Prepared by Professor Eliathamby Ambikairajah, Head of
School
Electrical Engineering and Telecommunications
University of New South Wales (UNSW), Australia
Modified and Delivered by Dr. T. Thiruvaran
Chapter 2
Chapter 2: Digital Signal Processing Fundamentals ........................ 20
2.1 Introduction........................................................................... 20
2.2 Overview of a DSP System .................................................. 21
2.3 Analogue to Digital Conversion Process.............................. 23
2.4 Quantisation and Encoding................................................... 25
2.5 Continuous-Time Fourier Transform (FT) ........................... 31
2.6 Sampling of continuous-time signal ..................................... 35
2.6.1 The Ideal Sampling Operation ........................................ 36
2.7 Aliasing ................................................................................. 37
2.8 Digital-to-Analogue Conversion (D/A) – Signal recovery .. 45
2.8.1 Reconstruction Filter ...................................................... 48
2.9 Summary ............................................................................... 49
Chapter 2: Problem Sheet 2
Chapter 2
19
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
20
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
x(t) – incoming analogue signal
s(t) – desired analogue signal
n(t) - noise
Analogue Signal
Processor
(e.g. Low-Pass Filter)
Magnitude
Analogue Input Signal:
x(t) = s(t) + n(t)
Analogue Output Signal:
sˆ ( t ) 
s (t)
Magnitude response
of a Low Pass filter
0
frequency
Figure 2.1 A general description of analogue systems whose input and output
are in analogue form
A digital signal processing system in Figure 2.1 provides an
alternative method for processing the analogue signal.
Chapter 2
21
Analogue prefilter
or antialiasing filter
dB
x(t)
Analogue to Digital
converter
-3 dB
0
xa(t)
Analogue
input signal
A/D
Converter
x[n]
Digital
Signal
Processor
Digital to analogue
dB
converter
s[n]
D/A
0
converter
-3 dB
s(t)
Analogue
output signal
f
fc 
f
fs
2
Lowpass filtered
signal
fs
2
Sampling
frequency (fs)
Discrete-time
signal
Reconstruction filter
(analogue filter) same as
the pre-filter
Figure 2.1: A general process of converting analogue signals into
digital signals and back to analogue form.
Name
Anti-aliasing filter
( fc 
fs
)
2
Function
To band-limit the analogue input signal prior to
digitisation to reduce aliasing (see Section 2.7)
Analogue-to-digital
converter
To convert analogue input signal into digital output
Digital Signal Processor
(heart of the system)
To process the digital signal according to the pre-defined
rules
Digital-to-analogue
converter
To convert the digital input signal into analogue output
signal by interpolating
Reconstruction filter
To smooth out the output of D/A converter
To remove unwanted high frequency components
f
( fc  s )
2
signal by sampling ( f s 
1
)
T
Table 2.1 Description of blocks contained in a general DSP process
The digital signal processor may implement one of the several
DSP algorithms, for example digital filtering (low-pass filter)
mapping the digital input signal x[n] into digital output signal
s[n].
Chapter 2
22
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.
Analogue-to-Digital Converter
Band-limited
Analogue Input
Signal
xa(t)
Sample & Hold
fs 
1
T
B-bits Quantiser
Encoder
2B
B bits
Logic
Circuit
1
T = sampling period
2B
x[n]
t
T
Quantisation
Levels
Sample & Hold
output signal
0
Digital Output
Signal
Analogue
signal
T
t
Figure 2.3 Analogue to digital conversion process
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.
Chapter 2
23
•
If the voltage varies from –mp to mp and the number of levels are L then
Note: There are two types of quantizer: Mid riser and mid tread. If nothing is mentioned assume
mid tread.
Chapter 2
24
Example: 4-bit (B = 4) A/D converter (bipolar)
digital
5
0101
0100
4
0011
3
0010
2
0001
1
0
Input-output characteristic
(two’s complement notation)
-1
1111
-2
1110
-3
1101
-4
1100
-5
1011
of
Analogue signal
4-bit
quantiser
(linear)
2.4 Quantisation and Encoding
Before conversion to digital, the analogue sample is assigned one
of 2B values (see Figure 2.4). This process, termed quantization,
introduces an error, which cannot be removed.
Chapter 2
25
Quantisation level
7
6
5
4
3
2
1
1
2 3 4
5 6
7 8
9
n – sampling
instant
Quantisation level
7
6
5
4
3
2
1
LSB 1 0 1 0 0 0 1 0
1 1 1 1 0 0 0 0
MSB 0 1 1 1 1 1 1 1
1 3-bit encoder output
1
0
Figure 2.4: Quantisation of discrete-time signals
Example
A 12 bit A/D converter (bipolar) with an input voltage range of
10V will have a least significant bit (LSB) of
20V
12
2
1
mV  4.9mV (resolution or quantisation step size, ∆𝑉)
+10V
0111 1111 1111
20
V  12
 4.9 mV
2 1
V 
Chapter 2
212 = 4096 levels
0V
Resolution
20
 4.9 mV
212 1
-10V
1000 0000 0000
26
For a B-bit A/D converter, the number of quantisation level is
2B, and the interval between levels, that is known as the
quantisation step size or resolution (V) is given by:
 V 
V
V
 B
2 1 2
B
The approximation holds when
when B is large (say B > 10 bits)
where V = resolution, B = number of bits and V = peak-to-peak
amplitude.
For a sine wave input of amplitude A, the quantisation step size
becomes
A
2A
-A
Quantisation Error (e):
level n+1
level n
v
∆V/2
∆V/2
level n-1
v: voltage at the
sampling instant
∆V
Quantisation error (e) =
∆V
sampling instant
For an A/D converter with bipolar signal inputs, the quantisation
error (e) is rounded up or down 
V
.
2
The quantisation error (e[n]) for each sample, is normally
assumed to be random and uniformly distributed in the interval
V
with zero mean.

Input analogue
2
Amplitude
signal
A/D output
signal
e[ n]  A  Aq
Actual
amplitude
Chapter
2
Quantized
amplitude
Quantisation
error
27
time
The probability density function of the error P(e) has the form as
shown below
The quantisation noise power or variance
 
V
2
2


V
2
1
V
V
2
2
e
 de 

V
2
is hence given by
Probability of quantisation
error is constant
 e P(e)de
2
e
 e2
P (e) 
1
V
V
2
1 e 
V  3   V
3
2
Hence, the quantisation noise power
V 2
 
12
2
e
for uniform quantisation
(Note: Uniform quantisation - all steps ( V ) are of equal size)
Chapter 2
28
Example
Signal-to-quantisation noise power ratio (SQNR) is defined as
SQNR 
1 N
x[n]2
Pin 

N
n 1
signal power
Pin
PN
noise power
1 N
e[n]2
and PN 

N
n 1
V 2
2
or  e 
12
N
 x [ n]
2
P
SQNR(dB)  10 log in  10 log n  1
N
P
N
e 2 [ n]

n 1
The dynamic range, R, of the signal is defined as
R  x[ n]max  x[ n]min
The quantisation step size of resolution
V 
R
L
V
is defined as
number of levels in the
quantiser
Pin
12 Pin L2
 SQNR( dB)  10 log
 10 log
( R / L) 2
R2
12
SQNR( dB)  10 log Pin  20 log L  10 log 12  20 log R
Chapter 2
29
Example:
A2
For the sine wave input, the average signal power is
, i.e.
2
2
 A  rms value


 2
The signal-to-quantisation noise power ratio (SQNR) in decibels
is
 A2

SQNR  10 log 2 2
 V

 12


A2


2
  10 log

B

 2A/ 2


12



2






 3 22B 

 10 log
2


SQNR = 6.02B + 1.76
dB
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
3
4
5
6
7
Levels
8
16
32
64
128
Thus, the signal-to-quantisation
approximately 6dB for each bit.
Chapter 2
30
SQNR
19.7 dB
25.3 dB
31.6 dB
37.7 dB
43.8 dB
noise
ratio
increases
Exercise:
Show that the input signal x(t) to quantisation noise ratio of a
linear A/D converter is given by
𝑆𝑄𝑁𝑅 = 10 log 𝑃𝑠 + 10.8 + 20 log 𝐿 − 20 log 𝑅
1
where 𝑃𝑠 is the signal power [𝑃𝑠 = ∑𝑁−1
𝑥 2 [𝑛]]; 𝐿 is the
𝑁 𝑛=0
number of quantisation levels and 𝑅 is the dynamic range of the
input signal. Using the above equation, show that for a B-bit
quantiser, 𝑆𝑄𝑁𝑅 = 6.02𝐵 + 1.76 (𝑑𝐵) if 𝑥 (𝑡) = 𝐴 cos(2𝜋𝑓𝑡)
2.5 Continuous-Time Fourier Transform (FT)
Recall:
The Fourier transform for non-periodic continuous-time signals
is a mathematical transformation to transform signals between
time domain and frequency domain, which has many applications
in engineering. Most signals of practical importance are nonperiodic. The FT pairs is given by
Fourier Transform (FT):
Inverse Fourier Transform (IFT):
The FT converts the time domain signal, x(t), into its frequency
domain representation, X ( ) and the IFT converts the frequency
domain representation, X ( ) , back into the time domain x(t).
Chapter 2
31
Fourier coefficients:
A0 
x(t ) 
  An cos(n0t )  Bn sin(n0t )
2 n 1
𝑇
2
𝑇
𝐴0 =
𝑇
𝐴𝑛 =
2
𝑇
2
𝑇
𝑥 (𝑡)𝑑𝑡
−𝑇
2
2
𝑥(𝑡) cos(𝑛𝜔0 𝑡) 𝑑𝑡
−𝑇
𝑇
𝐵𝑛 =
2
2
2
𝑥 (𝑡) sin(𝑛𝜔0 𝑡) 𝑑𝑡
−𝑇
2
Example:
Evaluate the Fourier transform of a rectangular pulse shown
below:
x(t)
A

X ( ) 

 x(t )e
 jt

2
dt 


2

2
 x(t )e

 jt
t
j
j

A   2
2
dt  
e

e


j 

2
X(w)
X ( )  A
sin 2

The FT is a continuous
function of frequency
Sinc function
2
2
Sinc function
4

Chapter 2
32
2


0
4

frequency (w)
 (t)
Example: Find the Fourier Transform of x(t) = (t).
1
(t)
t
( )
X(w)
1

X ( )    (t )e  jt dt  1
1

w
Example:

   e
FT e
j1t

j1t  jt
e

dt   e j (1  )t dt  2 (1   )

Using

  2 (   )
FT e
e
j1t
jt
the
properties:
dt  2 ( )

1
and (-1) = (1 - )
jω t
The magnitude spectrum of e 1 is show below:
|FT{ej1t}|
2
frequency,
1
Example:
 e j1t  e  j1t  1
1
j t
 j t
FT cos1t  FT 
  FT e 1  FT e 1
2
2

 2
    1      1 
The magnitude spectrum of cos1t is show below
 

Chapter 2
-1

33
1


frequency,
Chapter 2
34
Example:
Find the FT of 𝑥 (𝑡) = 𝑒 −𝑎𝑡 𝑢(𝑡) where a>0.
∞
𝑋(𝑤) =
∞
𝑒
−𝑎𝑡
𝑢(𝑡)𝑒
−𝑗𝑤𝑡
𝑑𝑡 =
−∞
𝑒
−(𝑎𝑡+𝑗𝑤𝑡)
0
1
𝑑𝑡 =
𝑎 + 𝑗𝑤
Converting to polar form, the magnitude and phase of (w) are
given by
1
|𝑋(𝑤)| = 2
(𝑎 + 𝑤 2 )1/2
𝑤
arg(𝑋(𝑤)) = − arctan ( )
𝑎
2.6 Sampling of continuous-time signal
If a continuous-time signal x(t) is sampled every T seconds, then
at the output of the Analogue to Digital Converter, as shown
below we obtain a discrete-time signal x[n]
x (t )
X ( )
A/D
converter
x[n]
 = T
X ( )
  2
fa
fs
where t = time; n = sample number; ω = analogue frequency; θ =
digital frequency; () = digital spectrum; () = analogue
spectrum.
The Discrete-Time Fourier Transform (DTFT) pair is given by
Discrete Time Fourier
Transform (DTFT):
Inverse Discrete Time Fourier
Transform (IDTFT):
where −𝜋 ≤ 𝜃 ≤ 𝜋
Chapter 2
35
Using Inverse Fourier Transforms in Continuous-time and
discrete-time domains, we can show that (see tutorial 2)
1 
2
X ( ) 
 X (  k ) .
T k 
T
2.6.1 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.
Convolution
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 a(f) at multiples of the sampling rate fs ( f s  1 or ws  2 ).
T
Chapter 2
36
T
2.7 Aliasing
Aliasing arises when a continuous-time signal is sampled at a rate
that is insufficient to capture the changes in the signal. If aliasing
occurs, the original continuous time signal cannot be recovered.
The Nyquist sampling theorem states that the sampling frequency,
𝑓𝑠 , should be at least twice the highest frequency, 𝑓𝑐 , contained in
the signal (𝑓𝑠 ≥ 2𝑓𝑐 ) to avoid aliasing.
Figure 2.5 illustrates the relationship between the digital spectrum
() and the analogue spectrum () for the case () = 0,   
T
or
f
f  s
2
.
Case 1: () = 0,
 

(sampling theorem holds)
T
()
Analogue spectrum
A
𝑓𝑠
2
-/T
frequency)
/T
 (Analogue
()
A/T
-3
-2
-
0
Digital spectrum

2
 (Digital frequency)
3
fs
Figure 2.5: Above: Frequency response of an analogue signal.
Below: Frequency response of the sampled analogue signal.
Note:

Chapter 2

T
corresponds to  =  (or
37
f 
fs
2
)
The digital spectrum is the same as the original analogue
spectrum and repeats at multiples of the sampling frequency fs
(refer to figure 2.6) as given by:
fk = f0+kfs
where k is an integer (k = 1, 2, 3, …) and f0 is the frequency
present in the fundamental region of the original analogue
spectrum. It is clear that the frequency fk is outside the
fundamental frequency range  f s  f 0  f s .
2
2
| X ( ) |
 2
 fs



0
fs
2
f0
fs
2
Fundamental region
Figure 2.6
Chapter 2
38
2
fs
digital frequency,
f1 analogue frequency, f
f1  f 0  1. f s
Case 2: ()  0,
 

T
, but () = 0,
 
3
2T
()
A


3


2T
T
T

3
2T
()
A/T
-3
-2

-
2

3
aliasing
Figure 2.7: 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.
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 2.7). Aliasing can only be avoided if the analogue signal
is band limited such that () = 0,     2f    f  f s . This
T
T
results in the familiar sampling theorem.
Note:
fs – sampling frequency, fa – analog frequency
f s corresponds to  = 
2
f s is the highest frequency that can be represented uniquely with a sampling rate f
s
2
fs
is called half the sampling frequency or folding frequency.
2
Digital frequency,   T  2
Chapter 2
fa
fs
39
2
Example: Suppose x(t) has the spectrum (f) as shown below.
Sketch the digital spectrum | ()| if the sampling frequency fs =
2 kHz.
|X(f)|
1
f (kHz)
-4
-3
-2
2
|X()|
3
4
1/T
-2 -
-3

 2
-2
2
3
Example: Consider the analogue signal
x(t) = 3 cos50t + 10 sin 300t – cos 100t
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.
Chapter 2
40
f (kHz)
Example : Consider x(t) = 10 sin 300t
fs  2  f = 300 Hz
xn  10 sin300nT 
 300
 10 sin
 fs
 10 sinn 
x[n]

n 

n
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
xt   10 sin300t    , where fs = 300Hz.
xn  10 sinn   
 10sinn cos   cosn sin  for n = 0,1,2,..

 10 cosn sin 
Since cos(n) = (-1)n ,
xn   1 10 sin 
n
If  = 0 or  = , the samples of the sinusoid taken at the Nyquist
rate are not all zero.
Example: Consider the analogue signal
xt   3 cos2000t   5 sin6000t   10 cos12000 t 
(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.
Chapter 2
41
(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 
fs
 2500
2
x(t) = 3cos(2  1000t) + 5sin(2  3000t) + 10cos(2  6000t)
 1000 
 3000 
 6000 
xn  3 cos 2
n   5 sin 2
n   10 cos 2
n
 5000 
 5000 
 5000 
 1 
 6 
 3 
 3 cos 2  n   5 sin 2 n   10 cos 2  n 
 5 
 5 
 5 
 1 
  2 
  1 
 3 cos 2  n   5 sin 2 1  n   10 cos 2 1  n 
 5 
  5 
  5 
 1 
 1 
2 

 3 cos 2  n   5 sin  2 n   10 cos 2  n 
5 

 5 
 5 
 1 
 2 
xn  13 cos 2  n   5 sin 2  n 
 5 
 5 
Second Method:
f s  5kHz 
fs
 2.5kHz
2
We have fk = f0 + kfs
f0 = fk – kfs can be obtained by subtracting from fk an integer
fs
fs


f

multiple of fs such that
.
0
2
2
Chapter 2
42
The frequency f1 = 1000 Hz is

fs
2
(= 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
  1000  
  2000  
  1000  
xn  3 cos 2 
n   5 sin 2  
n   10 cos 2 
n 
5000
5000
5000
 
 
 
 
 
 
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(2000t)-5sin(4000t)
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.
Exercise:
A digital communication link caries binary-coded words
representing samples of an input signal
x(t) = 5 cos(600t) + 4 cos (1800t) . The link is operated
at 10,000 bits/s and each input sample is quantised into 1024
different voltage levels.
(i)
What are the frequencies in the resulting discrete-time
signal x(n)?
(ii)
Determine the resulting discrete time signal x(n).
Chapter 2
43
Example:
Consider a signal, s(t), with spectrum satisfying the following
equation
||
||
|S(f)| = {3 − f 2kHz < f < 3kHz
0
otherwise
f – frequency in kHz. The signal s(t) is sampled uniformly with a
sampling frequency of 2kHz. Sketch the digital spectrum |S(θ)|,
if it is sampled at a sampling frequency of 2kHz.
|S(f)|
A
-3 -2
2
3
f(kHz)
fs = 2kHz, fk = fo + kfs
fo = fk − kfs , k = 0, ±1, ±2, …
For k = 1: fo = 2 − 1(2) = 0; fo = 3 − 1(2) = 1
For k = −1: fo = −2 + 1(2) = 0; fo = −3 + 1(2) = −1
k=-1
k=2
k=1
A
T
-2
Chapter 2
|S()|
- 0  2
(-1kHz) (1kHz)
44

2.8 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 2.8. 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.
Digital
Signal
Processor
y[n]
y(t)
Low pass
filter
D/A
8 or 12 bits
reconstruction filter
or smoothing filter
y[n]
n
t
T-Sampling period
T – Sampling period
Figure 2.8: Conversion process from digital signals to analogue signals.
The D/A shown in Figure 2.8 is referred to as a zero-order hold.
Chapter 2
45
By comparing its output yˆ (t ) 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.
h(t)
1
t
T
The corresponding frequency response is

 jt
T
e

T
H ( )   h(t )e  jt dt   e  jt dt  
 
  j  0 2

0
T
e
jT

2
The magnitude of H() is plotted in Figure 2.9.
|H()|
0
Figure 2.9: Magnitude response of a rectangular pulse.
Chapter 2
46

sin
T
2
T
2
In the frequency domain, the staircase action of the DAC
introduces a type of distortion known as the
distortion, where x 
T
.
2
sin x
or aperture
x
Y()
input to the D/A
-4
-3
-2
-
0

2
3

4
D/A output

The amplitude of the output signal spectrum is multiplied by the
sin x function, which acts like a lowpass filter, with the high
x
frequencies heavily attenuated. The sin x effect is due to the
x
holding action of the DAC and, in signal recovery, introduces an
amplitude distortion. For a zero-order hold, the function sin x
x
falls to about 4 dB at half the sampling frequency  f s  giving an
 2 
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 shape.
sin x
Chapter 2
47
2.8.1 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.
Note:
x(t)
(a) bit rate = fs  no of bits
= 8000 samples/sec  12 bits/sample
= 96000 bits/sec
12-bits A/D
(fs = 8,000 kHz)
x[n]
12
(b) In the case of Pulse Code Modulation (PCM), speech signals
are filtered to remove effectively all frequency components above
3.4 kHz and the sampling rate is 8000 samples per sec
Telephone speech
signal
x(t)
0-3.4 kHz
8-bits
(compressed PCM)
A/D
converter
8-bit persample
fs = 8000 Hz
(8000 samples/sec)
Bit rate (bits per second)
= sampling frequency  bits/sample
= 8000 samples/second  8-bits/sample
= 64,000 bits/sec
(c)
CD
16 bit
fs= 44.1
kHz
16 bit fs = 44.1kHz
Chapter 2
CD
Reader
16 bit
D/A
bit rate
=1644100 bits/sec
=0.7056 Mbits/sec
48
lowpass
filter
AMP
2.9 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.
 The definition of the Continouou-Time Fourier Transform (FT)
and its Inverse.
 Sampling of analog signals, in particular deriving the
mathematical relationship between the analog and digital
spectra.
 The definition of the Discrete-Time Fourier Transform (DTFT)
and its Inverse.
 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.
Chapter 2
49