SME
Communication Systems · Problem type
Eb/N₀ from SNR
Use this for “A 2 Mb/s signal arrives with a 10 dB SNR in a 1 MHz bandwidth. Find Eb/N₀.”
Chart
Eb/N₀ in dB: 10.0 − 3.0 = 7.0 dB
This is an example. Type your own values to start a new problem.
SNR
=
10
/
10
γb
=
SNR
×
B (MHz)
Rb (Mb/s)
γdB (dB)
=
10
log₁₀
γb
LB
Bandwidth over bit rate in dB
worked out (not typed) · dB
Formulas
SNR
=
10
SNRdB ÷ 10
10
=
10
10 ÷ 10
γb
=
SNR
×
B
Rb
5
=
10
×
1
2
γdB = 10 log₁₀(γb)
6.9897 = 10 log₁₀(5)
LB
=
10
log₁₀(
B
Rb
)
−3.0103
=
10
log₁₀(
1
2
)
Assumptions
•
The SNR is the signal power S over the noise power N measured in the bandwidth B, so N = N₀B.
•
Each bit gets the energy Eb = S ÷ Rb, so Eb/N₀ = SNR × B ÷ Rb.
•
A power ratio in dB is 10 log₁₀ of the ratio.
Step-by-step
WE KNOW
SNRdB = 10 dB, B = 1 MHz, Rb = 2 Mb/s
FIND
SNR as a ratio (SNR), Eb/N₀ as a ratio (γb), Eb/N₀ in dB (γdB), bandwidth over bit rate in dB (LB)
Step 1 · Find SNR as a ratio (SNR)
Use SNR = 10SNRdB ÷ 10
Undo the decibels: 10 to the dB over 10.
SNR
=
10
10 ÷ 10
SNR = 10
Step 2 · Find Eb/N₀ as a ratio (γb)
Use γb = SNR × B ÷ Rb
Eb/N₀
is
(
S
Rb
)
over
(
N
B
),
which
is
the
SNR
times
B
over
Rb.
γb
=
10
×
1
2
γb = 5
Step 3 · Find Eb/N₀ in dB (γdB)
Use γdB = 10 log₁₀(γb)
Eb/N₀ is an energy ratio, so 10 times its common log.
γdB = 10 log₁₀(5)
γdB = 6.9897 dB
Step 4 · Find bandwidth over bit rate in dB (LB)
Use LB = 10 log₁₀(B ÷ Rb)
In dB the product becomes a sum: Eb/N₀ is the SNR plus this term.
LB
=
10
log₁₀(
1
2
)
LB = −3.0103 dB
Check
Put the numbers back in. Both sides should match.
10
=
10
10 ÷ 10
✓
5
=
10
×
1
2
✓
6.9897 = 10 log₁₀(5) ✓
−3.0103
=
10
log₁₀(
1
2
)
✓
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