Question 2.11: Harmonic Analysis of a Square Wave. Find the Fourier series ......
Harmonic Analysis of a Square Wave. Find the Fourier series equivalent of the square wave in Fig. 2.31a, assuming that it has a peak value of 1 V.
We know by inspection, using (2.83) to (2.86), that the series will have only sines, with no even harmonics. Therefore, all we need are the even coefficients b_{n} from (2.82):
b_{n} \ = \ \frac{2}{T} \ \int_{0}^{T}{ \ f\left(t\right) \ \sin \ n\omega t \ dt} , \quad n \ = \ 1, 2, 3 \ . \ . \ . (2.82)
a_0 \ = \ 0 \ : \quad \text{when average value,} \ dc \ = \ 0 (2.83)
\text{no even harmonics:} \quad \text{when} \ f \ \left(t \ + \ \frac{T}{2} \right) \ = \ -f\left(t\right) (2.86)
Recall that the integral of a sine is the cosine with a sign change, so
Substituting ω = 2πf = 2π/T gives
Since this is half-wave symmetric, there are no even harmonics; that is, n is an odd number. For odd values of n,
\cos \ n\pi \ = \ \cos \ \pi \ = \ -1 \quad \text{and} \quad \cos \ 2n\pi \ = \ \cos \ 0 \ = \ 1That makes for a nice, simple solution:
b_{n} \ = \ \frac{4}{n\pi }That is,
b_{1} \ = \ \frac{4}{\pi } \ = \ 1.273, \quad b_{3} \ = \ \frac{4}{3\pi } \ = \ 0.424, \quad b_{5} \ = \ \frac{4}{5\pi } \ = \ 0.255 \ . \ . \ .So the series is
\text{(Square wave with amplitude of 1)} \ = \frac{4}{\pi } \ \left[ \ \sin \ \omega t \ + \ \frac{1}{3} \ \sin \ 3\omega t \ + \ \frac{1}{5} \ \sin \ 5\omega t \ + \ . \ . \ . \ \right]