Average Value of a Pulse Waveform

The average value of a pulse waveform can be determined using one of two methods. The first is the procedure outlined in Section 12.6, which can be applied to any alternating waveform. The second can be applied only to pulse waveforms since it utilizes terms specifically related to pulse waveforms; that is,
$$ V_{\text {av }}=( \text {duty cycle} )( \text {peak value} )+(1- \text {duty cycle} )\left(V_{b}\right) \tag{1} $$
In Eq. (1), the peak value is the maximum deviation from the reference or zero-volt level, and the duty cycle is in decimal form. Equation (1) does not include the effect of any tilt pulse waveforms with sloping sides.
Example 1: Determine the average value for the periodic pulse waveform of Fig. $ 1 $
Fig. 1: Example 1.
Solution: By the method of Section 12.6,
$$G=\frac{\text { area under curve }}{T} $$
$$T=(12-2) \mu \mathrm{s}=10 \mu \mathrm{s} $$
$$\begin{split} G&=\frac{(8 \mathrm{mV})(4 \mu \mathrm{s})+(2 \mathrm{mV})(6 \mu \mathrm{s})}{10 \mu \mathrm{s}}\\ &=\frac{32 \times 10^{-9}+12 \times 10^{-9}}{10 \times 10^{-6}} \\ &=\frac{44 \times 10^{-9}}{10 \times 10^{-6}}=\mathbf{4 . 4} \mathbf{~ m V}\end{split}$$
By Eq. (1),
$$V_{b}=+2 \mathrm{mV}$$
$$\text{Duty cycle} =\frac{t_{p}}{T}=\frac{(6-2) \mu \mathrm{s}}{10 \mu \mathrm{s}}=\frac{4}{10}=0.4 \text{(decimal form)}$$
$$\text{Peak value (from 0V reference)} =8 \mathrm{mV} $$
$$ \begin{split} V_{\mathrm{av}}&=( \text{duty cycle} )( \text{peak value} )+(1- \text{duty cycle} )\left(V_{b}\right) \\ &=(0.4)(8 \mathrm{mV})+(1-0.4)(2 \mathrm{mV}) =3.2 \mathrm{mV}+1.2 \mathrm{mV}=4.4 \mathrm{mV} \end{split}$$
as obtained above.