Sampling and the Sampling Theorem

Sampling and the Sampling Theorem

Andrew W. H. House

11 May 2004

1 Review of the Sampling Theorem

We previously stated the sampling theorem, which allows us to reconstruct a continuous-timesignal from its discrete-time samples under certain conditions.

The Sampling Theorem states that, for a continuous-time (CT) signal xc(t) with CT Fouriertransform Xc(ω), if Xc(ω) = 0 for |ω| > ωm and ωs > 2ωm then

xd(n) = xc(nTs) = xc(t)|t=nTs , n ∈ Z

completely represents xc(t) and xc(t) can be reconstructed from xd(n).

In other words, we need to sample fast enough to get more than 2 samples in each cycle ofxc(t)’s highest frequency. So, if Ts < 1

2Tm, then xc(nTs) represents xc(t).

2 The Sampling Theorem in Action

Let’s thing about applying the sampling theorem. Consider xc(t) = cos ωmt which hasfundamental period Tm.

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Sampling and the Sampling Theorem

Recall that the Fourier transform of Xc(ω) = F{xc(t)} = F{cos ωmt} so

Xc(ω) = πδ(ω − ωm) + πδ(ω + ωm) .

Obviously in this case, Xc(ω) = 0 for |ω| > ωm, so the signal xc(t) is band-limited.

Let’s choose a value for ωs so that ωs > 2ωm and therefore

2π

Ts

> 2 · 2π

Tm

or in other words Tm > 2Ts or Ts < 12Tm and then we can plot as below.

We can kind of see how the samples represent xc(t).

Now consider the plot below.

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Sampling and the Sampling Theorem

xx

xx

xx

x x x x x x xx

x x x x x x x x x x x xx

xx

xx

x x xx x

x x

o

o

o

o

oo

oo o o o o

oo

o o oo

o o o o o oo

o

o

o

oo

o oo

oo

o

o

Let’s assume that there are two sets of samples take, those marked with ‘x’, and thosemarked with ‘o’. We will also assume that in each case, ωs > 2ωm so our sampling rate ishigh enough.

Clearly, each set of samples will have different numerical values, since they record differentpoints on the original CT signal. Yet the sampling theorem suggests that the CT signal canbe reconstructed from either of the sets of samples.

How can this be? It seems counter-intuitive that different sets of samples will reproduce thesame signal. But that is what happens because the samples are used to recover the lowestfrequency signal. That basically means that there is only one possible signal that could haveproduced a given set of samples, so long as the assumptions of the sampling theorem weretrue.

3 Sampling Processes

We have seen, at least in a qualitative sense, how samples can represent a CT signal providedthat we follow the rules of the sampling theorem. At this point, we should consider how wecan get samples – after wall, without being able to sample real-world CT signals, there’s notmuch point to this course. That is, we need to know how to get samples from a CT signal.

?x (t)c x (n)d

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Sampling and the Sampling Theorem

There are many methods to get samples, some of which are covered in the following subsec-tions.

3.1 Sample-and-Hold Circuit

The sample-and-hold circuit (sometimes called zero-order hold) converts the held analogvalues to digital values during the hold interval Ts, as show in the following diagram.

Essentially the circuit latches values of the analog signal during the hold intervals, producinga series of stepped samples.

3.2 Pulse Amplitude Modulation (PAM) Sampling

This type of sampling multiplies a CT signal xc(t) with a pulse signal p(t) to get a series ofpulses that map certain portions of the original CT signal.

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Sampling and the Sampling Theorem

In the pulse train, note that the area of each pulse is 1 – that is, its height can be consideredto be ∆, and its width 1/∆.

When the signal is modulated (multiplied) with the pulses, we get xp(t) = xc(t) · p(t) shownbelow.

Each pulse of the modulated signal has an area that is approximately equal to the value ofthe original signal at the middle of the pulse width, which is also the value of the DT samplewe would ideally have. This is illustrated in the above diagram.

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Sampling and the Sampling Theorem

Each modulated pulse above can be integrated to get its area, as follows.

xd(n) =

∫ nTs+δ

nTs−δ

xp(t)dt ≈ xc(nTs)

Thus, the area of each modulated pulse is roughly the value of the sample. So this is a goodapproximation, but is not exactly what we want.

But, what happens when the width of the pulse narrows, as 1∆→ 0? Our pulse train

p(t) becomes an impulse train, with each pulse being a Dirac delta, δ(t). This suggests atheoretical approach to sampling, which we explore in the next section.

4 Ideal Impulse Train Sampling

Consider our pulse train p(t) from the previous section, with the width of the pulse going to0 (and thus the height going to infinity).

As1

∆→ 0, pulse train → impulse train.

So the impulse train is written as

p(t) =∞∑

n=−∞

δ(t− nTs)

which is a series of shifted deltas.

So, using this p(t), what is xp(t) = xc(t) · p(t) now?

If we recall that ∫ +Ts/2

−Ts/2

p(t)dt =

∫ +Ts/2

−Ts/2

δ(t)dt = 1

then we get the following.

If our original CT signal is

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Sampling and the Sampling Theorem

and our impulse train is

then we get a set of samples shown below.

So now

xd(n) =

∫ nTs+Ts2

nTs−Ts2

xp(t)dt = xc(nTs)

a.k.a. train to samples. Note that the above is exactly equal, not approximately as withPAM sampling before.

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Sampling and the Sampling Theorem

Therefore, this is our model of ideal sampling, which is the kind of sampling that the samplingtheorem means.

x (t)px (t)c x (n)d

p(t)

Convert impulsetrain to samples

Model:

This model above represents the basic concept of a C/D converter, illustrated below.

x (t)c x (n)dC/D

Ts

Example 4.1: Basic Sampling Theorem Exercises

Consider a CT signal xc(t) with a Fourier transform Xc(ω) that is sampled withideal sampling, with Ts = 10−4 seconds. For each of the following scenarios,determine whether the sampling theorem guarantees that xc(t) can be recoveredfrom its samples.

1. What if Xc(ω) = 0 for |ω| > 5000π?

Since Ts = 10−4 seconds, ωs = 2π/Ts = 2π/10−4 = 20000π or fs = 1/Ts =10000 Hz. In this scenario, ωm = 5000π = 2π · (2500).

Is (ωs = 20000π) > (2ωm = 10000π)?

Yes! Therefore, we can guarantee recovery of xc(t).

2. What if Xc(ω) = 0 for |ω| > 10000π?

In this case, ωm = 10000π.

Is (ωs = 20000π) > (2ωm = 20000π)?

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Obviously not. In this case, ωs = 2ωm which does not satisfy the require-ments of the sampling theorem, which requires strictly greater than, notgreater than or equal. Therefore, we cannot guarantee recovery of xc(t).

3. What if Xc(ω) = 0 for |ω| > 15000π?

Here, ωm = 15000π.

Is (ωs = 20000π) > (2ωm = 30000π)?

Obviously not. Therefore, we cannot guarantee recovery of xc(t).

4. What if |Xc(ω)| = 0 for |ω| > 5000π?

This question isn’t quite so obvious. Can we figure out, from the giveninformation, whether the signal is band-limited? And if so, can we figureout whether the sampling was fast enough?

Consider that since Xc(ω) = |Xc(ω)| ejΘ(ω), we can be sure that Xc(ω) = 0for |ω| > 5000π since in that range, Xc(ω) = 0 · ejΘ(ω) = 0.

So signal is band-limited, and ωm = 5000π.

Is (ωs = 20000π) > (2ωm = 10000π)?

Yes. Therefore, we can guarantee recovery of xc(t).

5. What if <e{Xc(ω)} = 0 for |ω| > 5000π?

Again, this is not an obvious question. Consider this representation ofXc(ω).

Xc(ω) = <e{Xc(ω)}+ j=m{Xc(ω)}

Here, we know that the real part is zero outside of the giving band limitsbut we know nothing about the imaginary part. Thus, we can’t say for surewhether it will be zero or non-zero.

Since we don’t know for sure whether Xc(ω) = 0 for |ω| > 5000π, we cannotguarantee recovery.

6. What if Xc(ω) ∗Xc(ω) = 0 for |ω| > 15000π?

This is another tricky situation. What can we figure out about Xc(ω)?

Thinking back to previous signals courses, recall that

Xc(ω) ∗Xc(ω)

l Fourier Transform

2πxc(t)xc(t)

and we can illustrate what happens in this scenario.

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Sampling and the Sampling Theorem

In this case, the band limits of the convolved transforms of Xc(ω) are twicethose of Xc(ω) itself.

So Xc(ω) = 0 for |ω| > 7500π, so ωm = 7500π.

Is (ωs = 20000π) > (2ωm = 15000π)?

Yes. Therefore, we can guarantee recovery of xc(t).

Additionally, for self-practice, you could identify the range of values of ωs foreach scenario that would allow the signal to be recovered.

5 Analysis of Ideal Sampling

In the preceding sections, we have shown our model of ideal sampling, in which the CT signalxc(t) is modulated by an impulse train p(t) to produce xp(t). Naturally, we want to knowwhether xp(t) really represents xc(t) uniquely and completely.

Since the sampling theorem talks about band-limited signals, it makes sense to do somefrequency analysis. First, we’ll do a quick review of the Fourier transform.

Aside: Review of Fourier Transform

Recall the Fourier transform Xc(ω) of CT signal xc(t) is given by:

Xc(ω) = F{xc(t)} =

∫ +∞

−∞xc(t)e

−jωtdt, ω ∈ R

This has the inverse transform given by:

xc(t) = F−1{Xc(ω)} =1

2π

∫ +∞

−∞Xc(ω)ejωtdω

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The above is true provided that xc(t) satisfies some conditions (such as beingfinite energy) or if xc(t) is periodic with a non-standard Fourier transform.

So, for our frequency analysis, let us assume our sampling function is

p(t) =+∞∑

n=−∞

δ(t− nTs)

and that xc(t) is any CT signal that has a CT Fourier transform.

Consider xp(t), which is still a CT signal as well.

xp(t) = xc(t) · p(t)

= xc(t) ·

[+∞∑

n=−∞

δ(t− nTs)

], n ∈ Z

=

[+∞∑

n=−∞

xc(t)δ(t− nTs)

], n ∈ Z

=

[+∞∑

n=−∞

xc(nTs)δ(t− nTs)

], n ∈ Z

Note that the values of xc(nTs) are the samples.

Now we analyze in the frequency domain. Recall that,

x(t) · y(t)

l Fourier Transform12π

X(ω) ∗ Y (ω)

so

Xp(ω) = F{xc(t) · p(t)}

=1

2πXc(ω) ∗ P (ω)

but that leads to the question, “What is P (ω)?” That is, what is the Fourier transform ofP (ω)?

P (ω) = F

{∑n

δ(t− nTs)

}We know that p(t) is periodic, with period Ts, so P (ω) is a non-standard transform basedon its Fourier series.

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Sampling and the Sampling Theorem

Aside: Fourier Series

For a periodic signal x(t) with period To, there is a Fourier series with coefficients

ak =1

To

∫To

x(t)ejkωotdt with ωo =2π

To

and therefore a non-standard Fourier transform, shown here.

X(ω) = 2π+∞∑

k=−∞

akδ(ω − kωo)

So, based on our knowledge of p(t) and Fourier transforms and series,

P (ω) = 2π+∞∑

k=−∞

akδ(ω − kωs) where ωs =2π

Ts

where the values of ak are determined by

ak =1

Ts

∫ Ts2

−Ts2

p(t)e−jkωstdt

=1

Ts

∫ Ts2

−Ts2

δ(t)e−jkωstdt

=1

Ts

e−jkωst

∣∣∣∣t=0

=1

Ts

which gives us a final expression for P (ω), shown and illustrated below.

P (ω) =2π

Ts

+∞∑k=−∞

δ(ω − kωs) with ωs =2π

Ts

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Sampling and the Sampling Theorem

Now that we know P (ω), we know

Xp(ω) =1

2πXc(ω) ∗ P (ω)

=1

2πXc(ω) ∗ 2π

Ts

+∞∑k=−∞

δ(ω − kωs)

=1

Ts

+∞∑k=−∞

Xc(ω) ∗ δ(ω − kωs)

which gives

Xp(ω) =1

Ts

+∞∑k=−∞

Xc(ω − kωs)

since x(t) ∗ δ(t− to) = x(t− to).

So Xp(ω) is a summation of scaled and shifted versions of Xc(ω). This is an important result.

Now, let’s consider an xc(t) and assume the sampling theorem conditions are met, so thatXc(ω) = 0 for |ω| > ωm and ωs > 2ωm.

So Xc(ω) =

{anything |ω| < ωm

0 |ω| > ωm

An example of a possible Xc(ω) is shown in the following figure.

With the Xc(ω) shown above, Xp(ω) would be as illustrated below, since it is made of shiftedand scaled replicas of Xc(ω):

Xp(ω) = . . .+1

Ts

Xc(ω+2ωs)+1

Ts

Xc(ω+ωs)+1

Ts

Xc(ω)+1

Ts

Xc(ω−ωs)+1

Ts

Xc(ω−2ωs)+ . . .

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Sampling and the Sampling Theorem

We have non-overlapping replicas of Xc(ω) so that we can recover Xc(ω) from Xp(ω) bydefining a reconstruction filter, Hr(ω), defined and illustrated as follows.

Hr(ω) =

{Ts

−ωs

2< ω < ωs

2

0 |ω| > ωs

2

Then, if the recovered signal Xr(ω) = Xp(ω) · Hr(ω) then Xr(ω) = Xc(ω) and thereforexr(t) = xc(t). Thus xc(t) is recovered from its samples!

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Sampling and the Sampling Theorem

The reconstruction occurs because it ignores all but one of the replicas (the lowest frequencyone) and undoes the scaling of Xc(ω) that occurs in Xp(ω) for that replica.

So this is how the reconstruction works when the sampling theorem conditions are met.What happens if the conditions aren’t met? We’ll do two examples to explore that scenario.

Example 5.1: xc(t) is band-limited but undersampled

Suppose we have a CT signal xc(t) with Fourier transform Xc(ω), that is band-limited. However, suppose we sample at less than the Nyquist rate. That is, letωs 6> 2ωm.

Let Xc(ω) be defined as in the figure below.

Also, recall the value of Xp(ω).

Xp(ω) =1

Ts

+∞∑k=−∞

Xc(ω − kωs)

With ωs 6> 2ωm, Xp(ω) would be as shown below. Note that the replicas overlap,and thus add, in places. The added version is shown by a dotted line where itdiffers from the non-adding parts.

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Sampling and the Sampling Theorem

Let’s try to recover Xc(ω) using Hr(ω) by applying it to Xp(ω).

Hr(ω) =

{Ts |ω| < ωs

2

0 |ω| > ωs

2

So now Xr(ω) = Hr(ω) ·Xp(ω).

The above Xr(ω) is clearly not the same as Xc(ω), and so xr(t) 6= xc(t) and theoriginal signal is not recovered. This is called aliasing due to undersampling.

Also, as an item of note, in this case, if ωs = 2ωm exactly, we still would havebeen able to recover Xc(ω). But that is only because, in this case, Xc(ω) = 0at the band-limits, and thus there is no interference. That’s why the samplingtheorem can only guarantee recovery for ωs > 2ωm, since ωs = 2ωm depends onthe values of Xc(ω).

In the prior example, since the sampling rate was too low to meet the conditions of thesampling theorem, we had overlapping replicas which added together and thus obscured theoriginal value. This is called aliasing.

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Sampling and the Sampling Theorem

Example 5.2: sin function

Let xc(t) = sin ωot and thus

Xc(ω) =π

jδ(ω − ωo)−

π

jδ(ω + ωo)

It is clear that Xc(ω) is band-limited to ωm = ωo.

Let us consider the possible scenarios for the sampling theorem.

1. When ωs > 2ωm,

We have non-overlapping replicas, and a simple reconstruction filter Hr(ω)as seen previously can reconstruct. This is not surprising, since we haveobeyed the rules of the sampling theorem.

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Sampling and the Sampling Theorem

2. When ωs < 2ωm,

Here, our replicas overlap. When we apply the reconstruction filter, we donot recover the same signal. In fact, in this case, xr(t) = sin−(ωs − ωo)t =− sin (ωs − ωo)t which is quite different from xc(t) = sin ωot. Under thissampling scheme, xr(t) is an alias of xc(t).

3. When ωs = 2ωm,

In this case, even if weird stuff didn’t happen due to the undefined boundaryof the reconstruction filter, the replicas clearly cancel each other out. Thus,the recovered signal will have no frequency components. xr(t) will be a DCsignal, which is clearly wrong.

We see aliasing again in this example. Aliasing can be avoided by following therequirements of the sampling theorem.

Signals and Systems, 2E: Chapter 7.0 to 7.3, page(s) 514–534.Signal Processing First: Chapter 4-1, 4-5, page(s) 71–79, 93–94.

We’ll conclude with an example covering ideal sampling, before looking at the concept ofaliasing in a bit more detail.

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Sampling and the Sampling Theorem

Example 5.3: Ideal Sampling

Given that

p(t) =∞∑

n=−∞

δ (t− nTs) and Ts =1

6000,

for each of the following signals, plot Xp(ω) which is the Fourier transform ofxp(t) = xc(t)·p(t), then determine what happens when that signal is reconstructedby the ideal reconstruction filter Hr(ω) with cut-off frequency ωs/2, as definedbelow.

Hr(ω) =

{Ts |ω| < ωs

2

0 |ω| > ωs

2

1. Case 1: xc(t) = sin 4000πtπt

Here, we can use the CT Fourier Transform tables in the textbooks to figureout what Xc(ω) will be.

xc(t) = sin 4000πtπt

l Fourier Transform

Xc(ω) =

{1 |ω| < 4000π0 |ω| > 4000π

This is clearly band-limited, with ωm = 4000π = 2π · 2000 and Tm =1/2000. Since (ωs = 2π/Ts = 12000π) > (2ωm = 8000π), this meets therequirements of the sampling theorem, and so no aliasing should occur.Therefore, Xp(ω) is simply a set of shifted, scaled replicas of Xc(ω), asshown below.

Xp(ω) =1

Ts

+∞∑k=−∞

Xc(ω − kωs)

This means that the ideal reconstruction filter will successfully recover theoriginal signal. In other words, Xr(ω) = Xp(ω) ·Hr(ω) = Xc(ω).

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Sampling and the Sampling Theorem

2. Case 2: xc(t) = 1 + cos 2000πt + sin 4000πt

Here, we can use the CT Fourier Transform tables in the textbooks to figureout what Xc(ω) will be.

xc(t) = 1 + cos 2000πt + sin 4000πt

l Fourier Transform

Xc(ω) = 2πδ(ω) + πδ(ω − 2000π) + πδ(ω + 2000π)

+π

jδ(ω − 4000π)− π

jδ(ω + 4000π)

This is also clearly band-limited, with ωm = 4000π = 2π · 2000 and Tm =1/2000. Since (ωs = 2π/Ts = 12000π) > (2ωm = 8000π), this meets therequirements of the sampling theorem, and so no aliasing should occur.Therefore, Xp(ω) is simply a set of shifted, scaled replicas of Xc(ω), asshown below.

Xp(ω) =1

Ts

+∞∑k=−∞

Xc(ω − kωs)

This means that the ideal reconstruction filter will successfully recover theoriginal signal. In other words, Xr(ω) = Xp(ω) ·Hr(ω) = Xc(ω).

More examples involving aliasing will follow in the next section.

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