Question

Find (a) the average and (b) rms value for the Saw-tooth voltage of peak value $V_0$ as shown in the figure.

Answer 1

As the equation of the Saw-tooth wave shown in the figure will be
$V=\frac{2V_0}{T}t-V_0=V_0(\frac{2ty}{T}-1)$
So (a) $V_{av}=\frac{{\int }_0^{T/2}Vdt}{{\int }_0^{T/2}dt}=\frac{2}{T}{\int }_0^{T/2}{V}_0(\frac{2ty}{T}-1)dt=\left|\frac{V_0}{2}\right|$
and (b) $V_{rms}={\left[\frac{{\int }_0^{T/2}Vdt}{{\int }_0^{T/2}dt}\right]}^{1/2}=\frac{V_0}{\sqrt{3}}$