Question

Determine the average and rms values for the function, y(t)=25+10sino it over the time periods (a) 0 to 0.1 sec and (b) 0 to 1/3 sec. Discuss which case represents the long-term behavior of the signal (Hint, consider the period of the signal).

Answer 1

Answer:

Explanation:

AVERAGE: the average value is given as

$\frac{1}{0.1} \int\limits^\frac{1}{10} _0 {25+10sint} \, dt = \frac{1}{0.1} [ 25t- 10cos\ t]_0^{0.1}$

=$\frac{1}{0.1} ([2.5-10]-10)=-175$

RMS= $\sqrt{\int\limits^\frac{1}{3} _0 {y(t)^2} \, dt }$

$y(t)^2 = (25 + 10sin \ t)^2 = 625 +500sin \ t + 10000sin^2 \ t$

$\frac{1}{\frac{1}{3} } \int\limits^\frac{1}{3} _0 {y(t)^2} \, dt =3[ 625t -500cos \ t + 10000(\frac{t}{2} - \frac{sin2t}{4} )]_0^{\frac{1}{3} }$

=$3[[\frac{625}{3} - 500 + 10000(\frac{1}{6} - 0.002908)] + 500] = 2845.92$

therefore, RMS = $\sqrt{2845.92} = 53.3$