Question

Determine the Fourier series expansion for full wave rectified sine wave i. (details inside)?

Answer 1

See below

Explanation:

Once rectified, it is even , so you only need the cosine series. Note it now has period `2L = 2 pi`:

Part (i)

• `a_n = 1/L int_0^(2L) f(x) cos ((n pi x)/L) dx`

`implies a_n = 5/pi int_0^(2 pi) \ sin (theta/2) cos (n theta) \ d theta`

Using:

`(sin ( X + Y) + sin (X - Y) )/2 = sin X cos Y`

`= 5/(2pi) int_0^(2 pi) \ sin (1/2+n) theta + sin (1/2 - n) theta \ d theta`

`= 5/(2pi) [ \ - 2/(1+2n) cos (1/2+n) theta - 2/(1-2n)cos (1/2 - n) theta ]_0^(2 pi)`

`=- 5/pi[ ( \ 1/(1+2n) cos (1/2+n) 2pi + 1/(1-2n)cos (1/2-n)2pi ) - (1/(1+2n) + 1/(1-2n)) ] = square`

• `cos (1/2 - n) 2pi = - cos 2 n pi = cos (1/2 + n) 2pi`

• `- cos 2 n pi = -1, qquad n in mathbb Z`

`square =- 5/pi ( \ (-2) ( 1/(1+2n) + 1/(1-2n))`

• `1/(1+n) + 1/(1-n) = 2/(1 - 4 n^2)`

`square = 20/pi \ 1/(1 - 4 n^2) = a_n`

• `implies a_0 = 20/pi`

Finally, from the definition:

`f(theta) = a_0/2 + sum_(n = 1)^(oo) a_n cos n theta`

`implies f(theta) = 10/ pi ( 1 - 2/3 cos theta - 2/15 cos 2 theta - 2/35 cos 3 theta ....)`

More formally:

`f(theta) = 10/pi (1 + sum_(n = 1)^(oo) \ 2/(1 - 4 n^2) cos n theta ) qquad triangle`

Part (ii)

From `triangle`:

`f(pi) = 5 = 10/ pi ( 1 - 2/3 cos (pi) - 2/15 cos (2 pi) - 2/35 cos (3 pi) ...)`

`implies 5pi = 10 ( 1 + 2/3 - 2/15 + 2/35 ...)`

Or:

`implies 5pi = 20 ( 1/2 + 1/3 - 1/(3(5)) + 1/(5(7)) ...)`