In this video we're going to implement a
digital filter.
Now, I'm assuming that we have a signal.
That we have run through an A2D converter,
it's been sampled appropriately.
And now we're have it inside the computer
and now we want to do some signal
processing on that signal.
The most common thing to do of course is
to filter it and we have to figure out how
to actually implement additional filter.
Implementation here, of course, means
software and we have many things we can do
in software.
You can do esentially anything you want,
but we need something that will wind up,
giving us a linear shifting during filter,
and the technique I'm going to talk about
today is the difference equation approach.
This is a quite general approach, and will
allow us to have any kind of filter we
want low pass, high pass, band pass.
We're going to talk though about some
filter categories that come up in the
digital world that this kind of
categorization doesn't come up in analog
filters at all.
So it turns out to be very important for
the digital filters to understand this
category, and we're going to develop new
input, output relationships both in the
time domain and in the frequency.
And we'll see how that works.
Well first of all we have to define what a
difference equation is.
So, we have for example a, what we want to
do is to implement a linear shift
invariant system.
And as always we are going to let x be our
input.
And y would be our output.
And the big question is going to be how do
you actually build filters?
How do you implement them for digital
signals?
Don't have circuits anymore you're inside
the computer.
It's all done in software.
So the technique we're going to use is one
called the difference equation.
And a difference equation is quite simple.
You have here the output and you can see
that the output depends on previous
outputs, the one just before it and the
one that was [unknown] samples away.
You multiply them by the constants and you
add them all up.
Then you take the current input, the
previous input when q back in time,
multiply them by their own constants, add
them up, take the sum of all that, and
that is your current output.
So the output now.
Equals the previous output times A1, plus
the 1P times ago times AP, plus B0 times
the current input, et cetera.
So, you can see, it's a pretty simple
thing, and it turns out we will only need
to know these What are known as filter
coefficients.
The a's and the b's.
And how many of them there are.
That's the p, and q.
Once we have those and on the filter
coefficients, we can do anything we want.
Literally, you can do some very fancy
filters, even ones that go beyond
bandpass, low-pass, high-pass, that kind
of thing.
Now I want to point out, that this is an
explicit input-output formula.
What I mean by that.
Not only does this equation, the
difference equation, specify the kind of
filter we want through its coefficients.
But this is a way you could implement it.
You can actually program this up.
You set up a table consisting of the
previous ys that you need.
The previous xes that you have.
Input comes in.
You stick it into this formula.
Multiply those things you stored in tables
by constants.
You've got the previous outputs.
Multiply them by their constants.
Add em all up, then now you've got the
current output.
For the next in, output value to compute,
you Put this over that place in the table.
Put this one over here, etcetera.
Do the same thing to the xes.
And you have now implemented a digital
filter.
You can actually, it's a very simple
programming exercise.
Just consistent multiplies and additions.
And turns out it's a very, very easy way
to implement a wide variety of folders.
Alright.
So let's see what the output is for a very
simple filter here.
For a very simple input.
So I only have one previous output there
that enters into the difference equation,
just the previous one.
And I'm just going to take the current
input.
Input.
So, the current input gets multiplied by
b, the previous output gets multiplied by
a.
Add those two things up, and you get the
output of the, the filter.
I do emphasize, this is linear and is
shifting variant, system.
The simple input I'm going to use is just
a, a unit sample.
So, it's a, a very, very simple input, I
mean one value is not here.
We'll learn in this video that this is a
very important input to know the output
for.
So, what do we do?
Well, what I like to do to point out how
the difference equation works, is to
construct a table.
So, what I'm going to do is have the in,
the time index.
The input value at that time.
And the output value of that time.
And we want to fill in the right hand
column of the table.
Now let's start at n equals minus 1.
So, we know that at this time the input is
zero, because that's that value there.
The question is now, what's y at minus 1?
What the difference equation says.
Is that, that is equal to a times y of n
minus 2 plus b times x at minus 1.
And we know that's zero.
But what's y of minus 2?
Well, that y of minus 2 from the
difference equation of course depends on y
of minus 3.
Y minus 3 depends on Y minus 4.
Going on, and on, and on back in time.
Well, how do we get around this?
Have these values been specified?
And there is an assumption that we make
and [inaudible] is that all sigmas if
minus infinity are 0.
So, with that and the fact that the input
is 0 for all these values back in negative
time, we know that Y of minus 2 is equal
to 0.
Well, that, of course, gives us that the
output at time N equals minus 1 is 0.
And now lets let that input come in, and
now we're going to talk about what happens
at n equals zero.
We know that the input is a 1 there.
And what we're going to do is to find y of
n.
We're going to multiply that by b.
We're going to multiply that by a.
Add them up and of course you get just b.
And that also applies to the next value.
We now, again, all we do is multiply that
value by B and that value by A.
Add them up.
That's exactly what the difference
equation says to do.
And, of course, we just get A times B.
And to go into the next moment in time and
all succeeding ones, I think it's pretty
clear that what happens, for example at
time 2, if we multiply that by B and that
times A we're just going to get a B
squared and it's, kind of think, pretty
obvious that at, at any time N.
As long as it's greater than or equal to
0, the output is b times a to the n.
So, when I put in a unit sample, my output
is this signal for n positive and I can
write that very succinctly Using the unit
step[INAUDIBLE] in my notation here.
So, unit step, of course, is equal to zero
for all previous values.
And then negative values.
And then it's equal to 1.
Full values [inaudible] and that are zero
are positive.
And that's a very concise way of using
this to write down this.
This output we're going to find this
happens all the time and is very, very
useful.
Okay, so what does this output look like?
So there it is our difference equation our
input and we have computed that for an
output.
Let's plot it.
For various values of a and d.
Now, I think you can see that the
coefficient b here amounts to a gain.
If I change b all it's going to do, is
make the signal bigger or smaller.
So in all of these plots I've said b equal
to 1.
To simplify things and here's my favorite
value for filtered coefficient one half
and you can see.
The output filter here for this unit
sample input is a one half to the end and
it just decays.
And in fact, it decays exponentially.
If you have n equals minus a half, it's
also decaying exponentially.
But now, every other, the odd value
samples alternate in sine.
They're negative, the pos-.
The ones for the even ones are positive.
And now, he has something that goes up
and.
Up and down, and has a very different
character than when a is positive.
Another interesting value here to play
with is when a is greater than 1.
And when you do that, does, the output
here is going to be 1 point 1 to the nth.
And it's going to grow.
And this turns out to be, not be terribly
useful.
If you think about it for a second.
This part here.
All these previous outputs, when you're
out in here, are going to keep getting
bigger and bigger and bigger.
And swamp out whatever values you had for
the input.
And basically, the filter ignores the
input.
Good, the n gets big enough.
So what we have is that we had better have
the absolute value of a to be less than 1
or we don't have a filter that's
interesting.
It will [unknown] of a outside that range
to.
And the sample response will just take
off, and the filter has a mind of its own,
as it were.
Okay.
So, lets think about what this filter is.
We know how to compute it.
We can plot it, but what kind of filter is
it?
And that of course means that we have to
hop into the frequency domain.
Well, to determine what the filter is,
we're going to do exactly the same thing
we did with analog circuits.
I'm going to assume the input is a complex
exponential, at some frequency f, having a
complex amplitude of x.
And, we're going to assume that, the
output has the same form, complex
exponential in, complex exponential out.
And we're going to see if this assumption
is correct by seeing if we can determine
an output amplitude that will satisfy the
difference eq [inaudible].
So I'm just going to stick in this these
assumptions for x, y and x.
And what you get is something that looks
like this, and notice this intermediate
term in here, y of n minus 1.
When you stick that in for the complex
exponential.
A factor of e to the minus j two pi f
comes out and we're left with our friend
the complex exponential.
And that's what I used when I wrote this
equation.
But I think you can see right away that
what happens is the complex exponentials
cancel And sure enough, if I bring the AY
term to the other side, factor out the Y
and divide by 1 minus A, I have a solution
for capital Y that works.
As long as the ratio of Y and X is equal
to this.
I stick in a complex exponential I get out
a complex exponential.
Well, we call this ratio a transfer
function.
So we can, have used this technique to
find that the transfer function has this
form.
Well as we've seen already before, the b
here just amounts to the gain or a simple
filter.
So, whatever number you put there just
supplies a gain for the transfer function.
For the filtering those affected by a,
what kind of filter it is, how good a
filter it is.
Is all determined by A because that's
where the terms involving frequency are.
Alright so, what we've seen is that the
output is equal to the transfer function
times the complex amplitude of the input
Times the complex exponential, output's
exponential N, complex exponential N.
Well, what does this, transfer function
look like?
So, here are some plots and, for various
values of A.
And again, just like I did before, I set B
equal to 1 because that's just the game,
we don't need to worry about it too much.
Okay.
So, for my favorite filter value here, an
A of a half, you can see that it starts
out and kind of gets small.
So, I guess you'd call that a low pass
folder.
It's not a very good one.
And what are these values here at the low
and high frequencies here?
At F equals zero, you're putting in F
equals 0 into this formula, you get 1 over
1 minus.
Hey.
Stick in f equals one half, well, e to the
minus j two pi f times a half, well that's
b to the minus j pi, and if you know
that's minus one, And so what we get for
the value at frequency half, is 1 over 1
plus a.
So this value is something like 2 thirds,
and this value up here is 2.
So it doesn't have much difference in the
game at high and low frequencies.
However, if you stick in a Bigger value.
A like .9.
Remember, we have to keep the values of a
in magnitude between 0 and 1.
So, A is a pretty big value for A.
And you can see, we get a much nicer
filter.
And it is low pass.
So what we have found out here is that for
a positive, that implies low pass open/g.
So if you now think about the, negative
values of a You get a high pass
[inaudible].
And, if I use a value of A that's minus
point 9 then you get something that's
symmetric and gives us a very nice high
pass[INAUDIBLE].
So, a greater than 0 but less than 1.
I get a low pass filter.
If I get, have an a less than zero and
greater than minus 1, I get a high-pass
filter.
Okay.
So, what happens in general?
How do we find the transfer function for a
difference equation in general?
And you do exactly what we've just gone
through.
Set x equal to a complex exponential.
And assume that the output is the same.
And I think it's pretty easy to see.
It's all going to work.
And what you get for a transfer function.
Always looks like the following.
You have a numerator term, which consists
of the input terms from the difference
equation.
You are going to have a denominator that
consists of all the terms that involved
the previous outputs.
The coefficients of the filter are staring
you in the face from the difference
equation.
Very, very easy to write this down.
That's not very difficult at all.
I do note that these minus signs come
about because of the way you have to solve
things.
Remember we have to bring these terms to
the other side when we found the transfer
function.
So the Ds all come in with plus signs.
The As enter downstairs with minus signs.
And this we can use to do what's called
filter design.
So, what you want is a way of choosing
these, filter coefficients a and b in such
a way that you get the kind of filter you
want.
Where there are software packages that do
this.
It's not nearly as hard as it might be in
the analogue world if you're stuck with
circuits.
You can use software packages.
And they will you tell them what kind of
filter you, you tell the program what kind
of filter you want.
And they'll pass let's say cut off
frequency of .2.
And you specify how good a filter it is
how sharply it goes to 0 from 2 and it
will pop out filter coefficients that you
need.
It's not very hard at all.
Well, let me point out something that is
different than the analog world.
And this is a special filter, special kind
of filter.
And what makes it special is that there's
no term here that talks about the previous
outputs.
That's right.
So, this filter, the difference equation
only depends on the input.
And we want to figure out how this works.
Well, I'm going to do the same thing I did
before and assume I have a unit sample
input and I'm going to build my table.
Well it turns out to be a whole lot easier
than it was before.
Since this output here only depends on the
previews, inputs, there's no concern about
what y minus 2 is.
So we were [unknown] n minus 1 what this
filter does, is it takes together We all
the in the input values here.
Adds them up.
Divide by the number of terms.
And puts out the answer.
Well of course that's 0.
Well if we now encompass that unit sample.
We look here.
We go back, q returns, add them all up
divide by q.
Well it's pretty clear that you get 1 over
q.
And if you go to time 2, we're now right
here, so what the filter does is it takes
that guy you.
N K and this previous adds them up divide
by q well you're only getting this innate
sample in here everything else is 0 so
we[UNKNOWN] for q.
And we can keep going with this I think
it's pretty clear we still get answer 1
over q.
Until we get to the time Q minus 1.
And that turns out to be right here.
So, you look at this last term, respond to
this, x of q minus 1 minus q plus 1, and
that's x of 0.
So this is going to be the last time in
which our filter grabs the input at timing
equals zero.
And of course, we get 1 over q for that.
Now, beyond that point, the filter is
sitting out there.
And it grabs only these inputs, adds those
up.
They're all 0 so you get 0.
So, for all subsequent times the output is
just, just 0.
So, for our special filter, which has this
difference equation.
We put in the unit sample.
The output basically looks like a pulse.
And that's what it is.
It's a pulse.
So here I plotted for specific value q and
pretty easy to implement and see what's
going on.
Let me point out something about this
filter that makes it special.
What is this filter doing?
It's taking q values of the input, adding
them up, and dividing by q, the number of
terms.
What would you call that?
In, standard, non-techno plots .
You take Q values together, add them up,
divide by the number of terms, and that's
called averaging.
So, this filter is special in the sense
that it averages it's input over Q values
and that's.
What the output is.
Sometimes this is called running average
because you keep doing this every N and
whichever N you pick, you're averaging
this, then you're averaging that, then
you're averaging that, etc.
And what kind of filter is the averager?
Remember we want to implement our And
software our filter with a difference
equation, but what kind of filter is it?
Well, what I'm going to do is find that
transfer function.
And remember, the general form is that
when you only have, the inputs, what you
do is your going to get a transfer
function that just has that in the
numerator.
So, for the transfer function here is
going to be, so we're going to get a 1
over q, times 1 plus e to the minus j 2
pie f plus E to the minus j, 2 pie f times
q minus 1, and all the [inaudible]
coefficients are 1 over q.
Well, we've seen this sum before.
And that should remind you of the digital
sin, and there it is.
So there's our digital sink.
There's over q, and we get this linear
phase term that we know comes from adding
up all of these terms.
Well, what does that look like?
And what we see is that our averager.
Is a low pass filter.
This is a very important concept to have.
Averager is equivalent to low pass filter.
The cutoff frequency if you will is at 1
over q.
This is a plot for q as follows.
And so if I come up with a filter with a
bigger value of q, I average more terms
the gain at the origin is still 1, but the
cutoff frequency is down here and the
ripples get smaller actually.
So you can add a better low pass filter By
averaging more terms we'll also, the
cutoff frequency moves to lower
frequencies.
And I think intuitively this is exactly
what averaging does.
It removes high frequencies, and leaves
only low frequency variations.
And I think it makes intuitive sense that
averaging.
Could be just a low pass filter.
And I think it's very neat that, in,
electrical engineering.
We can actually implement averaging.
And think about it as a filter.
Especially on a computer.
Well, and now I want to do something
special and point out why the unit sample
response is so important.
So, here's our linear shipped invariance
system and it turns out I'm not going to
worry about whether it's implemented with
a difference equation or not.
Is just linear in shift and variant.
What I want to do is for an input I want
to use a unit sample.
And I'm going to refer to the output for
that special input, as little h of n.
So in notation, a little concise notation.
If the input signal is a unit sample,
we're going to call that special output
little h of n.
Okay?
Now, because this filter is shift
invariant, if I delay that unit sample, by
n at the input, the output is just going
to be a delayed version of the output,
little h, okay, a very important fact to
realize.
For what's coming up.
Well as we've seen, you can think about
every signal in discrete time as a super
position of unit samples.
So, the unit sample at the origin, and
it's aplitude is x0.
Sample it time one, it's amplitude is x
sub 1.
Here's x of minus 2, and that's just a
unit-sample of time minus at time.
What am I saying, minus here should be
plus.
So unit-sample This is a unit sample whose
amplitude is exa plus 2, exa plus 1.
And I can write that concisely, that
expresses the super position.
Well, if I stick a super position and
signals into a linear shift and variance
system, what do I get out?
I get a super position of the applets to
each.
And therefore we can immediately write
that the output for our input x expressed
as a super position of unit samples is
given by, the, a super position of delayed
unit sample responses.
That's what little h is it's called unit
sample response.
Well, in fact we now have a very general
way of expressing the input output
relationship for a linear system.
You give me.
The input signal.
You give me the unit sample response.
I can use this sum to compute, what the
output is.
You don't need the difference equation,
this is an alternative.
It's usually not a efficient way of
computing the output.
The difference equation is much, much,
much better but for theoretical reasons as
you' re about to see this is important
input output relationship.
Well I want to find the transfer function
cause it should be as we've seen for a
linear shifting variant system.
The, we should be able to find the, the 4A
transformer.
The output is related to the 4A
transformer, the input.
And for that N, I'm going to compute the
DTFT of the output, which means I take the
output signal, multiply it by a complex
exponential and add them up.
I do the same thing to the other side.
And what I've done is, since I'm summing
over n here.
I've moved that sum on in-, inside, and do
it first.
'Because there's only 1 term that depends
on n.
Well, what's this?
What is that?
That, to me, looks like.
The DTFT of a delayed signal.
And so I know what that is.
That's the same as the DTFT of little h
times a linear phase.
Well, what do you call the DTFT of little
h?
You call it big H.
So I think you see where this is going.
Now, we're going to get to the transfer.
Function.
There's what we're going to call the
transfer function in just a second.
And so there's the linear phase that has
to do with that delay.
If I pull this term out of the sum on m.
It doesn't depend on m.
I now see that I've got the dtft.
Of the input.
And I arrive at a very nice result.
That the DTFT of the output equals the
DTFT of the input times the transfer
function.
But what the new thing is here.
Is that this, is the forier transform, of
the unit sample response.
So transfer function here.
Unit sample response here.
And they are Fourier, they constitute a
Fourier Transform pair.
So, this is a way of finding the transfer
function.
That's why finding the end sample response
in all those examples, what you did for
difference equations, was so important.
It's because I know that if I can computer
the DTFT of that unit sample response,
I've got the transfer function right away.
Very, very nice.
Well, let's talk about those filter
categories I mentioned.
What we like to think about digital
filters as falling into one of two
Classes.
The FIR and IIR.
And what does these acronyms mean?
Well, it turns out, the FIR case.
It turns out it is finite.
There's a word missing in the acronym,
duration.
Impulse, response.
That's what FIR stands for.
And the impulse is a, is another name for
unit-sample.
Example.
So to keep things short, this has been
adopted over, since the beginning of
digital-signal processing.
And so what that means, as we've seen, is
that the difference equation has no terms
that depend on previous outputs.
Only consists of terms that add together,
previous out, input values.
When you have a difference equation that's
a little bit more general, that includes
those output values, in addition to the
input values, we know that the unit sample
response as we've seen and our example is
infinitely long.
Remember, our, when we had one term our
unit sample response, what looked like A
to the N, was lasted forever.
Even though the unit sample, which is only
one sample long, yet the output lasts
forever.
So, that's what IIR stands for, it stands
for infinite duration impulse response.
So, these categories depend entirely on
the behavior of h.
Well, so one little thing I want to do is
compute the duration of the output for a
finite duration input in the FIR case.
And this turns out to be really important.
So, lets assume that my input is something
very simple.
Is three samples long.
Not very interesting, but will serve the
purpose for this example.
And my little h is something that's going
down, looks like that, and then it's 0.
Okay?
Well, what's the output going to be?
We know by the superposition principle
that all we do is found the output for
each of these inputs and then add them up.
So the first one gives that for an output
and goes through zero.
The next one gives that for an output.
[unknown] And this one gives that output,
[unknown] and then we add them up.
Well how long is this imput, I mean sorry
is this output?
The imput went to time N minus 1.
This end, goes to output.
Q minus 1, it's duration is q, they're q
non 0 values.
And this time right here, is n plus q
minus 2.
So that means it has duration n plus q
minus 1.
Duration of the output equals the duration
of the input ti-, plus the duration of the
unit sample response, q, minus 1.
So this little minus 1 here.
Sort of a bookkeeping kind of thing but
that's important to get this down as
you're going to see we're going to be talk
about this later.
So, back to the IIR case.
If you have a finite duration input the
output has an infinite duration in
general.
So, there's no real calculations to do
here, but this is very important as we'll
see in, in another video.
Well another property of FIR filters is
very important, has to do with the phase.
You can have a linear phase and we seen
that already for our averager.
But in the IIR case, it always has a non
linear phase.
And, let me bring up the phase plots for
our filters we've been talking about.
So, here's our linear phase coming out of
an FIR averager.
And what does linear phase mean?
[inaudible].
Linear phase means delay.
So what this means when you have a
transfer function at, with a linear phase.
Is that every frequency.
Maybe filtered, this amplitude is changed
but every frequency, no matter what it is,
experiences this same delay.
When you have a non-linear phase you're
talking about a frequency here and a
frequency here, well for this 1, it
experiences that, but since it's
non-linear.
These two frequencies experience different
delays.
Well, why is this important?
This has to do with the subtleties in ways
signals look.
So it turns out your ear is very
insensitive to fa-, to fe- Phase.
If you had the same filter implemented
with a FIR filter or done with an IIR
filter and listened to the result you
couldn't tell them apart.
You can phase shift the input, the speech
input, music input it doesn't really
matter as much as you want.
With non-linear phase, linear phase you
cannot with your ears tell the original
from the phase shifted input.
We just can't hear, your ear doesn't care.
However, if you reply a non-linear phase
shift to an image.
And filter that, it will really mess it
up.
It will be entirely gibberish in most
cases.
So for image processing they like to stick
with FIR filters, because you can design
them to have linear phase in which
everything just gets shifted over.
But [unknown] phase for an image is very,
very.
Very bad.
And you'll see that as you take other in,
or advanced digital signal processing.
Okay, so, digital filters.
Very interesting and very important
classes are described by difference
equations.
And now only is this the way of describing
the filter, this is a way of implementing.
Now it turns out that the difference
equation is not always the most efficient.
And we're going to talk about this in
another video.
It's not the best way implementing a
digital filter.
What I mean by that is, it is not as fast.
Clearly you don't want to wait for your.
Your results to come in, you want your
filter to run very quickly.
Well it turns out in terms of me getting
to the details of counting the
multiplications and additions it takes to
find the output for a given input.
It turns out the difference equation works
fine, but it isn't the best.
And we're going to talk about that in
another video.
So again the input output relationship for
general of linear filter can be expressed
at either the time domain or in the
frequency domain.
Time domain is this.
Frequency domain is given by our old
friend the transfer function kind of
relationship and we can do either specify
how what the filter does.
Either way I find using the transfer
function is more useful when you're trying
to figure out what kind of filter it is
for example.
What the new thing is that the unit-sample
response and that transfer function are
Fourier transform pairs.
So, if you can find this, you can find the
transfer function very easily.
It's not hard at all, you just find.
Fine with the DTFT.
And in succeeding videos we're going to
talk about efficient ways of implementing
filters.
And it turns out many of those cases wind
up being implementing filters in the
frequency domain.
Kind of a not obvious thing, and we'll
talk about that coming up.