SME
Communication Systems · Problem type
BPSK bit error rate
Use this for “A BPSK link runs at Eb/N₀ = 9.6 dB. What is its bit error rate?”
Chart
X is normal with μ = 0 and σ = 1.
P(X ≥ 4.2709) = < 0.0001.
x = 4.2709 is 4.2709 standard deviations above the mean: z = 4.2709.
This is an example. Type your own values to start a new problem.
γb
=
10
/
10
x
=
2
×
Pb
=
Q(
x
)
Formulas
γb
=
10
γdB ÷ 10
9.12
=
10
9.6 ÷ 10
x
=
√
2 × γb
4.271
=
√
2 × 9.12
Pb
=
Q(
x
)
9.736 ×
10
-6
=
Q(4.271)
Assumptions
•
Coherent BPSK in additive white Gaussian noise, with a matched-filter receiver.
•
Q(x) is the area of the standard normal curve past x: the chance the noise pushes a bit across the threshold.
•
A power ratio in dB is 10 log₁₀ of the ratio, so γb = 10dB ÷ 10.
Step-by-step
WE KNOW
γdB = 9.6 dB
FIND
Eb/N₀ as a ratio (γb), distance to the threshold in noise SDs (x), bit error rate (Pb)
Step 1 · Find Eb/N₀ as a ratio (γb)
Use γb = 10γdB ÷ 10
Undo the decibels: 10 to the dB over 10.
γb
=
10
9.6 ÷ 10
γb
=
10
0.96
γb = 9.12011
Step 2 · Find distance to the threshold in noise SDs (x)
Use x = √(2 × γb)
The
two
signal
points
sit
at
±√Eb,
each
√Eb
from
the
threshold,
with
noise
of
SD
√(
N₀
2
):
√
2Eb ÷ N₀
SDs.
x
=
√
2 × 9.12011
x
=
√
18.24
x
=
4.27086
Step 3 · Find bit error rate (Pb)
Use Pb = Q(x)
A
bit
is
wrong
when
the
noise
carries
it
past
the
threshold:
the
normal
tail
past
x
.
Pb = Q(4.27086)
Pb
=
9.736 ×
10
-6
Check
Put the numbers back in. Both sides should match.
9.12011
=
10
9.6 ÷ 10
✓
4.27086
=
√
2 × 9.12011
✓
9.736 ×
10
-6
=
Q(4.27086)
✓
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