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Digital Communication Vector Space conceptPowerPoint Presentation

Digital Communication Vector Space concept

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- Signal Space
- Inner Product
- Norm
- Orthogonality
- Equal Energy Signals
- Distance
- Orthonormal Basis
- Vector Representation
- Signal Space Summary

2

Signal Space

S(t)

S=(s1,s2,…)

- Inner Product (Correlation)
- Norm (Energy)
- Orthogonality
- Distance (Euclidean Distance)
- Orthogonal Basis

3

Distance, d

- For equal energy signals

- =-1 (antipodal)

- =0 (orthogonal)

- 3dB “better” then orthogonal signals

12

- Modulation
- BPSK
- QPSK
- MPSK
- QAM
- Orthogonal FSK
- Orthogonal MFSK
- Noise
- Probability of Error

18

Binary antipodal signals vector presentation

- Consider the two signals:

The equivalent low pass waveforms are:

20

The cross-correlation coefficient is:

The Euclidean distance is:

Two signals with cross-correlation coefficient

of -1 are called antipodal

22

Their complex-valued correlation coefficients are :

and the real-valued cross-correlation coefficients are:

The Euclidean distance between pairs of signals is:

25

28

30

MPSK

31

Multi-amplitude Signal

Consider the M-ary PAM signals

m=1,2,….,M

Where this signal amplitude takes the discrete values (levels)

m=1,2,….,M

The signal pulse u(t) , as defined is rectangular

U(t)=

But other pulse shapes may be used to obtain a narrower signal spectrum .

33

Clearly , this signals are one dimensional (N=1) and , hence, are represented by the scalar components

M=1,2,….,M

The distance between any pair of signal is

M=2

0

M=4

0

Signal-space diagram for M-ary PAM signals .

34

The minimum distance between a pair signals hence, are represented by the scalar components

35

Multi-Amplitude MultiPhase signals hence, are represented by the scalar components QAM Signals

A quadrature amplitude-modulated (QAM) signal

or a quadrature-amplitude-shift-keying (QASK) is represented as

Where and are the information bearing signal amplitudes of the quadrature carriers and u(t)= .

36

QAM signals are two dimensional signals and, hence, they are represented by the vectors

The distance between a pair of signal vectors is

k,m=1,2,…,M

When the signal amplitudes take the discrete values

In this case the minimum

distance is

37

For an represented by the vectorsM - ary QAM Square Constellation

In general for large M - adding one bit requires 6dB more energy to maintain same d .

41

Binary orthogonal signals represented by the vectors

Consider the two signals

Where either fc=1/T or fc>>1/T, so that

Since Re(p12)=0, the two signals are orthogonal.

42

The equivalent lowpass waveforms: represented by the vectors

The vector presentation:

Which correspond to the signal space diagram

Note that

43

M-ary Orthogonal Signal lowpass signals is

Let us consider the set of M FSK signals

m=1,2,….,M

This waveform are characterized as having equal energy and cross-correlation coefficients

45

First, we observe that =0 when and .

Since |m-k|=1 corresponds to adjacent frequency slots ,

represent the minimum frequency separation between adjacent signals for orthogonality of the M signals.

47

For the case in which ,the FSK signals and .

are equivalent to the N-dimensional vectors

=( ,0,0,…,0)

=(0, ,0,…,0)

Orthogonal signals for M=N=3

signal space diagram

=(0,0,…,0, )

Where N=M. The distance between pairs of signals is

all m,k

Which is also the minimum distance.

48

Orthogonal FSK and . (Orthogonal Frequency Shift Keying)

51

ORTHOGONAL MFSK and .

53

How to and . generatesignals

55

NOISE and .

61

- White Gaussian Noise and . has energy in every dimension

65

Exrecise and .

Probability of Error for Binary SignalingThe two signal waveforms are given as

These waveforms are assumed to have equal energy E and their equivalent lowpass um(t), m=1,2 are characterized by the complex-valued correlation coefficient ρ12 .

66

The optimum demodulator forms the decision variables and .

Or,equivalently

And decides in favor of the signal corresponding to the larger decision variable .

67

Lets see that the two expressions yields the same probability of error .

Suppose the signal s1(t) is transmitted in the interval 0tT . The equivalent low-pass received signal is

Substituting it into Um expression obtain

Where Nm, m=1,2, represent the noise components in the decision variables,given by

68

And . probability of error .

The probability of error is just the probability that the decision variable U2 exceeds the decision variable u1 . But

Lets define variable V as

N1r and N2r are gaussian, so N1r-N2r is also gaussian-distributed and, hence, V is gaussian-distributed with mean value

69

And variance probability of error .

Where N0 is the power spectral density of z(t) .

The probability of error is now

70

Distance, d as

- For equal energy signals

- =-1 (antipodal)

- =0 (orthogonal)

- 3dB “better” then orthogonal signals

72

It is interesting to note that the probability of error P2 is expressed as

Where d12 is the distance of the two signals . Hence,we observe that an increase in the distance between the two signals reduces the probability of error .

73

74 is expressed as

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