Question 23.13: Find the Fourier series representation of the function with ......
Find the Fourier series representation of the function with period T=\frac{1}{50} given by
f(t)= \begin{cases}1 & 0 \leqslant t<0.01 \\ 0 & 0.01 \leqslant t<0.02\end{cases}Step-by-Step
The function f (t) is shown in Figure 23.14. Note that f (t) is defined to be zero between t = 0.01 and t = 0.02. This means we need only consider 0 ≤ t < 0.01. Using Equations (23.3)-(23.5) we find
\begin{aligned} & a_0=100 \int_0^{0.02} f(t) \mathrm{d} t=100 \int_0^{0.01} 1 \mathrm{~d} t+100 \int_{0.01}^{0.02} 0 \mathrm{~d} t \\ & =100[t]_0^{0.01}=1 \\ & \begin{aligned} a_n=100 \int_0^{0.01} \cos 100 n \pi t \mathrm{~d} t & =100\left[\frac{\sin 100 n \pi t}{100 n \pi}\right]_0^{0.01} \\ & =0 \\ b_n=100 \int_0^{0.01} \sin 100 n \pi t \mathrm{~d} t & =100\left[\frac{-\cos 100 n \pi t}{100 n \pi}\right]_0^{0.01} \\ & =-\frac{1}{n \pi}(\cos n \pi-\cos 0) \end{aligned} \end{aligned}Noting that \cos n \pi=(-1)^n we find
b_n=\frac{1}{n \pi}\left(1-(-1)^n\right)If n is even b_n=0 . \text { If } n \text { is odd } b_n=\frac{2}{n \pi}. Therefore the Fourier series representation of f (t) is
f(t)=\frac{1}{2}+\frac{2}{\pi}\left(\sin 100 \pi t+\frac{\sin 300 \pi t}{3}+\frac{\sin 500 \pi t}{5}+\cdots\right)The average value of the waveform is \frac{1}{2} This is the zero frequency component or d.c. value. We note that in this example only odd harmonics are present.
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