U. Glaeser

The EER is the BER divided by the number of bit errors in the dominant error event. In Eq. (34.15) Q
refers to the well-known Q function defined by
Signal Gain (KPD) of the Phase Detector
Using the definition of KPD, for a constant sampling phase error τ,
© 2002 by CRC Press LLC
(34.16)
E{ tsˆ ( k)s(
k)
+ n( k)sˆ(
k)
}
E{ n( k)sˆ(
k)
}
KPD = ----------------------------------------------------------- = E{ sˆ ( k)s(
k)
} + ------------------------------ (34.17)
t
t
Consider E{ n( k)sˆ(
k)
}, where sˆ ( k)
is a linear function of the data bits, which can be realistically assumed
to be uncorrelated with the noise n(k). Therefore, this term is zero and as we should expect, the noise
does not contribute to the mean phase detector output. Thus,
(34.18)
If d is uncoded, hence white, with zero mean, E{ d( k– b)d(
k– c)
}
Consequently, the KPD is
= sd if b = c and is 0 if b ≠ c.
(34.19)
where it is assumed that slope table ouput is based on fewer than C 1 + C 2 + 1 terms to reduce the
summation to be from b = −B 1 to B 2. We note that the KPD values obtained here are equivalent to the
slopes of the f(τ) versus τ curve plotted in Fig. 34.24.
Output Noise of the Phase Detector
Computing the autocorrelation,
Q( x)
KPD = E{ sˆ ( k)s(
k)
} =
E n o k
1
= -----
2p
B 2
∑
∫
x
Because sˆ ( k)
is a filtered version of d(k), which is uncorrelated with n, n and sˆ are uncorrelated. Therefore,
(34.20)
With d being uncoded (hence white) and zero mean, E{ d(k– b)d(k+
l– b′) } = sd if b′ = b + l and
2
0 if b′ ≠ b + l. Also assuming, Rn(l) = snd[] l , i.e., n to be white, we need to consider only l = 0 in which
case we have b = b′. In that case,
∞
C 2 −1
∑
b=−B1 c=−C1 2
KPD = sd y 2
--exp
⎛ ⎞
⎝ 2 dy
⎠
y( b)y(
c)E{
d( k– b)d(
k– c)
}
B 2
∑
b=−B 1
y 2 ( b)
{ ( )no( k+ l)
} = E{ n( k)sˆ(
k)n(
k+ 1)sˆ(
k+ l)
}
{ ( )no( k+ l)
} = E{ n( k)n(
k+ l)
}E{ sˆ ( k)sˆ(
k+ l)
}
E n o k
B 2
∑
= Rn()E l { sˆ ( k)sˆ(
k+ l)
}
B 2
∑
E{ sˆ ( k)sˆ(
k+ l)
} = y( b)y(
b′ )E{ d( k– b)d(
k+ l– b′ ) }
E n o k
b=−B1 b′=−B1 { ( )no( k+ l)
} =
sn 2 sd
B 2
∑
2 2
d[] l y ( b)
b=−B 1
2
2
(34.21)