sdr.ebn0_to_snr(ebn0: ArrayLike, bps: int, rate: int = 1, sps: int = 1) → NDArray[float_]

Converts from \(E_b/N_0\) to \(S/N\).

\[ \frac{S}{N} = \frac{E_b}{N_0} \frac{k}{n} \log_2 M \frac{f_{sym}}{f_s} \]

Parameters:
ebn0: ArrayLike

Bit energy \(E_b\) to noise PSD \(N_0\) ratio in dB.

bps: int

Bits per symbol \(\log_2 M\), where \(M\) is the modulation order.

rate: int = 1

Code rate \(r = k/n\), where \(k\) is the number of information bits and \(n\) is the number of coded bits.

sps: int = 1

Samples per symbol \(f_s / f_{sym}\).

Returns:

The signal-to-noise ratio \(S/N\) in dB.

Examples

Convert from \(E_b/N_0 = 5\) dB to \(S/N\) for a 4-QAM signal with \(r = 2/3\) and 1 sample per symbol.

In [1]: sdr.ebn0_to_snr(5, 2, rate=2/3, sps=1)
Out[1]: 6.249387366082999

Convert from \(E_b/N_0 = 10\) dB to \(S/N\) for a 16-QAM signal with \(r = 1\) and 4 samples per symbol.

In [2]: sdr.ebn0_to_snr(10, 4, rate=1, sps=4)
Out[2]: 10.0