RF Toolkit ยท Proakis Digital Comms
Calculator SNR / Eb/N0
Convert between SNR, Eb/N0, Es/N0 and compare against standard modulation thresholds. Theoretical Shannon capacity included.
SNR (dB) = Eb/N0 + 10 x log10(R/B) ยท Shannon C = B x log2(1 + SNR_lin)
Standard Thresholds (AWGN, BER 10^-5, uncoded)
| Modulation | Bits / symbol | Eb/N0 (BER 10^-5) | SNR min (20 MHz) |
|---|---|---|---|
| BPSK | 1 | 9.6 dB | -3.4 dB |
| QPSK | 2 | 9.6 dB | -0.4 dB |
| 16QAM | 4 | 13.4 dB | 6.6 dB |
| 64QAM | 6 | 17.8 dB | 12.6 dB |
| 256QAM | 8 | 22.1 dB | 18.1 dB |
With LTE turbo channel coding rate 1/3: reduce by ~5-6 dB. With 5G NR LDPC: reduce by ~6-7 dB. Source: Proakis "Digital Communications" 5th edition.
Frequently Asked Questions
What is the difference between SNR and Eb/N0?
SNR (Signal to Noise Ratio) is the ratio between signal power and noise power within a given bandwidth. Eb/N0 is the ratio between energy per bit (Eb) and noise spectral density (N0). The relationship is: SNR = (Eb/N0) + 10 x log10(R/B), where R is the bit rate in bps and B is the bandwidth in Hz. Eb/N0 is independent of bandwidth, making it the standard metric for comparing modulation-coding performance.
What Eb/N0 is needed to achieve a BER of 10^-5?
Typical thresholds (AWGN, BER 10^-5) for common modulations (without channel coding): BPSK / QPSK: 9.6 dB; 16QAM: 13.4 dB; 64QAM: 17.8 dB; 256QAM: 22.1 dB. With LTE turbo coding rate 1/3: ~5-6 dB reduction. With 5G NR LDPC: ~6-7 dB reduction. Reference: Proakis 'Digital Communications' 5th ed.
How do SNR and throughput relate?
The Shannon limit gives the maximum throughput: C = B x log2(1 + SNR_linear), where C is in bps, B in Hz, and SNR is a power ratio. In practice, real systems achieve 60-80% of this limit thanks to adaptive modulation (OFDM) and LDPC coding. Example: SNR 20 dB = ratio 100, C = 20 MHz x log2(101) = 133 Mbps theoretical, i.e. ~100 Mbps in practice.
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