Sampling of Continuous Time Signalseeweb.poly.edu/~yao/EE3054/Ch12.3_sampling.pdf · Sampling of Continuous Time Signals ... map amplitude values into a set of discrete values Q:

EE3054

Signals and Systems

Sampling of Continuous Time Signals

Yao Wang

Polytechnic University

Some slides included are extracted from lecture presentations prepared by McClellan and Schafer

4/17/2008 © 2003, JH McClellan & RW Schafer 2

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LECTURE OBJECTIVES

� Concept of sampling

� Sampling using periodic impulse train

� Frequency domain analysis

� Spectrum of sampled signal

� Nyquist sampling theorem

� Sampling of sinusoids

Two Processes in A/D

Conversion

� Sampling: take samples at time nT

� T: sampling period;

� fs = 1/T: sampling frequency

� Quantization: map amplitude values into a set of discrete values

� Q: quantization interval or stepsize

xc(t) x[n] = xc(nT) $[ ]x n

Quanti-

zation

Sampling

Sampling

Period

T

Quantization

Interval

Q

∞<<−∞= nnTxnx ),(][

pQ±

)]([][ˆ nTxQnx =

0 0.2 0.4 0.6 0.8 1

-1

-0.5

0

0.5

1

T=0.1Q=0.25

Analog to Digital

Conversion

A2D_plot.m

How to determine T and Q?

� T (or fs) depends on the signal frequency range� A fast varying signal should be sampled more frequently!

� Theoretically governed by the Nyquist sampling theorem

� fs > 2 fm (fm is the maximum signal frequency)

� For speech: fs >= 8 KHz; For music: fs >= 44 KHz;

� Q depends on the dynamic range of the signal amplitude and perceptual sensitivity� Q and the signal range D determine bits/sample R

� 2R=D/Q

� For speech: R = 8 bits; For music: R =16 bits;

� One can trade off T (or fs) and Q (or R)� lower R -> higher fs; higher R -> lower fs

� We only consider sampling in this class

SAMPLING x(t)

� SAMPLING PROCESS� Convert x(t) to numbers x[n]

� “n” is an integer; x[n] is a sequence of values

� Think of “n” as the storage address in memory

� UNIFORM SAMPLING at t = nTs

� IDEAL: x[n] = x(nTs)

C-to-Dx(t) x[n]

Sampling of Sinusoid

SignalsSampling above

Nyquist rate

ωs=3ωm>ωs0

Reconstructed

=original

Sampling under

Nyquist rate

ωs=1.5ωm<ωs0

Reconstructed

\= original

Aliasing: The reconstructed sinusoid has a lower frequency than the original!

Nyquist Sampling Theorem

� Theorem:� If x(t) is bandlimited, with maximum frequency fb(or

ωb =2π fb)

� and if fs =1/ Ts > 2 fb or ωs =2π / Ts >2 ωb

� Then xc(t) can be reconstructed perfectly from x[n]= x(nTs ) by using an ideal low-pass filter, with cut-off frequency at fs/2

� fs0 = 2 fb is called the Nyquist Sampling Rate

� Physical interpretation:� Must have at least two samples within each cycle!


Sampling Using Periodic Impulse

Train

x[n] = x(nTs )

FOURIER

TRANSFORM

of xs(t) ???


Periodic Impulse Train

∑∞

−∞=

−=

n

snTttp )()( δ


Impulse Train Sampling

xs (t) = x(t) δ (t − nTs )n=−∞

∞

∑ = x(t)δ (t − nTs )n=−∞

∞

∑

xs(t) = x(nTs)δ(t −nTs)n=−∞

∞

∑


Illustration of Samplingx(t)

x[n] = x(nTs )

∑∞

−∞=

−=n

sss nTtnTxtx )()()( δ

n

t


Sampling: Freq. Domain

EXPECT

FREQUENCY

SHIFTING !!!

∑∑∞

−∞=

∞

−∞=

=−=k

tjkk

n

sseanTttp

ωδ )()(

∑∞

−∞=

=k

tjkk

seaω

How is the

spectrum of xs(t)

related to that of

x(t)?


Fourier Series Representation

of Periodic Impulse Train

ωs =2π

Ts∑∑∞

−∞=

∞

−∞=

=−=k

tjkk

n

sseanTttp

ωδ )()(

s

T

T

tjk

s

kT

dtetT

as

s

s1

)(1

2/

2/

== ∫−

− ωδFourier Series


FT of Impulse Train

∑∑ ∑∞

−∞=

∞

−∞=

−=↔=−=

k

s

sn k

tjk

s

s kT

jPeT

nTttp s )(2

)(1

)()( ωωδπ

ωδ ω

s

sT

πω

2=

Frequency-Domain Analysis:

Using Fourier Series

∑ ∑∞

−∞=

=−=

n k

tjk

s

sse

TnTttp

ωδ1

)()(

)()()( tptxtxs =

xs (t) = x(t)1

Tsk=−∞

∞

∑ ejkωst =

1

Tsx(t)

k=−∞

∞

∑ ejkωst

Xs ( jω) =1

TsX( j(ω

k=−∞

∞

∑ − kωs ))

ωs =2π

Ts

Frequency-Domain Analysis:

Using Multiplication-

Convolution duality

∑∑ ∑∞

−∞=

∞

−∞=

−=↔=−=

k

s

sn k

tjk

s

s kT

jPeT

nTttp s )(2

)(1

)()( ωωδπ

ωδ ω

x(t)p(t) ⇔1

2πX( jω )∗ P( jω)

( )( )∑

∑∞

−∞=

∞

−∞=

−=

−==

k

s

s

k

s

s

s

kjXT

kjXT

jPjXjX

ωω

ωωδωπ

πωω

πω

1

)(*)(2

2

1)(*)(

2

1)(


Frequency-Domain

Representation of Sampling

Xs ( jω) =1

TsX( j(ω

k=−∞

∞

∑ − kωs ))

“Typical”

bandlimited signal


Aliasing Distortion

� If ωs < 2ωb , the copies of X(jω) overlap, and we have aliasing distortion.

“Typical”

bandlimited signal

Original signal

Sampling

impulse train

Sampled signal

ωs>2 ωm

Sampled signal

ωs<2 ωm

(Aliasing effect)

Frequency Domain

Interpretation of Sampling

The spectrum of the

sampled signal includes

the original spectrum and

its aliases (copies) shifted

to k fs , k=+/- 1,2,3,…

The reconstructed signal

from samples has the

frequency components

upto fs /2.

When fs< 2fm , aliasing

occur.


Reconstruction: Frequency-Domain

)()()(

so overlap,not do )(

of copies the,2 If

ωωω

ω

ωω

jXjHjX

jX

srr

bs

=

>

Hr ( jω )

Nyquist Sampling Theorem

� Theorem:� If x(t) is bandlimited, with maximum frequency fb(or

ωb =2π fb)

� and if fs =1/ Ts > 2 fb or ωs =2π / Ts >2 ωb

� Then xc(t) can be reconstructed perfectly from x[n]= x(nTs ) by using an ideal low-pass filter, with cut-off frequency at fs/2

� fs0 = 2 fb is called the Nyquist Sampling Rate

� Physical interpretation:� Must have at least two samples within each cycle!

Sampling of Sinusoid

Signals: Temporal domainSampling above

Nyquist rate

ωs=3ωm>ωs0

Reconstructed

=original

Sampling under

Nyquist rate

ωs=1.5ωm<ωs0

Reconstructed

\= original

Aliasing: The reconstructed sinusoid has a lower frequency than the original!

Sampling of Sinusoid:

Frequency Domain

f0-f0 fs-fs fs+f0

fs-f0-fs+f0-fs -f0

f0-f0 0

f0-f0 fs-fs fs+f0

fs-f0-fs+f0-fs -f0

fs/2-fs/2

0

Spectrum of

cos(2πf0t)

No aliasing

fs >2f0

fs -f0 >f0

Reconstructed

signal: f0

With aliasing

f0<fs <2f0 (folding)

fs -f0 <f0

Reconstructed signal: fs -f0

f0-f0

fs-fs

fs+f0f0-fs-f0+fs-fs -f0

With aliasing

fs <f0 (aliasing)

f0-fs <f0

Reconstructed signal: fs -f0

0

0

More examples with

Sinusoids


SAMPLING GUI (con2dis)

Strobe Movie

� From SP First, Chapter 4, Demo on “Strobe Movie”

How to determine the necessary

sampling frequency from a signal

waveform?

� Given the waveform, find the shortest ripple, there should be at least two samples in the shortest ripple

� The inverse of its length is approximately the highest frequency of the signal

Tmin

Fmax=1/Tmin

Need at least two

samples in this

interval, in order not

to miss the rise and

fall pattern.

Sampling with Pre-Filtering

Pre-Filter Periodic

H (f) Sampling

x(t) x’(t) xd(n)

Sampling

period T

• If fs < 2fb, aliasing will occur in sampled signal

• To prevent aliasing, pre-filter the continuous signal so that fb<fs/2

• Ideal filter is a low-pass filter with cutoff frequency at fs/2

(corresponding to sync functions in time)

•Common practical pre-filter: averaging within one sampling interval

Summary

� Sampling as multiplication with the periodic impulse train

� FT of sampled signal: original spectrum plus shifted versions (aliases) at multiples of sampling freq.

� Sampling theorem and Nyquist sampling rate

� Sampling of sinusoid signals� Can illustrate what is happening in both temporal and freq.

domain. Can determine the reconstructed signal from the sampled signal.

� Need for prefilter

� Next lecture: how to recover continuous signal from samples, ideal and practical approaches

Readings

� Textbook: Sec. 12.3.1-12.3.2, 4.1-4.3

� Oppenheim and Willsky, Signals and Systems, Chap. 7.

� Optional reading (More depth in frequency

domain interpretation)

Documents

Sampling of Continuous Time Signalseeweb.poly.edu/~yao/EE3054/Ch12.3_sampling.pdf · Sampling of Continuous Time Signals ... map amplitude values into a set of discrete values Q: