9.5 - Receiver design
-
0:01 - 0:04Hi, and welcome to module 9.5 of digital
signal processing. -
0:04 - 0:09In this module, we will touch briefly on
some topics in receiver design. -
0:09 - 0:12A lot of things, unfortunately, happen to
the signal while it's traveling through -
0:12 - 0:15the channel.
The signal picks up noise, we have seen -
0:15 - 0:18that already.
It also gets distorted because the -
0:18 - 0:22channel will act as some sort of filter
that is not necessarily all pass and -
0:22 - 0:26linear phase.
Interference happens too, that might be -
0:26 - 0:30parts of the channel that we thought were
usable, and they're actually not, so the -
0:30 - 0:34receiver really has to deal with a copy
of the transmitted signal that is very, -
0:34 - 0:40very far from the idealized version we
have seen so far. -
0:40 - 0:45The way receivers, especially digital
receivers, can cope with the distortions -
0:45 - 0:49and noise introduced by the channel, is
by implemented adaptive filtering -
0:49 - 0:53techniques.
Now we will not have the time to go into -
0:53 - 0:56very many details about adaptive signal
processing, again these are topics that -
0:56 - 1:01you will be able to study in more
advanced signal processing classes. -
1:01 - 1:06But I think it is important to give you
an overview of things that have to happen -
1:06 - 1:10inside a receiver, inside your ADSL
receiver for instance. -
1:10 - 1:14So you can enjoy this high data rates
that are available today. -
1:14 - 1:18And the first technique that we will look
at is adaptive equalization and then we -
1:18 - 1:22will look at some very simple timing
recovery that it used in practice in -
1:22 - 1:28receivers.
Let's begin with a blast from the past. -
1:28 - 1:33[NOISE] Those of you that are a little
bit older will certainly have recognized -
1:33 - 1:38this sound as the obligatory soundtrack
every time you used to connect to the -
1:38 - 1:44internet.
And indeed this is the sound made by a -
1:44 - 1:49V34 modem that was the standard dial up
connection device seen in the 90s until -
1:49 - 1:55the early 2000.
Now if you have ever used a modem, you've -
1:55 - 2:01heard the sound and you probably wondered
what was going on, so we're going to -
2:01 - 2:07analyze what we just heard from the
graphical point of view. -
2:07 - 2:11If we look at the block diagram for the
receiver once again, what we're going to -
2:11 - 2:15do is we're going to plot the base band
complex samples as points on the complex -
2:15 - 2:20blank.
So we're going to take B, r of n has the -
2:20 - 2:26as the horizontal coordinate and b, i of
n as the vertical coordinate. -
2:26 - 2:30And before we do so let's just look for a
second at what happens inside the -
2:30 - 2:35receiver when the signal at the input is
a simple sinusoid, like cosine of omega c -
2:35 - 2:41plus omega zero n.
We are demodulating this very simple -
2:41 - 2:46signal with the two carriers, the cosine
of omega c n and sine of omega c n. -
2:46 - 2:51And then we're filtering the result with
a low pass field. -
2:51 - 2:54So if we work out this formula with
standard trigonometric identities, we can -
2:54 - 3:00always express for instance the product
of two cosine functions as the sum. -
3:00 - 3:04Of the cosine of the sum of the angles
plus the cosine of the difference of the -
3:04 - 3:06angles.
And same for the product of cosine and -
3:06 - 3:10sine.
So if we do that, we get four terms, two -
3:10 - 3:17of which have a frequency that will fall
outside of the pass band of the filter h. -
3:17 - 3:20So when we apply the filter to these
terms, we're left only with cosine of -
3:20 - 3:27omega 0 n plus j sine of omega 0 n.
Which is of course e to the j omega 0 n. -
3:27 - 3:31So when the input to the receiver is a
cosine, the points in the complex -
3:31 - 3:36baseband sequence will be points around
the circle and a difference between two -
3:36 - 3:43successive points is the angle omega 0.
The reason why we might be called to -
3:43 - 3:46demodulate a simple sinusoid is because
the receiver will send what are called -
3:46 - 3:50pile tones, simple sinusoids that are
used to probe the line and gauge the -
3:50 - 3:55response of the channel at particular
frequencies. -
3:55 - 3:58So with this in mind, let's look at the
slow-motion analysis of the base band -
3:58 - 4:02signals.
Samples, when the input is the audio file -
4:02 - 4:06we just heard before.
So lets start with a part that goes like -
4:06 - 4:10this[SOUND].
This signal contains several sinusoids, -
4:10 - 4:15that we can see here in the plot.
And the sinusoids also contain abrupt -
4:15 - 4:18phase reversal, meaning that, at some
given points in time The phase of the -
4:18 - 4:23sinusoid is augmented by pi.
You can see this as this small explosions -
4:23 - 4:28in the circular pattern in the plot.
These phase reversals, are used by the -
4:28 - 4:31transmitter and the receiver, as time
markers, to estimate the propagation -
4:31 - 4:35delay of the signal, from source to
destination. -
4:35 - 4:39The next part goes like this.
[NOISE]. -
4:39 - 4:43And this is a training sequence.
The transmitter sends a sequence of known -
4:43 - 4:47symbols, namely the receiver knows the
symbol that are transmitted. -
4:47 - 4:50And so the receiver can use this
knowledge to train an equalizer to undo -
4:50 - 4:54the affects of the channel.
The last part is the data transmission -
4:54 - 4:57proper, the noisy part if you want, of
the audio file. -
4:57 - 5:01And the interesting thing is that the
transmitter and receiver perform a -
5:01 - 5:05handshake procedure, using a very low bit
rate QAM transmission using only four -
5:05 - 5:11points, therefore two bits per symbol.
To exchange the parameters of the real -
5:11 - 5:14data transmission that is going to
follow. -
5:14 - 5:16The speed, the constellation size, and so
on. -
5:16 - 5:20Using the four point QAM constellation in
the beginning ensures that, even in very -
5:20 - 5:24noisy conditions, transmitter and
receiver can exchange their vital -
5:24 - 5:29information.
So even from this simple qualitative -
5:29 - 5:33description of what happens in a real
communication scenario, we can see that -
5:33 - 5:38the task that the receiver is saddled
with is very complicated. -
5:38 - 5:41So it's a dirty job, but a receiver has
to do it. -
5:41 - 5:45And a receiver has to cope with four
potential sources of problem. -
5:45 - 5:50Interference.
The propagation delay, so the delay -
5:50 - 5:53introduced by the channel.
The linear distortion introduced by the -
5:53 - 5:56channel.
And drifts in internal clocks between the -
5:56 - 6:01digital system inside the transmitter and
the digital system inside the receiver. -
6:01 - 6:05So when it comes to interference the
handshake procedure, and the line probing -
6:05 - 6:08pilot tones are used in clever ways to
circumvent the major sources of -
6:08 - 6:13interference.
We will see some example later on when we -
6:13 - 6:16discuss ADSL.
The propagation delay, is tackled by a -
6:16 - 6:21delay estimation procedure, that we will
look at in just a second. -
6:21 - 6:25The distortion to this by the channel is
compensated using adaptive equalization -
6:25 - 6:29techniques, and we will see some examples
of that as well. -
6:29 - 6:33And clock drifts are tackled by timing
recovery techniques that in and of -
6:33 - 6:38themselves are quite sophisticated and
therefore we leave them to more advanced -
6:38 - 6:42classes.
Graphically, if we sum up the chain of -
6:42 - 6:46events that occur between the
transmissions of the original digital -
6:46 - 6:51signal and the beginning of the
demodulation of the received signal. -
6:51 - 6:57We have a digital to analog converter and
a transmitter, this is transmitter part -
6:57 - 7:01of the chain that operates with a given
sample period Ts. -
7:01 - 7:06This generates an analog signal which is
sent over a channel. -
7:06 - 7:08We can represent the channel for the time
being as a linear filter in the -
7:08 - 7:13continuous time domain.
With frequency response d of j omega. -
7:13 - 7:17At the input of the receiver, we have a
continuous time signal, s hat of t. -
7:17 - 7:23Which is a distorted and delayed version
of the original analog signal. -
7:23 - 7:28We will neglect noise for the time being.
This signal is sampled by an a to d -
7:28 - 7:32converter that operates at a period t
prime of s. -
7:32 - 7:36And we obtain the sequence of samples
that will be input to the modulator. -
7:36 - 7:38So this is the receiver part of the
chain. -
7:38 - 7:42We have to take into account the
distortion introduced by the channel, and -
7:42 - 7:46we have to take into account the
potentially time varying discrepancies in -
7:46 - 7:49the clocks between the transmitter and
the receiver. -
7:49 - 7:53These two systems are geographically
remote and there is no guarantee that the -
7:53 - 7:56two internal clocks that're used in the A
to D and D to A converters are -
7:56 - 8:00synchronized or run exactly at the same
frequency. -
8:00 - 8:03Let's start with problem of delay
compensation. -
8:03 - 8:07To simplify the analysis we'll assume
that the clocks at transmitter and -
8:07 - 8:10receiver are synchronized and
synchronous. -
8:10 - 8:14So T prime of s is equal to T s, and the
channel acts as a simple delay. -
8:14 - 8:17So the received signal is simply a
delayed version of the transmitted -
8:17 - 8:21signal.
Which implies that the frequency response -
8:21 - 8:24of the channel is simply e to the minus j
omega d. -
8:24 - 8:27So, the channel introduces a delay of d
seconds. -
8:27 - 8:31We can express this in samples in the
following way. -
8:31 - 8:35We write d as the product of the sampling
period, times b plus tau. -
8:35 - 8:38Where b is an integer.
And tau is strictly less than one-half in -
8:38 - 8:41magnitude.
So b is called the bulk delay because it -
8:41 - 8:45gives us an integer number of samples of
delay at the receiver and tau is the -
8:45 - 8:51fractional delay.
So the fraction of samples introduced by -
8:51 - 8:56the continuous time delay of d.
So, how do we compensate for this delay? -
8:56 - 9:00Well, the bulk delay is rather easy to
tackle. -
9:00 - 9:03Imagine the transmitter begins
transmission by sending just an impulse -
9:03 - 9:06over the channel.
So, the discreet time signal is this one, -
9:06 - 9:10it's just a delta and a 0.
It gets sent to D-to-A converter. -
9:10 - 9:13And the converter will output a
continuous time signal that looks like an -
9:13 - 9:17interpolation function like the sinc.
And like all interpolation functions, it -
9:17 - 9:21will have a maximum peak at zero that
corresponds to the known zero sample. -
9:21 - 9:24This signal gets transmitted over the
channel. -
9:24 - 9:27And it gets to the receiver after a delay
d that we can estimate for instance by -
9:27 - 9:32looking at the displacement of the peak
of the interpolation function. -
9:32 - 9:35The receiver converts this into a
discrete time sequence. -
9:35 - 9:38Now in the figure here it looks as if the
sample incidence at the transmitter and -
9:38 - 9:42receiver are perfectly aligned.
Now this is not necessarily the case -
9:42 - 9:45because the starting time for the
interpolator at the transmitter and the -
9:45 - 9:49sampler at the receiver are not
necessarily synchronous. -
9:49 - 9:53But any difference in starting time can
be integrated into the propagation delay -
9:53 - 9:56as long as the sampling periods are the
same. -
9:56 - 10:01So with this, all we need to do at the
receiver is to look for the maximum value -
10:01 - 10:06in the sequence of samples.
Because of the shape of the interpolating -
10:06 - 10:10function, we know that the real maximum
will be at most half a sample in either -
10:10 - 10:14direction of the location, of the maximum
sample value. -
10:14 - 10:17So, add the receiver to offset the bulk
delay, we will just set the nominal time -
10:17 - 10:21n equal to 0, to coincide with the
location of the maximum value of the -
10:21 - 10:25sample sequence.
Now of course we need to compensate for -
10:25 - 10:29the fractional delay, so we need to
estimate tau. -
10:29 - 10:31And to do that we'll use a different
technique. -
10:31 - 10:35Let me add in passing, that in real
communication devices, of course we're -
10:35 - 10:38not using impulses to offset the bulk
delay. -
10:38 - 10:41Because impulses are full-band signals
and so they would be filtered out by the -
10:41 - 10:46passband characteristic of the channel.
The trick is to embed discontinuities in -
10:46 - 10:51pilot tones and to recognize this
discontinuities at the receiver. -
10:51 - 10:55As we have seen in the animation at the
beginning of this module, we use phase -
10:55 - 10:59reversals, which are abrupt
discontinuities in sinusoids, to provide -
10:59 - 11:03a recognizable instant in time for the
receiver to latch on. -
11:03 - 11:08Okay, so what about the fractional delay?
Well, for the fractional delay, we use a -
11:08 - 11:12sinusoid instead of a delta, so we build
a bass band signal, which is simply a -
11:12 - 11:17complex exponential at a known frequency,
omega 0. -
11:17 - 11:21This will be converted to a real signal
before being sent to the D-to-A -
11:21 - 11:24converter, and so what we transmit
actually is cosine of omega c, the -
11:24 - 11:30carrier frequency, plus the pilot
frequency omega 0, times n. -
11:30 - 11:35The receiver will receive a delayed
version of this, which contains the delay -
11:35 - 11:39now in sample, and fraction of sample, b
plus tau. -
11:39 - 11:43After we demodulate this cosine you
remember we got a complex exponential and -
11:43 - 11:48we can also compensate already for the
bulk delay which we know. -
11:48 - 11:52So for an integer number of sample b we
obtain a base band signal at b of n which -
11:52 - 11:58is e to the j omega and minus tau.
Since we know the frequency of omega 0 we -
11:58 - 12:02can just multiply this quantity by e to
the minus j omega 0 n and obtain e to the -
12:02 - 12:09minus j omega 0 tau, which is a constant.
And which we can invert given that we -
12:09 - 12:12know the frequency omega 0.
And so now we have an estimate for both -
12:12 - 12:17the bulk delay and the fractional delay.
Now we have to bring back the signal to -
12:17 - 12:20the original timing.
The bulk delay is really no problem. -
12:20 - 12:24It's just an integer number of samples.
What creates a problem is the fractional -
12:24 - 12:28delay because that will shift the peaks
with respect to the sampling intervals. -
12:28 - 12:33So if we want to compensate for the bulk
delay we need to compute subsample values -
12:33 - 12:38and in theory to do that we should use a
sinc fractional delay namely a filter -
12:38 - 12:45with impulse response sinc of n plus tau.
In practice however, we will use a local -
12:45 - 12:49interpolation and this is a very
practical application of the Lagrange -
12:49 - 12:53interpolation technique that we saw in
module 6.2. -
12:53 - 12:56So graphically the situation is like so,
we have a stream of samples coming in. -
12:56 - 13:01And for each sample, we want to compute
the subsample value with a distance of -
13:01 - 13:06tau from the nearest sample interval.
And we want to only use a local -
13:06 - 13:12neighborhood of samples to estimate this.
Now, you remember from module 6.2. -
13:12 - 13:15The Lagrange approximation works by
building a linear combination of Lagrange -
13:15 - 13:18polynomials weighed by the samples of the
function. -
13:18 - 13:22So, as per usual, we choose the sampling
interval equal to 1, so that we lighten -
13:22 - 13:26the notation.
We have a continuous time function x of -
13:26 - 13:30t, and we want to compute x of n plus
tau, with tau less than one half in -
13:30 - 13:35magnitude.
So we have samples of this function at -
13:35 - 13:40integers, n, and the local Lagrange
approximation around n, is given by this -
13:40 - 13:45linear combination of Lagrange
polynomials. -
13:45 - 13:49Weighted by the samples of the functions
around the approximation point. -
13:49 - 13:54So we use the notation x L of n and t.
N is the center point and t is the value -
13:54 - 14:00from the center point at which we want to
compute the approximation. -
14:00 - 14:03And the Lagrange polynomials are given by
this formula here, which is the same as -
14:03 - 14:08in module 6.2.
So the delayed compensated input signal -
14:08 - 14:11will be set equal to the Lagrange
approximation at tau. -
14:11 - 14:14So let's look at an example.
Assume that we want a second order -
14:14 - 14:17approximation.
So we pick N equal to 1 and we will have -
14:17 - 14:23three Lagrange polynomials.
And so, we will need to use three samples -
14:23 - 14:28of the sequence to compute interpolation.
These three polynomials will be centered -
14:28 - 14:32in n minus 1 and in n plus 1 and scaled
by the values of the samples at these -
14:32 - 14:36locations.
And finally, we will sum the poll numbers -
14:36 - 14:39together and compute their value in n
plus tau. -
14:39 - 14:42So, we start with the first one, which is
centered in n minus 1. -
14:42 - 14:47And, like all interpolation polynomials,
its value is 1 in n minus 1, and 0, at -
14:47 - 14:53other integer values of the argument.
The second polynomial will be centered in -
14:53 - 14:57n, and the third polynomial will be
centered in n plus 1. -
14:57 - 15:00When we sum them together, we obtain a
second order curve that goes through the -
15:00 - 15:03points, that interpolates the three
points, and then we can compute the -
15:03 - 15:08approximation as the value of this curve
in n plus tau. -
15:08 - 15:12Now the nice thing about this approach,
is that if we look at the approximation, -
15:12 - 15:17if we take the Lagrange approximation
around n. -
15:17 - 15:21We can define a set of coefficients, d
tau of k, which are the values of each -
15:21 - 15:27Lagrange polynomial in tau.
So d tau of k, are 2 N plus 1 values, the -
15:27 - 15:33form, the coefficients, of an FIR filter.
And we can compute the value of the -
15:33 - 15:37Lagrange approximation simply as the
convolution of the incoming sequence with -
15:37 - 15:42this interpolation filter.
So for example, if these are the three -
15:42 - 15:47Lagrange polynomials for n equal to 1, we
can compute these polynomials for t equal -
15:47 - 15:54to tau, where tau is the fractional delay
that we estimated before. -
15:54 - 15:59And we will obtain three coefficients,
like here, for instance, is an example -
15:59 - 16:03for tau equal to 0.2.
Three coefficients that give us an FIR -
16:03 - 16:06filter, and then we can just simply
filter the samples coming into the -
16:06 - 16:11receiver with this filter, to compensate
for the fractional delay. -
16:11 - 16:13So again, the algorithm is, estimate the
fractional delay. -
16:13 - 16:17The bulk delay is no problem, again.
Compute the 2 N plus 1 Lagrangian -
16:17 - 16:20coefficients and filter it with the
resulting FIR. -
16:20 - 16:24The added advantage of this strategy is
that if the delay changes over time for -
16:24 - 16:27any reason, all we need to do is to keep
the estimation running and update the FIR -
16:27 - 16:32coefficients as the estimation changes
over time. -
16:32 - 16:36Okay, now that we know how to compensate
for the propagation delay introduced by -
16:36 - 16:39the channel.
Let's go see the rechannel it with an -
16:39 - 16:42arbitrary frequency response D j of
omega. -
16:42 - 16:47And the transmission chain goes from the
pass band signal s of n, discreet time, -
16:47 - 16:52into a D-to-A converter, analog signal s
of t, it gets filtered by the channel, -
16:52 - 17:00gives us hat s of t, which is sampled at
the receiver to give us. -
17:00 - 17:05A received fast band signal, hat s of n.
But now we have seen in the previous -
17:05 - 17:08module that this block diagram can be
converted into an all digital scheme -
17:08 - 17:12where our band pass signal s of n gets
filtered by the discrete time equivalent -
17:12 - 17:17of the channel.
And, gives us a filtered version of the -
17:17 - 17:22bandpass signal, as would appear inside
the receiver. -
17:22 - 17:26So the problem now, is that we would like
to undo the effects of the channel, on -
17:26 - 17:30the transmitted signal.
And the classic way to do that, is to -
17:30 - 17:35filter the received signal hat s of n, by
a filter E, that compensates for the -
17:35 - 17:40distortion or the filtering introduced by
the channel. -
17:40 - 17:45So the target is that the output of the
filtering operation gives us a signal hat -
17:45 - 17:50s e of n, which is equal to the
transmitted signal. -
17:50 - 17:53How do we do that?
In theory, it would be enough to pick a -
17:53 - 17:57transfer function for the filter E, which
is just the reciprocal of the equivalent -
17:57 - 18:02transfer function of the channel.
But the problem is that we don't know the -
18:02 - 18:05transfer function of the channel in
advance because each time we transmit -
18:05 - 18:08data over the channel, this transfer
function may change. -
18:08 - 18:13And also, even while we're transmitting
data, the transfer function might change -
18:13 - 18:16because it is a physical system that
might be subject to. -
18:16 - 18:20Drifts and modifications.
So what do we do? -
18:20 - 18:26We need to use adaptive equalization.
So the filter that compensates for the -
18:26 - 18:30distortion introduced by the channel is
called an equalizer. -
18:30 - 18:35And what we want to do Is to change the
filter in time, so change the filter -
18:35 - 18:39coefficients in a DPS realization as a
function of the error that we obtain when -
18:39 - 18:48we compare the output of the filter with
the signal that we would like to obtain. -
18:48 - 18:52In our case the signal that we would like
to obtain is the transmitted signal. -
18:52 - 18:57And so we take the received signal, we
filter it with the equalizer. -
18:57 - 19:01We look at the result.
We take the difference, with respect to -
19:01 - 19:04the original signal, and we use the
error, which should be zero in the ideal -
19:04 - 19:09case, to drive the adaptation of the
equalizer. -
19:09 - 19:14But wait, how do we get the exact
transmitted signal at the receiver? -
19:14 - 19:17Well, we use two tricks.
The first one is boot strapping. -
19:17 - 19:23The transmitter will send a prearranged
sequence of symbols to the receiver. -
19:23 - 19:26So let's call the sequence of symbols a t
of n. -
19:26 - 19:32This gets modulated and generates a pass
band signal s of n. -
19:32 - 19:36Now at the receiver the sequence a t of n
is known. -
19:36 - 19:42And the receiver has an exact copy of the
modulator, of the transmitter, inside of -
19:42 - 19:46itself.
So the transmitter can generate locally, -
19:46 - 19:50an exact copy of the pass band signal s
of n. -
19:50 - 19:54And so, for the bootstrapping part of the
adaptation, we actually have an exact -
19:54 - 19:59copy of the transmitted pass band signal
that we can use to drive the adaptation -
19:59 - 20:04of the coefficients.
The training sequence is just long enough -
20:04 - 20:07to bring the equalizer to a workable
state. -
20:07 - 20:10For the handshake procedure that we saw
in the video before, for instance. -
20:10 - 20:14This would correspond to the moment where
the receiver starts demodulating the four -
20:14 - 20:18point QIM.
At that moment, the receiver will switch -
20:18 - 20:21strategy and implement a data driven
adaptation. -
20:21 - 20:26The thing works like this.
The received signal gets equalized, gets -
20:26 - 20:31demodulated and then the slicer will
recover the sequence of transmitted -
20:31 - 20:34symbols.
Since the receiver has a copy of the -
20:34 - 20:39transmitter inside of itself, it can use
the sequence of transmitted symbol. -
20:39 - 20:42To build a local copy of the transmitted
signal. -
20:42 - 20:47Now of course, errors might happen in the
slicing process, and so this local copy -
20:47 - 20:52is not completely error-free.
But the assumption is that the equalizer -
20:52 - 20:56is doing already enough of a good job to
keep the number of errors in this -
20:56 - 21:02sequence sufficiently low.
So that the difference, with respect to -
21:02 - 21:06the received signal, is enough to refine
the adaptation of the equalizer, and -
21:06 - 21:12especially to track the time varying
conditions of the channel. -
21:12 - 21:14What we have seen, is just a qualitative
overview of what happens inside of a -
21:14 - 21:18receiver.
And there're still so many questions that -
21:18 - 21:22we would have to answer to be thorough.
For instance, how do we carry out the -
21:22 - 21:25adaptation of the coefficients in the
equalizer? -
21:25 - 21:30How do we compensate for different clock
rates in geographically diverse receivers -
21:30 - 21:35and transmitters?
How do we recover from the interference -
21:35 - 21:41from other transmission devices, and how
do we improve the resilience to noise? -
21:41 - 21:46The answers to all those questions
require a much deeper understanding of -
21:46 - 21:51adaptive signal processing, and hopefully
that'll be the topic of your next signal -
21:51 - 21:54processing class.
- Title:
- 9.5 - Receiver design
- Description:
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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.
Claude Almansi edited English subtitles for 9.5 - Receiver design | ||
Claude Almansi edited English subtitles for 9.5 - Receiver design | ||
Claude Almansi commented on English subtitles for 9.5 - Receiver design | ||
Claude Almansi edited English subtitles for 9.5 - Receiver design | ||
Claude Almansi added a translation |