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

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We are talking about digital [br]communication systems, and in this

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

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

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upsampling. [br]And we will see how upsampling will allow

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

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the channel. [br]Remember that our assumption is that the

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signal generated by the transmitter is a [br]wide sequence.

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

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

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so that it fits on the band allowed by [br]the channel.

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The way we do this is by using multirate [br]techniques.

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

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digital signal. [br]One way to do this is to interpolate the

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digital signal into a continuous time [br]signal.

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And then resample the interpolation at a [br]different sampling rate.

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

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

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

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

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

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

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

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

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times sequence here. [br]Graphically, assume this is our discrete

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times signal. [br]The interpolation to continuous time will

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

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

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

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

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

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interplay signal is given by the standard [br]sync interplay formula.

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

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

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

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centered at fractional intervals, so n [br]over big K.

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In the frequency domain, the process [br]looks like so.

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

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

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

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

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equivalent to multiplying nyquist [br]frequency by k.

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So we make it bigger and we move it over [br]here.

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And when we plot the result in digital [br]spectrum.

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

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contraction of the frequency spectrum by [br]a factor of K.

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

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

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

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domain? [br]Well the idea is that we need to increase

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

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

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

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

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

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

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

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between. [br]So for example, for k equal to 3, the

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

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zero. [br]Then we put two zeros, then we put x1,

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then we put two zero, then we put x2, and [br]then we put two zeroes.

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

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before. [br]And what we are doing, we're simply

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

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of the upsample sequence is rather easy [br]to compute.

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

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

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

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

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

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

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

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

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

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

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

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the interval from minus 3 pi to 3 pi back [br]onto the minus pi, pi interval.

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

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

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

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

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

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

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

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

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

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

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

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

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

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So now, the resulting sequence is simply [br]the convolution of the upsampled sequence

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with zeroes. [br]And the impulse response of the filter

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

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

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

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

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

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

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

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upsampled sequence we can always recover [br]the original sequence by downsampling,

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

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

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

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

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interpolation properties that we have [br]seen in module 6.

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

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

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

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interpolation. [br]We have the pesky problem of aliasing,

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

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

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

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

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imposes that we only use frequencies [br]between Fmin and Fmax.

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

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

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will be our nyquist frequency in the [br]analog domain.

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

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

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that W. [br]Now, pick the sample frequency so that

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

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

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

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as an integer multiple of the positive [br]bandwidth.

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

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With this choice when we translate the [br]analogue specifications into digital

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

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

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

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

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

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

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of symbols. [br]From the user data bitstream and then we

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are introducing the zeros between [br]samples.

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So, we are not introducing extra [br]information.

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

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

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

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

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

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

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the transmitter. [br]User data comes in as a bit stream.

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

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The mapper will create, the random [br]sequence of symbols.

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

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

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

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

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

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This frequency is the center frequency of [br]the channel's band.

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

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being transmitted over the channel. [br]Graphically, assume these are the

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specifications dictated by the channel [br]and translated to the digital domain.

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So here we have the positive bandwidth, [br]and the negative bandwidth.

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

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

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

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

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

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

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

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

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

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

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

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

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no longer a discontinuity. [br]But it is actually a smooth transition

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

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

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

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

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the sync. [br]Another good property of the raise

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

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upsampling. [br]And the final selling pointing, is that

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

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

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