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CHAPTER 4: DIGITAL TRANSMISSION SYSTEM
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BLOCK DIAGRAM
A/D
CONVERTER
LINE
CODING
PULSE
TXN
Analog
signalOutput
PCM
signal
Bit
stream
Pulse Analog ModulationPAM, PWM, PPM
Pulse Digital ModulationPCM
- Sampling- Quantizing
- Encoding - Binary code
- Grey code
DM
NRZ
RZ
Manchester
Etc.
ASK
FSK
PSK
MultiplexingFDM
TDM
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In any digital transmission system, it will involve the analog todigital conversion since most of baseband signals are in analogform.
Therefore, digital modulation is required in order to convert theanalog signal to a digital form.
The advantages of digital system compared to the analogsystem: Easy to regenerate using the regenerative repeaters.
Less noise and distortion (high SNR).
Easy to process and to store QP, DSP, RAM, ROM, etc.
Small size, low cost.
Low power.
INTRODUCTION
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However, there are some disadvantages of digital system:
Require wider bandwidth.
Digital signal detection needs synchronization
Require a coder to convert the analog signal to a digital signal.
Might need partial changes to the available analog channels.
In order to transmit the analog signal in digital form, we need to
convert the signal into pulses.
Thus, a periodic pulse train is required as a carrier signal.The carrier pulses will be changed by the instantaneous value of
the analog modulating signal yield apulse modulation.
INTRODUCTION
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Pulse modulation includes many different methods of converting
information into pulse form for transferring them from a source to a
destination.
It can be categorized into two:
Pulse analog modulation
PAM Pulse Amplitude Modulation
PWM Pulse Width Modulation
PPM Pulse Position Modulation
Pulse digital modulation
PCM Pulse Code Modulation
DM Delta Modulation
INTRODUCTION
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Advantages of pulse modulation:
Noise immunity
Inexpensive digital circuitry
Can be time-division multiplexed with other pulse modulated signals.
Digital pulse stream can be stored
Error detected and correction is easily implemented
Disadvantages of pulse modulation:
Special encoding and decoding technique may be necessary to increasetransmission rates, thus making the pulse stream more difficult to recover.
INTRODUCTION
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SAMPLING THEOREM
Any analog signals can be digitized through sampling and quantization.
The sampling rate must be sufficiently large so that the analog signal
can be reconstructed from the samples with sufficient accuracy.
The sampling theorem is the basis for determining the proper sampling
rate for a given signal, has a deep significance in signal processing and
communication theory.
Generally, it is a process of changing the analog signal to a discrete
form without losing any information.It is the first part that has to be done in the process of analog to digital
conversion.
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SAMPLING THEOREM
Sampling theorem is very important in signal analysis, processing andtransmission.
It allows us to replace a continuous-time signal by a discrete sequence
of numbers.The continuous-time signal is sampled, and sample values are used tomodify certain parameters of a periodic pulse train.
We may vary the amplitudes, widths or positions of the pulses inproportion to the sample values of the signal.
Accordingly we have three types ofpulse analog modulation; Pulse Amplitude Modulation (PAM)
Pulse Width Modulation (PWM)
Pulse Position Modulation (PPM)
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SAMPLING THEOREM
9
Sampling process can be done by multiplying the information
signal, vm(t) with the sampling signal (pulse signal), s(t) as given
by:
The sampling signal, s(t) is shown:
Where,
)()()( tstvtv ms y!
X Ts
A
amplitudepulse
intervalsampling
timesampling
!
!
!
A
Ts
X
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Sampling frequency, fs is given by:
It must be sufficiently large so that no information will loss oraliasing
occurs during the transmission.
Therefore, information signal must be sampled at least twice of the
maximum frequency of the information signal.
This rate is known as Nyquist rate and it is given by:
SAMPLING THEOREM
s
s
Tf
1!
frequencymax.
frequencysamplingwhere,2
max
max
!
!u
f
fffss
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If the information signal is sampled at Nyquist rate (fsu 2fmax),thus the
spectrum can be shown below:
Iffs < 2fmax:
SPECTRUM OF THE SAMPLED SIGNAL
fm
fs-f
mfs
2fs
fs+f
m2f
s-f
m2f
s+f
m
V
fm
fs-f
mfs
2fs
fs+f
m
2fs-f
m
2fs+f
m
Valiasing
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1. Two signals have their time for one full cycle of 0.01s and 0.02srespectively. Calculate the sampling frequency, fs required foreach signal.
2. Determine the Nyquist rate for the following signal:a)
b)
Answer:
1. 200Hz 2. a) 5000Hz b) 200Hz
EXAMPLES
ttv
tttv
m
m
T
TT
200sin200)(
5000cos3000cos5.2)(
!
!
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