The half cycle of an alternating singnal is shown in fig. It increases uniformly form zero at `0^(@) to F_(m) at (alpha)^(@)` and decrease uniformly from `(F_m) at 180^(@)`

The average value of the singnal is

Options

A. `((pi+alpha)F_(m))/(pi)`
B. `((pi-alpha)F_(m))/(pi)`
C. `((pi+alpha)/(3 pi))F_(m)`
D. `((2pi+alpha)/(2 pi))F_(m)`

Answer

B

Explanation

For finding the average value m, we would find the total area of the trapezium and divide it by `(pi)`.
Area`=2 xx Delta OAE + rectangle ABDE`
`=2xx(1//2) xx F_(m0alpha +(pi - 2 alpha) F_(m0=(pi-alpha)F_(m)`
Average value `=(pi-alpha) F_(m)//(pi)`
rms value: From similar triangles , we get
`(y)/(thata) =(F_m)/(alpha)` or `(y^2)=(F_(m)^(2))/(alpha^(2)) thata^(2)`
This gives the equation of the single over the two triangles OAE and DBC. The single reamins constant over the angle `alpha` to `(pi-alpha)`, i.e., over an angluar distance of `(pi - alpha)-alpha = (pi-2 alpha)`.
Sum of the square
`(2 f_(m)^(2))/(alpha^2) F_(m)^(2)(pi -(4 alpha)/(3))`
The mean value of the squares
`=(1)/(pi) F_(m)^(2)(pi -(4 alpha)/(3))= F_(m)^(2)(1 -(4 alpha)/(3pi))`
rms value `=F_(m) sqrt((1 -(4 alpha)/(3pi)))`.