Editing Power (physics) (section)

==Peak power and duty cycle==
[[File:peak-power-average-power-tau-T.png|thumb|right|In a train of identical pulses, the instantaneous power is a periodic function of time. The ratio of the pulse duration to the period is equal to the ratio of the average power to the peak power. It is also called the duty cycle (see text for definitions).]]

In the case of a periodic signal <math>s(t)</math> of period <math>T</math>, like a train of identical pulses, the instantaneous power <math display="inline">p(t) = |s(t)|^2</math> is also a periodic function of period <math>T</math>.  The ''peak power'' is simply defined by:
<math display="block">P_0 = \max [p(t)].</math>

The peak power is not always readily measurable, however, and the measurement of the average power <math>P_\mathrm{avg}</math> is more commonly performed by an instrument.  If one defines the energy per pulse as
<math display="block">\varepsilon_\mathrm{pulse} = \int_0^T p(t) \, dt </math>
then the average power is
<math display="block">P_\mathrm{avg} = \frac{1}{T} \int_0^T p(t) \, dt = \frac{\varepsilon_\mathrm{pulse}}{T}. </math>

One may define the pulse length <math>\tau</math> such that <math>P_0\tau = \varepsilon_\mathrm{pulse}</math> so that the ratios
<math display="block">\frac{P_\mathrm{avg}}{P_0} = \frac{\tau}{T} </math>
are equal. These ratios are called the ''duty cycle'' of the pulse train.