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

Step-by-step explanation:

AVERAGE: the average value is given as

`(1)/(0.1) \int\limits^(1)/(10) _0 {25+10sint} \, dt = (1)/(0.1) [ 25t- 10cos\ t]_0^(0.1)`

=
`(1)/(0.1) ([2.5-10]-10)=-175`

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

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

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

=
`3[[(625)/(3) - 500 + 10000((1)/(6) - 0.002908)] + 500] = 2845.92\\`

therefore, RMS =
`√(2845.92) = 53.3`