Islamic University of Gaza

Faculty of Engineering Electrical Engineering Department

Fall-2011

Communication II Lab (EELE 4170)

Lab#3 Sampling and Quantization

Objectives

When you have completed this assignment, you will:

− understand the concept of sampling, aliasing, quantization and reconstruction.

− demonstrate the effect of aliasing on the reconstruction process.

Introduction

Digital communications has proved to be a very efficient means of transporting

speech, music, video, and data over different kinds of media. These media include satellite,

microwave, fiber-optic, coaxial, and cellular channels. One special advantage that digital

communication holds over analog communication is in the superior handling of noise in the

channel.

Baseband signals such as speech, music, and video are naturally occurring analog signals.

Hence, the processes of analog to digital (A/D) conversion at the transmitter and digital to

analog conversion (D/A) at the receiver are integral sections of the entire communication.

Theory

(1) Sampling

In order to store, transmit or process analog signals using digital hardware, we must

first convert them into discrete-time signals by sampling.

The processed discrete-time signal is usually converted back to analog form by

interpolation, resulting in a reconstructed analog signal xr(t).

The sampler reads the values of the analog signal xa(t) at equally spaced sampling instants.

The time interval Ts between adjacent samples is known as the sampling period (or

sampling interval). The sampling rate, measured in samples per second, is fs =1/Ts.

𝑥[𝑛] = 𝑥𝑎(𝑛𝑇𝑆) , n=…, -1, 0, 1, 2, ….

Also it possible to reconstruct 𝑥𝑎(𝑡) from its samples: 𝑥𝑎(𝑡)= 𝑥(t𝐹𝑆)

Figure 3.1 Sampling and Reconstruction process

Sampling Theorem

The uniform sampling theorem states that a bandlimited signal having no spectral

components above fm hetez can be determined uniquely by values sampled at uniform

intervals of : 𝑇𝑠 ≤1

2𝑓𝑚

The upper limit on Ts can be expressed in terms of sampling rate, denoted fs=1/Ts. The

restriction, stated in term of the sampling rate, is known as the Nyquist criterion. The

statement is: 𝑓𝑠 ≥ 2𝑓𝑚

The sampling rate 𝑓𝑠 = 2𝑓𝑚 is also called Nyquist rate .The allow Nyquist criterion is a

theoretically sufficient condition to allow an analog signal to be reconstructed completely

from a set of a uniformly spaced discrete-time samples.

Impulse Sampling

Assume an analog waveform x(t) ,as shown in figure 3.2(a), with a Fourier transform,

X(f) which is zero outside the interval (− fm < f < fm ) ,as shown in figure 3.2(b). The

sampling of x(t) can be viewed as the product of x(t) with a periodic train of unit impulse

function xδ(t), shown in figure 3.2(c) and defined as

𝑥𝛿 𝑡 = 𝛿(

∞

𝑛=−∞

𝑡 − 𝑛𝑇𝑠)

where 𝑇𝑠 is the sampling period and δ(t) is the unit impulse or Dirac delta function .Let us

choose 𝑇𝑠 = 1/2𝑓𝑚 , so that the Nyquist criterion is just satisfied. The sifting property

of the impulse function states that

𝑥 𝑡 𝛿 𝑡 − 𝑡𝑜 = 𝑥 𝑡𝑜 𝛿 𝑡 − 𝑡𝑜

Using this property, we can see that 𝑥𝑆 𝑡 , the sampled version of x(t) shown in figure

3.2(e), is given by

𝑥𝑆 𝑡 = 𝑥(𝑡)𝑥𝛿 (𝑡) = 𝑥(𝑡)𝛿(

∞

𝑛=−∞

𝑡 − 𝑛𝑇𝑠) = 𝑥(𝑛𝑇𝑠)𝛿(

∞

𝑛=−∞

𝑡 − 𝑛𝑇𝑠)

Using the frequency convolution property of the Fourier transform we can transform the

time-domain product 𝑥(𝑡)𝑥𝛿 (𝑡) to the frequency-domain convolution 𝑓 ∗ 𝑋𝛿 𝑓 ,

𝑋𝛿 𝑓 =1

𝑇𝑠 𝛿(

∞

𝑛=−∞

𝑓 − 𝑛𝑓𝑠)

where 𝑋𝛿 𝑓 is the Fourier transform of the impulse train 𝑥𝛿 (𝑡) . Notice that the Fourier

transform of an impulse train is another impulse train; the values of the periods of the two

trains are reciprocally related to one another. Figure 3.2(c) and (d) illustrate the impulse

train 𝑥𝛿 (𝑡) and its Fourier transform 𝑋𝛿 𝑓 , respectively.

We can solve for the transform 𝑋𝑆 𝑓 of the sampled waveform:

𝑋𝑆 𝑓 = 𝑋 𝑓 ∗ 𝑋𝛿 𝑓 = 𝑋 𝑓 * 1

𝑇𝑠 𝛿(∞

𝑛=−∞ 𝑓 − 𝑛𝑓𝑠)

𝑋𝑆 𝑓 =1

𝑇𝑠 𝛿(

∞

𝑛=−∞

𝑓 − 𝑛𝑓𝑠)

Figure 3.2 Sampling theorem using the frequency convolution property of the Fourier

transform

We therefore conclude that within the original bandwidth, the spectrum 𝑋𝑆 𝑓 is, to within

a constant factor (1/Ts) , exactly the same as that of x(t) in addition, the spectrum repeats

itself periodically in frequency every fs hertz.

Aliasing

Aliasing in Frequency Domain

If fs does not satisfy the Nyquist rate, 𝑓𝑠 < 2𝑓𝑚 , the different components of 𝑋𝑆 𝑓

overlap and will not be able to recover x(t) exactly as shown in figure 3.3(b). This is

referred to as aliasing in frequency domain.

Figure 3.3 Spectra for various sampling rates. (a) Sampled spectrum fs > 2𝑓m

(b) Sampled spectrum fs < 2𝑓m

Aliasing in time domain

As we shall presently see, the frequency content of the analog signal xa(t) has to be

limited before attempting to sample it. This means that xa(t) cannot change too fast;

consequently, the average value over the interval nTs −Δ to nTs+Δ is a good approximation

of the ideally sampled value xa(t) of nTs .

𝑡 = 𝑛𝑇𝑆 , 1

𝐹=

1

𝑓

1

𝐹𝑆 , 𝐹 = 𝑓𝐹𝑆 , Ω = ω𝐹𝑆

where, F : is the frequency of x(t) , f : is the frequency of x[n]

Note that the range of the frequency variable F for continuous-time sinusoids are

−∞ < 𝐹 < ∞ , − ∞ < Ω < ∞

However, the situation is different for discrete-time sinusoids

−1

2< 𝑓 <

1

2 , − 𝜋 < ω < 𝜋

Discrete- time sinusoid whose radian frequencies are separated by integer multiples of 2π

are identical. Sequence that result from a sinusoid with a frequency |w| > π is identical to a

sequence obtained from a sinusoidal with frequency : -π ≤ w ≤ π .

But for continuous-time sinusoids result in distinct signals for Ω or F in the range :

-∞< F <∞ or -∞< Ω <∞

Figure 3.4 shows a simple example .The solid line describes a 0.5Hz continuous-time

sinusoidal signal and the dash-dot line describes a 1.5 Hz continuous time sinusoidal signal.

When both signals are sampled at the rate of Fs =2 samples/sec, their samples coincide, as

indicated by the circles. This means that x1[nTs] is equal to x2[nTs] and there is no way to

distinguish the two signals apart from their sampled versions. This phenomenon is known

as aliasing.

Figure 3.4 Two sinusoidal signals are indistinguishable from their sampled versions

𝑥1 𝑛𝑇𝑆 = cos 2𝜋𝐹1𝑛𝑇𝑆 = cos 2𝜋 ×1

2 × 𝑛 ×

1

2 = cos 0.5𝜋𝑛

𝑥2 𝑛𝑇𝑆 = cos 2𝜋𝐹2𝑛𝑇𝑆 = cos 2𝜋 ×3

2 × 𝑛 ×

1

2 cos 1.5𝜋𝑛 = cos 1.5𝜋𝑛 − 2𝜋𝑛

= cos −0.5𝜋𝑛 = cos −0.5𝜋𝑛 = cos 0.5𝜋𝑛 𝑥1 𝑛𝑇𝑆 = 𝑥2 𝑛𝑇𝑆

Reconstruct 𝑥1 𝑡 𝑎𝑛𝑑 𝑥2 𝑡

𝑥1 𝑡 = 𝑥 𝑡𝐹𝑆 = cos 0.5𝜋 𝑡𝐹𝑆 = cos 0.5𝜋 𝑡 × 2 = cos 2𝜋 × 0.5 × 𝑡 = 𝑥2 𝑡

On the other hand,

𝑭𝒂𝒍𝒊𝒂𝒔𝒆𝒅 = 𝑭𝒃𝒂𝒔𝒆 + 𝒌 𝑭𝑺 , 𝒌 = ±𝟏, ±𝟐, ….

𝒇𝒂𝒍𝒊𝒂𝒔𝒆𝒅 = 𝒇𝒃𝒂𝒔𝒆 + 𝒌 , 𝒌 = ±𝟏, ±𝟐, ….

where 𝐹𝑎𝑙𝑖𝑎𝑠𝑒𝑑 𝑎𝑛𝑑 𝑓𝑎𝑙𝑖𝑎𝑠𝑒𝑑 are outside the fundamental frequency range:

−𝐹𝑆

2 ≤ 𝐹𝑏𝑎𝑠𝑒 ≤

𝐹𝑆

2 and −

1

2 ≤ 𝑓𝑏𝑎𝑠𝑒 ≤

1

2

Avoiding Aliasing

In order to avoid aliasing we must restrict the analog signal xa(t) to the frequency

range, Fo less than Fs/2. Thus, every sampler must be preceded by an analog low-pass

filter, known as an anti-aliasing filter; with a cut-off frequency fc equals fs /2.

This means that the high frequency artifacts of the continuous-time signal will be lost. This

is the cost of managing aliasing. In practice, the anti-aliasing filter will have a cutoff

frequency at roughly 90% of

Figure 3.5 Avoiding aliasing in signal processing

Quantization

The second stage in the A/D process is amplitude quantization, where the sampled

discrete-time signal x(n), is quantized into a finite set of output levels . The quantized signal

can take only one of L levels, which are designed to cover the dynamic range:

xmin ≤ x(n) ≤ xmax

The step size or resolution of the uniform quantizer is given as:

Δ = 𝑥𝑚𝑎𝑥 − 𝑥 𝑚𝑖𝑛

L − 1

The step size can be either integer or fraction and is determined by the number of levels L.

For binary coding, L is usually a power of 2, and practical values are 256 (=2 ^8) or greater.

The difference between the actual analog value and quantized digital value is

called quantization error. This error is due to rounding.

and,

A continuous time signal, such as voice, has a continuous range of amplitudes and therefore

its samples have a continuous amplitude range i.e. they are only discrete in time not in

amplitude. In other words, within the finite amplitude range of the signal, we find an

infinite number of amplitude levels. It is not necessary in fact to transmit the exact

amplitude of the samples. Any human sense (the ear or the eye), as ultimate receiver, can

detect only finite intensity differences. This means that the original continuous time signal

may be approximated by a signal constructed of discrete amplitudes selected on a minimum

error basis from an available set. Clearly, if we assign the discrete amplitude levels with

sufficiently close spacing we may take the approximated signal practically indistinguishable

from the original continuous signal.

Amplitude quantization is defined as the process of transforming the sample amplitude

m(nTs) of a message signal m(t) at time t=nTs into a discrete amplitude v(nTs) taken from a

finite set of possible amplitudes.

Quantization error : ( ) ( ) ( )q qe n x n x n

2 2error

Practical Parts:

Part 1: Aliasing in Time Domain

The signal 𝑥 𝑡 = cos 2𝜋𝐹𝑜𝑡 can be sampled at the rate 𝐹𝑆 =1

𝑇𝑆 to yield

𝑥(𝑛) = 𝑐𝑜𝑠 2𝜋𝐹𝑜

𝐹𝑆 𝑛

Aliasing will be observed for various 𝐹𝑜

a) Let 𝑭𝑺=10 kHz and 𝑭𝒐=1 kHz. Compute and plot x[n] using stem.

n=0:50;

Fs=10000;

Fo=1000;

xn=cso(2*pi*(Fo/Fs)*n)

figure(1)

stem(xn)

b) Use subplot to plot x(t) for 𝐹𝑜=300 Hz and 700 Hz.

Fs=10000;

t=0:0.0001:50/Fs;

Fo=[300,700];

for i=1:2;

subplot(2,1,i)

xt=cos(2*pi*Fo(i)*t)

plot(t,xt)

end

Comment on the following words for the result figure in (b):

(a) The period: .

(b) The rate of oscillation: .

.

c) Use subplot to plot x[n] for 𝐹𝑜=300 Hz and 700 Hz.

Fs=10000;

n=0:50;

Fo=[300,700];

for i=1:2;

subplot(2,1,i)

xn=cos(2*pi*(Fo(i)/Fs)*n)

stem(xn)

end

Comment on the following words for the result figure in (c):

(a) Aliasing: :

(b) The period: .

(c) The rate of oscillation: .

d) Repeat (b) and (c) for 𝐹𝑜=9700 Hz and 9300 Hz.

e) Repeat (b) and (c) for 𝐹𝑜=10300 Hz and 10700 Hz.

Part 2: Aliasing in Frequency Domain

a) Construct the simulink model shown in Figure 4.6.

b) Input a sine wave of Fo=5 Hz frequency from the signal generator. Choose a

sampling time of 0.01 sec for the pulse generator with pulse duration of 50 % of the

sampling period. Choose the cut-off frequency of 120π for the Butterworth low-pass filter.

c) Start the simulation for 1sec and notice all visualizers(included them in your report).

d) Repeat steps 2 and 3 for a sine wave with Fo = 95 Hz. Comment.

e) Repeat steps 2 and 3 for a sine wave with Fo = 105 Hz. Comment.

Figure 3.6 Simulink model to implement aliasing in frequency domain

Part 3: Quantization:

a) Write a Matlab function Y = uquant(X,L) which will uniformly quantize an input

array X (either a vector or a matrix) to L discrete levels.

function y=uquant(x,n)

del=((max(max(x))-(min(min(x)))))/(n-1);

r=(x-min(min(x)))/del;

r=round(r);

y=r*del+min(min(x));

b) Use this function to quantize an analog sinusoidal signal with L=4 and 32 levels.

Compute the SQR for each level.

t=0:.001:1;

y=2*sin(2*pi*t)

figure(1)

subplot(311)

plot(y)

q1=uquant(y,4)

subplot(312)

plot(q1)

q2=uquant(y,32)

subplot(313)

plot(q2)

Ps=mean(y.^2);

Pq1=mean(q1.^2);

Pq2=mean(q2.^2);

SQR1=Ps/Pq1;

SQR2=Ps/Pq2;

Example: Quantized x=2sin (2pi*t) using 16 levels.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

0 20 40 60 80 100 1200

5

10

15

0 20 40 60 80 100 1200

5

10

15

0 20 40 60 80 100 120-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

0 20 40 60 80 100 1200

0.5

1

1.5

2

2.5

3

3.5

4

0 20 40 60 80 100 1200

0.5

1

1.5

2

2.5

3

3.5

4

Mapping to

[ 0 to 15 ]

Round to the

nearest level.

Re-mapping to

[ -2 to 2 ]

Shift to

+ve region

[ o to 4]

x-min(min(x))

Divide by del

c) Audio quantization, use uquant function to quntize an audio signal at different levels and compare the result quantized signals by hearing them.

clc

clear all

[y,fs]=wavread('speech_dft.wav');

sound(y,fs)

for b=1:7;

L=2.^b;

yQ=uquant(y,L);

pause

b

sound(yQ,fs);

end

d) Construct the simulink model shown in Figure 3.6.

Input a sine wave of Fo=1 Hz frequency from the signal generator. Choose a sampling

time of 0.1 sec for the pulse generator with pulse duration of 50 % of the sampling

period. Choose the quantization interval of 0.5 for the Quantizer.

Start the simulation and notice all visualizers(included them in your report).

Figure 4.7 Simulink model for sampling and quantization

Image Quantization

Use uquant function to quantize the office_4.jpg image (you can found it in the direction c:\Program Files\matlab\toolbox\images\imdemosimage\office_4.jpg)

to 7, 4, 2 and 1 b/pel. Observe and print the output images for each level.

Note you must first convert the image from RGB to gray image.

Audio Quantization

Form results in part 3(c) answer the following questions:

(a) Describe the change in quality as the number of b/sample is reduced? (b) Do you think 4 b/sample is acceptable for telephone systems? (c) Plot the SQNR versus number of bits .Generate this curve by computing the SQNR for

7, 6, 5,..., 1 bits/sample. Also display this curve in semi-log(dB). (d) Plot the distortion versus number of bits in linear and semi-log scales.

Exercise 1

Exercise 2

Given a sinusoidal waveform with a frequency of 100 Hz,

(a) Write a MATLAB program to quantize the x(t) using 4-bit quantizers to obtain the

quantized signal xq, assuming that the signal range is from -5 to 5 volts.

(b) Plot the original signal, quantized signal, and quantization error, respectively. (c) Calculate the SQNR due to quantization using the MATLAB program.

Check for your answer.

Note:

Your report should include the following:

1- All Matlab program and its results with a short comment

on each result.

2- Answer any internal questions in practical parts.

3- Solve all lab exercises.

Exercise 3