91Ó°ÊÓ

The given curve is the absolute value of a sine function

$\text{V}\left(\text{t}\right)\text{=}\left|\text{Asin}\left(\frac{\text{2Ï€³Ù}}{\text{Ï„}}\right)\right|$, where $\text{Ï„=}\frac{\text{1}}{\text{60}}$and A=100

A Fourier series is that a sum that be a periodic function as a sum of sine and cosine waves. It can be written as

$\text{f}\left(\text{x}\right)\text{=}\frac{{\text{a}}_{\text{0}}}{\text{2}}\text{+}\underset{\text{n=1}}{\overset{\mathrm{∞}}{∑}}\left[{\text{a}}_{\text{n}}\text{cos}\frac{\text{²ÔÏ€³æ}}{\text{L}}{\text{+b}}_{\text{n}}\text{sin}\frac{\text{²ÔÏ€³æ}}{\text{L}}\right]$

The corresponding Fourier coefficients are

$\begin{array}{l}{\text{a}}_{\text{0}}\text{=}\frac{\text{1}}{\text{L}}\underset{\text{2L}}{∫}\text{f}\left(\text{x}\right)\text{dx}\\ {\text{a}}_{\text{n}}\text{=}\frac{\text{1}}{\text{L}}\underset{\text{2L}}{∫}\text{f}\left(\text{x}\right)\text{cos}\frac{\text{²ÔÏ€³æ}}{\text{L}}\text{dx}\\ {\text{b}}_{\text{n}}\text{=}\frac{\text{1}}{\text{L}}\underset{\text{2L}}{∫}\text{f}\left(\text{x}\right)\text{sin}\frac{\text{²ÔÏ€³æ}}{\text{L}}\text{dx}\end{array}$

The coefficients are equal to

$\begin{array}{l}{\text{a}}_{\text{0}}\text{=2}\frac{\text{2}}{\text{Ï„}}\underset{\text{0}}{\overset{\raisebox{1ex}{$\text{Ï„}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}{∫}}\text{Asin}\frac{\text{2Ï€³Ù}}{\text{Ï„}}\text{dt}\\ \text{=-}\frac{\text{4A}}{\text{Ï„}}{\left[\frac{\text{Ï„}}{\text{2Ï€}}\text{cos}\frac{\text{2Ï€³Ù}}{\text{Ï„}}\right]}_{\text{0}}^{\raisebox{1ex}{$\text{Ï„}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\\ \text{=-}\frac{\text{2A}}{\text{Ï€}}\left(\left(\text{-1}\right)\text{-1}\right)\\ \text{=}\frac{\text{4A}}{\text{Ï€}}\end{array}$

Solve for the value of a_n

$\begin{array}{l}{\text{a}}_{\text{n}}\text{=}\frac{\text{2}}{\text{Ï„}}\left[\underset{\text{0}}{\overset{\raisebox{1ex}{$\text{Ï„}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}{∫}}\text{Asin}\frac{\text{2Ï€³Ù}}{\text{Ï„}}\text{cos}\frac{\text{2Ï€²Ô³Ù}}{\text{Ï„}}\text{dt-}\underset{\raisebox{1ex}{$\text{Ï„}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}{\overset{\text{Ï„}}{∫}}\text{Asin}\frac{\text{2Ï€³Ù}}{\text{Ï„}}\text{cos}\frac{\text{2Ï€²Ô³Ù}}{\text{Ï„}}\text{dt}\right]\\ \text{=}\frac{\text{2}}{\text{Ï„}}\left[\frac{\text{A}}{\text{2}}\underset{\text{0}}{\overset{\raisebox{1ex}{$\text{Ï„}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}{∫}}\left(\text{sin}\frac{\text{2Ï€³Ù}\left(\text{n+1}\right)}{\text{Ï„}}\text{+sin}\frac{\text{2Ï€³Ù}\left(\text{1-n}\right)}{\text{Ï„}}\right)\text{dt-}\frac{\text{A}}{\text{2}}\underset{\raisebox{1ex}{$\text{Ï„}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}{\overset{\text{Ï„}}{∫}}\left(\text{sin}\frac{\text{2Ï€³Ù}\left(\text{n+1}\right)}{\text{Ï„}}\text{+sin}\frac{\text{2Ï€³Ù}\left(\text{1-n}\right)}{\text{Ï„}}\right)\text{dt}\right]\\ \text{=}\frac{\text{A}}{\text{Ï„}}{\left[\text{-}\frac{\text{Ï„}}{\text{2Ï€}\left(\text{n+1}\right)}\text{cos}\frac{\text{2Ï€³Ù}\left(\text{n+1}\right)}{\text{Ï„}}\text{-}\frac{\text{Ï„}}{\text{2Ï€}\left(\text{1-n}\right)}\text{cos}\frac{\text{2Ï€³Ù}\left(\text{1-n}\right)}{\text{Ï„}}\right]}_{\text{0}}^{\raisebox{1ex}{$\text{Ï„}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\\ \text{-}\frac{\text{A}}{\text{Ï„}}{\left[\text{-}\frac{\text{Ï„}}{\text{2Ï€}\left(\text{n+1}\right)}\text{cos}\frac{\text{2Ï€³Ù}\left(\text{n+1}\right)}{\text{Ï„}}\text{-}\frac{\text{Ï„}}{\text{2Ï€}\left(\text{1-n}\right)}\text{cos}\frac{\text{2Ï€³Ù}\left(\text{1-n}\right)}{\text{Ï„}}\right]}_{\raisebox{1ex}{$\text{Ï„}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}^{\text{Ï„}}\end{array}$

Solving, further

$\begin{array}{l}\text{=}\frac{\text{A}}{\text{2Ï€}}\left[\frac{\text{1}}{\text{n+1}}\left(\text{1-}{\left(\text{-1}\right)}^{\text{1+n}}\text{-}{\left(\text{-1}\right)}^{\text{1+n}}\text{+1}\right)\text{+}\frac{\text{1}}{\text{1-n}}\left(\text{1-}{\left(\text{-1}\right)}^{\text{1-n}}\text{-}{\left(\text{-1}\right)}^{\text{1-n}}\text{+1}\right)\right]\\ \text{=}\frac{\text{1+}{\left(\text{-1}\right)}^{\text{n}}}{\text{Ï€}}\text{A}\left[\frac{\text{1}}{\text{1+n}}\text{+}\frac{\text{1}}{\text{1-n}}\right]\\ \text{=}\frac{\text{4A}}{\text{Ï€}}\frac{\text{1}}{{\text{1-n}}^{\text{2}}}\end{array}$

Where$\text{n=2k,k}∈\mathrm{ℕ}$

Solve for the value of

$\begin{array}{l}{\text{b}}_{\text{n}}\text{=}\frac{\text{2}}{\text{Ï„}}\left[\underset{\text{0}}{\overset{\raisebox{1ex}{$\text{Ï„}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}{∫}}\text{Asin}\frac{\text{2Ï€³Ù}}{\text{Ï„}}\text{sin}\frac{\text{2Ï€²Ô³Ù}}{\text{Ï„}}\text{dt-}\underset{\raisebox{1ex}{$\text{Ï„}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}{\overset{\text{Ï„}}{∫}}\text{Asin}\frac{\text{2Ï€³Ù}}{\text{Ï„}}\text{sin}\frac{\text{2Ï€²Ô³Ù}}{\text{Ï„}}\text{dt}\right]\\ \text{=}\frac{\text{A}}{\text{Ï„}}\left[\underset{\text{0}}{\overset{\raisebox{1ex}{$\text{Ï„}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}{∫}}\left(\text{cos}\frac{\text{2Ï€³Ù}\left(\text{1-n}\right)}{\text{Ï„}}\text{-cos}\frac{\text{2Ï€³Ù}\left(\text{1+n}\right)}{\text{Ï„}}\right)\text{dt-}\underset{\raisebox{1ex}{$\text{Ï„}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}{\overset{\text{Ï„}}{∫}}\left(\text{cos}\frac{\text{2Ï€³Ù}\left(\text{1-n}\right)}{\text{Ï„}}\text{-cos}\frac{\text{2Ï€³Ù}\left(\text{1+n}\right)}{\text{Ï„}}\right)\text{dt}\right]\\ \text{=}\frac{\text{A}}{\text{Ï„}}{\left[\frac{\text{Ï„}}{\text{2Ï€}\left(\text{1-n}\right)}\text{sin}\frac{\text{2Ï€³Ù}\left(\text{1-n}\right)}{\text{Ï„}}\text{-}\frac{\text{Ï„}}{\text{2Ï€}\left(\text{1+n}\right)}\text{sin}\frac{\text{2Ï€³Ù}\left(\text{1+n}\right)}{\text{Ï„}}\right]}_{\text{0}}^{\raisebox{1ex}{$\text{Ï„}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\\ \text{-}\frac{\text{A}}{\text{Ï„}}{\left[\frac{\text{Ï„}}{\text{2Ï€}\left(\text{1-n}\right)}\text{sin}\frac{\text{2Ï€³Ù}\left(\text{n+1}\right)}{\text{Ï„}}\text{-}\frac{\text{Ï„}}{\text{2Ï€}\left(\text{1+n}\right)}\text{sin}\frac{\text{2Ï€³Ù}\left(\text{1+n}\right)}{\text{Ï„}}\right]}_{\raisebox{1ex}{$\text{Ï„}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}^{\text{Ï„}}\end{array}$

Solving, further

$\begin{array}{l}\text{=}\frac{\text{A}}{\text{2π}}\left[\frac{\text{sin}\left(\text{π}\left(\text{1-n}\right)\right)}{\text{1-n}}\text{-}\frac{\text{sin}\left(\text{2π}\left(\text{1-n}\right)\right)}{\text{1-n}}\text{+}\frac{\text{sin}\left(\text{π}\left(\text{1-n}\right)\right)}{\text{1-n}}\right]\\ \text{=}\frac{\text{A}}{\text{2π}}\left({\text{πδ}}_{\text{1n}}{\text{-2πδ}}_{\text{1n}}{\text{+πδ}}_{\text{1n}}\right)\\ \text{=0}\end{array}$

Thus the voltage function is equal to

$\begin{array}{l}\text{V}\left(\text{t}\right)\text{=}\frac{\text{2A}}{\text{Ï€}}\text{+}\frac{\text{4A}}{\text{Ï€}}\underset{\text{neven}}{\overset{\mathrm{∞}}{∑}}\frac{\text{1}}{{\text{1-n}}^{\text{2}}}\text{cos}\frac{\text{2Ï€²Ô³Ù}}{\text{Ï„}}\\ \text{=}\frac{\text{200}}{\text{Ï€}}\left[\text{1+2}\underset{\text{neven}}{\overset{\mathrm{∞}}{∑}}\frac{\text{cos}\left(\text{120Ï€²Ô³Ù}\right)}{{\text{1-n}}^{\text{2}}}\right]\end{array}$