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Data Transmission
EECE 542 Fall 2003
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Time/Frequency Relationships The relationship between time and
frequency domain representation of signalsis defined by Fourier analysis.
Unmodulated (non-sinusoidal) signals havetheir frequency domain spectra centeredabout 0 Hz. (i.e. baseband transmission)
General rule: A faster (shorter period) signal in the time
domain results in a wider (larger bandwidth)signal in the frequency domain
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Ex: A random sequence of 0s
and 1s
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Baseband Data Transmission
Most physical layer transmission systems rely on
baseband transmission.
Almost exclusively use a type of cable or fiber
Supports only one current transmission
No parallel transmissions on the same wire unless
multiple wires are used (both tx and rx)
Exception: some fiber optic systems
Transmission involves a mapping of binary data to
analog waveforms.
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Baseband Data Reception
Line components typically block the transmission
in the vicinity of 0 Hz (DC).
The received signal is first filtered and amplifiedto reduce the effects of noise and line attenuation.
Correct decisions on the data being a 0 or 1
requires knowledge of the bit transition edges or
boundaries.
Requires a bit clock which is not typically sent with the
data
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Bit Synchronization
A bit (data) clock must be generated at thereceiver for the data being received.
The generation of this clock and thealignment (phase adjustment) of its edgeswith the edges of the received data isperformed by a bit synchronizer.
A bit synch is basically a Phase Lock Loop(PLL)
PLLs work best if bit transitions occur atmost if not all data bit boundaries
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Line Coding Data embedded in a layer 2 frame may
easily contain long strings of 0s or 1s
Few bit transitions for the PLL to work well Line coding is the translation of the binary
data into a new digital stream
Good line coding schemes guarantee bit
transitions The spectral shape of the transmission is often
affected
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Types of Line Coding Unipolar
Polar
NRZ (Nonreturn to Zero)
RZ (Return to Zero)
Biphase
Bipolar
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Unipolar Line Coding Simple
Binary 1 = high voltage
Binary 0 = low (zero) voltage
Properties
No edge transitions when the original data
doesnt change
No change in the spectral shape (still has DC
component)
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Unipolar cont.
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NRZ (Nonreturn to Zero) Coding A type of polar (two non-zero voltage levels)
coding
Removes the DC component NRZ-I
NRZ-L 0 -> positive (or neg.) voltage
1 -> negative (or pos.) voltage NRZ-I
0 -> voltage remains the same
1 -> causes an inversion in the voltage
creates bit transitions in long strings of 1s (but not 0s)
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NRZ Cont.
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RZ (Return to Zero) Coding Another type of polar encoding
The first half of each bit is mapped as in
NRZ-L
The second half of each bit is set to 0 volts
Guarantees bit transitions
Removes the DC component The width of the transmitted pulse is cut in
half so the spectral bandwidth increases
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RZ Cont.
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Biphase Coding
Another type of polar
Like RZ, transitions are created in the middle of
the bit periods Most common methods used in LANs
Manchester
Middle transition = o if bit =1, q if bit = 0
Ethernet
Differential Manchester
Middle transition always present, but a transition at thebeginning of a bit only occurs if the bit = 0
Token Ring
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Manchester & Diff. Manchester
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Line Coding Spectra
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Block Coding
Enhances the performance of line codingwhile also introducing some error-detecting
capability Based on substituting a block of n bits for a
block of m bits, where n > m
A dictionary contains the mapping. Someof the n-bit blocks are not used in the one-to-one mapping
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Block Coding cont.
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Block coding subsitution(m=4, n=5)
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Block coding cont.
Errors can be detected if the received n-bitword is invalid
Also called mBnB coding
Used in some of the newer Ethernetstandards
100B
ase-TX (2-wire twisted pair) 100Base-FX (Fiber)
1000Base-T (2-wire Gigabit Ethernet)
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RF Transmission Not baseband
Requires modulation
The placement of data onto a cosinusoidal signal
Multiple bits may be mapped into one
modulation symbol
Baud rate = modulation symbol rate
Traditional schemes:ASK, FSK, PSK, QAM
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ASK Amplitude Shift Keying Susceptible to channel degradations
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FSK Frequency Shift Keying
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PSK Phase Shift Keying
BPSK: bit rate = baud rate, 0 or 180 deg. phase
QPSK: bit rate = 2 * baud rate, [45, 135, 225, 315] deg.
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QAM Quadrature Amplitude
Modulation Combined ASK and PSK
Higher-order modulation scheme that
lowers the symbol rate
More susceptible to noise and nonlinearities
Used in most modern phone modems
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8-QAM
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Multiplexing Transmission resources are usually limited
in either time, frequency, or both
Normally two separate signals cannot sharethe same time and frequency space
As multiple users or segments become
necessary, a method of sharing the theseresources is critical
Multiplexing allows this sharing
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FDM Frequency Division
Multiplexing The frequency channel is divided and each
user receives one portion of the spectrum
Requires at least one non-baseband signal
Guard bands are used to limit the effect of
adjacent channel interference (ACI)
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FDM cont.
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FDM cont.
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Time-division Multiplexing
(TDM) Dividing by time
Supports any combination of baseband and
modulated signals
Two types of TDM:
Synchronous TDM
Asynchronous TDM
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Synchronous TDM Each user (1, 2, n) is allocated a time slot
A frame consists of one full cycle of a time
slot from every user
Requires framing bits for time slot
synchronization
Inefficient if data is not always being sent
by ALL users
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Synch. TDM Cont.
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Asynchronous TDM m time slots for n users, m < n
Time slots are not reserved for each user
Scans user input lines for available data
Tries to fill all time slots during each frame
Requires addressing overhead for correct
de-multiplexing
Typically more efficient that synch. TDM
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Asynch. TDM cont.
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Limits on Data Throughput Nyquist Bit Rate
Noiseless, bandlimited channels
Bit Rate (bps) = 2 x B x log2(L) L = # of signal levels used to represent the data
B = frequency bandwidth available (Hz)
Shannons Capacity Theorem
Bandlimited channels with noise
C (bps) = B x log2(1 + SNR)
SNR = signal-to-noise ratio of the channel
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Nyquist Example A noiseless channel with a 5 kHz bandwidth
and binary transmission (2 levels) can
deliver:
Bit Rate = 2 x 5000 x log2(2) = 10,000 bit/sec.
If transmission using 4 bits/symbol is used
(16 levels) thenBit Rate = 2 x 5000 x log2(16) = 40,000 bit/sec.
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Shannon Capacity Example A modem operating over a telephone line
has a maximum useful bandwidth of about
3400 Hz (300 Hz to 3700 Hz). Themaximum SNR of the channel is 39 dB.
What is the maximum capacity?
First, un-dB the SNR: SNR = 10^(39/10) = 7943
C = 3400 x log2(1+7943) = 44 kbps