Question

Given a periodic rectangular waveform of frequency 1 kHz, symmetrical about t=0 and having a pulse width of 500 µs and amplitude 5 V. The Fourier series is ___________

(a) 2.5 + 3.18 cos (2π × 10^3 t) – 1.06 cos (6π × 10^3 t)

(b) 2.5 – 3.18 cos (2π × 10^3 t) – 1.06 cos (6π × 10^3 t)

(c) 2.5 + 3.18 cos (2π × 10^3 t) + 1.06 cos (6π × 10^3 t)

(d) 2.5 – 3.18 cos (2π × 10^3 t) + 1.06 cos (6π × 10^3 t)

I have been asked this question in an international level competition.

I'm obligated to ask this question of Trigonometric Fourier Series in chapter Fourier Series of Signals and Systems

Answer 1

The correct answer is (a) 2.5 + 3.18 cos (2π × 10^3 t) – 1.06 cos (6π × 10^3 t)

For explanation: Period, T = \(\frac{1}{f} =  \frac{1}{1×10^3} = 10^{-3}s\)

Ratio, \(\frac{τ}{T} = \frac{500 × 10^{-6}}{10^{-3}} \) = 0.5

Now, \(f (t) = \frac{Aτ}{T} + 2 \frac{Aτ}{T} \frac{sin⁡ \frac{πτ}{T}}{\frac{πτ}{T}} cos⁡[\frac{2πt}{τ}] + \frac{2Aτ}{T} \frac{sin⁡ \frac{2πτ}{T}}{\frac{2πτ}{T}} cos⁡ \frac{4πt}{T}\)

= 5(0.5) + 2(5) (0.5)\(\frac{sin⁡0.5π}{0.5π}\) cos⁡(2π × 10^3 t) + 2(5) (0.5) \(\frac{sin⁡π}{π}\)  cos⁡(4π × 10^3 t)

= 2.5 + 3.18 cos (2π × 10^3 t) – 1.06 cos (6π × 10^3 t).