1497-Chapter4

Chapter 4Sampling of Continuous-Time Signals

Der-Feng Tseng

Department of Electrical EngineeringNational Taiwan University of Science and Technology

(through the courtesy of Prof. Peng-Hua Wang of National Taipei University)

February 19, 2015

Der-Feng Tseng (NTUST) DSP Chapter 4 February 19, 2015 1 / 100

http://homepage.ntust.edu.tw/dtseng/http://www.ee.ntust.edu.tw/http://www.ntust.edu.tw/http://homepage.ntust.edu.tw/dtseng/

Outline

1 4.1 Periodic Sampling

2 4.2 Frequency Domain representation

3 4.3 Reconstruction

4 4.4 DT Processing of Signals

5 4.5 CT Processing of Signals

6 4.6 Changing Sampling Rate

7 4.7 Multirate Signal Processing

8 4.8 A/D and D/A

9 4.9 Oversampling A/D and D/A


http://homepage.ntust.edu.tw/dtseng/

About The figures

All the figures are from Discrete-Time Signal Processing, 2e, byOppenheim, Schafer, and Buck,Prentice Hall, Inc.



4.1 Periodic Sampling



Concept

Figure 4.1 Block Diagram of an ideal C/D converter

The sequence of samples x[n] is obtained from a continuous-timesignal xc(t) according to

x[n] = xc(nT ), < n <

T is the sampling period. fs = 1/T is the sampling frequency.s = 2fs = 2/T is the sampling frequency in radians per second.

The system is an ideal continuous-to-discrete-time (C/D) converter.



Figure 4.2

(a) C/D converter = modulate by s(t) + impulse to sequence

(b) sampling by two rates

(c) output sequences



4.2 Frequency Domain representation



Derivation

Let s(t) be the periodic impulse train

s(t) =

n=

(t nT ), S(j) =2

T

k=

( ks)

The sampled signal xs(t) from a continuous-time signal xc(t) is

xs(t) = xc(t)s(t) = xc(t)

n=

(t nT )

=

n=

xc(nT )(t nT )

The Fourier transformation of xs(t) is

Xs(j) =1

2Xc(j) S(j) =

1

T

k=

Xc(j( ks))



Figure 4.3

Suppose xc(t) is bandlimited at = N , the replicas of Xc(j) do notoverlap if s N > N or s > 2N



Recovery

Xc(j) can be recovered by Xs(j)Hr(j) where Hr(j) is an ideallowpass filter with gain T and cutoff frequency c withN < c < s N .



Alias

If s 2N , the copies of Xc(j) overlap, and Xc(j) cant berecovered by lowpass filtering. The reconstructed signal byXr(j) = Xs(j)Hr(j) is distorted. This is refered to aliasdistortion.

Nyquist Sampling Theorem. Let xc(t) be a bandlimited signal withXc(j) = 0 for || N . Then xc(t) is uniquely determined by itssamples x[n] = xc(nT ), n = 0,1,2, . . . , if

s =2

T 2N .

N is commonly referred to as the Nyquist frequency.The minimal sampling frequency 2N is called the Nyquist rate.



Example

Let xc(t) = cos 0t. With no aliasing, xr(t) = cos 0t. With aliasing,xr(t) = cos(0 0)t.



Discrete-Time FT

The sampled signal

Xs(j) =

n=

xc(nT )ejTn

Since x[n] = xc(nT ) and X(ej) =

n=

x[n]ejn, it follows that

Xs(j) = X(ej)

=T= X(ejT )

Since X(ejT ) =1

T

n=

Xc(j( ks)), we have

X(ej) =1

T

n=

Xc

(

j

(

T

2k

T

))

,



Example 4.1

xc(t) = cos(4000t) (2000 Hz).

Sampling period T = 1/6000, s = 12000 (6000 Hz)

x[n] = cos(4000Tn) = cos0n where 0 = 2/3.

Xc(j) = ( 4000) + ( + 4000)



Example 4.2

xc(t) = cos(16000t) (8000 Hz).


x[n] = cos(16000Tn) = cos0n where 0 = 8/3 = 2/3.



Example 4.3

xc(t) = cos(4000t) (2000 Hz).


x[n] = cos(4000Tn) = cos0n where 0 = 2/3.



4.3 Reconstruction



Derivation

Sampled sequence

xs(t) =

n=

x[n](t nT )

Reconstruction by a lowpass filter with cutoff frequencyN < c < s N .

xr(t) =

n=

x[n]hr(t nT )

A common choice c = s/2 = /T

hr(t) =sin(t/T )

t/T

xr(t) =

n=

x[n]sin[(n T )/T ]

(t nT )/T



Illustration

xr(t) =

n=

x[n]hr(t nT )



Illustration

hr(t) =sin(t/T )

t/TDer-Feng Tseng (NTUST) DSP Chapter 4 February 19, 2015 20 / 100


Interpolation

hr(nT ) = 0, xr(mT ) = x(mT )



Bandlimited Reconstruction

Ideal discrete-to-continuous-tine (D/C) converter

Xr(j) =

n=

x[n]Hr(j)ejTn = Hr(j)X(jT )



4.4 DT Processing of Signals



Concept

x[n] = xc(nT )

X(ej) =1

T

k=

Xc

(

j

(

T

2k

T

))

yr(t) =

n=

y[n]sin[(n T )/T ]

(t nT )/T

Yr(j) = Hr(j)Y (jT ) =

{

TY (jT ), || < /T,

0, otherwise.



LTI System

If the discrete-time system is LTI, we have Y (ej) = H(ej)X(ej)

The spectrum of the output signal

Yr(j) = Hr(j)H(ejT )X(ejT )

= Hr(j)H(ejT )

1

T

k=

Xc

(

j

(

T

2k

T

))

If Xc(j) = 0 for || /T , then

Yr(j) =

{

H(ejT )Xc(j), || < /T,

0, || /T.

We have Yr(j) = Heff(j)Xc(j) where

Heff(j) =

{

H(ejT ), || < /T,

0, || /T.



Illustration

Heff(j) =

{

H(ejT ), || < /T,

0, || /T.



Example 4.4

H(ej) =

{

1, || < c,

0, c < || .



Example 4.5

The ideal continuous-time differentiator yc(t) =ddtxc(t)

The corresponding frequency response Hc(j) = j

The effective frequency response Heff(j) = j for || < /T .

The discrete-time have frequency response

H(ej) =j

T, || <

The corresponding impulse response

h[n] =n cos n sinn

n2T=

{

0, n = 0cos n

nT, n 6= 0.



Example 4.5



Example 4.6

xc(t) = cos(0t) x[n] = cos(0n) where 0 = 0T

X(ejT ) =

T( 0) +

T( + 0), for || /T

X(ej) = ( 0) + ( + 0)

The output of the digital differentiator

Y (ej) = H(ej)X(ej) =j

T[( 0) + ( + 0)]

=j0

T( 0)

j0

T( + 0)]

The output of the D/C converter

Yr(j) = Hr(j)Y (ejT ) = TY (ejT )

= T

[

j0

T(T 0T )

j0

T(T + 0T )

]

= j0( 0) j0( + 0)

yr(t) = j012e

j0t j012e

j0t = 0 sin(0t)



Impulse Invariance

If h[n] = hc(nT ) H(ej) = 1T Hc(j

T ), thus

h[n] = Thc(nT ) H(ej) = Hc(j

T), ||



Example 4.7

Ideal lowpass filter

Hc(j) =

{

1, || < c

0, || c.

The impulse response of the continuous-time system

hc(t) =sin(ct)

t

The impulse response of the discrete-time system by impulse invariantis

h[n] = Thc(nT ) = Tsin(cnT )

nT=

sin(cn)

n



Example 4.8

The ideal impulse response of the continuous-time system

hc(t) = Aes0tu(t)

The impulse response of the discrete-time system by impulse invariantis

h[n] = Thc(nT ) = TAes0nTu(nT ) = ATes0nTu[n]

The z-transform is

H(z) =AT

1 es0nz1

Alias occurs!



4.5 CT Processing of Signals



Concept

Suppose

xc(t) =

n=

x[n]sin[(t nT )/T ]

(t nT )/T

yc(t) =

n=

y[n]sin[(t nT )/T ]

(t nT )/T

We have

Xc(j) = TX(ejT ), Yc(j) = Hc(j)Xc(j), Y (e

j) =1

TYc(j

T)

Hc(j) = H(ejT )



Example 4.9

The desired systemH(ej) = ej

for implementing y[n] = x[n] (no formal meaning).

The continuous system

Hc(j) = H(ejT ) = ejT

andy(t) = xc(tT )



Example 4.10

The desired system

H(ej) =1

M + 1

sin[(M + 1)/2]

sin(/2)ejM/2

M is even y[n] = w[n M/2]M is odd y[n] = w[n] followed by a continuous-time delay ofMT/2.



4.6 Changing Sampling Rate



Motivation

Given x[n] = xc(nT ), we want to find x[n] = xc(nT

) where T 6= T .

One approach is to reconstruct xc(t) from x[n], then sample xc(t) byT .

Not desirable, because reconstruction, A/D and D/Aare not ideal.

We are interested in the method that involve only discrete-timeoperations.



Reduction by M

If T = MT , we have the (sampling rate) compressor

xd[n] = x[nM ] = xc(nMT )

If Xc(j) = 0 for || N , then xd[n] is an exact representation ofxc(t) if /T

= /(MT ) N .The sampling rate can be reduced by a factor of M

if the original sampling rate was at least M times the Nyquist rate, orif the bandwidth of the sequence is first reduced by a factor of M bydiscrete-time filtering.

In general, we the operation of reducing the sampling rate (includingany prefiltering) will be called the downsampling.



Reduction by M

What is the Fourier transform of xd[n] ?

x[n] = xc(nT )

X(ej) =1

T

k=

Xc

(

j 2k

T

)

xd[n] = xc(nMT ), r = i+ kM

Xd(ej) =

1

MT

r=

Xc

(

j 2r

MT

)

=1

M

M1

i=0

[

1

T

k=

Xc

(

j

(

2i

MT

2k

T

))

]

=1

M

M1

i=0

X(ej(/M2i/M))



Reduction by M



Downsampling



Increase by L

T = T/L, we want to obtain

xi[n] = xc(nT) from x[n] = xc(nT ).

This is called the upsampling.

xi[n] = x[n/L] = xc(nT/L) for n = 0,L,2L, . . .

(Sampling rate) expander

xe[n] =

{

x[n/L], n = 0,L,2L, . . .

0, otherwise.=

k=

x[k][n kL]

Xe(ej) =

k=

x[k]ejLk = X(ejL)



Increase by L

T = T/L, we want to obtain

xi[n] = xc(nT) from x[n] = xc(nT ).

This is called the upsampling.

xi[n] = x[n/L] = xc(nT/L) for n = 0,L,2L, . . .



Increase by L



Interpolator

Ideal lowpass filter, gain L, cutoff frequency /L

hi[n] =sin(n/L)

n/L

xi[n] =

k=

x[k]sin[(n kL)/L]

(n kL)/L

We can use other interpolator instead of the ideal lowpass filter

linear interpolation

hlin[n] =

{

1 |n|/L, |n| L

0, otherwise

Hlin(ej) =

1

L

[

sin(L/2)

sin(/2)

]



Linear Interpolator



Noninteger Factor



Example 4.11



4.7 Multirate Signal Processing



Introduction

Multirate signal processing refers in general to utilizing upsampling,down sampling, compressor, and expanders to increase the efficiencyof signal-processing systems.

For example, if T = 1.01T , we can first interpolate by L = 100 usinga lowpass filter, that cutoff at c = /101. then decimate byM = 101.

Require large amounts of computation for each outputsample.



Downsampling Identities

These two systems are equivalent



Downsampling Identities

In (b), Xb(ej) = H(ejM )X(ej) and

Y (ej) =1

M

M1

i=0

Xb(ej(/M2i/M))

=1

M

M1

i=0

H(ej(2i))X(ej(/M2i/M))

= H(ej)1

M

M1

i=0

X(ej(/M2i/M))

= H(ej)Xa(ej)



Upsampling Identities

These two systems are equivalent



Upsampling Identities

In Fig.(a),

Y (ej) = Xa(ejL) = X(ejL)H(ejL)

In Fig.(b), Xb(ej) = X(ejL), thus

Y (ej) = H(ejL)Xb(ej)



Polyphase Decom of Signal

Decompose h[n] into

h[n] =

M1

k=0

hk[n k],

where hk[n] =

{

h[n + k], n = integer multiple of M,

0, otherwise.

ek[n] = h[nM + k] = hk[nM ]



Polyphase Decom of Signal



Polyphase Decom of Filter

H(z) =

M1

i=0

Ek(zM )zk



Polyphase Decimation



Polyphase Interpolation



4.8 A/D and D/A



Introduction



Prefiltering: Ideal

Antialiasing filter

Haa(j) =

{

1, || < c < /T,

0, || > c.

Heff(j) Haa(j)H(ejT )

Need sharp-cutoff antialiasing filters. Drawbacks

May account for a major part of the cost. Difficult and expensive to implement. Highly nonlinear phase response.



Prefiltering: Oversampling

Simple antialiasing filter Haa(j) has a gradual cutoff with significantattenuation at MN .

Oversampling at MN .

Downsampling by M with sharp cutoff antialiasing digital filter.



Prefiltering: Oversampling



A/D Conversion

Sampling-and-hold

Analog-to-digital: quantization



A/D: Sample-and-hold

The output of the ideal sample-and-hold system is

x0(t) =

n=

x[n]h0(t nT ) = h0(t)

n=

x[n](t nT )

h0(t) is the impulse response of the zero-order-hold system

h0(t) =

{

1, 0 < t < T,

0, otherwise.



A/D: Concept

Practical A/D Converter

Conceptual A/D Converter



A/D: Quantizer

The quantizer is a nonlinear system whose purpose is to transform theinput sample x[n] into one of a finite set of prescribed values.

x[n] = Q(x[n])

Uniform quantizers: quantization levels are spaced uniformly.

Linear quantizers: quantization levels are of linear progression.

Bipolar quantizers: both positive and negative samples can bequantized.



A/D: Quantization

Uniform, linear, bipolar

The input sample is rounded to the nearest quantization level.

8 (even) quantization levels No quantization level at zero amplitude. An equal number of positive and negative quantization

level.Der-Feng Tseng (NTUST) DSP Chapter 4 February 19, 2015 78 / 100


A/D: Quantization

Let Xm be the full-scale level of the A/D converter. Eachquantization level is coded by B + 1 bits.For a bipolar quantizer, the step size of the quantizer is

=2Xm2B+1

=Xm2B

If the signal amplitude exceeds the full-scale value, some samples areclipped.Der-Feng Tseng (NTUST) DSP Chapter 4 February 19, 2015 79 / 100


A/D: Quantization Error

Quantization error defined as

e[n] = x[n] x[n]

We can analyze the quantization error statistically by assuminge[n] is a stationary random process.e[n] and x[n] are uncorrelated.e[n] is a white-noise process.The pdf of e[n] is uniformly.



Example 4.12




/2 < e[n] < /2

Variance of e[n]: 2e =2

12 =22BX2m

12

Let 2x be the rms value of the signal. The signal-to-noise ratio (SNR)

SNR = 10 log102x2e

= 6.02B+ 10.8 20 log10Xmx

Increase B by 1 bit, SNR increases by 6 dB. Halve x, SNR decreases by 6 dB.




If the signal amplitude has a Gaussian distribution, only 0.064% ofthe samples would have an amplitude greater than 4x.

We may set Xm = 4x, then

SNR 6B 1.25dB

The signal amplitude should be carefully matched to the full-scaleamplitude of the A/D converter.



D/A Conversion

Ideal reconstruction

xr(t) =

n=

x[n]sin[(t nT )/T ]

(t nT )/T

Practical reconstruction

xDA(t) =

n=

x[n]h0(t nT )

=

n=

x[n]h0(t nT ) +

n=

e[n]h0(t nT )

= x0(t) + e0(t)Der-Feng Tseng (NTUST) DSP Chapter 4 February 19, 2015 84 / 100


D/A Conversion

The spectrum of x0(t) is

X0(j) =

[

1

T

k=

Xa (j( 2k/T ))

]

H0(j)

Compensated reconstruction filter

Hr(j) =Hr(j)

H0(j)

The frequency response of the zero0order hold filter is

H0(j) =2 sin(T/2)

ejT/2

The compensated reconstruction filter is

Hr(j) =

{

T/2sin(T/2)e

jT/2, || < /T

0, || > /T.



D/A Conversion



4.9 Oversampling A/D and D/A



Oversampled A/D

Assume that xa(t) is a zero-mean, WSS random process with PSDxaxa(j) and autocorrelation function xaxa(). xaxa(j) isbandlimited to N .

Sampling rate 2/T = M 2N . M is called the oversamplingratio.

The decimation filter is an ideal lowpass filter with c = /M .

We want to analyze the quantization error.



Oversampled A/D

Additive noise model.

We want to determine the ratio of signal power E{x2da[n]} toquantization-noise power E{x2de[n]}. The ratio is a function ofquantization step and the oversampling ratio M .

Original signal xa(t), PSD is xaxa(j), power is

E{x2a(t)} =1

2

N

N

xaxa(j)d



Oversampled A/D

Analysis of signal powerAfter C/D, x[n] = xa(nT ), PSD and power are

xx(ej) =

{

1T xaxa(j/T ), || < /M

0, /M <

E{x2[n]} =1

2

xx(ej)d = E{x2a(t)}

After downsampling by M , PSD and power are

xdaxda(ej) =

1

Mxx(e

j/M )

E{x2da[n]} =1

2

xdaxda(ej)d

=1

2

1

Mxx(e

j/M )d = E{x2[n]} = E{x2a(t)}

We have E{x2da[n]} = E{x2[n]} = E{x2a(t)}



Oversampled A/D

Analysis of noise power

Since variance of e[n] is 2e = 2/12, we have ee[m] =

2e[m] and

ee(ej) = 2 for || < .

After ideal lowpass filter, the noise power is

E{e2[n]} =1

2

/M

/M

2ed =2eM

After downsampling, the noise power does not changed.

E{x2de[n]} =2eM

=2

12M=

1

12M

(

Xm2B

)2

Let Pde = E{x2de[n]}, we have

B = 1

2log2 M

1

2log2 12

1

2log2 Pde + log2 Xm

For example, M = 4, one less bit to achieve a desired accuracy.



Oversampled A/D



Noise Shaping A/D

If we want to reduce the number of bits from 16 to 12, we needM = 44 = 256.

The concept of noise shaping is to modify the A/D conversionprocedure so that the PSD of the quantization noise is not uniform,but rather, is shaped such that most of the noise power is outside|| < /M .



1st-Order Noise Shaping A/D

Denote the transfer function from x[n] to y[n] as Hz(z)

Yx(z) =1

1 z1[

X(z) z1Yx(z)]

Yx(z) = X(z), Hx(z) = 1

Denote the transfer function from e[n] to y[n] as He(z)

Ye(z) = E(z) +

(

1

1 z1

)

[

z1Ye(z)]

Ye(z) = (1 z1)E(z), He(z) = 1 z

1




e[n] = e[n] e[n 1]

ee(ej) = 2e |He(e

j)|2 = 2e [2 sin(/2)]2

Pda = E{x2da[n]} = E{x

2[n]} = E{x2a(t)}IfM is sufficient large such that sin(/2M) /2M , thequantization-noise power is

Pde =1

2

2

12M

[2 sin(/2M)]2d 1

36

22

M3

B = 32 log2 M + log2(/6) 12 log2 Pde + log2Xm




Equivalent saving in quantizer bits relative to M = 1Direct quantization

B = 1

2log2M

1

2log2 12

1

2log2 Pde + log2Xm

First-order noise shaping

B = 3

2log2M + log2(/6)

1

2log2 Pde + log2Xm



2nd-Order Noise Shaping A/D

He(z) = (1 z1)2

ee(ej) = 2e [2 sin(/2)]

4



Oversampling D/A



Oversampling D/A



4.1 Periodic Sampling4.2 Frequency Domain representation4.3 Reconstruction4.4 DT Processing of Signals4.5 CT Processing of Signals4.6 Changing Sampling Rate4.7 Multirate Signal Processing4.8 A/D and D/A4.9 Oversampling A/D and D/A

1497-Chapter4

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