Digital Communication Vector Space concept. Signal space. Signal Space Inner Product Norm Orthogonality Equal Energy Signals Distance Orthonormal Basis Vector Representation Signal Space Summary. Signal Space. S(t). S=(s1,s2,…). Inner Product (Correlation) Norm (Energy)
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Digital
Communication
Vector Space
concept
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Signal space
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S(t)
S=(s1,s2,…)
3
ONLY CONSIDER SIGNALS, s(t)
T
t
Energy
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Similar to Vector Dot Product
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Example
A
T
t
A
2A
A/2
t
T
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Similar to norm of vector
A
T
A
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A
T
A
Y(t)
B
Similar to orthogonal vectors
T
8
X(t)
{
T
Y(t)
T
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Correlation Coefficient
1 1
=1 when x(t)=ky(t) (k>0)
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Example
Y(t)
X(t)
10A
A
t
t
A
T
T/2
7T/8
Now,
shows the “real” correlation
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(antipodal signals)
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PSK (Orthogonal Phase Shift Keying)
(Orthogonal if
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QAM
BPSK
QPSK
BFSK
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The equivalent low pass waveforms are:
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The vector representation is –
Signal constellation.
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The crosscorrelation coefficient is:
The Euclidean distance is:
Two signals with crosscorrelation coefficient
of 1 are called antipodal
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The equivalent low pass waveforms are:
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The vector representation is:
Or in complexvalued form as:
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Their complexvalued correlation coefficients are :
and the realvalued crosscorrelation coefficients are:
The Euclidean distance between pairs of signals is:
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The minimum distance dmin corresponds to the case which
 mk =1
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(00)
(10)
(11)
(01)
*
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X(t)
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(00)
(10)
(11)
(01)
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Exrecise
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Consider the Mary 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 .
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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
Signalspace diagram for Mary PAM signals .
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The minimum distance between a pair signals
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A quadrature amplitudemodulated (QAM) signal
or a quadratureamplitudeshiftkeying (QASK) is represented as
Where and are the information bearing signal amplitudes of the quadrature carriers and u(t)= .
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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
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d
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d
QAM=QASK=AMPM
Exrecise
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M=256
M=128
M=64
M=32
M=16
M=4
+
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For an M  ary QAM Square Constellation
In general for large M  adding one bit requires 6dB more energy to maintain same d .
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Consider the two signals
Where either fc=1/T or fc>>1/T, so that
Since Re(p12)=0, the two signals are orthogonal.
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The equivalent lowpass waveforms:
The vector presentation:
Which correspond to the signal space diagram
Note that
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We observe that the vector representation for the equivalent lowpass signals is
Where
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Let us consider the set of M FSK signals
m=1,2,….,M
This waveform are characterized as having equal energy and crosscorrelation coefficients
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The real part of is
0
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First, we observe that =0 when and .
Since mk=1 corresponds to adjacent frequency slots ,
represent the minimum frequency separation between adjacent signals for orthogonality of the M signals.
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For the case in which ,the FSK signals
are equivalent to the Ndimensional 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.
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“0”
“1”
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All signals are orthogonal to each other
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0 T 2T 3T 4T 5T 6T
+
0 T 2T 3T 4T 5T 6T
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0 T 2T 3T 4T 5T 6T
+
0 T 2T 3T 4T 5T 6T
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0 T 2T 3T 4T 5T 6T
+
0 T 2T 3T 4T 5T 6T
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IQ Modulator
+
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IQ Modulator
Pulse shaping filter
+
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T
T
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We write
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Exrecise
The 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 complexvalued correlation coefficient ρ12 .
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The optimum demodulator forms the decision variables
Or,equivalently
And decides in favor of the signal corresponding to the larger decision variable .
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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 lowpass received signal is
Substituting it into Um expression obtain
Where Nm, m=1,2, represent the noise components in the decision variables,given by
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And .
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 N1rN2r is also gaussiandistributed and, hence, V is gaussiandistributed with mean value
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And variance
Where N0 is the power spectral density of z(t) .
The probability of error is now
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Where erfc(x) is the complementary error function, defined as
It can be easily shown that
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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 .
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M=256
M=128
M=64
M=32
M=16
M=4
+
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