Transcript Slide 1

Chapter 5 – Signal Encoding and
Modulation Techniques
1/21
Analog Data, Digital Signal
 Digitization is conversion of analog data into
digital data which can then:
 be transmitted using NRZ-L (digital signal)
 be transmitted using code other than NRZ-L
 be converted to analog signal by using modulation
techniques (ASK, PSK, FSK)
signal
2/21
Codec (Coder-decoder)
 Analog to digital conversion done using a
codec (coder-decoder). Two techniques:
1. Pulse Code Modulation (PCM)
2. Delta Modulation (DM)
3/21
Pulse Code Modulation (PCM)
Sampling Theorem:
“If a signal is sampled at regular intervals at a rate
higher than twice the highest signal frequency, the
samples contain all information in original signal”
Samples can be used to reconstruct the original
signal
e.g., 100-4000Hz voice data, requires 2*4000=8000
sample per sec
These are analog samples, called Pulse
Amplitude Modulation (PAM) samples
To convert to digital, each of these analog
samples must be assigned a binary code
4/21
Pulse Code Modulation (PCM) Example
 The signal is assumed to be band-limited with bandwidth B
 The PAM samples are taken at a rate of 2B, or once every
Ts=1/(2B) seconds
 Each PAM sample
is quantized into
one of 16 levels
 Each sample is
then represented
by 4 bits.
 8 bits→256 level
→better quality
 4000Hz voice→
(8000sample/s)*
8bits/sample=
64Kbps
5/21
Pulse Code Modulation (PCM)
Block Diagram
 By quantizing the PAM samples, the resulting signal is an
approximation of the original one
 This effect is known as quantization error or quantization
noise
 The Signal-to-Noise-Ratio (SNR) for quantizing noise:
SNRdB  6.02n  1.76 dB,
n : numberof bits
6/21
Linear Versus Non-Linear Encoding
 Linear Encoding (uniform quantization):
 Equally spaced quantization steps
 Lower amplitude values are relatively more distorted
 Non-Linear Encoding (non-uniform quantization):
 Non-equally spaced quantization steps
 Large number of quantization steps for signals with low amplitude, and smaller
number of quantizing steps for signals with large amplitude
7/21
Companding (Compressing-Expanding)
 Instead of non-linear encoding, use companding+linear
encoding
 Companding gives more gain to weak signals than to strong
signals on the input. At output, the reverse operation is
performed
X
Compressing
Y
Expanding
X
Norm al value( X )
com pressed(Y  10* log10 X )
Expanded( X  10Y /10 )
X  10
Y  10* log10 10  10
X  1010 /10  10
X  100
Y  10* log10 100  20
X  1020 /10  100
X  1000
Y  10 * log10 1000 30
X  1030 /10  1000
X  10000
Y  10 * log10 10000 40
X  1040 /10  10000
8/21
Delta Modulation (DM)
 An analog input is approximated by a staircase function that moves up or
down by one quantization level () at each sampling interval (Ts).
 A 1 is generated if the staircase function is to go up during the next interval;
a 0 is generated
otherwise.
 The staircase
function tracks
the original
waveform
9/45
Delta Modulation Operation
 For transmission:
 the analog input is compared
to the most recent value of the
approximating staircase function.
 If the value of the analog input
exceeds that of the staircase
function, a 1 is generated;
otherwise, a 0 is generated.
 Thus, the staircase
is always changed in the
direction of the input signal.
Staircase
 For reception:
 The output of the
DM process is therefore a binary
sequence that can be used at the
receiver to reconstruct the
staircase function.
10/21
Pulse Code Modulation (PCM)
Versus Delta Modulation (DM)
DM has simplicity compared to PCM
DM has worse SNR compared to PCM
PCM requires more bandwidth
eg., for good voice reproduction with PCM
 want 128 levels (7 bit) & voice bandwidth 4khz
 need 8000 sample/s x 7bits/sample = 56kbps
PCM is more preferred than DM for analog
signals
11/21
Analog Data, Analog Signal
Modulate carrier signal with analog data (voice)
Why modulate analog signals?
 higher frequency can give more efficient transmission
 permits frequency division multiplexing (chapter 8)
Types of modulation
 Amplitude Modulation (AM)
 Frequency Modulation (FM)
 Phase Modulation (PM)
analog data
m(t )
Modulator
modulated signal
s (t )
s (t )
Demodulator
m(t )
carrier signal
Ac cos(2f c t   )
12/21
Amplitude Modulation (AM)
AM is the simplest form of analog modulation
Used in AM radio with carrier 0.535MHz  f c  1.605MHz
Used also in analog TV broadcasting
Analog data modulates a carrier signal
Mathematically, the AM wave can be expresses as
s (t )  [1  na x(t )] cos(2f c t )
where
na x(t )  m(t ) : input data signal
0  na  1 : Modulationindex
f c : carrier frequency
13/21
Time Domain description of AM Signal
 Derive an expression for the AM wave if the input signal:
m(t )  na cos(2f mt ),
s(t )  [1  m(t )] cos(2f c t )
f m  f c
 [1  na cos(2f m t )] cos(2f c t )
 Envelope of AM signal:
[1  na cos(2f m t )]
max . when cos(2f m t )  1
min . when cos(2f m t )  1
Amax 1  na

Amin 1  na
Amax  Amin
na 
Amax  Amin
na *100% : % Modulation
11/45
Frequency Domain description of AM Signal
The Double SideBand Transmitted Carrier (DSBTC):
s(t )  [1  na cos(2f m t )] cos(2f c t )
 cos(2f c t )  na cos(2f m t ) cos(2f c t )
1
employingthe trigonomet
ric identity: cos( ) cos( )  [cos(   )  cos(   )]
2
na
s(t )  cos(2f c t )  [cos(2 ( f c  f m )t )  cos(2 ( f c  f m )t )]
2
S( f )
Carrier
Upper Side Band
(USB)
Lower Side Band
(LSB)
0
fc  fm
fc
fc  fm
f
15/21
Frequency Domain description of AM Signal
 Consider a voice signal m(t) with a
bandwidth that extend from 300Hz
to 3000Hz being modulated on
a 60 KHz Carrier

Bandwidth B  3KHz
carrier f c  60KHz
The resultingsignal contains:
Upper SideBand: 60.3KHz  63KHz
Lower SideBand: 57KHz  59.7 KHz
Carrier at 60KHz
11/45
Variations of AM signal
Double Side Band Transmitted carrier (DSBTC)
wast of power as the carrier is transmitted with the side
bands
wast of bandwidth as both upper and lower side bands are
transmitted (each side band contains the complete spectrum
of the message signal m(t) ): Transmitted bandwidth=BT=2B
Double Side Band Suppressed Carrier (DSBSC)
Less power is required as no carrier is transmitted
wast of bandwidth as both upper and lower side bands are
transmitted: Transmitted bandwidth=BT=2B
Single Side Band (SSB)
Less power is required as no carrier is transmitted
Less bandwidth as one side band is transmitted: BT=B
17/21
Angle Modulation
 Frequency Modulation (FM) and Phase Modulation (PM) are
special cases of angle modulation
 Used in FM radio with carrier 88MHz  f c  108MHz
 The angle modulated signal is expressed as:
s(t )  Ac cos(2f c t   (t ))
 Phase Modulation (PM):
 (t )  n p m(t )
- Example: find s(t) if
 (t )  n p cos(2f m t )
where
n p : phase Modulationindex

s(t )  Ac cos[2f c t  n p cos(2f m t )]
m(t ) : input messagesignal

Max. phase deviation n p Am  n p
Max. phase deviation n p Am
where Am is the max . of m(t )
18/21
Frequency Modulation (FM)
 The angle modulated signal is expressed as:
s(t )  Ac cos(2f c t   (t ))
 FM when:
- Example: find s(t) if
d (t )
 n f m(t )
dt
n f : frequencyModulationindex
d (t )
  n f sin(2f m t )
dt
The frequencyat any time:
  (t )   n f sin(2f m t ) dt
d
[ 2f c t   (t )]
2f i (t ) 
dt
 f i (t )  f c  n f m(t ) / 2


nf
2f m
cos(2f m t )


Max. freq. deviationF  n f Am / 2 s (t )  cos[2f t  n f cos(2f t ) ]
c
m
2f m
where Am is the max . of m(t )
19/21
Transmitted Bandwidth for AM, PM and FM
 Transmitted bandwidth for AM:
BT  2 B
where B is the messagesignalbandwidth
 Transmitted bandwidth for PM and FM:
BT  2(   1) B
where
n p Am

   n f Am

 2B
for PM
for FM
 Thus, both PM and FM require greater bandwidth than AM
20/21
AM, PM, FM
21/21