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5.3.a The spectrum of interpolated signals
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來自 École Polytechnique Fédérale de Lausanne 的課程
数字信号处理
306 個評分
從本節課中
Module 5: Sampling and Quantization
與講師見面
Paolo PrandoniLecturer
School of Computer and Communication Science
Martin VetterliProfessor
School of Computer and Communication Sciences
The ingredients of sinc interpolation are a discrete-time signal
xn with DTFT capital X of e to the omega.
An interpolation interval Ts which will set
the distance between samples that we interpolate.
And the sinc function properly scaled to have zero crossings at multiples of Ts.
The result will be a smooth, continuous time signal, x of t and
the question is what is the spectrum of x of t?
Recall key facts about the sinc functions, so five Ts, sinc of T, or
T S, T S is the sampling interval.
Pi over capital omega N, is fully transformed, we know it's a rect function,
properly scaled, and omega N is just the reverse, it's pi over T S.
It is much nicer to see this in pictures so on the top you have the time domain
function, these one at the origin and that all multiples of Ts it has zero crossings.
And the fourier transform of 5T is the box function between minus omega N to omega N.
And properly scaled and [INAUDIBLE] to pi over capital omega N.
The sinc interpolation, we have seen before.
You take the sequence xn, you multiply each term,
xn with a sinc function, centered at location n times Ts.
And the result x(t) is simply the sum of all these terms.
So what is the spectrum of this interpolated signal?
So let's compute it.
Fourier transform capital X j omega Which is interval of x of t,
multiplied by e to the minus j omega t.
We write out the expression for xt, as the interpolation of the sequence xn.
Then we inter change the order of summation and integral.
We can take xn outside, so inside we have sinc of t minus nts over ts.
Then we remember of course that the fourier transform of sinc is
direct function.
But it's also sinc shifted by nts.
This gives a phase factor e to the minus jnts omega which is a variant.
Let's analyze this formula in detail.
Capital X of omega fourier transform of our sinc interpolated signal is
the summation here of Xn times the rec and e to the- j and ts omega.
The rect function does not depend on the small n so we can take it in front.
We are left on the right side with the summation
from minus infinity plus x n e to the minus j pi over
capital omega capital n times omega as a variable times n.
Now we recognize an old friend.
This is a discrete time fourier transform of xn.
The sequence rescaled by a two times capital Omega capital N.
Now this DTFT is periodic and the periodic is two times capital Omega capital N.
However,it's multiplied by direct function which takes
away all the repetition except the central one.
That's what we see in the last equation.
So that fully transforms of X of T is equal to the scale version of
the DTFT of the sequence between minus capitol Omega and non capital Omega.
And NDT's zero otherwise.
It is best to look at this in a figure.
So on top you have the DTFT of the sequence xn.
This is a 2 pi periodic function.
We show the central period for minus pi 2 pi.
The fourier transform of x of t,
the interpolated version with sampling integral of ts is a scaled version.
And it lays between minus capital omega m, and capital omega n.
It has a same shape as a DTFT, simply it is rescaled along
the horizontal axis to fit on the interval of lengths, 2 omega n.
What happens if we change the sampling interval?
So for example, let's space the samples twice apart, then the spectrum will be
sharpened, and it will leave between minus omega n over 2 and omega n over 2.
Or if we take a sampling interal,
which is half of the one that we had seen at the beginning ts over 2.
Then it will broaden the spectrum which now lives between minus 2 omega n and
2 omega n.
The intuition is simple.
If the samples are spread apart as the function becomes slower.
If the samples are closer together the signal becomes faster.
Let us summarize this important point.
So when you pick the interpolation period ts the interpolated signal has
only got n bandlimited spectrum where omega n is pi over ts.
So with fast interpolation when ts is small the spectrum will be wide.
And when you do slow interpolation so ts is large.
Then the spectrum will be narrow.
In some sense it's like changing the speed of a record player.
We have seen that starting with a sequence in l2 of z
some interpolation with sampling interval of TS, we get X of T, and
band limited single, namely band limited we only got M.
This function X of T is uniquely specified by the sequence XM.
So an intriguing question is to find out if we can take samples of XT and
get the sequence again that these uniquely specified.
And so we have a one to one relationship between sequences and
bandlimited function.
The answer to this question is actually positive, and
it's given by some sampling theorems that we will see in a moment.
