Transcript KLKSK - Universitas Udayana

KLKSK
Pertemuan III
Analog & Digital Data
Shannon Theorem
xDSL
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Analog and Digital Signaling of
Analog and Digital Data
• Analog signals: represents data with
continuously varying electromagnetic
wave
• Digital signals: represents data with
sequence of voltage pulses
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Data and Signals
• Usually use digital signals for digital data
and analog signals for analog data
• Can use analog signal to carry digital data
– Modem
• Can use digital signal to carry analog data
– Compact Disc audio
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Analog & Digital Transmission
Analog Signal
Digital Signal
Two alternatives: (1) signal
Analog data are encoded
Analog occupies the same spectrum as
the analog data; (2) analog data using a codec to produce
Data are encoded to occupy a different a digital bit stream
portion of spectrum
Digital Digital data are encoded using a
Data modem to produce analog signal
Data and Signal
Two alternatives: (1)
signal consists of two
voltage levels to represent
the two binary values; (2)
digital data are encoded to
produce a digital signal
with desired properties
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Analog Signals Carrying Analog
and Digital Data
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Digital Signals Carrying Analog
and Digital Data
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Analog & Digital Transmission
Analog
Transmission
Analog
Signal
Digital
Signal
Digital Transmission
Is propagated through
amplifiers; same
treatment whether
signal is used to
represent analog data
or digital data
Assumes that the analog signal represents
digital data. Signal is propagated through
repeaters; at each repeater, digital data are
recovered from inbound signal and used to
generate a new analog outbound signal
Not used
Digital signal represents a stream of 1s and
0s, which may represent digital data or may
be an encoding of analog data. Signal is
propagated through repeaters; at each
repeater, stream of 1s and 0s is recovered
from inbound signal and used to generate a
new digital outbound signal
Treatment of Signals
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Channel Capacity
• Data rate
– In bits per second
– Rate at which data can be communicated
• Bandwidth
– In cycles per second of Hertz
– Constrained by transmitter and medium
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Nyquist Bandwidth
• If rate of signal transmission is 2B then signal
with frequencies no greater than B is sufficient to
carry signal rate
• Given bandwidth B, highest signal rate is 2B
• Given binary signal, data rate supported by B Hz
is 2B bps
• Can be increased by using M signal levels
• C= 2B log2M
• M is the of discrete signal or voltage levels
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Shannon Capacity Formula
• Consider data rate, noise and error rate
• Faster data rate shortens each bit so burst
of noise affects more bits
– At given noise level, high data rate means
higher error rate
• Signal to noise ratio (in decibels)
• SNRdb=10 log10 (signal/noise)
• Capacity C=B log2(1+SNR)
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• Consider a voice channel being used, via
modem, to transmit digital data. Assume a
bandwidth of 3100 Hz. A typical value of
S/N for a voice-grade line is 30 dB; or a
ratio of 1000:1
• Thus
• C = 3100 log2(1 + 1000) = 30,894 bps
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• Shannon proved that if the actual
information rate on a channel is less than
the error-free capacity, then it is
theoretically possible to use a suitable
signal code to achieve error – free
transmission through the channel
• Unfortunately Shannon formula does not
suggest a means for finding such codes
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• Eb/No : ratio of signal energy per bit to noise
power density per hertz
• Consider a signal, digital or analog, that contains
binary digital data transmitted at a certain bit rate
R
• 1 W = 1 J/s
• The energy per bit in a signal Eb = STb
• S : the signal power
• Tb : the time required to send one bit
• The data rate R = 1/Tb
• (Eb/No) = (S/R)/No = S/(kTR)
• Eb/No = S – 10log R + 228.6 dBW – 10 logT
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• Ex. For binary phase-shift keying, Eb/No =
8.4 dB is required for a bit error rate of 10-4
(probability of error = 10-4). If the effective
noise temperature is 290oK (room
temperature) and the data rate is 2400
bps, what received signal level is
required?
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• Ans.
• 8.4 = S (dBW) – 10 log 2400 + 228.6 dBW
– 10 log 290
• = S (dBW) – (10) (3.38) + 228.6 –
(10)(2.46)
• S = -161.8 dBW
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