When we incorporate additive noise into our channel model, so
that
rt=α
s
i
t+nt
r
t
α
s
i
t
n
t
, errors can creep in. If the transmitter sent bit
0 using a BPSK
signal set, the integrators' outputs in the matched filter
receiver would be:
∫nT(n+1)Trt
s
0
td
t
=αA2T+∫nT(n+1)Tnt
s
0
td
t
t
n
1
T
n
T
r
t
s
0
t
α
A
2
T
t
n
1
T
n
T
n
t
s
0
t
(1)
It is the quantities containing the noise terms that cause
errors in the receiver's decision-making process. Because they
involve noise, the values of these integrals are random
quantities drawn from some probability distribution that vary
erratically from bit interval to bit interval. Because the noise
has zero average value and has an equal amount of power in all
frequency bands, the values of the integrals will hover about
zero. What is important is how much they vary. If the noise is
such that its integral term is more negative than
αA2T
α
A
2
T
, then the receiver will make an error, deciding that
the transmitted zero-valued bit was indeed a one. The
probability that this situation occurs depends on three factors:
-
Signal Set Choice — The difference
between the signal-dependent terms in the integrators' outputs
(equations Equation 1) defines how
large the noise term must be for an incorrect receiver
decision to result. What affects the probability of such
errors occurring is the energy in the difference of the received signals in
comparison to the noise term's variability. The signal-difference energy equals
∫0Ts 1t−s 0t2dt
t
T
0
s 1
t
s 0
t
2
For our BPSK baseband signal set, the difference-signal-energy term is
4α2A4T2
4
α
2
A
4
T
2
.
-
Variability of the Noise Term — We
quantify variability by the spectral height of the white noise
N02
N0
2
added by the channel.
-
Probability Distribution of the Noise
Term — The value of the noise terms relative
to the signal terms and the probability of their occurrence
directly affect the likelihood that a receiver error will
occur. For the white noise we have been considering, the
underlying distributions are Gaussian. Deriving the following expression for the probability the
receiver makes an error on any bit transmission is complicated but can be found at (Reference) and (Reference).
pe=Q∫0Ts 1t−s 0t2dt2N0=
Q2α2A2T
N
0
for the BPSK case
pe
Q
t
T
0
s 1
t
s 0
t
2
2
N0
Q
2
α
2
A
2
T
N
0
for the BPSK case
(2)
The term
A2T
A
2
T
equals the energy expended by the transmitter in sending the
bit; we label this term
E
b
E
b
. We arrive at a concise expression for the
probability the matched filter receiver makes a bit-reception
error.
p
e
=Q2α2
E
b
N
0
p
e
Q
2
α
2
E
b
N
0
(3) shows how the receiver's
error rate varies with the signal-to-noise ratio
α2
E
b
N
0
α
2
E
b
N
0
.
Derive the probability of error expression for the modulated
BPSK signal set, and show that its performance identically
equals that of the baseband BPSK signal set.
The noise-free integrator outputs differ by
αA2T
α
A
2
T
, the factor of two smaller value than in the
baseband case arising because the sinusoidal signals have
less energy for the same amplitude. Stated in terms of
E
b
E
b
, the difference equals
2α
E
b
2
α
E
b
just as in the baseband case.
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