Learn the fundamentals of digital signal processing theory and discover the myriad ways DSP makes everyday life more productive and fun.

Loading...

Do curso por École Polytechnique Fédérale de Lausanne

Processamento Digital de Sinais

293 classificações

Learn the fundamentals of digital signal processing theory and discover the myriad ways DSP makes everyday life more productive and fun.

Na lição

Module 5: Sampling and Quantization

- Paolo PrandoniLecturer

School of Computer and Communication Science - Martin VetterliProfessor

School of Computer and Communication Sciences

Remember the raw-sampling of an arbitrary signal.

So you have a continuous time signal Xc of (t) sampled every Ts

seconds to give a sequence x[n].

Which is equal to the continuous time signals at

multiples of the sampling intervals TS.

In Fourier Transform domain we have a spectra of the continuous time

signal Xc(j capital omega).

And at the output we have a discrete time Fourier Transform of the sequence,

that's X(e to the j omega).

What is that going to be in general?

And how is it going to be related to the input spectrum?

The key idea is the following.

Pick a sampling interval Ts for a frequency omega N = to pi over Ts.

Start with omega zero, which is smaller than omega N.

The frequency, which is the maximum frequencies that can be

faithfully represented with a sampling system.

So the input is e to the j omega zero T is sampled with Ts.

It gives an output, e to the j omega nought, Ts times n,

that's the sequence resulting from the sampling operation.

Then, let's add 2 omega N to the input frequencies.

So now the frequency's (omega 0 + 2 omega N).

It is sampled at the output, it is (omega 0 + 2 omega N) times Ts times n.

We expand this product, then we see that the second term becomes 2 pi n.

But (2 pi n) raised to the power e to the j is equal to one and

therefore, we have the same discreet time sequence as before.

So we do no see this higher frequency complex exponential,

it simply looks like the lower frequency exponential omega 0.

So in general, if we have two frequencies, A times the complex exponential

omega 0 + B times the complex exponential (omega 0 + 2 times omega N),

sampled, it gives you (A + B) e to the j omega 0Tsn.

So the upper frequency one has been folded back or

aliased back onto the lower frequency one and we simply see the sum of these two.

And so the aliased part has been shifted back to the non-aliased one.

So what is the spectrum of a raw-sampled signal in general?

Well, let's look at the sample x[n], it is xc, at location (nTs).

We can write this as the inverse Fourier Transform of the spectrum capital Xc,

multiplied by the e to the (j omega) nTs.

Because that's the location we want to evaluate the inverse Fourier Transform.

Frequencies that are 2 omega N apart will be aliased.

We have seen this specifically on an example just a couple of slides before.

So we can split the integration interval into pieces of size 2 omega N.

And this we do here.

So x[n], this inverse Fourier Transform is now as a sum of

the pieces from (2k- 1) omega N to (2k + 1) omega N.

With the Fourier Transform here multiplied by the complex exponential at

nTs Let's look this on a picture.

So everything at location 2 omega N will be folded back to the origin.

What is that?

2 omega N plus epsilon will be folded back to epsilon.

What is that?

2 omega N minus epsilon gets folded back to minus epsilon.

It's the same symmetrically around.

Minus 2 omega N, it's the same from 4 omega N, and minus 4 omega N, et cetera.

So we can go back to our formula.

We do a change of variables so

that the integral is between minus omega N and omega N.

This changes the frequency variable inside of Xc.

Then we also note that adding multiples of 2k omega N to the frequency,

doesn't change the frequency, that's the famous aliasing effect.

We can interchange summation and integral to end up with 1 over 2 pi,

the integral, [the summation of the spectrum Xc and

its shifts by multiples of 2k omega N].

And then at the right end, we still have the integration variable for

the inverse Fourier Transform.

Let's define a periodized spectrum, so X tailed c is

the summation of Xc, and it's shifted versions by even multiples of omega N.

Therefore, the sample x[n] is integral between minus omega N and

omega N of the periodized spectrum, x tilde, multiplied by e to the j omega nTs.

Let's do the following change of variable.

Small omega is equal to capital omega times Ts.

This will change the integration boundaries from minus omega N to

omega N into minus pi 2 pi.

It changes the frequency inside the X tilde,

and it changes exponent to j omega n, which is more convenient.

So we recognize now that x[n] is the inverse

discrete time Fourier Transform of 1 over Ts,

this perodized spectrum X tilde c, rescaled by Ts.

An important point to note is that this periodized spectrum

rescaled is actually 2 pi periodic, so it's a valid DTFT.

Since this is the IDTFT, it means that the DTFT of the sequence,

X of e to the omega, is simply up to a scale factor of 1 over Ts,

the summation of Xc and its shifts by multiples of 2 pi over Ts.

A picture is worth 1,000 equations, maybe, so let us look at the spectrum and

how it evolves.

So we start with capital Xc.

It's a spectrum which is inside the boundary minus omega N to omega N,

doesn't cover the entire interval.

Then, the periodized version puts repetition spectras at 2 omega N,

4 omega N, etcetera.

This is now a periodic spectrum of period 2 capital omega N.

So spectrum of X of (e to the j omega) is a scaled version of this,

it is 2 pi periodic, and it is shown here at the bottom in blue.

Let's take a bandlimited signal that is bandlimited to omega 0 equal to omega N.

This is the limit of the Nyquist sampling frequency.

It will be periodic of period 2 omega N which is equal to 2 omega 0.

So the repetitions will just touch each other.

It is therefore the usual periodic spectrum.

And now we look at the spectrum of X of (e to the j omega).

It fills the entire interval, minus pi to pi.

What happens when omega 0 is bigger than omega N?

Well we know trouble is going to come because aliasing will happen.

So we see that the spectrum is beyond the central interval of minus omega,

and 2 omega N.

So the repetitions will overlap with each other.

The result is still 2 omega N periodic spectrum.

We see it has a funny shape now.

It's not the original shape, because of the overlaps.

And when we look at the spectrum of x of (e to the j omega), so blue spectrum,

it indeed does not resemble the original spectrum,

because an entire interval has been messed up by aliasing.

Let us look at a non-bandlimited signal.

So this is a Gaussian type of signal, which has decay, but

goes well beyond the central interval, minus omega N to omega N.

It is after sampling the periodized version is 2 omega N periodic of course.

But there are overlaps all over.

And it has, in the center of the interval,

a shape that resembles the original spectrum but then it has overlap.

And so the blue spectrum, which is between minus pi and

pi resembles the original spectrum around the origin 0,

but is different from the original at the boundary of the interval

O Coursera proporciona acesso universal à melhor educação do mundo fazendo parcerias com as melhores universidades e organizações para oferecer cursos on-line.