9.2 - Controlling the bandwidth

From the official description of 9.. videos:

Welcome to Week 8 of Digital Signal Processing.

This week's module is about digital communication systems and this is where it all comes together; from complex-valued signals, to spectral analysis, to stochastic processing, sampling and interpolation: everything plays a role in the design and implementation of a digital modem. Digital communications is an extremely vast and fascinating topic and it is arguably the pinnacle achievement of DSP in the sense that it's the domain where the most extraordinary quantitative progress has been made thanks to the digital paradigm. The fact that MOOCs such as this one are available to such an incredibly vast audience is just one of the tangible results of digital communication systems. It is only fitting, therefore, to devote the last module of our class to this subject.

We will start with the basics of data modulation and demodulation and we will progress to describing how your ADSL box works by way of its direct predecessor, the voiceband modem that spearheaded the Internet revolution by allowing for the first time the delivery of substantial data rates in the home.

Hi, and welcome to module 9.2 of digital
signal processing.

We are talking about digital
communication systems, and in this

module, we will talk about how to fulfil
the bandwidth constraint.

The way that we're going to do this is by
introducing an operation called

upsampling.
And we will see how upsampling will allow

us to fit the spectrum generated by the
transmitter onto the band allowed for by

the channel.
Remember that our assumption is that the

signal generated by the transmitter is a
wide sequence.

And therefore, it's spiral spectral
density will be full band.

What we need to do is to shrink the
support of it's spiral spectral density

so that it fits on the band allowed by
the channel.

The way we do this is by using multirate
techniques.

In multirate, the goal is to increase or
decrease the number of samples of a

digital signal.
One way to do this is to interpolate the

digital signal into a continuous time
signal.

And then resample the interpolation at a
different sampling rate.

However, we want to avoid the transition
to discrete time, and we want to perform

this aritficial change of sampling rate
entirely in the digital domain.

Let's consider the up sampling operation,
which is really what we're interested in.

And let's look at how to do this, going
through an interpolation and resampling

operation first.
So we have a discrete time signal here,

we interpolate with the given period Ts.
We obtain a continuous time signal and

then we sample this continuous time
signal with a period that is k times

smaller than the original interpolation
sample, and we obtain another discrete

times sequence here.
Graphically, assume this is our discrete

times signal.
The interpolation to continuous time will

give us this and the resampling with a
smaller sampling period will give us a

higher density of samples on the same
curve so that the resulting upsampled

signal would look like this.
Now, we are interpolating to continuous

time, so the choice of the sampling
period is completely arbitrary for

simplicity as per usual, which is Ts
equal to 1, and here we have that the

interplay signal is given by the standard
sync interplay formula.

When we resample, with the period that is
1 over k, k times smaller than the

original, we are taking samples of the
interplayed function at n over big K.

And the result Is an interpolation
formula, where the sync function now, is

centered at fractional intervals, so n
over big K.

In the frequency domain, the process
looks like so.

Imagine we have a discreet time signal,
whose spectrum is limited to 3 pi over 4.

We interpolate it to continuous time, and
we get an analog spectrum that looks like

this, where our nyquist frequency is
omega n, equal to pi over Ts.

When we resample, with the sampling
period which is k times smaller, that is

equivalent to multiplying nyquist
frequency by k.

So we make it bigger and we move it over
here.

And when we plot the result in digital
spectrum.

We map, as per usual, the nyquist
frequency to pi, which corresponds to a

contraction of the frequency spectrum by
a factor of K.

If here, we choose k, which is equal to
3.

We get that what was the highest
frequency of the spectrum, 3 pi over 4,

now becomes pi over 4.
Can we do this completely in the digital

domain?
Well the idea is that we need to increase

the number of samples by a factor of K.
And obviously the sample sequence will

have to coincide with the original values
when the index of the up sample sequence

is a multiple of K.
There are several reasons why this is so,

but probably the most intuitive one is
that if we then discard the extra

samples, we should be able to obtain the
original sequence again.

So for lack of a better strategy, we can
start by building a sequence where we put

the original samples, every K samples and
then we put zeros everywhere else in

between.
So for example, for k equal to 3, the

upsample sequence, say this is m equal to
0, will be equal to x0 and m equal to

zero.
Then we put two zeros, then we put x1,

then we put two zero, then we put x2, and
then we put two zeroes.

We can see this in the time domain start
with the same sequence that we showed

before.
And what we are doing, we're simply

introducing zeroes between each stamp.
With this choice, the Fourier transform

of the upsample sequence is rather easy
to compute.

We just write out the standard DTFT
formula, but now here, we remember that

xu of m will be equal to zero every time
that m is not a multiple of K.

And so with this, we can simplify the sum
and use only the known zero terms.

And we get the sum from n that goes to
minus infinity to plus infinity of x of

n, which is our original sequence, that
multiplies e to the minis j, omega nK.

And so, this is simply a scaling of the
frequency axis by a factor of K.

Graphically, we can plot the digital
spectrum and we know that now, since

we're multiply the frequency access be a
factor of K, there will be a shrinkage of

the frequency access like this.
But we should never forget that the

digital spectrum is 2 pi periodic.
So lets plot this explicitly, for minus 5

pi to 5 pi.
If we choose k equal to 3, we're mapping

the interval from minus 3 pi to 3 pi back
onto the minus pi, pi interval.

And when we do that, we get something
that is very close to what we obtained

going through the analog domain.
In the sense that this frequency here is

again pi over 4.
But, we have extra copies that have crop

in the main frequency interval.
Now, we know what to do in this cases, we

apply some drastic low pass filter to get
rid of them.

We choose an ideal low-pass filter with
cutoff frequency pi over K, because this

is where the original pi in the frequency
spectrum would be mapped to.

And this will get rid of the extra
copies, and leave us with a spectrum that

is identical to what we obtain using an
interpolator followed by a sampler.

So now let's look at the procedure back
in the time domain.

So the first step is to insert K minus 1
zeros after each sample, followed by an

ideal low-pass filter.
And we choose the cutoff frequency for

the filter to be pi over K as we saw in
the previous graph.

So now, the resulting sequence is simply
the convolution of the upsampled sequence

with zeroes.
And the impulse response of the filter

that, with this cutoff frequency will be
simply sink of n over K.

And if we work out the convolution sum,
we have this summation here.

But again, we remember that of these
terms, only 1 every K will be non zero.

So we replace i with mK, and we sum over
m.

And we get the sum for m that goes to
minus infinity to plus infinity of x of

m, sync of n over K minus m.
Which is exactly the same formula we got

using an interpolator and a sample.
As we've mentioned before, if we have an

upsampled sequence we can always recover
the original sequence by downsampling,

which means we keep only one sample out
of k and throw away the rest.

Now, this is obvious in the case of an
upsample sequence where we just introduce

K minus 1 zeroes every sample, but it is
also true for a filtered sequence where

we used an ideal filter.
Or any other filter that fulfills the

interpolation properties that we have
seen in module 6.

In other words, we want impulse response
to be equal to 1 for n equal to 0, and to

be equal to 0, for all multiples of K.
In general, downsampling is a more

complex operation that upsampling, just
like sampling is more complicated than

interpolation.
We have the pesky problem of aliasing,

because we are throwing away information.
We will not develop the properties of the

downsampling operator in detail because
we will not need it in the following.

But you're encouraged to read about
multirate signal processing in the book.

So let's go back to the bandwidth
constraint, as you remember the channel

imposes that we only use frequencies
between Fmin and Fmax.

We also want to translate these
requirements into the digital domain.

So we choose a sampling frequency and Fs
over two, half of our sampling frequency

will be our nyquist frequency in the
analog domain.

Now, here's a new trick.
Compute the positive bandwidth.

Namely, the width of the channel
bandwidth on the positive axis, and call

that W.
Now, pick the sample frequency so that

two things happen.
First of all, the sample frequency will

have to be at least twice the maximum
frequency that we can use in the channel

to avoid aliasing.
But then, we choose the sample frequency

as an integer multiple of the positive
bandwidth.

So we say that Fs is equal to KW for K a
positive integer.

With this choice when we translate the
analogue specifications into digital

domain and remember the formula is always
the same, 2 pi F over Fs.

What happens is that the bandwidth in
visual domain will be 2 pi W divided by

Fs which is equal to 2 pi over K and so.
We can simply upsample the symbol

sequence by k so that it's bandwidth will
move from two pi to two pi over k and

therefore, it's width will fit on the
band allowed for on the channel.

Now, upsampling does not change the data
rate, because we're creating a sequence

of symbols.
From the user data bitstream and then we

are introducing the zeros between
samples.

So, we are not introducing extra
information.

So, we produce and transmit W symbols per
second and then we have sampled that by K

to achieve a sample rate which is equal
to the sample of frequency.

W, therefore is the fundamental data rate
of the system and sometimes it's called

the baud rate of the system.
A golden rule for digital communication

systems is that the baud rate will be
equal to the positive bandwidth allowed

by the channel.
So here is our revised baud diagram for

the transmitter.
User data comes in as a bit stream.

The Scrambler makes sure that we have a
random sequence of symbols.

The mapper will create, the random
sequence of symbols.

We upsample the sequence by K.
We filter this with a low pass filter

with cutoff frequency pi over K, and we
obtain base band signal b of n, which is

centered in 0, and extends from minus pi
over K to pi over K.

Now, we need to move this base band
signal, to the pass band of the channel.

And to do so, we modulated with a cosine
carrier.

This frequency is the center frequency of
the channel's band.

This pass band signal s of n can now be
converted to the analog domain, before

being transmitted over the channel.
Graphically, assume these are the

specifications dictated by the channel
and translated to the digital domain.

So here we have the positive bandwidth,
and the negative bandwidth.

The sequence of symbols generated by the
mapper, is a wide sequence and therefore,

inspires spectral density Pa of e to the
j omega, is a full band signal.

Now when we upsample this sequence by a
factor of k, we reduce its spectral

support, to a baseband signal that goes
from a minus pi over K, 2pi over K.

And then, we modulate this with a cosine
carrier, to fit it onto the bands that

are available on the channel.
As a final note, since we are developing

a completely digital transmission system,
we will probably want to use FIR filters

in its implementation.
Now, we know that the sync filter that we

have used in the upsampling operator Will
be a notoriously difficult filter to

approximate with an FIR.
So what is used in practice is another

type of filter called a raised cosine.
The frequency response of the raised

cosine is shown here in this picture, and
you can see that the transition band is

no longer a discontinuity.
But it is actually a smooth transition

from past band to star band as a matter
of fact a raised cosine has a parameter

that you can tune to have an even gentler
transition then.

Now, the raised cosine remains an ideal
filter because you can see it is constant

over the pass band, and the stop band.
But it is much easier to approximate than

the sync.
Another good property of the raise

cosine, is that it fulfills the
interpolation property that we need to do

upsampling.
And the final selling pointing, is that

the impulse response, it can be shown
decays as 1 over n cubed.

So even short FIR approximations can get
a very good response.

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