Digital communications - week 1 1 · 2010. 10. 8. · Digital communications - week 2 12 How large is the bandwidth W [Hz] for a given information bit rate Rb [bps]? Digital communications

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].


Equally likely information bits is desired and common in practice.Why?


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=600064-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, we need to find R(f) for a given M and a 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 QAMM-ary PSKM-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 GSM!


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


All frequencies are equally disturbed.


Gaussian probabilitydistribution:

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


Very usefultables!


Sent message (k bits): Decidedmessage:



ML receiver when M=2.


Example:

zo(t)=0, z1(t) rectangular with amplitude A and T=Tb.

Rb=400 kbps, A^2/No= 70 dB

Pb=?

D^2=A^2/Rb

D^2/(2No)=(A^2/No)*(0.5/Rb)=12.5

Table 3.1 on Page 182: Pb=Q(3.536)=2.3*10^-4

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 especiallygood 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 important!



Example: M equal energy orthogonal signals (FSK, or PPM).

D^2=2E for all pairs.

Union bound = (M-1)Q( )++++++++++++++++++++++++++++

The coefficients in general:













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!

Digital communications - week 1 1 · 2010. 10. 8. · Digital communications - week 2 12 How large is the bandwidth W [Hz] for a given information bit rate Rb [bps]? Digital communications

Documents