multiplication

The passage around this formula

An always-on system’s design target is not peak throughput but the average power drawn between the (usually rare) events that matter, which can be written as a simple duty-cycle model: $\bar{P} = D \cdot P_{\mathrm{active}} + (1-D) \cdot P_{\mathrm{standby}}$. where D is the fraction of time spent actively computing and $P_{\mathrm{standby}}$ is the power drawn the rest of the time. For a keyword detector or a vibration monitor, D is typically small and falling, so $\bar{P}$ is dominated by $P_{\mathrm{standby}}$ almost regardless of how fast the active phase runs — which is exactly why this category optimises the idle floor first and peak throughput second, the opposite priority from a dedicated accelerator sized for a known, recurring, higher-duty…

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