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>> Introduction
Digital transmission
Digital
is the transmittal of digital signals
Transmission between two points in a communication
system.
Advantages and Disadvantages of Digital >> Pulse Modulation
Pulse modulation
• Cost Effective - Integration of Switching and
Transmission Equipment consist of sampling analog information
• Digital Signal Regeneration “blocks” noise signals and then converting those samples into
• Performance Monitoring and Error Correction discrete and transporting the pulses from a
• All messages are 0s and 1s source to a destination over a physical
transmission medium.
Cost Incurred for Analog to Digital to Analog Conversion
Need Synchronization - A Common Clock
Increased Bandwidth Requirements
2. Pulse position modulation (PPM)
Predominant methods of pulse modulation
the position of a constant width pulse within a
prescribe timeslot is varied according to the amplitude of
1.Pulse width modulation (PWM) the sample of the analog signal.
is sometimes called pulse duration modulation
(PDM) or pulse length modulation (PLM), as the width of
the constant amplitude pulse is varied proportional to the
3. Pulse Amplitude Modulation (PAM)
amplitude of the analog signal at the time the signal is the amplitude of a constant width, constant
sampled. PWM and PPM are used in special purpose position pulse is varied according to the amplitude of
communication systems mainly for the military but are the sample of the analog signal. PAM is used as an
seldom used for commercial digital transmission intermediate form of modulation with PSK, QAM, and
systems. PCM.
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4. Pulse Code Modulation
the analog signal is sampled and then converted Analog signal
to a serial n-bit binary code for transmission. Each code
has the same number of bits and requires the same Sample pulse
length of time for transmission. PCM is by far the most
prevalent form of pulse modulation.
PWM
PPM
PAM
PCM
Simplex PCM Transmission System
Band pass filter
Limits the frequency of the analog input signal to the
standard voice-band frequency of 300 Hz to 3000 Hz.
Sample and Hold
periodically samples the analog input signal and converts
those samples to a multilevel PAM signals.
Analog to Digital Converter (ADC)
converts the PAM samples to parallel PCM codes, which are
converted to serial binary data in the parallel to serial converter
and then outputted onto the transmission line as serial digital pulses.
PCM Sampling
Transmission Line Repeaters
Natural Sampling
are placed at prescribed distances to regenerate the digital
pulses
Serial to Parallel Converter
converts serial pulses received from the transmission line to
parallel PCM codes
Digital to Analog Converter (DAC)
converts the parallel PCM codes to multilevel PAM signals.
Hold Circuit
is basically a low pass filter that converts the PAM signals
back to its original analog form
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Dynamic Range
Flat- Top Sampling
Dynamic Range (DR)
is the ratio of the largest possible magnitude that can be decoded by the
digital to analog converter in the receiver
DR = Vmax / Vmin
Where:
DR = dynamic range (unit less)
Vmin = the quantum value (resolution)
Vmax = max voltage magnitude that can be discerned by
the DAC’s in the receiver
It can be written as:
DR = V max / resolution To solve for the number of bits (n) necessary to produce a DR, convert to
logs:
DR is generally expressed as a dB value: n log 2 = log (DR + 1)
DR = 20 log V max / V min
The number of bits used for a PCM code depends on the n = log 2 (DR + 1) / log 2
dynamic range. The relationship between DR and the
number of bits in a PCM code is: For a DR of 3, a PCM code with two bits is required. DR can be expressed
on dB as:
2n – 1 ≥ DR
DR(dB) = 20 log (V max / V min)
And for the minimum number of bits:
2n – 1 = DR DR(dB) = 20 log (2n – 1)
Where: n = number of bits in a PCM code, excluding the sign bit
DR = absolute value of DR Where: n = number of bits
Sample problem: Solution:
for a PCM system with the following parameters, determine a) fs = 2 fa b) 46 dB = 20 log (V max / V min)
= 2 (4 khz) 2.3 = log (V max / V min)
a) min. sample rate
fs = 8 khz 102.3 = V max / V min
b) min. no. of bits used in PCM code
199.5 = DR
c) resolution
d) quantization error
Given: c) resolution = V max / 2n -1
d) Qe = resolution / 2
max analog input frequency = 4khz = 2.55 / (256 -1)
= 0.01 / 2
max decoded voltage at the receiver = ±2.25 V resolution = 0.01 V
Qe = 0.005 V
min. dynamic range = 46 dB
