Minimization of error probability in transceivers - Signal Processing and Optimization for Transceiver Systems - page 517

Using Eq. (16.90) we have

1

E − B − 1 / 2

df ( E ) 2 A

√ 2 × 0 . 5 × × −

1
E 2

√ π e − 0 . 5 A 2 ( E −B ) ×

= −

d E

√ 2 π e − 0 . 5 A 2 ( E −B ) 1

B − 1 / 2

A 1

E 2

=

E −

call this g ( E )

The sign of d 2 f ( E ) /d E 2 is identical to that of dg ( E ) /d E . Now,

e − 0 . 5 A 2 ( E −B ) 1 B − 1 / 2

0 . 5 A 2
E 2 ×

dg ( E ) 1

E 2

=

E − d E

1
E −

B − 3 / 2

0 . 5
E 2

1
E 2

e − 0 . 5 A 2 ( E −B ) ×

+

2 e − 0 . 5 A 2 ( E −B ) 1 B − 1 / 2

1
E 3

−

E −

B − 1 / 2 A 2 + 1

.

e − 0 . 5 A 2 ( E −B ) 1 B − 1

0 . 5
E 4

=

E − E −

− 4 E

This is nonnegative, or equivalently P br ( E ) is convex, if and only if

A 2 + 1 B − 1

E −

− 4 E≥ 0 ,

which can be rewritten as

E 2 − a 1 E + a 2 ≥ 0 , (16 . 91)

where a 1 and a 2 are positive numbers given by

a 1 = 3+ A 2 B
4 B

A 2
4 B .

, 2 =

Figure 16.9 shows a plot of the left-hand side of Eq. (16.91). From the plot
it is clear that there exists a range 0 ≤E≤E t such that Eq. (16.91) holds

(since a 2 > 0). This concludes the proof. Notice that Eq. (16.91) also holds
in the range

E≥E 0 , where E 0 is the second zero-crossing.