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Naemahussein
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Tyro
Asked: 2021-03-27T19:23:12+00:00 In: ,

Fourier Analysis

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What is the Fourier analysis of the shown curve

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  2. acedstud
    This answer was edited.

    The given periodic function can be rewritten as

    f(t)=\begin{cases} 1 & \text{ if } 0<t<0.2 \\ 3 & \text{ if } 0.2<t<0.4 \end{cases}
    f(t+0.4)=f(t)

    The Fourier series expansion of a function f(t) with period, T, is given by f(t)=\dfrac{a_0}{2}+\sum_{n=1}^{\infty}{\left\{a_n\cos{\dfrac{2n\pi t}{T}}+b_n\sin{\dfrac{2n\pi t}{T}}\right\}}, where a_0=\dfrac{2}{T}\int_{0}^{T}f(t)dt, a_n=\dfrac{2}{T}\int_{0}^{T}f(t)\cos{\dfrac{2n\pi t}{T}}dt, and b_n=\dfrac{2}{T}\int_{0}^{T}f(t)\sin{\dfrac{2n\pi t}{T}}dt

    Here, T=0.4, so we have: f(t)=\dfrac{a_0}{2}+\sum_{n=1}^{\infty}{\{a_n\cos{5n\pi t}+b_n\sin{5n\pi t}\}}.

    Where a_0=\dfrac{2}{T}\int_{0}^{T}f(t)dt=5\int_{0}^{0.4}f(t)dt

    =5\left(\int_{0}^{0.2}dt+\int_{0.2}^{0.4}3dt\right)=5\left([t]_0^{0.2}+[3t]_{0.2}^{0.4}\right)

    =5(0.2+0.6)=4.

    a_n=\dfrac{2}{T}\int_{0}^{T}f(t)\cos{\dfrac{2n\pi t}{T}}dt=5\int_{0}^{0.4}f(t)\cos{5n\pi t}dt

    =5\left(\int_{0}^{0.2}\cos{5n\pi t}dt+\int_{0.2}^{0.4}3\cos{5n\pi t}dt\right)

    =5\left(\left[\dfrac{\sin{5n\pi t}}{5n\pi}\right]_0^{0.2}+3\left[\dfrac{\sin{5n\pi t}}{5n\pi}\right]_{0.2}^{0.4}\right)

    =0        [\because \sin{n\pi}=0].

    b_n=\dfrac{2}{T}\int_{0}^{T}f(t)\sin{\dfrac{2n\pi t}{T}}dt=5\int_{0}^{0.4}f(t)\sin{5n\pi t}dt

    =5\left(\int_{0}^{0.2}\sin{5n\pi t}dt+\int_{0.2}^{0.4}3\sin{5n\pi t}dt\right)

    =5\left(-\left[\dfrac{\cos{5n\pi t}}{5n\pi}\right]_0^{0.2}-3\left[\dfrac{\cos{5n\pi t}}{5n\pi}\right]_{0.2}^{0.4}\right)

    =-\dfrac{\cos n\pi}{n\pi}+\dfrac{1}{n\pi}-\dfrac{3\cos2n\pi}{n\pi}+\dfrac{3\cos n\pi}{n\pi}

    =\dfrac{2\cos n\pi}{n\pi}-\dfrac{2}{n\pi}

    =0 (when n is even), and \dfrac{-4}{n\pi} (when n is odd).

    Therefore, the Fourier series is given by f(t)=2-\dfrac{4}{\pi}\left\{\sin5\pi t+\dfrac{1}{3}\sin15\pi t+\dfrac{1}{5}\sin25\pi t+... \right \}.

     

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