Digital communications - week 1 1e4svecka-6-Ht1-2012-SUMMARY.pdfDigital communications - week 4 40 The union bound is an upper bound and it is especially good at “high” signal-to-noise

Digital communications - week 1 1


Note: T=pulse duration, Tb=bit time, Ts=symbol time=k*Tb

___________________________________ t | new signal sent | new signal sent | new signal sent | 0 Ts 2Ts 3Ts

A new signal alternative is sent every Ts. Ts = the signaling (or symbol) time.

A signal alternative carries k information bits, and

The signaling (or symbol) rate = Rs = 1/Ts [symbols/s].


Average signal energy and signal power


M=2: “Antipodal signals are better then orthogonal signals”



4-ary PSK


4-ary FSK (orthogonal)



16-ary QAM with Gray-coding


Orthogonal Frequency-Division Multiplex (OFDM)

OFDM = the sum of N orthogonal QAM signals. Example: N=6000 64-ary QAM in each QAM signal Then an QFDM signal carries 36000 bits! How can this be built? An OFDM signal alternative?


The bandwidth W of a signal is the width of the frequency interval where most of the signal energy (or power) is located.

________________________________________ f [Hz] | W |


How large is the bandwidth W [Hz] for a given information bit rate Rb [bps]?



G(f+fc) is left shift. G(f-fc) is right shift.



Consequently, to find the bandwidth we need to find R(f) for the given set of M signal alternatives.


Remember Appendix D!

Useful for: M=2 and equally likely antipodal signals!



Bandpass case.


VERY USEFUL!


Consequently, Table 2.1 can be used also in this M-ary PAM case!


Consequently, Table 2.1 can also be used for: M-ary QAM M-ary PSK M-ary bandpass PAM



Consequently, the bandwidth efficiency is bad for large M!


A practical implementation is therefore:

General bandpass:



Common challenge in both Wireless and Wireline applications!

Remember the training bits in the GSM-example!


This will cause overlapping signals unless Tb is increased to 3 s!


The noise power is equally distributed over all frequencies.


Gaussian probability distribution:

What is the probability that the output noise is above a critical level A (“bit-error”)?


Very useful tables!


Sent message (k bits): Decided message:



ML receiver when M=2.


FUNDAMENTAL RESULT!


How much received energy per bit is required for a given Pb?

d2 measures energy efficiency: the larger the better.




A “typical” problem formulation.

Consequences:

Note! The received signal power Pz decreases with communication distance.


The union bound is an upper bound and it is especially good at “high” signal-to-noise ratios. In that case it is also easy to calculate!

Assume 4-PAM: Then 3 different distances exist.

So, the minimium Euclidean distance is very important!















How large is ISI? Is it too large? Can we make ISI=0?








How can we get better Pb than binary antipodal signals? PAM, PSK, QAM, PWM, PPM, FSK? Uncoded: memoryless, i.e. no dependency between sent signal alternatives. Coding: In a clever way introduce memory (dependency, redundancy) between the sent signal alternatives! The memory can be used by the receiver to significantly reduce Pb!



Adaptive coding and modulation!

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Digital communications - week 1 1e4svecka-6-Ht1-2012-SUMMARY.pdfDigital communications - week 4 40 The union bound is an upper bound and it is especially good at “high” signal-to-noise