Signal-to-Noise Ratio of a Noise Schedule

Problem Statement

Turn a cumulative-alpha schedule into the two quantities that actually describe "how hard is step $t$": the signal-to-noise ratio and its logarithm.

Background

Under the closed form $x_t = \sqrt{\bar{\alpha}_t}\,x_0 + \sqrt{1-\bar{\alpha}_t}\,\varepsilon$, the signal power is $\bar{\alpha}_t$ and the noise power is $1-\bar{\alpha}_t$. Hence

$\mathrm{SNR}(t) = \frac{\bar{\alpha}_t}{1 - \bar{\alpha}_t}, \qquad \lambda_t = \log \mathrm{SNR}(t)$

Two facts make this the right coordinate system for diffusion:

• $\bar{\alpha}_t = 0.5$ is exactly the half-signal point: $\mathrm{SNR} = 1$ and $\lambda_t = 0$.
• $\lambda_t$ decreases roughly linearly over the useful part of a schedule, which is why continuous-time formulations parameterize by log-SNR rather than by $t$, and why two schedules with the same $\lambda$ curve are the same model.

Your Task

Implement:

def snr_and_logsnr(alpha_bars):

Return (snr, log_snr), two 1-D NumPy arrays the same length as alpha_bars.

Input Format

• alpha_bars: 1-D NumPy array with every entry strictly inside (0, 1).

Output Format

A tuple (snr, log_snr) of NumPy arrays.

Sample

ab = np.array([0.9, 0.5, 0.1])
snr, ls = snr_and_logsnr(ab)
print(np.round(snr, 4).tolist())
print(np.round(ls, 4).tolist())

At $\bar{\alpha} = 0.5$ signal and noise power are equal, so SNR = 1.0 and log SNR = 0.0; at 0.9 the SNR is 9, and at 0.1 it is one ninth.