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

May 21, 2018

See below

#### Explanation:

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

Part (i)

• ${a}_{n} = \frac{1}{L} {\int}_{0}^{2 L} f \left(x\right) \cos \left(\frac{n \pi x}{L}\right) \mathrm{dx}$

$\implies {a}_{n} = \frac{5}{\pi} {\int}_{0}^{2 \pi} \setminus \sin \left(\frac{\theta}{2}\right) \cos \left(n \theta\right) \setminus d \theta$

Using:

$\frac{\sin \left(X + Y\right) + \sin \left(X - Y\right)}{2} = \sin X \cos Y$

$= \frac{5}{2 \pi} {\int}_{0}^{2 \pi} \setminus \sin \left(\frac{1}{2} + n\right) \theta + \sin \left(\frac{1}{2} - n\right) \theta \setminus d \theta$

$= \frac{5}{2 \pi} {\left[\setminus - \frac{2}{1 + 2 n} \cos \left(\frac{1}{2} + n\right) \theta - \frac{2}{1 - 2 n} \cos \left(\frac{1}{2} - n\right) \theta\right]}_{0}^{2 \pi}$

$= - \frac{5}{\pi} \left[\left(\setminus \frac{1}{1 + 2 n} \cos \left(\frac{1}{2} + n\right) 2 \pi + \frac{1}{1 - 2 n} \cos \left(\frac{1}{2} - n\right) 2 \pi\right) - \left(\frac{1}{1 + 2 n} + \frac{1}{1 - 2 n}\right)\right] = \square$

• $\cos \left(\frac{1}{2} - n\right) 2 \pi = - \cos 2 n \pi = \cos \left(\frac{1}{2} + n\right) 2 \pi$

• $- \cos 2 n \pi = - 1 , q \quad n \in m a t h \boldsymbol{Z}$

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

• $\frac{1}{1 + n} + \frac{1}{1 - n} = \frac{2}{1 - 4 {n}^{2}}$

$\square = \frac{20}{\pi} \setminus \frac{1}{1 - 4 {n}^{2}} = {a}_{n}$

• $\implies {a}_{0} = \frac{20}{\pi}$

Finally, from the definition:

$f \left(\theta\right) = {a}_{0} / 2 + {\sum}_{n = 1}^{\infty} {a}_{n} \cos n \theta$

$\implies f \left(\theta\right) = \frac{10}{\pi} \left(1 - \frac{2}{3} \cos \theta - \frac{2}{15} \cos 2 \theta - \frac{2}{35} \cos 3 \theta \ldots .\right)$

More formally:

$f \left(\theta\right) = \frac{10}{\pi} \left(1 + {\sum}_{n = 1}^{\infty} \setminus \frac{2}{1 - 4 {n}^{2}} \cos n \theta\right) q \quad \triangle$

Part (ii)

From $\triangle$:

$f \left(\pi\right) = 5 = \frac{10}{\pi} \left(1 - \frac{2}{3} \cos \left(\pi\right) - \frac{2}{15} \cos \left(2 \pi\right) - \frac{2}{35} \cos \left(3 \pi\right) \ldots\right)$

$\implies 5 \pi = 10 \left(1 + \frac{2}{3} - \frac{2}{15} + \frac{2}{35} \ldots\right)$

Or:

$\implies 5 \pi = 20 \left(\frac{1}{2} + \frac{1}{3} - \frac{1}{3 \left(5\right)} + \frac{1}{5 \left(7\right)} \ldots\right)$