Question 9.1: (a) Derive a Fourier series to represent the voltage wavefor......
(a) Derive a Fourier series to represent the voltage waveform shown in Fig. 9.2(a), a square wave with amplitude ± 1 V, and period T seconds, by representing one period as an even function.
(b) Repeat (a) using an odd function to represent one period.
(c) Compare the results.
(d) If the period of the square wave is 1 second, plot the sums of each of the two series derived in (a) and (b) above, against time, t, showing that the original square wave is reproduced approximately.
Part (a)
Figure 9.2(b) shows one complete period, of duration T, chosen to be an even function, defined as a function where x(t) = x(-t) . For analysis purposes this period will be assumed to extend from t = -T/2 to t = T/2. Since the waveform has zero mean value, a_{0} = 0, in this case.
The coefficients a_{n} (n=1, 2, 3, . . . , ∞) are given by Eq. (9.8), which, using Eq. (9.2), ω_{0} = (2π / T), can be written as:
The coefficients b_{n} (n = 1, 2, 3, . . . , ∞) are given by Eq. (9.10), which can similarly be written as:
b_{n} = \frac{2}{T} \int_{-T/2}^{T/2}{x(t)\sin n\omega_{0}t\cdot dt} (n = 1, 2, 3, . . . , \infty ) (9.10) \\ b_{n} = \frac{2}{T} \int_{-T/2}^{T/2}{x(t)\sin n\omega_{0}t\cdot dt} = \frac{2}{T} \int_{-T/2}^{T/2}{x(t)\sin n \left(\frac{2\pi}{T}\right) t\cdot dt} (B)It can be seen from Fig. 9.2 that the values of a_{n}, given by Eq. (A), are zero for even values of n, and the values of b_{n}, given by Eq. (B), are all zero. Therefore, the complete Fourier series consists only of cosine terms with odd values of n:
x(t) = a_{1} \cos \left(\frac{2\pi}{T} \right) t + a_{3} \cos 3\left(\frac{2\pi}{T}\right)t + a_{5} \cos 5\left(\frac{2\pi}{T} \right) t + a_{7} \cos 7\left(\frac{2\pi}{T} \right) t + . . . (C)From Eq. (A), the numerical value of a_{1} is given by:
a_{1} = \frac{2}{T} \int_{-T/2}^{T/2}{x(t)\cos \left(\frac{2\pi}{T} \right)t\cdot dt} = 4\cdot \frac{2}{T}\int_{0}^{T/4}{\cos \left(\frac{2\pi}{T} \right)t\cdot dt} = 4\cdot \frac{2}{T} \cdot \frac{T}{2\pi }\left[\sin \left(\frac{2\pi}{T} \right)t \right]^{T/4}_{0}= \frac{4}{\pi } (D)In the same way, it can be shown that
a_{3} = – \frac{1}{3}\cdot \left(\frac{4}{\pi }\right) ; a_{5}= \frac{1}{5} \cdot \left(\frac{4}{\pi }\right) ; a_{7} = -\frac{1}{7}\cdot \left(\frac{4}{\pi }\right) + . . .and Eq. (C) becomes
noting that the terms are alternately positive and negative.
Part (b)
Alternatively, the single period to be analysed can be taken as the odd function shown in Fig. 9.2 at (c), an odd function being defined mathematically as one where x(t) = -x(-t). It can be seen that, in this case, all values of a_{n} are zero, as are values of b_{n}, when n is an even number. The complete series therefore consists only of sine terms with odd values of n, and from Eq. (B):
noting that, in this case, the terms are all positive.
Part (c)
Comparing the two series in Eqs (E) and (H), it can be seen that the amplitude components are the same, but the phases are apparently different. This is due to the fact that the period chosen as an even function, shown in Fig. 9.2 at (b), leads that chosen as an odd function, in Fig. 9.2 at (c), by T/4 seconds, which is equivalent to a phase shift of π/2 radians at a frequency of (2π/T) rad./s; or (3π/2) radians at frequency 3(2π/T) rad/s and so on. Thus the magnitudes of the terms in the two series are identical, but their apparent phases naturally depend upon the arbitrary point in the waveform chosen as t = 0.
Part (d)
Plotting x(t), as given by either Eq. (E) or Eq. (H), versus t, should reproduce the original waveform. Taking T equal to 1 second, Fig. 9.3(a) is a plot of Eq. (E), using only the first four terms of the series. Figure 9.3(b) is a similar plot, from Eq. (H), also using only the first four terms, and it can be seen that the the waveforms reproduced are identical, apart from the shift of T/4 seconds horizontally, equal to 0.25 s. in this case, as would be expected. The square wave is only approximately reproduced, due to the small number of terms included. However, this improves as the number of terms is increased, as shown in Fig. 9.3(c), which is a repeat of Fig. 9.3(b), using seven terms.
