Question 5.17: Find the average and rms values of the sinusoid and sawtooth......

As noted previously, the sinusoid has an average value of zero. The sawtooth clearly has a positive average value. By geometry, the net area under one cycle of the sawtooth waveform is V_{A}T_{0}/2, so its average value is (1/T_{0})(V_{A}T_{0}/2) = V_{A}/2. To obtain the rms value of the sinusoid we apply Eq. (5–34) as

V_{rms} =\sqrt{\frac{(V_{A})^2}{T_{0}}\int_{0}^{T_{0}}\sin^2(2\pi t/T_{0})dt}

=\sqrt{\frac{(V_{A})^2}{T_{0}}\left[\frac{t}{2}-\frac{\sin(4\pi t/T_{0})}{8 \pi /T_{0}}\right]_{0}^{T_{0}}}=\frac{V_{A}}{\sqrt{2}}

V_{rms} =\sqrt{\frac{1}{T} \int^{t+T}_{t} [v(t)]^2dt}         (5–34)

For the sawtooth the rms value is found as:

V_{rms} =\sqrt{\frac{1}{T_{0}}\int_{0}^{T_{0}}(V_{A}t/T_{0})^2dt}=\sqrt{\frac{(V_{A})^2}{T_{0}^3}\left[\frac{t^3}{3}\right]_{0}^{T_{0}}}=\frac{V_{A}}{\sqrt{3}}