Chapter 2

25

Chapter 2

Chapter 2: Digital Signal Processing Fundamentals ...................... 26

2.1 Introduction ...................................................................... 26

2.2 Overview of a DSP System ............................................... 27

2.3 Analogue to Digital Conversion Process ............................ 29

2.4 Quantisation and Encoding ............................................... 31

2.5 Sampling of Analogue Signal ............................................ 38

2.5.1 The Ideal Sampling Operation ..................................... 40

2.6 Aliasing ............................................................................ 41

2.7 Digital-to-Analogue Conversion (D/A) – Signal recovery .. 54

2.7.1 Reconstruction Filter ................................................... 57

2.7.2 Ideal D/A Converter (Sinc Interpolation)...................... 58

2.8 Summary .......................................................................... 61

Chapter 2: Problem Sheet 2

Chapter 2

26

Chapter 2: Digital Signal Processing

Fundamentals

2.1 Introduction

Digital Signal Processing (DSP) is a rapidly developing

technology for scientists and engineers. In the 1990s the digital

signal processing revolution started, both in terms of the

consumer boom in digital audio, digital telecommunications and

the wide used of technology in industry.

Due to the availability of low cost digital signal processors,

manufacturers are producing plug-in DSP boards for PCs,

together with high-level tools to control these boards. There are

many areas where DSP technology is now being used and the

current proliferation of such technology will open up further

applications.

In the medical field, DSP systems are widely utilized for

recording data analysis and the interpretation of ECG signals.

Audiologists and speech therapists are exposed to DSP systems

for both testing a person’s level of hearing and subsequently

DSP hearing aid filtering.

The professional music industry uses spectrum analysers, digital

filtering, sampling conversion filters etc and is one of the

biggest users and exploiters of DSP technology.

In summary, DSP is applied in the area of control and power

systems, biomedical engineering, instrumentation (test and

measurement), automotive engineering, telecommunications,

mobile communication, speech analysis and synthesis, audio and

Chapter 2

27

video processing, seismic, radar and sonar processing and neural

computing.

There are many advantages to using DSP techniques for variety

of applications, these include:

- high reliability and reproducibility

- flexibility and programmability

- the absence of component drift problem

- compressed storage facility

DSP hardware allows for programmable operations. Through

software, one can easily modify the signal processing functions

to be performed by the hardware. For all these reasons, there has

been vast growth in DSP theory & applications over the past

decade.

2.2 Overview of a DSP System

An analogue signal processing system is shown in Figure 2.1, in

which both the input signal and output signal are in analogue

form

A digital signal processing system in Figure 2.2 provides an

alternative method for processing the analogue signal.

Analogue signal

processor (e.g. low-

pass filter) Analogue

Input Signal

Analogue

Output Signal

s(t) x(t) = s(t) + n(t)

signal noise

Figure 2.1: A general description of analogue systems whose

input and output are in analogue form.

Chapter 2

28

Note:

- The A/D converter converts the analogue input signal into

a digital form.

- The D/A converter converts the processed signal back into

analogue form.

- The reconstruction filter smooths out the outputs of the

D/A and removes unwanted high frequency components.

- The analogue input filter is used to band-limit the analogue

input signal prior to digitisation to reduce aliasing (see

later).

- The heart of the system in Figure 2.2 is the digital signal

processor which may be based on a DSP chip such as

Texas instruments TMS 320C60.

The digital signal processor may implement one of the several

DSP algorithms, for example digital filtering (low-pass filter)

mapping the input x[n] into the output s[n].

Digital signal processor implies that the input signal must be in a

digital form before it can be processed.

s[n]

xa(t)

dB 3 dB x[n]

Digital

Signal

Processor

analogue prefilter

or antialiasing filter analogue to digital

converter

Lowpass filtered

signal

x(t)

2

sf Sampling

frequency

Tf s

1

discrete-time

signal

D/A

converter

Digital to

analogue

converter

dB s(t)

2

sf

reconstruction

filter (analogue

filter) same as the

pre-filter

Figure 2.2: A general process of converting analogue signals into

digital signals and back to analogue form.

A/D

Converter

Chapter 2

29

2.3 Analogue to Digital Conversion Process

Before any DSP algorithm can be performed, the signal must be

in a digital form. The A/D conversion process involves the

following steps:

- The signal (Band-limited) is first sampled, converting the

analogue signal into a discrete-time signal

- The amplitude of each sample is quantised into one of 2B

levels (where B is the number of bits used to represent a

sample in the A/D converter)

- The discrete amplitude levels are represented or encoded

into distinct binary words each of length B bits.

A practical representation of the A/D conversion process is

shown in Figure 2.3.

A/D converter

2B

1

close & open the

switch at fs Hz

Analogue Signal

(bandlimited)

xa(t)

2.1.1.1.1.1.1 T

Logic circuit B bits

x[n]

digital

output

Sample & Hold Quantiser encoder

Figure 2.3: Analogue to digital conversion process

Chapter 2

30

Sample and hold (S/H) takes a snapshot of the analogue signal

every T sec and then holds that value constant for T secs until

the next snapshot is obtained.

Example:

S/H output

t

xa(t)

t

Tfs

1

2.1.1.1.1.1.2 T

Input signal

Figure 2.4: An example of “sample and hold” process

to convert analogue signals into digital signals.

Sampled Signal n

0V

-6V

-12V

12V x[n]

6V

T = sampling

period

Analogue

Signal

Samples

Figure 2.5: An example of sampling analogue signals

to discrete-time signals. The sampling period is T.

Chapter 2

31

Example: 4-bit (B = 4) A/D converter (bipolar)

Input-output characteristic of 4-bit quantiser (linear)

(two’s complement notation)

2.4 Quantisation and Encoding

Before conversion to digital, the analogue sample is assigned

one of 2B values (see Figure 2.6). This process, termed

quantization, introduces an error, which cannot be removed.

5

4

3

2

1

digital

-1 1111

-2 1110

-3 1101

-4 1100

-5 1011

0101

0100

0011

0010

0001

analogue

Chapter 2

32

A 12 bit A/D converter (bipolar) with an input voltage range of

10V will have a least significant bit (LSB) of

mVmVV

9.412

20

12

(resolution)

Resolution (step-size)

mVV

V 9.412

2012

1000 0000 0000

12 bits

0111 1111 1111

12 bits

+10V

-10V

212

levels = 4096

6

1

2

3

4

5

sampling instants

Quantisation

Level

1

2

3

4

5

6

1

0

0

0

0

1

1

0

1

0

1

0

1

1

0

3 bits code output

Quantisation

Level 3-bit A/D

Converter

(Unipolar)

encoder output

Figure 2.6: Quantisation of discrete-time signals

LSB

Chapter 2

33

Note:

For an A/D converter with Binary digits the number of

quantisation level is 2B, and the interval between levels, that is

the quantisation step size (V) – resolution is given by

BB

VVV

212

V-full scale range of the A/D converter with bipolar signal

inputs. The maximum quantisation error, for the case where the

values are rounded up or down 2

V .

For a sine wave input of amplitude A, the quantisation step size

becomes

The quantisation error (e[n]) for each sample, is normally

assumed to be random and uniformly distributed in the interval

2

V with zero mean.

qAAne ][

level n+1

level n

level n-1

sampling instant

v ∆V

∆V/2

∆V/2 ∆V

Quantisation error =2

V (one

half of an LSB)

= 4.9 mV / 2 = 2.45 mV

B

AV

2

2

2A A

-A

actual amplitude quantized amplitude

Chapter 2

34

The probability density function of the error P(e) has the form as

shown below

The quantisation noise power or variance 2

e is hence given by

2

2

22 )(

V

V

e deePe

2

2

21

V

V

deeV

Hence,

12

22 Ve

for uniform quantisation

V

1

2

V

2

V

VeP

1)(

Probability of quantisation

error is constant

VeP

1)(

Chapter 2

35

(Note : Uniform quantisation - all steps are of equal size)

Signal-to-quantisation noise power ratio (SQNR) is defined as

N

in

P

PSQNR

12

][1

][1

22

1

2

1

2

Vor

neN

PandnxN

P

e

N

n

N

n

Nin

N

n

N

n

N

in

ne

nx

dBSQNR

P

PdBSQNR

1

2

1

2

][

][

log10)(

log10)(

The dynamic range of the signal is defined as

minmax ][][ nxnxR

The quantisation step size of resolution V is defined as

L

RV

2

2

2

12log10

12

)/(log10)(

R

LP

LR

PdBSQNR inin

RLPdBSQNR in log2012log10log20log10)(

signal power

noise power

number of levels in the

quantiser

Chapter 2

36

Example:

For the sine wave input, the average signal power is 2

2A, i.e.

2

2

A rms value

The signal-to-quantisation noise power ratio (SQNR) in decibels

is

2

23log10

12

2/2

2log10

12

2log10

2

2

2

2

2

B

BA

A

V

A

SQNR

The SQNR increases with the number of bits, B. In many DSP

applications, an A/D converter resolution between 12 and 16 bits

is adequate.

Number of Bits Levels SQNR

3 8 18.7 dB

4 16 25.3 dB

5 32 31.6 dB

6 64 37.7 dB

7 128 43.8 dB

Thus, the signal-to-quantisation noise ratio increases

approximately 6dB for each bit.

SQNR = 6.02B + 1.76 dB

Chapter 2

37

Example:

Consider the ramp x(t) = t over (0, 3). For a sampling interval

of 0.2s and number of levels, L = 6, the sampled signal,

quantized (by rounding) signal, and error signal are

x[n]={0, 0.2, 0.4, 0.6, 0.8, 1.0, 1.2, 1.4, 1.6, 1.8, 2.0, 2.2, 2.4, 2.6, 2.8, 3.0}

xQ[n] = {0, 0.0, 0.5, 0.5, 1.0, 1.0, 1.0, 1.5, 1.5, 2.0, 2.0, 2.0, 2.5, 2.5, 3.0, 3.0}

e[n] = {0, 0.2, -0.1, 0.1, -0.2, 0.0, 0.2, -0.1, 0.1, -0.2, 0.0, 0.2, -0.1, 0.1, -0.2, 0}

where {e[n]=x[n]-xQ[n]}

Compute the SQNR.

Method 1:

dBne

nx

P

PSQNR

N

ini 2.22

3.0

6.49log10

][

][log10log10

2

2

Method 2:

dBnxN

SQNR

NandRwithRLPSQNR

N

n

s

ins

7.213log206log208.10][1

log10

163log20log208.10log10

1

0

2

Method 3:

If x(t) forms a period of a periodic signal with T=3, we can also

find inP and

sSQNR as

dBSQNR

dttxT

P

s

in

583.213log206log2012log103log10

3)(1 2

Note: iSQNR and

sSQNR differs because sSQNR is a statistical

estimate. The larger the number of samples N, the less iSQNR

and sSQNR differ.

Chapter 2

38

2.5 Sampling of Analogue Signal

Suppose that an analogue signal x(t) is sampled every T seconds,

then at the output of the sampler we obtain a discrete-time signal

x(n) = x(t)|t = nT.

deXtx tj)(2

1)( Inverse Fourier Transform

t=nT

deXnx nTj)(2

1

deXnx

enxX

jn

n

jn

)(2

1

)(

Discrete Time Fourier

Transform (DTFT):

Inverse Discrete Time

Fourier Transform

(IDTFT):

deXtx

dtetxX

tj

tj

)(2

1)(

)()(

Fourier Transform:

Inverse Fourier Transform:

A/D

Sampling

freq.(T

fs

1 )

x[n]

X()

x(t)

X()

= T

s

a

f

f 2

(2.1)

Chapter 2

39

Now ejnT

is a periodic function of period 2. Equation (2.1)

becomes

TdeT

kXT

TdeT

kXT

deXnx

k

nTj

nTT

kj

k

k

k

k

nTj

)2

(1

2

1

)2

(1

2

1

)(2

1

)2

(

2

2

Let T =

de

TkX

Tnx nj

k

)2

(1

2

1 …

We have

deXnx nj

)(2

1 …

Inverse Fourier Transform for discrete signal

By comparing (2.2) and (2.3), We obtain

It is seen that X() is periodic with period 2. The digital

spectrum is a repetition of the analogue spectrum.

k T

kXT

X )2

(1

)(

- ,

Digital spectrum Analogue spectrum

(2.2)

(2.3)

(2.4)

Chapter 2

40

2.5.1 The Ideal Sampling Operation

An analogue signal multiplied by a periodic impulse train results

in a train of impulses that match the values of the analogue

signal at the sampling instants.

Multiplication of the analogue signal and the ideal impulse train

results in the convolution of their respective spectra.

The spectrum of the sampled signal x[n] thus consists of replicas

of Xa(f) at multiples of the sampling rate fs (T

wT

f ss

2or

1 ).

Tw

Tf ss

2or

1

Chapter 2

41

2.6 Aliasing

Figure 2.7 illustrates the relationship between the digital

spectrum X() and the analogue spectrum X() for the case

X() = 0, T

or

2

sff .

Case 1: X() = 0, T

(sampling theorem holds)

Note: T

corresponds to = (or

2

sff )

The digital spectrum is the same as the original analogue

spectrum and repeats at multiples of the sampling frequency fs.

X()

A

Figure 2.7: Above: Frequency response of an analogue signal.

Below: Frequency response of the sampled analogue signal.

-/T /T (Analogue frequency)

X()

A/T

-3 -2 - 0 2 3 (Digital frequency)

2

sf fs

Analogue spectrum

Digital spectrum

𝑓𝑠

2

Chapter 2

42

Case 2: X() 0, T

, but X() = 0,

T2

3

T2

3

T

T

T2

3

If the sampling frequency, fs is not sufficiently high, the

spectrum centred on fs will fold over or alias into the base band

frequencies (Figure 2.8). Equation (2.4) tells us that aliasing can

only be avoided if the analogue signal is band limited such that

X() = 0, T

22 sff

Tf

.

X()

A

Figure 2.8: Above: Frequency response of an analogue signal whose highest

frequency component is larger than the sampling frequency.

Below: Frequency response of the sampled analogue signal.

The overlapped region represents aliasing.

-3 -2 2

3 -

2

3 2 3

X()

A/T

aliasing

Chapter 2

43

This results in the familiar sampling theorem. The minimum

sampling frequency T

1 for which equation

holds is called the Nyquist frequency.

Note: It should be noted that even if X() is not strictly band

limited, that it has some negligible energy outside T

, a small

enough T can be chosen so that the overlap of the components of

the summation in the above equation is below a prescribed level.

This is important when a sampling frequency is to be selected

for a particular signal.

Aliasing Examples

Example: Suppose x(t) has the spectrum X(f) as shown

below. Sketch the digital spectrum |X()| if the sampling

frequency fs = 2 kHz.

k T

kXT

X )2

(1

)(

|X(f)|

f (kHz)

2 4 -4 -2 3 -3

1

f (kHz)

|X()|

2 -2

-3

-

3

1/T

2 -2

Chapter 2

44

Example: Consider the analogue signal

x(t) = 3 cos50t + 10 sin 300t – cos 100t

What is the Nyquist rate for this signal?

The frequencies present in the signal above are

f1 = 25Hz; f2 = 150 Hz; f3 = 50 Hz

Hence, fmax = 150 Hz

fsampling > 2 fmax = 300 Hz

The Nyquist rate is fN = 2 fmax = 300 Hz.

Note: Consider x(t) = 10 sin 300t

fs 2 f = 300 Hz

n

nf

nTnx

s

sin10

300sin10

300sin10

We are sampling the analogue sinusoid at its zero-crossing

points and hence we miss the signal completely. The situation

will not occur if the sinusoid is offset by some phase (here). In

such case we have

ttx 300sin10 and sf

T1

, where fs = 300Hz.

sincos10

sincoscossin10

sin10

n

nn

nnx

for n = 0,1,2,..

n

x[n]

Chapter 2

45

Since cos(n) = (-1)n , sin101

nnx

If = 0 or = , the samples of the sinusoid taken at the

Nyquist rate are not all zero.

Note: x(t) = A cos(2f0t) is a continuous-time sinusoidal signal

22

2cos)(

0

0

ss

s

ff

f

nf

fAnx

On the other hand, if the sinusoids,

tfAtx k2cos

[where fk = f0 + kfs , k = 1, 2, 3, ….]

are sampled at a rate fs, it is clear that the frequency fk is outside

the fundamental frequency range 22

0ss f

ff

; consequently the

sampled signal is

knnf

fA

nf

kffA

nf

fAnx

s

s

s

s

k

22cos

)(2cos

2cos

0

0

nf

fAnx

s

02cos

(2.5)

(2.6)

(2.7)

Chapter 2

46

which is identical to the discrete-time signal in equation (2.5). If

we are given a sequence x[n] there is an ambiguity as to which

continuous-time signal x(t) these values represent. We can say

the frequencies fk = f0+kfs are indistinguishable from the

frequency f0 after sampling and hence they are aliases of f0.

Note:

Example: Consider the analogue signal

ttttx 12000cos106000sin52000cos3

(a) What is the Nyquist rate for this signal?

The frequencies existing in the analogue signal are:

f1 = 1 kHz; f2 = 3 kHz; f3 = 6 kHz

Thus fmax = 6 kHz and according to the sampling theorem,

fs > 2 fmax = 12 kHz

The Nyquist rate is = 12 kHz.

fs – sampling frequency

2

sf corresponds to =

2

sfis the highest frequency that can be represented uniquely

with a sampling rate fs

2

2fis called half the sampling frequency or folding frequency.

sf

fT 2

Chapter 2

47

(b) Assume now that we sample this signal x(t) using a

sampling rate fs = 5 kHz (samples/sec). What is the discrete-time

signal obtained after sampling?

First Method:

fs = 5000Hz 25002sf

x(t) = 3cos(2 1000t) + 5sin(2 3000t) + 10cos(2 6000t)

nnn

nnn

nnn

nnnnx

5

12cos10

5

22sin5

5

12cos3

5

112cos10

5

212sin5

5

12cos3

5

62cos10

5

32sin5

5

12cos3

5000

60002cos10

5000

30002sin5

5000

10002cos3

nnnx

5

22sin5

5

12cos13

The same result can be obtained using equation (2.6).

Second Method:

kHzf

kHzf ss 5.2

25

We have fk = f0 + kfs

f0 = fk – kfs can be obtained by subtracting from fk an integer

multiple of fs such that 22

0

ss ff

f .

Chapter 2

48

The frequency f1 = 1000 Hz is 2

sf (= 2500 Hz) and thus it is not

affected by aliasing.

However, the other two frequencies f2 & f3 are above the folding

frequency and they will be changed by the aliasing effect.

f2' = f2 – 1 fs = 3000 – 5000 = -2 kHz

f3' = f3 – 1 fs = 6000 – 5000 = 1 kHz

This is agreement with the result obtained before.

(c) What is the analogue signal y(t) we can reconstruct from the

samples if we use ideal interpolation.

Since only frequency components at 1 kHz and 2 kHz are

present in the sampled signal, the analogue signal we can

recover is,

y(t)=13cos(2000t)-5sin(4000t)

which is obviously different from the original signal x(t).

The distortion of the original analogue signal was caused by the

aliasing effect, due to the low sampling rate used.

nnnnx

5000

10002cos10

5000

20002sin5

5000

10002cos3

Chapter 2

49

Example: An analogue signal x(t) = sin(480t)+3sin(720t) is

sampled 600 times per second.

(a) Determine the Nyguist sampling rate for x(t)

(b) Determine the folding frequency (or half the sampling

frequency)

(c) What are the frequencies, in radians, in the resulting

discrete time signal x[n]?

(d) If x[n] is passed through an ideal D/A converter what is

the reconstructed signal y(t)?

(a) x(t) = sin(2 240t) + 3sin(2 360t)

f1 = 240 Hz f2 = 360 Hz

fmax = 360 Hz FNyquist = 2 fmax = 720 Hz

(b) fs = 600 Hz ffold or 2

sf = 300 Hz

(c) First Method:

n

nn

nn

nn

nntxnxnTt

5

4sin2

5

4sin3

5

4sin

5

42sin3

5

4sin

5

6sin3

5

4sin

600

3602sin3

600

2402sin)(

Chapter 2

50

Second Method:

We have fk = f0 + kfs and therefore f0 = fk – k fs

22

0

ss ff

f

f1 = 240 Hz and is 2

sf (= 300 Hz) not affected by aliasing

f2 = 360 Hz and is 2

sf (= 300 Hz) affected by aliasing

Aliased frequency f0 = fk – k fs

= 360 – 1 600

= -240 Hz

n

nnnx

5

4sin2

600

2402sin3

600

2402sin

(d)

nny

600

2402sin2)(

600

n

f

nnTt

s

tty 480sin2)(

Chapter 2

51

Note: Fourier transform – analogue signal

An analogue signal xa(t) = cos(2500t) is sampled at a rate of

fs=4kHz. Determine the resulting analogue and digital

magnitude spectra.

)(22

1)(2

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2)}{cos(

11

)()(

1

11

11

11

11

dtedte

dteedtee

eFTeFT

eeFTtFT

tjtj

tjtjtjtj

tjtj

tjtj

)()(

)()(

11

11

)()()}{cos( 111 tFT

)()( 11

)(2

dte tj

Chapter 2

52

Hence, Fourier transform of xa(t) is )500()500()( fffX a

as shown below. Note that fmax = 500Hz < fs/2 = 2000Hz, so this

obeys the sampling theorem. Sampling produces copies

(aliases) of the analogue spectrum centred at multiples of 2.

Chapter 2

53

Example: An analogue signal )()( tuetx at

a is sampled.

Determine the resulting analogue and digital magnitude spectra.

Fourier transform tables gives us fja

fX a2

1)(

, so

22 2

1|)(|

fafX a

, as shown below. Note that 0|)(| fX a at all

frequencies (i.e. )(txa is not bandlimited, or maxf , so that no

sampling frequency (no matter how high) can obey the sampling

theorem. Sampling produces copies (aliases – shown dotted) of

the analogue spectrum centred at multiples of 2 which sum to

produce the overall response (solid), and this overall response

suffers from aliasing distortion as dictated by the sampling

theorem. In order to correctly sample )(txa in practice, it would

first be need to be band-limited by low pass filtering )(txa with

some cut-off frequency sff 2/1max and then sampled at a

sampling rate of fs.

Chapter 2

54

2.7 Digital-to-Analogue Conversion (D/A) – Signal

recovery

The D/A conversion process is employed to convert the digital

signal into an analogue form after it has been digitally

processed. The reason for such conversion may be for example,

to generate an audio signal to drive a loudspeaker or to sound an

alarm. The D/A process is shown in Figure 2.9. A register is

used to buffer the D/A’s input to ensure that its output remains

the same until the D/A is fed the next digital input.

Note: The inputs to the D/A are series of impulses, while the

output of the DAC has a staircase shape as each impulse is held

for a time T sec.

The D/A shown in Figure 2.9 is referred to as a zero-order hold.

)(ˆ ty

n t

T-Sampling period T – Sampling period

Digital

Signal

Processor

D/A Low pass

filter

y[n]

8 or 12 bits

)(ˆ ty y(t)

y[n]

reconstruction filter

or smoothing filter

Figure 2.9: Conversion process from digital signals to analogue signals.

Chapter 2

55

By comparing its output )(ˆ ty and its input y[n], it is evident that

for each digital code fed into the D/A, its output is held for a

time T. The result is the characteristic staircase shape at the D/A

output.

The D/A output approximates the analogue signal by a series of

rectangular pulses whose height is equal to the corresponding

value of the signal pulse.

Just consider one pulse.

The corresponding frequency response is

2

2sin

2

2

2

]1[1

)()(

222

2

222

00

T

T

eT

Tj

eee

T

j

eeee

j

j

edte

dtethH

TjTjTj

Tj

TjTjT

jTj

TtjT

tj

tj

h(t)

1

T

t

otherwise

Ttth

0

01)(

Chapter 2

56

The magnitude of H() is plotted in Figure 2.10.

In the frequency domain, the staircase action of the DAC

introduces a type of distortion known as the x

xsin or aperture

distortion, where 2

Tx

.

|H()|

2

T

7

6 7

4 7

2 0 7

2 7

4 7

6

Figure 2.10: Magnitude response of a rectangular pulse.

x

xsin

Y()

input to the D/A

-4 -3 -2 - 0 2 3 4

)(ˆ y

D/A output x

xsin

Chapter 2

57

The amplitude of the output signal spectrum is multiplied by the

x

xsin function, which acts like a lowpass filter, with the high

frequencies heavily attenuated. The x

xsin effect is due to the

holding action of the DAC and, in signal recovery, introduces an

amplitude distortion.

For a zero-order hold, the function x

xsin falls to about 4 dB at

half the sampling frequency

2

sf giving an average error of

about 36.4%.

Aperture error can be eliminated by equalization. In practice this

can be achieved by first applying the signal, before converting it

to analogue, through a digital filter whose amplitude-frequency

response has a x

x

sinshape.

2.7.1 Reconstruction Filter

The output of the D/A converter contains unwanted high

frequency at multiples of the sampling frequency as well as the

desired frequency components. The role of the output filter is to

smooth out the steps in the D/A output thereby removing the

unwanted high frequency components. In general, the

requirements of the anti-imaging filter are similar to those of the

anti-aliasing filter.

Chapter 2

58

2.7.2 Ideal D/A Converter (Sinc Interpolation)

Consider just an impulse

If we apply the signal )(ˆ ty to the input of the filter, we obtain

n

nTtnyTt

Tt

tythty

)()(*)/(

)/sin(

)(ˆ*)()(

T Impulse not square pulses

as in the case of an non-

ideal D/A

Ideal D/A y[n] y(t)

Ideal Lowpass filter

y(t) ^

T

T

)(H T

1 h(t) h(t) )(tδ

)(H

T

T

tT

t

th

sin

)(

1

t

Chapter 2

59

y(t) can be written in this form.

nTt

TtnTtnyty

)/(

)/sin(*)()(

Using the property

)()(*)( 00 ttxtttx

we can obtain

nT

nTt

T

nTt

nyty)(

)(sin

)(

(2.8)

The original signal can be obtained by adding together an

infinite number of x

xsinpulses. The n

th

x

xsinpulse here is

shifted through a distance nT with respect to the origin and

multiplied (weighted) by a factor y[n]. This recovery process is

called interpolation. Figure 2.11 shows the implementation of

equation (2.8).

Chapter 2

60

The signal x(t) is reconstructed from the samples of nynx

by summation of weighted and shifted x

xsin pulse.

Note:

(a) bit rate = fs no of bits

= 8000 samples/sec 12 bits/sample

= 96000 bits/sec

(b) In the case of PCM, speech signals are filtered to remove

effectively all frequency components above 3.4 kHz and the

sampling rate is 8000 samples per sec

Bit rate (bits per second)

= sampling frequency bits/sample

= 8000 samples/second 8-bits/sample

= 64,000 bits/sec

x[n]

12

x(t) 12-bits A/D

(fs = 8,000

kHz)

0-3.4 kHz

8-bit persample

fs = 8000 Hz

(8000 samples/sec)

speech signal

x(t) 8-bits

(compressed PCM)

A/D

converter

t 2T T 0 -T

y(1)

y(2) y(0)

y(-1)

y(-2)

y(t) = x(t) original signal

Figure 2.11: Each discrete-time sample is multiplied by a shifted sinc function.

Summing these sinc functions will produce the original analogue signal.

Chapter 2

61

(c)

2.8 Summary

At the end of this chapter, it is expected that you should know:

A block diagram of the conversion from analog to digital and

back to analog form, including descriptions of the blocks

Analog to digital conversion, in particular amplitude

quantization and quantization error. Be able to calculate the

signal-to-quantization error ratio.

Sampling of analog signals, in particular deriving the

mathematical relationship between the analog and digital

spectra.

The sampling theorem and the Nyquist frequency.

How to demonstrate the effect of aliasing using sketched

magnitude spectrum plots.

Digital to analog conversion and the role of the reconstruction

filter. Show your understanding using both mathematical and

hand-sketched explanations.

Calculations of aliased frequencies: f0 = fk –kfs, where fk is the

frequency outside the Nyquist frequency.

bit rate

=1644100 bits/sec

=0.7056 Mbits/sec

CD

16 bit fs=

44.1 kHz

CD

Reader

16 bit

D/A

lowpass

filter

AMP

16 bit fs = 44.1kHz