DSP Lab Experiments - Digital Signal Processing: An Introduction with MATLAB and Applications - page 336

s o ( t )

1

t

0 T

s 1 ( t )

1

T

t

− 1

T/2

Fig. D.10 Orthogonal signals for binary transmission

Assuming that s 0 and s 1 are as shown in Fig. D.10 , and that s 0 was transmitted,
the outputs of the two matched filters at the time instant t = T are as follows:

r 0 ¼ Z r ð t Þ s 0 ð t Þ dt ¼ Z

T T

½ s 0 ð t Þþ n ð t Þ s 0 ð t Þ dt

0 0

¼ Z s 0 ð t Þ dt þ Z

T T

n ð t Þ s 0 ð t Þ dt ¼ E þ n 0

0 0

r 1 ¼ Z r ð t Þ s 1 ð t Þ dt ¼ Z

T T

½ s 0 ð t Þþ n ð t Þ s 1 ð t Þ dt

0 0

¼ Z s 0 ð t Þ s 1 ð t Þ dt þ Z n ð t Þ s 1 ð t Þ¼ 0 þ Z

T T T

n ð t Þ s 1 ð t Þ¼ n 1

0 0 0

Both n 0 and n 1 are Gaussian and have zero means. The variances of n 0 and
n 1 , r i 2 (i = [ {1, 2}), are given by:

r i ¼ g

2 E ;
where E is the energy of the signals s 0 and s 1 [Note that s 1 2 (t) = s 0 2 (t)].
Now if s 1 (t) was transmitted, then r 0 = n 0 , r 1 = E + n 1 with same statistics as
above.

Probability of Error

The matched filter compares r 1 and r 0 . It will decide that a ''0'' was transmitted if
r 0 [ r 1 , and that ''1'' was transmitted if r 0 \ r 1 . If ''0'' was transmitted, then error
will occur only if r 1 [ r 0 .
It can be shown that the above probability of error can be expressed as follows: