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PROFESSOR: So today I'm going to continue
with frequency response and filtering,
but also begin the story of spectral content of signals.
So our starting point is still something
you've seen before, namely the statement that for an LTI
system, a sinusoid into the system
gives you a sinusoid out at the same frequency,
but maybe shifted in phase and scaled in amplitude.
So a bit of terminology here just for general interest,
we refer to the exponential as an eigenfunction of the LTI
system, because the only effect the LTI
system has on it a scaling.
So an input to some kind of a mapping, which comes out
the same except for a scaling is referred
to as an eigenfunction, or an eigenvector
if you're talking about matrices.
So we say that the exponential-- the complex exponential
here is an eigenfunction of the LTI system.
Because when it comes through, it's just the same exponential,
but scaled by some number.
And that number is what we refer to as a frequency response,
right?
And we've seen that there's a simple expression for it.
And let me put that expression up,
because we're going to use it repeatedly.
The m here is irrelevant.
It can be any dummy index, because we're
summing over the m.
You can call it anything you want.
And I just should mention that there's
other notation for this object.
There are people who refer to it as--
well, it's often referred to as h of ej omega,
because actually, the way omega enters is always in the term e
to the j omega.
So this is-- if you want to think of it that way,
this is e to the j omega to the power minus m, right?
Well, let me just write it as 1 over 1 over e to the j omega m.
OK, so it's some function of e to the j omega.
And people will often write it this way.
And one of the advantages of this
is the notation right away tells you
that this object is periodic with period 2 pi.
Because if you were to increase omega
by an integer multiple of 2 pi in the numerator here,
you'd get the same argument again.
And therefore, h must be the same again.
So this notation has the value that it
keeps the periodicity front and center.
It also makes sense when you're developing
various other transforms.
There's something called a z transform, which we
won't deal with in this class.
But it's used a lot when dealing with discrete-time systems.
And the way that you get from the z transform
to this object is by making the substitution z equals ej omega.
So people will use this notation.
So the z transform uses z exactly where
we use e to the j omega.
But for our purposes, this is a much simpler notation.
It's just that we need you to remember when you see this
that we're talking about something
that's got period 2 pi.
And if you look at the definition, that becomes clear.
If you increase big omega here by any integer
multiple of 2 pi, you're going to get the same thing
back again.
There's another bit of notational confusion that
can arise, which is that people will sometimes
write little omega instead of big omega.
So that's also used.
So this is other notation, and it's notation
that we will try not to use, but you
might see vestiges of this when you look through old problems.
Because in some terms, we may have
used this notation, and some terms,
we may have used a little omega instead of a big omega.
But for our purposes, we'll stick to this.
OK, so when we say big omega, we're
thinking of it as an angle around the unit circle.
So if you've got the complex number here
at an angle big omega, this complex number
is e to the j omega, right?
So we're thinking of big omega as an angle, something
measured in radians, and it's different from little omega.
You can write the expression for the frequency response
in various ways.
So here, I've just used Euler's identity
to split that into a cosine and a sine,
and that's straightforward enough.
The sums are over infinite intervals.
And we talked last time about how stability of the system--
bounded input, bounded output stability of the system
will guarantee that those summations are well defined.
OK, now there's another name for this formula.
Basically, we've called it the frequency response.
But when you compute an h of omega using this formula,
another way to say what you're doing
is to say that you're taking the discrete-time Fourier transform
of the sequence h dot, OK?
So it's the discrete-times Fourier transform.
Again, that's just terminology for now.
We'll come to expand our view of it later.
But we've called it the frequency response so far,
because it describes how sinusoids or exponentials here
get to the output, but it's also referred to
as the discrete-time Fourier transform of the unit sample
response.
So you've got some time signal--
happens to be a unit sample response.
You compute an object through this formula
to get an h of omega.
That's the DTFT, OK?
Another thing we've already seen is
that knowing that you have an LTI system,
and that a cosine is a superposition
of complex exponentials, you can use the result
that we had so far to just describe
what happens to a cosine when it goes through the system.
So it's no longer a complex exponential.
It's a real signal of the kind that we're
more likely to work with.
And we've seen that the only thing that happens
is the cosine that went in gets scaled
in amplitude by an extra factor, which
is the magnitude of the frequency response.
And whatever phase it had, you get an extra phase,
which is the angle of the frequency response.
So actually, if you had an LTI system,
this is a good way to measure the frequency response
in the lab.
What you do is you take your system there,
excite it with a sinusoid.
In continuous-time, we know we can do that with an oscillator.
In discrete-time, you generate a sequence like this.
And then look to see what comes out of the system
and express it in this form, and you'll label the scale factor
there as the magnitude of the frequency response
and the extra phase angle as the phase angle of the frequency
response.
So it makes for a very systematic way
to probe a system and get at the frequency response.
Again, of a point that I've made before,
which is that when you do this probing,
you only need to vary big omega over the range minus pi to pi.
So when we write a frequency response, because h of omega
is periodic with period 2 pi, we only need to probe h of omega--
either the magnitude, that would be one plot, and the angle
would be another plot.
Both of these would be plotted from minus pi to pi.
Because outside of that range-- well,
you can see it already with the cosine.
If I added an integer multiple of 2 pi to omega 0,
I'm going to get an integer multiple of 2 pi added
into the argument of a cosine.
And I'm getting the same cosine back again.
And the reason that's the case is
because the n that's multiplying it here is an integer.
So in continuous-time, it doesn't
work quite the same way.
If I had a little omega 0t and I added multiple of 2 pi,
I wouldn't get the same argument back again.
So what's different here is that if I increase omega
0 by an integer multiple of 2 pi,
because I've got an integer, and outside that,
again, I end up adding an integer multiple of 2 pi,
and I'm back at the same cosine.
So frequency response for a discrete-time system
is always in the interval minus pi to pi.
It repeats periodically outside of that if you chose
to look at some other omega.
And I've said that already.
You actually heard the term frequency response
in all sorts of settings, I'm sure.
One setting in which it's used a lot is in describing,
for instance, the performance characteristics
of a loudspeaker.
So people will tell you how good their loudspeaker
is by showing you the frequency response of the speaker.
And what they're doing is they're
applying a sinusoidal voltage to the input
and looking at the sound pressure that comes out.
SPL here is sound pressure level.
This is measured in dB, so it's actually
a measurement of the ratio of the pressure
that you hear under certain standardized conditions
to a pressure which is taken as the lowest
audible pressure on the ear.
So there's a particular ratio there.
So what they'll do is they'll feed the loudspeaker
with 1 watt at 1,000 Hertz, so just a steady tone.
And then, a meter away from the speaker in an anechoic chamber,
they'll look to see what sound pressure they pick up
on a specialized sensor--
a detector, a microphone basically--
and that number in dB is what they'll represent.
And so typical speakers are--
have values in that kind of range.
Now, if you probe it at different frequencies
applying the same input voltage and looking at pressure,
you'll get varying pressure depending on the frequency
that you probe at.
So this is the frequency response of the speaker,
and if you go too low in frequency,
then you don't get much of a response.
If you go to high in frequency, you
don't get much of a response.
Now, of course, when you use the speaker,
you're not going to probe it with sines and cosines.
You're actually going to put more complicated sounds
in there.
So what you're really interested in
is how does the speaker behave to signals that
are combinations of cosines?
And again, we're using our model of the speaker as an LTI
system.
All bets are off if you drive your speaker so hard that you
get distortion and exercise all the nonlinearities there
or burn it out.
But if you're in a normal range, the speaker is acting linearly,
you can talk about its frequency response.
And what you're really interested in
is how does the speaker respond to linear combinations
of cosines?
And all of these various signals can be thought of as--
at least over reasonable time intervals--
as combinations of cosines appropriately chosen.
So if you hit a particular key on the piano,
you get a dominant note, but you'll get harmonics of that.
And that's what's going into your speaker.
So knowing how an LTI system responds to cosines
then puts you in a position to say
how it responds to combinations of cosines,
or signals that are combinations of cosines.
So the other part of the story that we're going to get
to-- and maybe even by the end of this lecture--
is we need a way to take a general signal
and represent it as a combination of cosines.
And that's what we refer to as the spectral content
of the signal.
So when we talk of exposing the spectral content of a signal,
as over here, what we're saying is
we're going to show you what combination of cosines it
takes to make up that signal.
And once you figure that out, and you have the frequency
response of your LTI system, you can say how
your system responds to that.
OK, so this theme runs through every stage
of what happens, actually, in communication.
Now, the example I've given you here
is one that you would typically probe
with a continuous-time oscillator in the lab.
And so there's some connections that you
might want to make between probing
with a continuous-time signal and probing
with a discrete-time sequence that comes
from sampling that signal.
But I'm going to leave you to look at that later or leave
your recitation instructors to pull that back,
or leave you to draw this up if you have a homework
problem that needs you to think about how continuous-time maps
to discrete-time.
But the basic point is the actual, physical speaker
you might probe with a cosine in continuous-time,
if you're generating that signal from a computer, what you'd
actually be sending to your amplifier
is a sequence of numbers.
And the frequency of the numbers that you would send,
this frequency is related to the frequency
of the continuous-time cosine that you
want in a very particular way.
So I'll leave you to chew on that.
But I don't want to spend time on that now.
OK, so let's spend a little time talking
about the properties of frequency response now
that we know why we would use it.
And this I've already said.
The value of the frequency response at--
some of this, by the way, you may have seen in recitation.
But it doesn't hurt to repeat.
The frequency response at frequency 0--
well, we've said if you've got e to the j omega sub
0 and some frequency omega sub zero going into a system h
omega, a system with frequency response h omega,
an LTI system with frequency response h omega-- all right,
I'm leaving out lots of words.
But frequency response doesn't make sense
unless you have an LTI system.
OK, so for what kind of input signal
would you be looking at omega equals 0?
DC, right-- a constant signal.
It's what the electrical engineers
call DC, which used to stand for direct current
but has now come to mean constant.
When we say a DC input, we just mean a constant input.
So if I pick omega sub 0 to be 0, then e to the j 0n,
well, that's just one for all time.
And so I'm feeding the system with a constant.
That's the slowest possible input that you can find.
It's a 0 frequency input.
And the amount that it's scaled by
is the number that you're going to plot here.
So whatever value you get is going
to end up being plotted there at omega equals 0.
And let's see, do we believe this other statement--
h0?
It's just a substitution in here.
If I put omega equals 0, it's a summation of all the hm's.
But there's another way to think of it also.
If you want to think of it in the time domain--
let's see.
I have an LTI system.
It's got some unit sample response.
And I'm feeding it with an input that's constant for all time.
It's actually constant at the value 1 for all time.
If you're thinking in terms of convolution, the flip slide
and dot product picture, what is the output at any time here?
You're going to draw out your unit sample response.
You're going to draw out your input, which is 1 for all time,
take one of them and flip it over, slide
it the appropriate amount over the other,
and then take the dot product.
Well, for every shift of this flipped and shifted input,
you're going to pick up all of the unit sample response.
So every time, you're going to get summation hm outside.
So if you fed an input that was DC at the value 1,
this is what the output will be at all times.
You can see that from the convolution picture.
So what's the frequency response at frequency 0?
What's the ratio of the output to the input--
the output amplitude to the input?
This is for all time.
So the input amplitude was 1 at each time.
The output amplitude was that.
And so that's the DC gain--
the DC gain of the system, or the frequency response at 0.
Let's say, so h0 is what's referred to as the DC gain.
What about high frequency?
So what's the highest frequency variation
that you can have with a discrete time sequence?
I've got a sequence here at the input.
We've seen what the slowest variation possible is.
It's something that's constant.
If you're talking about a discrete-time signal that
can only take values at integer times,
what's the highest frequency variation that you can get?
Just something that alternates in sine, right?
So you're going to have--
OK, so is this of the form e to the j omega
0n for some omega 0?
Is that a signal of exponential form?
Yes?
AUDIENCE: [INAUDIBLE]
PROFESSOR: Yeah, if you take omega not equal to pi,
this is just e to the j pi n.
In fact, you can take plus or minus pi.
So when you probe the system with an input
of this type, which is the highest frequency
input that you can probe with, what you're really
probing is what's the frequency response at this point?
You get the same value at minus pi or pi.
So these are the two extremes.
And then the frequency response, the rest of it
lies in between for other sorts of inputs.
Now, do you believe this other identity that I have up there?
Well, you can go back to the definition, set
big omega equal to pi or minus pi,
and you'll get an alternating sequence of 1's and minus
1's here.
And so that verifies that identity.
Or you can think in terms of convolution.
If I convolve a sequence like this with a system
with this unit sample response, what comes out at every time
is an alternating sum of the hm's, except the sine flips
from one time to the next.
And so, again, you can verify in the time domain
that that's actually the high frequency gain of the system,
OK?
Now, there's a bunch of other symmetry properties
of the frequency response that I think
in-- at least in some of the recitations you've done.
And the easiest way to see these symmetry properties
is to actually go back to the rewriting
I did of the frequency response in terms of sines and cosines.
This first term here I'm calling C of omega.
The second term here, the summation,
I'm calling S of omega.
So where would a statement like this come from?
Let's see.
For real h of n, that's the only kind of h of n
we're going to worry about in general.
We're going to talk about systems with real unit sample
responses.
if h is real, why would it be true
that the real part of the frequency response
is an even function of frequency?
Well, the real part of the frequency response
is this term, because the other term is the imaginary part.
So the real part of the frequency response
is this term.
And if I change big omega to minus omega,
the cosine doesn't change.
It's the same.
And therefore, the real part is even, OK?
So the real part of the frequency response
is an even function of omega.
The imaginary part, which is the minus S omega,
well if I change omega to minus omega, I flip the sine.
So that's an odd function of omega, and so on.
So you can go through these properties.
Whenever you're stuck trying to figure out a property,
this is the expression to go back to.
So rewrite the basic definition in this form,
and you'll understand a lot of this.
And again, you'll get practice in recitation
if you haven't done that already.
Another important property of--
that you encounter when you go from the time domain
to the frequency domain--
so remember, in the time domain, we
said that if you have an input here, you
convolve that input with h1 to get the output
of the intermediate point?
OK, so if I call the output of the intermediate point--
I should have done it there.
But here's h1.
If I call this w, this is x.
And then I go into a second system, h2.
And here's y.
OK, well w is equal to h1 convolved with x.
And y equals h2 convolved with w.
So that's this.
But I can put the parentheses any way I like for convolution,
right?
We've already established that property.
So the net effect of the cascade of systems
is the effect you'd get by having a single system LTI
with this unit sample response.
Now, if I think the frequency domain--
if I put e to the j omega n here, then
what comes out at the intermediate point?
At the intermediate point, I get h1 omega ej omega n.
All right, so that's wn.
But this is, again, an input of exponential form.
So what comes out of the second system when
I put this input into it?
So what's w of n-- sorry, what's y of n going to be?
Yeah?
AUDIENCE: [INAUDIBLE]
PROFESSOR: Yeah, so it's basically the second system's
frequency response scaling the exponential that
went into the second system, which is this.
So the net effect when I put ej omega in at the first spot
is at the output, I get the same ej omega n,
but scaled by the product of the two frequency responses.
So the nice thing here is that when
I'm describing a cascade of two systems,
if I describe the net effect in the time domain,
I've got to do a convolution of these two units' sample
responses.
If I think of it in the frequency domain,
I just have to take the product of the individual frequency
responses.
So the key observation here is that convolution
in the time domain maps to multiplication in the frequency
domain.
So if I wanted the DTFT of this--
if I wanted the discrete-time Fourier
transform of this result of a convolution,
I can find it by just multiplying
the individual DTFTs, all right?
So convolution in time maps to multiplication in frequency.
And this actually makes design much more easy,
because we're often cascading systems in this form.
And if you think in terms of frequency,
you can track a frequency component
through a cascade of such systems
just focusing on the frequency response of each system
as you go.
So here's an example.
Suppose we have a channel.
Let's say that it's a channel with an echo, so when I put--
let me actually draw it out here.
So I've got a channel here which I'm modeling as LTI.
And if I put in a unit sample function
here, so this has the value 1 at time 1,
suppose the channel is one that has some echoing in it.
So what I actually get out for this input
is the same delta of n plus 0.8 delta of n minus 1.
So there is a later arrival scaled by something
which corresponds to the echo.
So this must be the unit sample response of the channel.
What's the frequency response of the channel?
So if I call this h1 of n, what is h1 big of omega--
big omega-- h1 of big omega?
I don't have it up there, do I?
No.
Anyone?
Just from the definition.
Is the problem here that you don't quite
see what h1, 0 is, h1, 1, h1, 2, and so on?
If I asked you to plot this out, how would you plot it?
Yeah?
AUDIENCE: [INAUDIBLE]
PROFESSOR: Is it 1.8?
Where?
Where would you put the 1--
just over there?
Oh, you're talking about the frequency response.
Let's get the unit sample response first.
Let's sketch this.
What's your sketch of that?
AUDIENCE: At 0B1
PROFESSOR: At 0B1?
AUDIENCE: [INAUDIBLE]
PROFESSOR: OK, on 0 everywhere else--
that's the unit sample response.
OK, so what's the frequency response?
Well, we just plug it into the definition.
All the h's except the ones that argue in 0 on 1 are equal to 0.
So this is going to be 1 plus 0.8 to the minus j omega.
Is that what you said?
It was not quite what you said, right?
What you said was the number I'd get at omega
equals 0-- the DC gain of the system.
But the frequency response is that.
Let's just work backwards here.
So the frequency response is that.
Or if I wanted to write it--
we're going from that board to here--
h1 of omega, I can write it as a real plus imaginary part.
So it would be 1 plus 0.8 cosine omega.
This would be the real part.
Then I have a minus j--
sorry, 0.8 sine omega.
OK, so that's the frequency response--
some complex number with a real part and an imaginary part.
OK, and if I asked you to give it to me in magnitude and angle
form, you could do that.
It's just rearranging things.
So you'd-- the magnitude would be the square root of the sum
of squares of these two pieces.
And the angle would be the arctan of the ratio.
So I assume that you know how to do all of that.
And what you find-- actually, you
can see it in these expressions already.
Just as-- well, I didn't quite claim this earlier.
But the magnitude of the frequency response
will always be a real function of frequency-- sorry,
an even function of frequency.
And the phase will always be an odd function of frequency.
So if you're drawing the results of a computation like this
and you find that you don't have an even function
for the magnitude, then you know you've done something wrong.
So I'm not-- I'm going to sketch something here
which I'm not pretending is the magnitude of that.
I just want you to get the idea of what I mean by even.
It's going to be something that's symmetric in omega.
This is the magnitude.
And then, if I did the phase, the phase
is always going to be something that's
an odd function of frequency.
So if it's an odd function of frequency,
what's the value at 0 of the phase?
It's got to go through 0, right?
And so I might get--
well, what would it actually be?
It would be some shape.
I'm not pretending I have the right shape here.
But it's going to have an odd symmetry.
I'll leave you to figure out what it actually looks like.
So that's the frequency response of this echo channel.
So here's what I want you to do now.
At your receiver, build for me a filter that's
going to undo the distortion that the echo has produced.
So what I'd like is, I'd like an output,
after you've done your filtering, to be
exactly equal to the input.
So my question is, what should--
and my claim is you can do that with an LTI filter.
How would you describe that LTI filter?
What should that LTI filter be?
Yeah?
AUDIENCE: [INAUDIBLE]
PROFESSOR: Right, OK, so if you wanted the output
to be exactly equal to the input,
no matter what input was, you want a frequency response of 1
overall.
And the overall frequency response we know
is the product of the two individual ones.
And so we want h2 omega times h1 omega to be equal to 1.
And therefore, h2 should be 1 over h1.
So you can see here how things get a lot easier when you
think in the frequency domain.
If I had to do this in the time domain,
I would have had to say h2 convolved
with h1 has got to give me the unit sample function.
And I'll give you h1, now you've got to figure h2.
Well, you've got to go and work the convolution picture
backwards, which is doable for simple cases.
But this is much simpler.
So this shows that h2 should be 1 over h1.
Seems like a reasonable way to go.
And you can actually work the whole thing through.
But there's a problem with this.
And we've seen this in other settings
as well, which is something that works
fine in the noise-free case doesn't
work so well when you've got noise in your system.
So look at what this receiver filter is doing.
The receiver filter-- let's see, what is its magnitude?
How does the magnitude of the receiver filter
relate to the magnitude of the channel filter--
of the channel frequency response?
So this magnitude is a magnitude of 1 over h1.
Is that the same as 1 over magnitude of h1?
Is that how complex numbers work?
OK, right?
So look what happens.
Where the channel has a very low frequency response--
in other words, where the channel output is very
low for a sinusoidal input at that frequency,
the receiver filter is going to have a very high magnitude.
So the receiver filter is trying to boost up
whatever signal it sees in a frequency range
where the channel actually has very little output.
So what happens if I come and have a bit of noise
here where I'm receiving the signal?
Well, it's going to be very badly exaggerated
by the inverse filter.
So a little bit of noise here will
get accentuated at frequencies where
the frequency response of the receiver filter is large.
But that's precisely where the channel had
a very low frequency response.
And it's precisely where the output-- the channel--
has nothing interesting for me.
So my receiver filter ends up accentuating the noise.
OK, so yet again, we see that these sorts
of inversion operations may look nice on paper.
But if you don't take account of what noise does,
then you can run into trouble.
And the picture is very transparent
when you think in the frequency domain.
OK, some more practice with filters and cascade--
I think I'm going to leave you to work through this
in recitation, perhaps.
So I'll leave it on the slides.
But let's go to design of filters.
So now, we've seen one example of trying to design a filter--
the receiver filter-- to undo the distortion of the channel.
Here's another-- actually, I want that.
Here is another design problem that you
run into all the time, which is that you see a signal that's
got a whole bunch of frequencies mixed up in it,
and you want to exclude some of them.
So maybe you're looking for an audio signal.
You know that the combinations of sinusoids
that make up an audio signal are unlikely to go above--
whatever you want.
Pick your number-- 10 kilohertz, 20 kilohertz.
And so you want to exclude frequencies outside
of that range.
So you're very often in the position
of trying to build what's called an ideal low pass filter.
So here's an ideal low pass filter.
I'd like you to build for me a filter that
passes all frequencies in some range without distortion,
and that completely kills everything outside.
So let me call this the cutoff frequency.
So that's the h of omega I want.
And now my question is, how are you going to build this filter?
I want you to give me the unit sample
response that goes with it.
And you see a hint over here.
But can you tell me how you might go about that?
Not so obvious, right?
Because we've specified the filter characteristic
in the frequency domain, and now we want to find the h's
that go with it.
So what we're really looking for is a formula
that will give us the time domain signal in terms
of the frequency domain.
So we want to invert this somehow.
So what we're looking for is really
what's called the inverse DTFT.
And actually, if you've done Fourier series,
you've seen this trick before.
Because really, we're not far from Fourier series here.
It's just that the domains are a little different,
so maybe you don't recognize it.
Here, we've got a periodic something expressed
as a combination of sines and cosines,
or as a combination of exponentials.
And now, we want to invert that, OK?
If you thought of these as Fourier coefficients
for some periodic signal, and then went and looked up
whatever book you use for Fourier series,
you'd get the formula.
Because we're just trying to extract the Fourier
coefficients for this periodic signal.
But you can actually do it from scratch.
So if you think of multiplying both sides of this by,
let's say, e to the j omega n--
OK, so I'm going to multiply both sides.
So I've got e to the minus j omega m minus n now, right?
And I'm going to then integrate both sides over an interval
of length 2 pi--
any contiguous interval of length 2 pi.
It actually does matter because of the periodicity.
So I'll take any interval of length 2 pi
and I integrate both sides.
And I'll assume that I can hop this integral in there.
I'll assume my signal is well-behaved enough for that.
So here's what I end up getting.
On this right hand side, I get summation integral hm.
Oh, I should put a d omega there.
Sorry.
I've gotten casual with my integration.
So on this side, I have this integral.
On this side, I have that integral.
And if you work through this, out
of all this infinity of terms, there's
only one term that survives.
Because any term in which m is different from n
will have this exponential still sitting here.
This exponential is like a cosine plus a j sine,
or a cosine minus a j sine.
You're integrating it over an interval of 2 pi.
So any term here that has the exponential,
or has the sine or cosine in it, will disappear
under the integration.
The only term that survives is the one where m equals n.
And so what you discover is that this is
2 pi hn when you're all done.
I'm not going through the details here.
So here is the formula we wanted for the inverse DTFT.
Here's the inverse DTFT, OK?
I've forgotten my colored chalk today, but that'll do.
So if I gave you a filter characteristic like this
and asked you to find the unit sample response of the filter
that went with it, you would just have to plug
in the frequency response characteristic that I gave you
and solve for the h's.
I think I have a bunch of this on the slides.
This is what we just went through.
So let's do this now for the ideal low pass filter.
What is it that we do?
I've got the formula that I just derived for you there.
h is equal to 1 in the pass band of the filter,
and it's 0 outside of that.
So I set h equal to 1 in the pass band of the filter, which
is from minus omega C to plus omega C, and the rest of it
doesn't contribute anything.
And then, I just work out this integral.
And I've actually got to do it in two pieces.
For n not equal to 0, this is what I get.
For n equals 0, this is what I get.
If n was continuous, actually, you'd
say that this is the same expression as here,
because you just use L'Hopital's rule
and you'll get from here to here.
But since n is an integer, we've got
to be a little careful how we write it, OK?
So you can't really say you're going to use L'Hopital's rule
to see what this is in the limit of n going to 0,
because n takes integer values.
But if you work it out from scratch for n equals 0,
you'll see that you get a formula that's
consistent with using L'Hopital's rule.
OK, so this is a function that we'll see again and again when
we do filtering of this type, and it's
referred to as a sinc function.
So it's not S-I-N, but S-I-N-C. And if you plot it out,
this is what it is.
So it's got the oscilation that comes from the sine,
but it's got a reduction in amplitude that comes from the 1
over n.
So it's a signal that falls off as 1
over n with this kind of a characteristic.
Do you think it's a bounded input, bounded output
stable system?
What's your hunch?
Remember what it takes for a system to be stable?
The unit sample response has to be absolutely summable.
So if you take the absolute values here,
and sum from minus infinity to infinity,
you want to get something finite to call this stable.
Well, since this only falls off as 1 over n,
it turns out to not be stable.
So it's actually an extreme idealization
that is not bounded-- input bounded, output stable,
but it's close.
Just to go back when I showed you this filter characteristic
here, to give the cheap version of a low pass filter, what
we actually did was take the sinc function
and truncated to a finite interval.
And so what happens when you truncate it to a finite
interval is that instead of the sharp box-like shape
for the frequency response, you get a closer approximation
to it--
not exactly the ideal low pass filter, but maybe good enough.
The other thing that you might notice
if you're looking carefully is that I had a sinc that
was centered around 0 and even.
And now, I seem to have a causal version of the filter.
And I think I'll leave you in recitation to figure out
how you can go from the centered, non-causal filter
to a causal filter, and what that
does to phase and to frequency response magnitude.
So basically, I'll leave you to go through the details here.
But the key idea here is the inverse DTFT.
So now, I want to just take a slightly different perspective
on this formula that we derived.
We said we've got a frequency response, which we're
calling the DTFT of the signal h of n-- the unit sample
response.
We've got an inverse formula that
allows us to get the time signal from the frequency response.
But here's yet another way of looking at what
this formula is telling us.
This formula is saying, I can think of h of n
as being made up of a whole bunch of complex exponentials.
So you see that this is what we were looking for.
We were looking for a way to take a signal
and figure out its spectral content.
We want to know what complex exponentials,
or what sinusoids does it take to make that signal?
Well, we have a hint of that in this expression,
because this is saying, take the time domain signal.
I can think of it as being a combination.
Now, this is not a finite combination, it's a continuum.
But it is a combination of exponentials of the type
that we know to work with.
So this is actually giving us a spectral decomposition
of the unit sample response, where
the amount of e to the j omega n that it
takes to make up the signal is told to me by h of omega.
So the h of omegas are sort of the weights
that we use to combine these exponentials to get the signal.
So the idea for a spectral decomposition,
or for describing the spectral nature of a signal
is actually sitting there.
All we have to do is say, we'll use the same formulas,
but let's no longer restrict it to the unit sample
response of a system and the frequency
response of that system.
Let's use it for any signal--
so the same formulas, but now for any signal xn.
Give me any signal xn, I'll compute for you
this object, which is the DTFT of that signal, just
the same way I did for a frequency response.
So I'll compute the x of omega for you.
What's the significance of x of omega?
Well, it tells me in what combination
I have to wait the e to the j omega n's
to construct for you the signal.
So the x of big omega, the DTFT tells me
what the spectral content of the signal is.
If I plot that as a function of frequency,
it tells me how to assemble the signal out
of sums of sines and cosines.
So let's see here.
More specifically, what I would say is that the DTFT at omega
0--
omega sub 0 times d omega is the spectral content
of the signal in that particular interval.
And if I add up all those components over all frequencies
in this 2 pi range, I'll get the original signal
that I'm interested in.
So what we'll do next time is work with this idea
to see how it lets us think about signals through systems,
and how it enables us to do filtering in a systematic way.
All right, let's leave it at that for now.