Digital Signal Processing Notes

Chap 2: A/D Conversion and Multiplexing

ELE745 Digital Communications

Chapter 2 : A/D Conversion and Multiplexing

Lian ZhaoProfessor

Department of Electrical and Computer EngineeringRyerson University

1


Formatting and Digital Coding of Waveforms

• A lot of information sources generate analog (continuous-time) information. In order torepresent, process, store and transmit data using computers and networks, the analogdata has to be converted to a faithful digital representation.

• The original form of most communicated data is either textual or analog. It is needed tobe transformed to binary digits for transmission.

– Textual information is transformed by use of a coder, etc. ASCII (7-bit code).

– Analog signal is formatted using sampling, quantization and encoding.

• In this chapter, the well established concepts of sampling and quantization are reviewed.Representing a signal waveform using pulse code modulation will be discussed.

ChannelSampling Quantizing Encoding Modulator

A/D conversiona(n)a(t)

Source{a_q}

Figure 1: Digital Coding of Waveforms.

2


Chap. 2.1 Sampling

• Analog signals can be digitized through sampling and quantization.

• The sampling rate must be sufficiently high so that the analog signal can bereconstructed from the samples with sufficient accuracy.

• The sampling theorem is the basis for determining the proper sampling rate of a givensignal, and is state as If fm is the maximum frequency of the signal, then the signalhas to be sampled at a rate of at least fs samples per second, where fs = 2fm.

• In practice, sampling is performed by multiplying the signal in the time-domain with atrain of impulses where each pulse is separated by sampling time (a comb function).

3


x(t)

0

fs0

0 fsTs

Figure 2: Sampling process analyzed in Frequency domain.

4


Mathematically, Denote the original analog signal as x(t), sampling function (pulse train) asx!(t)

1. pulse train x!(t) =!!

n="! !(t! nTs) is periodic, and its FS are

Dn =1

Ts

" Ts/2

"Ts/2xp(t)e

"jn2"fst dt =1

Ts(fs =

1

Ts) (1)

Then the Fourier transform of x!(t) is

X!(f) =1

Ts

!#

n="!!(f ! nfs) (2)

2. In time domain, the sampled signal is

xs(t) = x(t) · x!(t) =!#

n="!x(t) · !(t! nTs) =

!#

n="!x(nTs)!(t! nTs)

5


3. In frequency domain, the sampled signal is

Xs(f) = X(f) "X!(f) = X(f) "$1

Ts

!#

n="!!(f ! nfs)

%

=1

Ts

!#

n="!X(f ! nfs) (3)

4. We observe that the spectrum of Xs(f) is a periodic function of the original spectrumX(f) scaled by a constant factor ( 1

Ts) and centered at nfs.

5. The original signal x(t) can be recovered without distortion by using a low pass filterwhen there is no spectrum overlap. Here comes the sampling theorem (fs # 2fm).

6. When sampling frequency is not high enough, there will be spectrum overlap, which iscalled aliasing.

6


f−W W

X(f)

0−fs−2fs fs 2fsOver sampling

Nyquist sampling

Under sampling

0−fs−2fs

−fs 0 fs

2fsfs

2fs−2fs

2fs0−2fs fs−fs

Figure 3: Spectrum of the sampled signal.

7


Figure 4: Overlapping spectra due to aliasing.

8


Figure 5: Illustration of aliasing.

9


Sampling Theorem

A band-limited signal of finite energy which has no frequency components higher than B Hz,is completely described by specifying the values of the signal at instants of time separated by1/2B seconds.

Discussion:

• The sampling rate of 2B samples per second for a signal bandwidth of B Hz is called theNyquist rate, and 1/2B is called the Nyquist interval.

• We can oversample, but cannot undersample.

• The sampling process discussed is for baseband (low-pass) signal.

• If a signal is not bandlimited, no matter how fast we sample the signal, the signal cannotbe accurately recovered. For example, a square waveform.

10


The distortion is due to fs < 2B (undersampling). This is called aliasing. When aliasingoccurs, {x(nTs)} cannot accurately recover the original signal x(t). To combat the aliasingerror in practice,

• Prior the sampling, we use an LPF pre-alias filter to attenuate those high frequencycomponent of the signal that are not essential to the information be carried by the signal.

• Taken account of the guard band of practical LP filter. Sample the signal at a rate slightlyhigher than the Nyquist rate (fs > 2B).

11


Ideal Signal Reconstruction

0−fs−2fs fs 2fs

0−fs−2fs 2fs

−W W

X(f)

fsfs−W

fc−fc

W < fc < fs−W

f

f

f

f

f−W W

X(f)

12


Ideal Signal Reconstruction – Analysis

(1) Frequency-domain description

Review that the spectrum of the sampled signal is

Xs(f) =1

Ts

!#

n="!X(f ! nfs)

The original signal can be recovered as

X(f) =1

fsXs(f) · rect(f/2B)

where

rect(f/2B) =

&'

(1 !B $ f $ B

0 o.w.

is called the ideal low pass filter.

13


(2) Time-domain description

x(t) = F"1[X(f)] =1

fsF"1[Xs(f) · rect(f/2B)] =

1

fsF"1[Xs(f)] " F"1[rect(f/2B)]

=1

fsxs(t) " (2Bsinc(2"Bt)) =

2B

fsxs(t) " sinc(2"Bt)

=2B

fs

$ !#

n="!x(nTs)!(t! nTs)

%" sinc(2"Bt)

=2B

fs

!#

n="!x(nTs)sinc[2"B(t! nTs)]

If fs = 2B, then

x(t) =!#

n="!x(nTs)sinc[2"B(t! nTs)] (4)

Equation (4) is an interpolation formula for reconstruction the original signal from thesequence of sample values x(nTs). The function sinc(2"Bt) is the interpolation function.

14


Figure 6: Ideal sampling (textbook Fig 2.6 P64).

15


Figure 7: Natural sampling (textbook Fig 2.8 P67).

16


Practical Sampling

Ideal sampling: !Ts =!!

n="! !(t! nTs).

Natural sampling: c(t) is a periodic sampling signal. Using Fourier Series analysis, we have

c(t) =A#

Ts

!#

n="!sinc(n"fs#)e

j2"nfst (5)

where # is the pulse width, and A is the pulse amplitude. The sampled signal can beexpressed as

xs(t) = x(t) · c(t) = A#

Ts· x(t)

!#

n="!sinc(n"fs#)e

j2"nfst (6)

In the frequency domain.

Xs(f) = F [xs(t)] =A#

Ts

!#

n="!sinc(n"fs#) · F [x(t)ej2"nfst]

=A#

Ts

!#

n="!sinc(n"fs#) ·X(f ! nfs)

17


We can observe that

• Using natural sampling, no distortion in receiving x(t) if fs > 2B.

• If A# = 1, natural sampling is equivalent to ideal sampling when n = 0.

18


Chap. 2.2 Quantization

Quantization is the process in which the amplitude of the sampled signal (discrete-timesignal) is limited to a predefined set of new voltage levels. As a result, for each sample therecould be a quantization error e(t), resulting due to differences in voltage between thesampled and the quantized values.

x(t) % quantizer % xq(t)

e(t) = xq(t)! x(t)

For uniform quantization,

stepsize q =Vpp

LVpp : peak-to-peak voltage;

L = 2N : quantization levels; N : quantization bits

Let e represent the quantization error of a data sample x,

e = xq ! x

19


The range of the quantization error would be

!q

2$ e $ q

2

For a particular signal sample, the quantization error depends on the value of the sample. Ingeneral, with a constant (fixed) peak-to-peak value,

quantization level & =' step size q ( =' quantization error (

On the other hand,

quantization level & =' the number of bits necessary to represent all the discrete level &

That is, if we want to reduce the quantization error, we need to use more bits to representeach sampled data.

20


If we consider the quantizer input as a random variable X , then the quantization error is alsoa random variable Q. Assume that Q is uniformly distributed with a pdf (probability densityfunction) as

f(e) =

&'

(

1q !q/2 $ e $ q/2

0 otherwise

then the mean of the quantization error e is

me =

" q/2

"q/2ef(e) de = 0

the variance of the quantization error, i.e., quantization noise power is

$2q = E[(e!me)2] = E[e2] =

" q/2

"q/2e2f(e) de =

" q/2

"q/2e2

1

qde =

q2

12

The quantizer’s performance can be described by the output signal-to-quantization noise ratio(SNR) defined as

SNR =ave. signal powerave. noise power

=$2X$2q

=$2X

q2/12

21


Example 1: What is the resulting SNR for a signal uniformly distributed on [-1,+1] whenuniform quantizer with 256 levels is employed?

Solution: The step size is the ratio of peak-to-peak value of the signal to the number ofquantization levels,

q =1! (!1)

256=

1

128Since signal X is uniformly distributed over [-1,+1], its pdf is

fX(x) =

&'

(

12 !1 $ x $ 1

0 otherwise

Yields E[X] = 0, and

$2X =

" !

"!x2fX(x)dx =

" 1

"1

1

2x2dx =

1

3

therefore

SNR =$2X

q2/12=

1/3

(1/128)2/12= 65536 ) 48.16(dB)

22


Example 2: For a full-scale sinusoid input signal, the signal power is

S =1

2

)Vpp

2

*2

=1

2

)Lq

2

*2

=L2q2

8

Therefore the signal to quantization noise ratio is

SNR =ave. signal powerave. noise power

=L2q2/8

q2/12=

3L2

2

when expressed in dB, then

SNR = 10 · log10)3L2

2

*= 20 · log10 L+1.76 = 20 · log10 2N +1.76 = 6.02N +1.76(dB)

23


Pulse Code Modulation (PCM)

PCM is a simple method of converting an analog signal into a digital signal (A/D conversion).Analog signal is characterized by the fact that its amplitude can take on any value over acontinuous range. This means that it can take on an infinite number of amplitude values. Ananalog signal is converted into a digital signal amplitude level by means of sampling andquantization, and followed by encoding of each quantized sample to a binary code. Based onthe number of levels, if a fixed length coding is used then that method is described as PCM

ChannelSampling Quantizing Encoding Modulator

A/D conversiona(n)a(t)

Source{a_q}

Figure 8: Digital Coding of Waveforms.

a(t): continuous-time continuous-amplitude analog signal.a(n): discrete-time continuous-amplitude discrete signal.aq: discrete-time discrete-amplitude digital signal.

24


Chap. 2.3 Pulse Code Modulation (PCM)

To discretize the amplitude of a(n), the most convenient way is to quantize the amplitude toL = 2N levels, using

log2 L = N log2 2 = N bits

i.e., each signal level can be represented by a N -bit word. For example, when N = 3, we canquantize the samples into codeword with 3-bits length to represent L = 8-level PCM signal.

The transmission of an L-level signal as an N -bit codeword (consisting of 0’s and 1’s with2-level) is known as Pulse-code Modulation (PCM).

25


0.51

21.5

2.53

3.54

−0.5−1

−1.5−2

−2.5−3

−3.5−4

1.3 3.6 2.3 0.7 −0.7 −2.4 −3.4 Natural sample value−3.5 Quantized sample value−2.5−0.50.52.53.51.5

5 7 6 4 3 1 0 Code numberPCM sequence000001011100110111101

t

x(t)

Figure 9: Natural sampling, quantized samples, and pulse code modulation.

26


PCM word size

• Quantization level: L; Number of bits each level: N , then

L = 2N N = log2(L) (7)

• How to determine N? It will depend on how much quantization distortion we are willingto tolerate. Let the magnitude of the quantization error, |e|, be specified as

|e| $ pVpp

while

|e|max =q

2=

Vpp

2LTherefore,

Vpp

2L$ pVpp

which implies

2N = L # 1

2pN # log2

)1

2p

*(8)

27


Pulse Amplitude Modulation (PAM)

PAM is one way to modulate information on to a sequence of pulses.

h(t) <=> H(f)

modulatory(t)

PAM signalg (t)

where

g!(t) =!#

n="!g(nTs) · !(t! nTs)

therefore

y(t) = g!(t) " h(t) =!#

n="!g(nTs) · !(t! nTs) " h(t) =

!#

n="!g(nTs) · h(t! nTs) (9)

TTs

t

PAM waveformsSamples

28


Remark:

• The information carried in the PAM signal resides in the amplitude value {g(nTs)}.

• The width of a PAM signal pulse (T ) is less than or equal to Ts. The ratio T/Ts < 100%

is called the duty cycle.

• The sampled values {g(nTs)} are amplitude continuous, therefore, PAM is an analogsignaling scheme.

• The information samples without any quantization are modulated on to pulses, theresulting pulse modulation can be called analog pulse modulation.

• When the information samples are first quantized, yielding symbols from an M-aryalphabet set, and then modulated on to pulses, the resulting pulse modulation is digitaland we refer to it as M-ary pulse modulation.

29


PCM vs. PAM

• PCM: use 2-level, easy to detect but need more bandwidth.

• PAM: use more amplitude levels, difficult to detect but save bandwidth.

30


Example (textbook p93): The information in an analog waveform, with maximum fre-quency fm = 3 kHz, is to be transmitted over an M -ary PAM system, where the number ofpulse level is M = 16. The quantization distortion is specified not to exceed ±1% of the Vpp

• What is the minimum number of bits/sample, or bits/PCM word that should be used indigitizing the analog waveform?

• What is the minimum required sampling rate, and what is the resulting bit transmissionrate?

• What is the PAM pulse or symbol transmission rate?

• If the transmission bandwidth (including filtering) equals 12 kHz, determine the band-width efficiency for this system.

1


Chap. 2.4 Multiplexing

Multiplexing is a method for simultaneous transmission of several signals over a commoncommunication channel.

fs1 s2 s3 fc s1 s2 s3

tT_s

fc: bandwidth of the channel capacity Ts: sampling interval

FDM TDM

• Frequency division multiplexing (FDM): to separate signals in frequency domain (forexample: AM/FM radio) and each signal occupies its frequency slot all the time.

#

n

fi $ fc

• Time division multiplexing (TDM): the signal from different information sources aremultiplexed in time domain, and each signal occupies all the bandwidth of the channelduring its time slot.

The pulse duration of the individual inputs has to be shortened from Ts to Tx = Ts/N ,

31


where Tx seconds is allocated to an individual sample during one sample interval. Thisresults in an expansion of the required bandwidth. That is TDM transmission requires awider transmission channel bandwidth.

Example: 10 voice signals each bandlimited to 3.2 kHz are sequentially sampled at 8 kHzand time multiplexed on one channel.

Sampling frequency fs = 8kHz # 2B(B = 3.2kHz), then the sampling intervalTs = 1/fs = 1/8000 = 125µs.

• For TDM, each voice signal can occupy to

Ts/N = 125/10 = 12.5µs

The bandwidth required to transmit the pulse is roughly 80 kHz.

• For FDM system, 8 kHz bandwidth is needed to transmit each signal, but 10 transceiversare needed for 10 signals.

32


33


34


35

Date post:	06-Dec-2015
Category:	Documents
Upload:	danial-zaman
View:	43 times
Download:	0 times

Digital Signal Processing Notes

Documents