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Tutorial 10: Fourier Series

Tutorial Question

  1. Find the fundamental period of the following function:

    a. cos⁡2x\cos 2 x

    b. sin⁡2πx\sin 2 \pi x

    c. cos⁡nx\cos n x

    d. sin⁡2πnxk\sin \frac{2 \pi n x}{k}

    Solution

    a. 2xp=2π⇒xp=2π2=π2x_p=2\pi\quad\Rightarrow\quad x_p=\frac{2\pi}{2}=\pi

    b. 2πxp=2π⇒xp=2π2π=12\pi x_p=2\pi\quad\Rightarrow\quad x_p=\frac{2\pi}{2\pi}=1

    c. nxp=2π⇒xp=2πnnx_p=2\pi\quad\Rightarrow\quad x_p=\frac{2\pi}{n}

    d. 2πnxk=2π⇒x=kn\frac{2\pi n x}{k}=2\pi\quad\Rightarrow\quad x=\frac{k}{n}

  2. Sketch the given function and find the Fourier Series.

    a. f(x)={−1;−π<x<02;0<x<π;p=2πf(x)=\left\{\begin{array}{rrr}-1 & ; & -\pi<x<0 \\ 2 & ; & 0<x<\pi\end{array} \quad ; p=2 \pi\right.

    b. f(x)={0−2<x<−1−2−1≤x<010≤x<101≤x<2;p=4f(x)=\left\{\begin{array}{lc}0 & -2<x<-1 \\ -2 & -1 \leq x<0 \\ 1 & 0 \leq x<1 \\ 0 & 1 \leq x<2\end{array} \quad ; p=4\right.

    c. f(x)={1;−1<x<0x0<x<1;p=2f(x)=\left\{\begin{array}{lr}1 & ;-1<x<0 \\ x & 0<x<1\end{array}\quad;p=2\right.

    Solution

    a.

    a0=1π∫−ππf(x)dx=1π∫−π0(−1)dx+1π∫0π2dx=1an=1π∫−ππf(x)cos⁡nxdx=1π∫−π0−cos⁡nxdx+1π∫0π2cos⁡nxdx=0bn=1π∫−ππf(x)sin⁡nxdx=1π∫−π0−sin⁡nxdx+1π∫0π2sin⁡nxdx=3nπ[1−(−1)n]f(x)=12+3π∑n=1∞1−(−1)nnsin⁡nx\begin{aligned} a_{0}&=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) d x=\frac{1}{\pi} \int_{-\pi}^{0}(-1) d x+\frac{1}{\pi} \int_{0}^{\pi} 2 d x=1 \\ a_{n}&=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos n x d x=\frac{1}{\pi} \int_{-\pi}^{0}-\cos n x d x+\frac{1}{\pi} \int_{0}^{\pi} 2 \cos n x d x=0 \\ b_{n}&=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin n x d x=\frac{1}{\pi} \int_{-\pi}^{0}-\sin n x d x+\frac{1}{\pi} \int_{0}^{\pi} 2 \sin n x d x\\&=\frac{3}{n \pi}\left[1-(-1)^{n}\right] \\\\ &f(x)=\frac{1}{2}+\frac{3}{\pi} \sum_{n=1}^{\infty} \frac{1-(-1)^{n}}{n} \sin n x \end{aligned}

    b.

    a0=12∫−22f(x)dx=12(∫−10−2dx+∫011dx)=−12an=12∫−22f(x)cos⁡nπ2xdx=12(∫−10(−2)cos⁡nπ2xdx+∫01cos⁡nπ2xdx)=−1nπsin⁡nπ2bn=12∫−22f(x)sin⁡nπ2xdx=12(∫−10(−2)sin⁡nπ2xdx+∫01sin⁡nπ2xdx)=3nπ(1−cos⁡nπ2)f(x)=−14+∑n=1∞[−1nπsin⁡nπ2cos⁡nπ2x+3nπ(1−cos⁡nπ2)sin⁡nπ2x]\begin{aligned} a_{0}&=\frac{1}{2} \int_{-2}^{2} f(x) d x=\frac{1}{2}\left(\int_{-1}^{0}-2 d x+\int_{0}^{1} 1 d x\right)=-\frac{1}{2} \\ a_{n}&=\frac{1}{2} \int_{-2}^{2} f(x) \cos \frac{n \pi}{2} x d x\\&=\frac{1}{2}\left(\int_{-1}^{0}(-2) \cos \frac{n \pi}{2} x d x+\int_{0}^{1} \cos \frac{n \pi}{2} x d x\right)=-\frac{1}{n \pi} \sin \frac{n \pi}{2} \\ b_{n}&=\frac{1}{2} \int_{-2}^{2} f(x) \sin \frac{n \pi}{2} x d x\\&=\frac{1}{2}\left(\int_{-1}^{0}(-2) \sin \frac{n \pi}{2} x d x+\int_{0}^{1} \sin \frac{n \pi}{2} x d x\right)=\frac{3}{n \pi}\left(1-\cos \frac{n \pi}{2}\right) \\ &f(x)=-\frac{1}{4}+\sum_{n=1}^{\infty}\left[-\frac{1}{n \pi} \sin \frac{n \pi}{2} \cos \frac{n \pi}{2} x+\frac{3}{n \pi}\left(1-\cos \frac{n \pi}{2}\right) \sin \frac{n \pi}{2} x\right] \end{aligned}

    c.

    a0=∫−11f(x)dx=∫−101dx+∫01xdx=32an=∫−11f(x)cos⁡nπxdx=∫−10cos⁡nπxdx+∫01xcos⁡nπxdx=1n2π2[(−1)n−1]bn=∫−11f(x)sin⁡nπxdx=∫−10sin⁡nπxdx+∫01xsin⁡nπxdx=−1nπf(x)=34+∑n=1∞[(−1)n−1n2π2cos⁡nπx−1nπsin⁡nπx]\begin{aligned} a_{0}&=\int_{-1}^{1} f(x) d x=\int_{-1}^{0} 1 d x+\int_{0}^{1} x d x=\frac{3}{2} \\ a_{n}&=\int_{-1}^{1} f(x) \cos n \pi x d x=\int_{-1}^{0} \cos n \pi x d x+\int_{0}^{1} x \cos n \pi x d x\\&=\frac{1}{n^{2} \pi^{2}}\left[(-1)^{n}-1\right] \\ b_{n}&=\int_{-1}^{1} f(x) \sin n \pi x d x=\int_{-1}^{0} \sin n \pi x d x+\int_{0}^{1} x \sin n \pi x d x=-\frac{1}{n \pi} \\ &f(x)=\frac{3}{4}+\sum_{n=1}^{\infty}\left[\frac{(-1)^{n}-1}{n^{2} \pi^{2}} \cos n \pi x-\frac{1}{n \pi} \sin n \pi x\right] \end{aligned}

  3. Even and Odd Functions. Sketch the given functions, f(x)f(x). Determine whether it is an even, odd or neither odd nor even. For part (a) and (b), find the appropriate Fourier Cosine or Fourier Sine Series.

    a. f(x)=∣x∣;−π<x<π;f(x)=f(x+2πn),n=integerf(x)=|x| ;-\pi<x<\pi ; f(x)=f(x+2 \pi n), n=\text{integer}

    b. f(x)=x;−1<x<1;f(x)=f(x+2n),n=integerf(x)=x ;-1<x<1 ; f(x)=f(x+2 n), n=\text{integer}

    c. f(x)=x2;−1<x<1;f(x)=f(x+2n),n=integerf(x)=x^{2} ;-1<x<1 ; f(x)=f(x+2 n), n=\text{integer}

    (Do not find the Fourier Series for 3c)

    d. f(x)=ex;−π<x<π;f(x)=f(x+2πn),n=integerf(x)=e^{x} ;-\pi<x<\pi ; f(x)=f(x+2 \pi n), n=\text{integer}

    (Do not find the Fourier Series for 3d)

    Solution

    a.

    Since f(x)f(x) is an even function, we expand in a cosine series:

    a0=2π∫0πxdx=πan=2π∫0πxcos⁡nxdx=2n2π[(−1)n−1].\begin{aligned} &a_{0}=\frac{2}{\pi} \int_{0}^{\pi} x d x=\pi \\ &a_{n}=\frac{2}{\pi} \int_{0}^{\pi} x \cos n x d x=\frac{2}{n^{2} \pi}\left[(-1)^{n}-1\right] . \end{aligned}

    Thus,

    f(x)=π2+∑n=1∞2n2π[(−1)n−1]cos⁡nxf(x)=\frac{\pi}{2}+\sum_{n=1}^{\infty} \frac{2}{n^{2} \pi}\left[(-1)^{n}-1\right] \cos n x

    b.

    It is an odd function → a0=an=0a_0=a_n=0

    bn=2L∫0Lf(x)sin⁡nπxLdx=2∫01xsin⁡nπxdx=2[−xnπcos⁡nπx+∫1nπcos⁡nπx]01=2[−xnπcos⁡nπx∣01+1(nπ)2sin⁡nπx∣01]=2[−1nπcos⁡nπ+0−0]=−2nπcos⁡nπ=−2nπ(−1)n\begin{aligned}b_n&=\frac{2}{L}\int_0^Lf(x)\sin\frac{n\pi x}{L}dx\\&=2\int_0^1x\sin n\pi xdx\\&=2\left[-\frac{x}{n\pi}\cos n\pi x+\int \frac{1}{n\pi}\cos n\pi x\right]_0^1\\&=2\left[\left.-\frac{x}{n\pi}\cos n\pi x\right|_0^1+\left.\frac{1}{(n\pi)^2}\sin n\pi x\right|_0^1\right]\\&=2\left[-\frac{1}{n\pi}\cos n\pi+0-0\right]\\&=-\frac{2}{n\pi}\cos n\pi\\&=-\frac{2}{n\pi}(-1)^n\end{aligned}f(x)=∑n=1∞bnsin⁡nπxL=2π[sin⁡πx−12sin⁡2πx+13sin⁡3πx−14sin⁡4πx+...]\begin{aligned}f(x)&=\sum_{n=1}^{\infty}b_n\sin\frac{n\pi x}{L}\\&=\frac{2}{\pi}\left[\sin\pi x-\frac{1}{2}\sin2\pi x+\frac{1}{3}\sin3\pi x-\frac{1}{4}\sin4\pi x+...\right]\end{aligned}

    c.

    Even function

    d.

    Neither odd or even

  4. Solve the following questions:

    a. Obtain the Fourier series for a periodic function f(t)f(t) with period 2π2 \pi :

    f(t)=t2,−π<t<πf(t)=t^{2}, \quad-\pi<t<\pi

    b. Obtain the Fourier series for a periodic function f(t)f(t) with period 2π2 \pi :

    f(t)=t,−π<t<πf(t)=t, \quad-\pi<t<\pi

    c. Differentiate the Fourier series in (a) to obtain f′(t)f^{\prime}(t)

    d. Find the Fourier series of f(t)=tf(t)=t by using result in part (c) and compare it with (b).

    Solution

    a. Since f(t)=t2f(t)=t^2 is an even function, bn=0b_n=0

    a0=1π∫0πt2dt=π33×1π=π23an=2π∫0πt2cos⁡ntdt=2π{(1nsin⁡nt(t2))∣0π−1n∫0π2tsin⁡ntdt}=2π(−2n){[−1ncos⁡nt(t)]0π+1n∫0πcos⁡ntdt}=−4nπ×(−1n)(πcos⁡nπ−0)=4n2cos⁡nπ=4n2(−1)n∴f(t)=t2=π23+4∑n=1∞(−1)nn2cos⁡nt\begin{aligned}a_0&=\frac{1}{\pi}\int_0^\pi t^2dt=\frac{\pi^3}{3}\times\frac{1}{\pi}=\frac{\pi^2}{3}\\a_n&=\frac{2}{\pi}\int_0^\pi t^2\cos ntdt\\&=\frac{2}{\pi}\left\{\left.\left(\frac{1}{n}\sin nt(t^2)\right)\right|_0^\pi-\frac{1}{n}\int_0^\pi2t\sin ntdt\right\}\\&=\frac{2}{\pi}\left(-\frac{2}{n}\right)\left\{\left[-\frac{1}{n}\cos nt(t)\right]_0^\pi+\frac{1}{n}\int_0^\pi\cos ntdt\right\}\\&=-\frac{4}{n\pi}\times\left(-\frac{1}{n}\right)\left(\pi\cos n\pi-0\right)\\&=\frac{4}{n^2}\cos n\pi=\frac{4}{n^2}(-1)^n\\\\\therefore f(t)&=t^2=\frac{\pi^2}{3}+4\sum_{n=1}^\infty \frac{(-1)^n}{n^2}\cos nt\end{aligned}

    b. a0=an=0a_0=a_n=0 since f(t)=tf(t)=t. Thus, it is an odd function.

    bn=2π∫0πtsin⁡ntdt=2π{[−1ncos⁡nt(t)]0π+1n∫0πcos⁡ntdt}=2π{−πcos⁡nπ+1n[1nsin⁡nt]0π}=2π{−πncos⁡nπ}=−2n(−1)n=2n(−1)n+1∴f(t)=t=2∑n=1∞(−1)n+1nsin⁡nt\begin{aligned}b_n&=\frac{2}{\pi}\int_0^\pi t\sin ntdt\\&=\frac{2}{\pi}\left\{\left[-\frac{1}{n}\cos nt(t)\right]_0^\pi+\frac{1}{n}\int_0^\pi\cos nt dt\right\}\\&=\frac{2}{\pi}\left\{-\pi\cos n\pi+\frac{1}{n}\left[\frac{1}{n}\sin nt\right]_0^\pi\right\}\\&=\frac{2}{\pi}\left\{-\frac{\pi}{n}\cos n\pi\right\}\\&=-\frac{2}{n}(-1)^n=\frac{2}{n}(-1)^{n+1}\\\\\therefore f(t)&=t=2\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}\sin nt\end{aligned}

    c.

    f′(t)=ddt(t2)=ddt(π23+4∑n=1∞(−1)nn2cos⁡nt)f′(t)=2t=−4∑n=1∞(−1)nnsin⁡(nt)\begin{alignat*}{3}f'(t)&=\frac{d}{dt}(t^2)&&=\frac{d}{dt}\left(\frac{\pi^2}{3}+4\sum_{n=1}^\infty \frac{(-1)^n}{n^2}\cos nt\right)\\f'(t)&=2t&&=-4\sum_{n=1}^\infty\frac{(-1)^n}{n}\sin(nt)\end{alignat*}

    d.

    f(t)=t=−2∑n=1∞(−1)nnsin⁡(nt)or2∑n=1∞(−1)n+1nsin⁡(nt)f(t)=t=-2\sum_{n=1}^\infty\frac{(-1)^n}{n}\sin(nt)\quad\text{or}\quad2\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\sin(nt)

    The answer obtained by differentiation is faster and the answer is the same as 4(b).

  5. Consider the following ODE which represents an undamped mass-spring system:

    116d2xdt2+4x=f(t)\frac{1}{16} \frac{d^{2} x}{d t^{2}}+4 x=f(t)

    where f(t)f(t) is a periodic function as shown in Fig. 1. Obtain a particular solution for the ODE.

    Solution

    We expand f(t)f(t) as an odd function.

    ∴f(t)=πt,0<t<1\therefore f(t)=\pi t,\quad0<t<1bn=2∫01πtsin⁡nπtdt=2π{[−tnπcos⁡nπt]01+1nπ∫01cos⁡nπtdt}=2π{−1nπcos⁡nπ}=−2ncos⁡nπ=2n(−1)n+1∴116d2xdt2+4x=∑n=1∞2n(−1)n+1sin⁡nπt\begin{aligned}b_n&=2\int_0^1\pi t\sin n\pi tdt\\&=2\pi\left\{\left[-\frac{t}{n\pi}\cos n\pi t\right]_0^1+\frac{1}{n\pi}\int_0^1\cos n\pi tdt\right\}\\&=2\pi\left\{-\frac{1}{n\pi}\cos n\pi\right\}\\&=-\frac{2}{n}\cos n\pi=\frac{2}{n}(-1)^{n+1}\\\\\therefore\frac{1}{16}\frac{d^2x}{dt^2}+4x&=\sum_{n=1}^\infty \frac{2}{n}(-1)^{n+1}\sin n\pi t\end{aligned}

    We assume a particular solution of the form

    xp(t)=∑n=1∞βnsin⁡nπtxp′(t)=∑n=1∞βn(nπ)cos⁡nπtxp′′(t)=∑n=1∞−βn(nπ)2sin⁡nπt\begin{aligned}x_p(t)&=\sum_{n=1}^\infty\beta_n\sin n\pi t\\x'_p(t)&=\sum_{n=1}^\infty \beta_n(n\pi)\cos n\pi t\\x''_p(t)&=\sum_{n=1}^\infty -\beta_n(n\pi)^2\sin n\pi t\end{aligned}∑n=1∞(−116n2π2βn+4βn)sin⁡nπt=∑n=1∞2n(−1)n+1sin⁡nπt−116n2π2βn+4βn=2n(−1)n+1\sum_{n=1}^\infty\left(-\frac{1}{16}n^2\pi^2\beta_n+4\beta_n\right)\sin n\pi t=\sum_{n=1}^\infty\frac{2}{n}(-1)^{n+1}\sin n\pi t\\-\frac{1}{16}n^2\pi^2\beta_n+4\beta_n=\frac{2}{n}(-1)^{n+1}βn=2n(−1)n+14−116n2π2=32n(−1)n+164−n2π2∴xp(t)=∑n=1∞32(−1)n+1n(64−n2π2)sin⁡nπt\begin{aligned}\beta_n&=\frac{\frac{2}{n}(-1)^{n+1}}{4-\frac{1}{16}n^2\pi^2}\\&=\frac{\frac{32}{n}(-1)^{n+1}}{64-n^2\pi^2}\\\\\therefore x_p(t)&=\sum_{n=1}^\infty \frac{32(-1)^{n+1}}{n(64-n^2\pi^2)}\sin n\pi t\end{aligned}