I have implemented a basic noise estimator using the minimum statistics method. Noise power estimate is obtained as a minimum of the short time power estimate within a window of subband power samples. In short, you take DFT to get amplitudes, square the amplitudes to get power, and then find minimum power across several frames, hoping that noise is the minimum.

However, the noise in this case is underestimated due to the fact that minimums are taken.

In the paper "Spectral Subtraction Based on Minimum Statistics" by Rainer, there is an $omin$ factor used to compensate for the underestimation, so that the final noise power $P_{final} = P_{estimated} \cdot omin$.

What I don't understand is the formula for $omin$ (marked under (12) in the paper):

$$omin = \frac{1}{E\left\{ P_{min} \right\}_{|\sigma^2(k)=1}} $$

"It follows that the bias of the minimum subband power estimate is proportional to the noise power and that the bias can be compensated by multiplying the minimum estimate with the inverse of the mean."

How is the $omin$ calculated?


2 Answers 2


The overestimation factor, $omin$, is given by: $$omin = \frac{1}{E\{P_{min}\}_{|\sigma^2(k)=1}}$$

$E\{P_{min}\}$ is the expected value of the minimum power estimate. According to the paper, this parameter has the following probability density function: $$f_{P_{min}}(y) = D \cdot \left(1-F_{P_x}(y)\right)^{D-1} \cdot f_{P_x}(y)$$

Where $D$ is the number of independent power estimates, $F_{P_x}(y)$ is the subband power estimate cumulative distribution function, and $f_{P_x}(y)$ is the subband power estimate probability density function.

The expected value of the minimum power estimate can be then computed from the minimum power estimate probability density function: $$E\{P_{min}\} = \int_{0}^{\infty} y \cdot f_{P_{min}}(y) \,dy$$


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