which is 1 is very small.
So if this distance is
small, the deviation
is that little triangle
which goes like t-square.
So for t small t-square is
very small compared to t,
I can ignore it,
and so the function
is going to start
going up like t,
and if you work out the details
it will eventually roll off.
Because as you shift
it further and further,
the one exponential is in the
tail of the other exponential,
so one of the exponentials
kills the other one,
and so the response
goes to zero.
If you're more
mathematically inclined, yes?
AUDIENCE: [INAUDIBLE]
you were saying
that if both functions
are left sided [INAUDIBLE]
right sided it always starts
out t equals 0 is always 0.
PROFESSOR: That's not quite
right because right sided just
means that the left is 0.
So if I tell you that a signal
is right sided, all I've said
is that all of the non-zero
values are on the right,
but I haven't told you whether
they're impulses or not.
Right sided says
something about the left.
The left is zero.
Kind of weird, so
if I'm right handed,
I might as well not have
a left hand when I'm
writing so the left is zero.
So right sided
signals have zeros.
The signals on the left of
right sided signals are zero,
but I haven't told you
what was on the right.
The right could be
an impulse or worse.
So I just inferred
some properties
of what this convolution
is going to look like.
If you were
mathematically inclined,
you could do it by math.
The math doesn't
look very different
from the math for
the discrete version.
You simply write this as a
function of tau rather than t.
This one as a function of
t minus tau rather than t.
Recognize that the u's cut
off parts of the integral.
This u is 1 only if
t is bigger than 0.
So that lops off the
t less than 0 part.
This lopped off the part bigger
than capital, bigger than t.
So that leaves the
integrals 0 to t.
Putting these two
together results
in the tau parts
killing each other,
so I'm left with only a t
part but the integrals on tau.
So just like the other
one, the integral
goes to the integral
of 1 over finite limit
0 to t, so the answer is t.
So the overall answer is
te to the minus tu of t
which is plotted here.
So the point of today
then is that this
is a different representation
for the way systems work.
It's often computationally
interesting.
This is the way.
This is a perfectly
plausible way
of doing discrete time
signal processing.
Represent the system
by a signal h of n,
and then compute the response to
any signal by convolving h of n
with that signal.
Perfectly reasonable
way to compute things.
Honestly, we'll find better
ways of computing things
by the end of the course.
The real reason for studying
convolution is conceptually,
you can think about
how a system ought
to work by thinking
about convolution,
and I want to show
an example of that
by thinking about systems
that I'm interested in.
I do work on hearing.
I do work with microscopes,
and we can regard a microscope
as an LTI system.
A linear time invariant
system, and convolution
is a very good way of thinking
about such optical systems.
So the idea is that
even the best microscope
gives blurry images, and that's
very fundamental physics.
It has to do with
the diffraction
limit of optical systems.
If you're interested
in that sort of thing,
take an optics course in
the physics department
or come to my lab
and do a [INAUDIBLE]..
So the idea is that even the
best microscopes in the world
are blurred.
We have the best microscopes
in the world in my lab,
and they fundamentally
blur things,
and we have to worry about that.
The blurring is
inversely related
to the numerical
aperture which has to do
with the size of the optic.
Big optics are good.
So if you imagine
that a target emits
a spherical wave of light.
So every point on the target
emits a spherical wave,
then there's some optic that's
collecting all of those waves,
, and relaying them back
to a different point.
The resolution of
the picture goes
with how many of those rays
the optic system picked up.
So if you make the
optics smaller,
the picture becomes blurrier.
If you make the
optic even smaller,
the picture becomes
even smaller,
and the way we think about
that is by convolution.
We think about the
microscope as an LTI system,
so we characterize it by
its point spread function.
We don't like to
use any words that
come from a different field.
Just like every other field,
we like to invent our own.
So in optics, the thing that
we will call a impulse response
is called a point
spread function.
It just says if you had
an ideal point of light,
what would the image look like?
It's exactly the
same as convolving,
so you can think of the blurry
image as the convolution
of the effect of the microscope,
the point spread function,
the impulse response
with the ideal target.
So then as you change
the size of the optic,
it changes the size of
the impulse response,
the point spread function.
Crummy optics, fat
point spread motions.
Fat point spread
functions, blurry pictures,
and so here's a picture
of how our system works.
This is a
representation to scale
of a tiny microscopic bead
a fraction of a micron.
So it's about six times
smaller than the image.
So this is an image taken
with our microscope system,
and you can see that
most of the energy
fits inside a region
about half a micron.
World's best microscope,
you can't do better
than this by physics.
This is using 500 nanometer
light, and the size of that
has to do with the length
scale of the light.
So you end up with this
particular microscope
not being able to make images
with less blurring than that.
That's the point
spread function.
Now of course, the point
spread function of a microscope
is three dimensional.
In this class, we're only
talking about 1-D time.
In a microscope is
3D, x, y, and z.
So the impulse response
has extent in x, y, and z.
So here is a picture taken
by Anthony Patire, who
was a student in my lab of
that same tiny little dot.
When the in-focus
plane shows the dot,
and as you go out
of focus, the dot
gets bigger, smearier,
and blurrier.
What you can do then is
assemble those pictures that
were taken one of the
time into a 3-D volume,
and that 3-D volume then
represents the point spread
function or the three
dimensional impulse
response of the microscope.
So the idea then is that
convolution is a very good way
to think about optical
systems because they
are very easy to relate
to the underlying physics.
The blurring is a direct result
of a fundamental property
of light.
The diffraction limit, and
you can very accurately
represent the effect
of the blurring
as convolving with a point
spread function and impulse
response.
Same sort of thing applies
to optics at any scale.
So going from the microscopic
to the rather macroscopic-- that
is to say the
universe and beyond.
We can think about the
Hubble Space Telescope.
Same thing.
Light, that's all that
matters, and here the issue
is-- the reason they wanted
to make a Space Telescope
is that there are two
principal sources of blurring
for a ground based telescope.
One is the atmosphere
blurring, because we're
looking through an atmosphere
and other is blurring because
of the property of the optic
elements in the telescope,
and it turns out
pretty easy to show
that the combined effect of
the atmosphere and the lenses
is the convolution of
the individual parts.
You can think about
atmospheric blurring
as convolving with the
atmosphere's point spread
function, and you can
think of the blurring
due to the microscope as
convolving with the blurring
function due to optics.
The combined is the
convolution of those
which means that if
you've got some amount
of atmospheric blurring
and a telescope made out
of 12 centimeter optics, then
the combined responses showed
here not very different
from each individual.
But if you made
a big telescope--
a one meter type
telescope-- you might be
expecting this much blurring.
But because of the atmosphere,
you get much more blurring.
So the deviation between
the small telescope,
and what you actually measure
is not very different.
The atmosphere makes
an enormous difference
when you start talking about
a high resolution telescope.
That's the reason for
putting it in space.
You get rid of the atmosphere.
The Hubble Space Telescope
was made principally
out of two big mirrors,
both were parabolic,
both were highly optimized,
both were enormous.
This is the main lens.
The lens is 2.4 meters,
about eight feet in diameter,
and it was
astonishing the thing.
So in order for a lens to work
perfectly like I illustrated,
it's important that
every reflection
remain in phase coherence.
So the length that
every ray travels
has to be precisely matched
to what it's supposed to be.
In the case of the Hubble,
they matched the surface.
The surface was controlled
to within 10 nanometers.
That's absurd.
So the blurring of my microscope
was about half a micron.
So 50 times worse.
The best I could see
with my microscope
is 50 times worse
than was required
for making this mirror.
It was absolutely astonishing
feat to make the mirror
and they made a mistake.
So when they put
this in space, they
were expecting to see
pictures like this.
This is a picture
of a distant star.
They were expecting the distant
star would look like this.
This is what they
actually measured,
and the reason was
that the feedback
system that they used
to grind the lens
made a mistake by 2.2 microns.
2.2 microns would have
been just barely resolvable
on my microscope, but barely.
So the hair?
That's about 100
microns in diameter.
They were off by 2.2
microns, and because of that,
it was like a complete disaster.
So that small error was enough
to make the images terrible.
So the solution-- it wasn't
very practical to ship up
a new lens, so they
shipped up eyeglasses.
The eyeglasses was
another transformation,
just like your eyeglasses.
Your eyeglasses work by changing
the point spread function that
is determined by your
retina and your lens
into a new point
spread function.
They shipped up
eyeglasses, and the result
of putting the
eyeglasses into Hubble
was to turn this into
that, and to give
some of the most dazzling
pictures we've ever had.
So the point is that
convolution is a complete way
of describing a system.
It's a very intuitive way
for certain kinds of systems,
and it's especially
useful for systems
like light based
systems where blurring
is a natural way of thinking
about the way the system works.
Have a good time.
See you tomorrow at 7:30.
