Question 17.2: Obtain the Fourier series for the periodic function in Fig. ......
Obtain the Fourier series for the periodic function in Fig. 17.7 and plot the amplitude and phase spectra.
The function is described as
f(t) = \begin{cases} t , & 0 < t < 1 \\ 0, & 1 < t < 2 \end{cases}
Since T = 2 , ω_{0} = 2 π / T = π . Then
a_{0} = \frac{1}{T} \int^{T}_{0}{f(t) dt} = \frac{1}{2} [ \int^{1}_{0}{t dt} + \int^{2}_{1}{0 dt} ] = \frac{1}{2} \frac{t^{2}}{2} \mid^{1}_{0} = \frac{1}{4} (17.2.1)
To evaluate a_{n} and b_{n} , we need the integrals in Eq. (17.15):
\int{\cos at dt} = \frac{1}{a} \sin at (17.15a)
\int{\sin at dt} = – \frac{1}{a} \cos at (17.15b)
\int{t \cos at dt } = \frac{1}{a^{2}} \cos at + \frac{1}{a} t \sin at (17.15c)
\int{t \sin at dt} = \frac{1}{a^{2}} \sin at – \frac{1}{a} t \cos at (17.15d)
a_{n} = \frac{2}{T} \int^{T}_{0}{f(t) \cos nω_{0} t dt }
= \frac{2}{2} [ \int^{1}_{0}{t \cos n π t dt } + \int^{2}_{1}{0 \cos n π t dt } ] (17.2.2)
= [ \frac{1}{n^{2} π^{2}} \cos n π t + \frac{t}{n π } n π t ] \mid^{1}_{0}
= \frac{1}{n^{2} π^{2}} (\cos n π – 1) + 0 = \frac{(-1)^{n} – 1}{n^{2} π^{2}}
since \cos nπ = (-1)^{n} ; and
b_{n} = \frac{2}{T} \int^{T}_{0}{f(t) \sin n ω_{0} t dt}
= \frac{2}{2} [ \int^{1}_{0}{t \sin nπt dt } + \int^{2}_{1}{0 \sin nπt dt } (17.2.3)
= [\frac{1}{n^{2}π^{2}} \sin n πt – \frac{t}{n π} \cos n πt] \mid^{1}_{0}
= 0 – \frac{\cos nπ }{n π} = \frac{(-1)^{n + 1}}{n π}
Substituting the Fourier coefficients just found into Eq. (17.3) yields
\boxed{f(t)=\underbrace{a_0}_{dc}+\underbrace{\sum_{n=1}^{\infty}\left(a_{n}\cos n\omega_{0}t+b_{n}\sin n\omega_{0}t\right)}_{ac}}(17.3)
f(t)=\frac{1}{4}+\sum_{n=1}^{\infty}\left[\frac{\left[(-1)^{n}-1\right]}{(n\pi)^{2}}\mathrm{cos}\,n\,\pi t+\frac{(-1)^{n+1}}{n\pi}\mathrm{sin}\,n\pi t\right]To obtain the amplitude and phase spectra, we notice that, for even harmonics, a_n=0,b_n=-1/n\pi, so that
A_{n}/\underline{\phi_{n}}=a_n -j b_n=0+j\frac{1}{n\pi} (17.2.4)
Hence,
\begin{array}{ll}A_{n}=|b_{n}|={\frac{1}{n\pi}},\qquad n=2,4,\ldots.&(17.2.5)\\\phi_{n}=90^{\circ},\qquad n=2,4,\ldots.\end{array}For odd harmonics, a_{n}= -2/(n^{2}\pi^{2}),b_{n}=1/(n\pi) so that
A_{n}/\underline{\phi_{n}}=a_{n}-j b_{n}=-\frac{2}{n^{2}\pi^{2}}-j\frac{1}{n\pi}\qquad\qquad(17.2.6)That is,
\begin{array}{l l}{{{ A}_{n}=\,\sqrt{a_{n}^{2}+b_{n}^{2}}=\sqrt{\frac{4}{n^{4}\pi^{4}}+\frac{1}{n^{2}\pi^{2}}}~~~~~}}&{{~~~~(17.2.7)}}\\ {{}}&{{}}&{{}}\\ {{=\frac{1}{n^{2}\pi^{2}}~\sqrt{4+n^{2}\pi^{2}}~~~~~n=1,3,\dots.}}\end{array}From Eq. (17.2.6), we observe that lies in the third quadrant, so that
\phi_{n}=180^{\circ}+\tan^{-1}{\frac{n\pi}{2}},\qquad n=1,3,\cdot\cdot\cdot (17.2.8)
From Eqs. (17.2.5), (17.2.7), and (17.2.8), we plot A_n and \phi_n for different values of n\omega_0=n\pi to obtain the amplitude spectrum and phase spectrum as shown in Fig. 17.8.
