Question 17.1: Determine the Fourier series of the waveform shown in Fig. 1......
Determine the Fourier series of the waveform shown in Fig. 17.1. Obtain the amplitude and phase spectra
The Fourier series is given by Eq. (17.3), namely,
f(t) = \underbrace{a_0}_{dc} + \underbrace{\sum\limits_{n=1}^{\infty }{(a_n \ cos \ n\omega_0 t + b_n sin \ n \omega_0 \ t)} }_{ac} \quad (17.3)f(t) = a_{0} + \sum\limits_{n= 1}^{∞}{(a_{n} \cos nω_{0} t + b_{n} \sin nω_{0} t)} (17.1.1)
Our goal is to obtain the Fourier coefficients a_{0} , a_{n} and b_{n} using Eqs. (17.6), (17.8), and (17.9). First, we describe the waveform as
a_0 = \frac{1}{T}\int_{0}^{T}{f(t)dt} \quad (17.6) \\ a_n = \frac{2}{T}\int_{0}^{T}{f(t) cos \ n \omega_0 t \ dt} \quad (17.8) \\ b_n = \frac{2}{T}\int_{0}^{T}{f(t) sin \ n \omega_0 t \ dt} \quad (17.9)\\f(t) = \begin{cases} 1 , & 0 < t < 1 \\ 0 , & 1 < t < 2 \end{cases} (17.1.2)
and f(t) = f(t + T) . Since T = 2, ω_{0} = 2π/T = π Thus,
a_{o} = \frac{1}{T} \int^{T}_{0}{f(t) dt} = \frac{1}{2} [\int^{1}_{0}{1 dt} + \int^{2}_{1}{0 dt} ] = \frac{1}{2} t \mid^{1}_{0} = \frac{1}{2} (17.1.3)
Using Eq. (17.8) along with Eq. (17.15a),
a_{n} = \frac{2}{T} \int^{T}_{0}{f(t) \cos nω_{0} t dt}= \frac{2}{2} [ \int^{1}_{0}{1 \cos nπt dt } + \int^{2}_{1}{0 \cos nπt dt} (17.1.4)
= \frac{1}{nπ} \sin nπt \mid^{1}_{0} = \frac{1}{nπ} [ \sin nπ – \sin (0) ] = 0From Eq. (17.9) with the aid of Eq. (17.15b),
\int{cos \ at \ dt = \frac{1}{a} sin\ at }\quad (17.15a) \\\int{sin \ at \ dt = -\frac{1}{a} cos\ at }\quad (17.15b)\\ b_{n} = \frac{2}{T} \int^{T}_{0}{f(t) \sin nω_{0} t dt}= \frac{2}{2} [\int^{1}_{0}{1 \sin nπt dt } + \int^{2}_{1}{0 \sin nπt dt}] (17.1.5)
= – \frac{1}{nπ} \cos nπt \mid^{1}_{0} =- \frac{1}{nπ} (\cos nπ – 1 ] , \cos n π = (-1)^{n} = \frac{1}{nπ} [ 1 – (-1)^{n} ] = \begin{cases} \frac{2}{nπ} , & n = odd \\ 0 , & n = even \end{cases}Substituting the Fourier coefficients in Eqs. (17.1.3) to (17.1.5) into Eq. (17.1.1) gives the Fourier series as
f(t) = \frac{1}{2} + \frac{2}{π} \sin πt + \frac{2}{3π} sin 3πt + \frac{2}{5π} sin 5πt + . . . (17.1.6)
Since f(t) contains only the dc component and the sine terms with the fundamental component and odd harmonics, it may be written as
f(t) = \frac{1}{2} + \frac{2}{π} \sum\limits^{∞}_{k = 1}{\frac{1}{n} \sin 5 πt} , n = 2k – 1 (17.1.7)
By summing the terms one by one as demonstrated in Fig. 17.2, we notice how superposition of the terms can evolve into the original square. As more and more Fourier components are added, the sum gets closer and closer to the square wave. However, it is not possible in practice to sum the series in Eq. (17.1.6) or (17.1.7) to infinity. Only a partial sum ( n = 1, 2,3 , … , N , where N is finite) is possible. If we plot the partial sum (or truncated series) over one period for a large N
as in Fig. 17.3, we notice that the partial sum oscillates above and below the actual value of f(t). At the neighborhood of the points of discontinuity (x= 0, 1, 2, … ), there is overshoot and damped oscillation. In fact, an overshoot of about 9 percent of the peak value is always present, regardless of the number of terms used to approximate f(t). This is called the Gibbs phenomenon.
Finally, let us obtain the amplitude and phase spectra for the signal in Fig. 17.1. Since a_{n} = 0
A _{n} = \sqrt{a^{2}_{n} + b^{2}_{n}} = |b_{n}| = \begin{cases} \frac{2}{nπ} , n = odd \\ 0 , n = even \end{cases} (17.1.8)
and
\phi_n = -tan^{-1} \frac{b_n}{a_n} = \begin{cases} -90 , \quad n = odd \\ 0 , \quad n = even \end{cases} \quad (17.1.9)and (17.1.9) The plots of A_n and \phi_n for different values of n \omega_0 = n \pi provide the amplitude and phase spectra in Fig. 17.4. Notice that the amplitudes of the harmonics decay very fast with frequency.
