91Ó°ÊÓ

In electric signals, a harmonious is defined as the signal content at a specific frequence, which is a multiple integral of the current frequence system or main frequence produced by the creators. With an oscilloscope, it's possible to observe a complex signal in the sphere of time.

The given sketch of an electrical signal is

The value of harmonic content of signal is to be found

From the given sketch, the harmonic function can be defined as

$$\begin{gathered} V\left(t\right)=\left\{\stackrel{˙}{A}t,t∈\left[0,\frac{\mathrm{τ}}{2}\right] \\ A,t∈\left[\frac{\mathrm{τ}}{2},\mathrm{τ}\right]\right. \end{gathered}$$

where,$$\begin{gathered} \stackrel{Ë™}{A}=12000 \\ A=100 \\ \mathrm{Ï„}=\frac{1}{60} \end{gathered}$$

First calculate the values of$a_0,a_n,b_n$

$\frac{a_0}{2}$is the mean value of the function which can be calculated as$\frac{a_0}{2}=75$

Calculate the other coefficients as

role="math" localid="1664300050030" $$\begin{gathered} a_n=\frac{2}{\mathrm{Ï„}}\left[{∫}_0^{\raisebox{1ex}{$\mathrm{Ï„}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\stackrel{.}{A}t\cos \frac{2\mathrm{Ï€²Ô³Ù}}{\mathrm{Ï„}}dt+{∫}_{\raisebox{1ex}{$\mathrm{Ï„}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\mathrm{Ï„}}\stackrel{.}{A}t\sin \frac{2\mathrm{Ï€²Ô³Ù}}{\mathrm{Ï„}}dt\right] \\ =\frac{2}{\mathrm{Ï„}}\left[\stackrel{.}{A}{\left(\frac{\mathrm{Ï„}t}{2\mathrm{Ï€²Ô}}\sin \frac{2\mathrm{Ï€²Ô³Ù}}{\mathrm{Ï„}}+\frac{{\mathrm{Ï„}}^2}{4{\mathrm{Ï€}}^2{\mathrm{n}}^2}\cos \frac{2\mathrm{Ï€²Ô³Ù}}{\mathrm{Ï„}}\right)}_0^{\raisebox{1ex}{$\mathrm{Ï„}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{\text{´¡Ï„}}{\text{2Ï€²Ô}}\text{sin}\frac{\text{2Ï€²Ô³Ù}}{\mathrm{Ï„}}{|}_{\raisebox{1ex}{$\mathrm{Ï„}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\mathrm{Ï„}}\right] \\ =\frac{2}{\mathrm{Ï„}}×\frac{{\mathrm{Ï„}}^2}{4{\mathrm{Ï€}}^2{\mathrm{n}}^2}\stackrel{.}{A}\text{((}-{\text{1)}}^{\text{n}}-\text{1)} \\ =-\frac{\text{200}}{{\displaystyle \mathrm{Ï€}}{}^{\text{2}}{\text{n}}^{\text{2}}} \end{gathered}$$

Now calculate value of b_n

_{$$\begin{gathered} b_n=\frac{2}{\mathrm{Ï„}}[{∫}_0^{\raisebox{1ex}{$\mathrm{Ï„}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\stackrel{.}{A}t\sin \frac{2\mathrm{Ï€²Ô³Ù}}{\mathrm{Ï„}}dt+{∫}_{\raisebox{1ex}{$\mathrm{Ï„}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\mathrm{Ï„}}\stackrel{.}{A}t\sin \frac{2\mathrm{Ï€²Ô³Ù}}{\mathrm{Ï„}}dt] \\ =\frac{2}{\mathrm{Ï„}}[\stackrel{.}{A}{\left(\frac{\mathrm{Ï„}t}{2\mathrm{Ï€²Ô}}\cos \frac{2\mathrm{Ï€²Ô³Ù}}{\mathrm{Ï„}}+\frac{{\mathrm{Ï„}}^2}{4{\mathrm{Ï€}}^2{\mathrm{n}}^2}\sin \frac{2\mathrm{Ï€²Ô³Ù}}{\mathrm{Ï„}}\right)}_0^{\raisebox{1ex}{$\mathrm{Ï„}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{\text{´¡Ï„}}{\text{2Ï€²Ô}}\text{sin}\frac{\text{2Ï€²Ô³Ù}}{\mathrm{Ï„}}{|}_{\raisebox{1ex}{$\mathrm{Ï„}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\mathrm{Ï„}}] \\ =\frac{2}{\mathrm{Ï„}}\left[\stackrel{.}{A}\frac{{\mathrm{Ï„}}^2}{4{\mathrm{Ï€}}^2{\mathrm{n}}^2}\text{((}-{\text{1)}}^{\text{n}}-\frac{A\mathrm{Ï„}}{2\mathrm{Ï€²Ô}}(1-{(-1)}^n)\right] \\ =-\frac{\text{100}}{{\displaystyle \mathrm{Ï€}\text{n}}} \end{gathered}$$}

$$\begin{gathered} \text{V(t)}=\frac{{\mathrm{a}}_0}{2}\text{+}\underset{\text{n - odd}}{\overset{∞}{∑}}{\text{a}}_{\mathrm{n}}\frac{\cos \text{(}\frac{\text{2Ï€²Ô³Ù}}{\mathrm{Ï„}}\text{)}}{\text{n}}+\underset{\text{n - odd}}{\overset{∞}{∑}}{\text{b}}_{\text{n}}\frac{\text{sin(}\frac{\text{2Ï€²Ô³Ù}}{\mathrm{Ï„}}\text{)}}{\text{n}} \\ \text{=75-}\frac{200}{2}\text{+}\underset{\text{n - odd}}{\overset{∞}{∑}}{\text{a}}_{\mathrm{n}}\frac{\cos \text{(120nt)}}{{\text{n}}^2}\text{-}\frac{100}{\mathrm{Ï€}}\underset{\text{n=1}}{\overset{∞}{∑}}{\text{b}}_{\text{n}}\frac{\text{sin(120nt)}}{\text{n}} \end{gathered}$$

Therefore, the hormonic content of signal of the given sketch

is$\text{V(t)}=\text{75-}\frac{200}{{\mathrm{Ï€}}^2}\underset{\text{n - odd}}{\overset{∞}{∑}}\frac{\text{³¦´Ç²õ(120Ï€²Ô³Ù)}}{{\text{n}}^{\text{2}}}-\frac{\text{100}}{\mathrm{Ï€}}\underset{\text{n = 1}}{\overset{∞}{∑}}\frac{\text{²õ¾±²Ô(120Ï€²Ô³Ù)}}{\text{n}}$