For fourier series of wave shown in Figure below, Select correct expression for f(t)
\(f(t)=\frac{A}{2}+\frac{2A}{\pi}\left[\sin(\omega_o t)+\frac{1}{3}\sin(\sin 3\omega_o t)+ \ldots \right]\)
The given problem involves finding the Fourier series for a periodic wave shown in the figure. To determine the correct expression for \( f(t) \), let's analyze the waveform and derive its Fourier series.
The waveform depicted is a square wave, which is periodic with period \( T \). The fundamental frequency is \( \omega_o = \frac{2\pi}{T} \). The Fourier series expression for a square wave is generally given as:
\(f(t) = \frac{A}{2} + \frac{2A}{\pi} \left[\sin(\omega_o t) + \frac{1}{3}\sin(3\omega_o t) + \frac{1}{5}\sin(5\omega_o t) + \ldots \right]\)
This series consists only of odd harmonics (i.e., terms like \(\sin(\omega_o t)\), \(\sin(3\omega_o t)\), etc.), which is typical for a square wave.
Now, let's compare the derived expression with the given options:
- Option 1: \(f(t)=\frac{A}{2}+\frac{2A}{\pi}\left[\sin(\omega_o t)+\frac{1}{3}\sin(\sin 3\omega_o t)+ \ldots \right]\)
- Option 2: \(f(t)=A+\frac{2A}{\pi}\left[\sin(\omega_o t)+\frac{1}{3}\sin(3\omega_o t)+ \ldots \right]\)
- Option 3: \(f(t)=\frac{A}{4}+\frac{A}{\pi}\left[\sin(\omega_o t)+\frac{1}{3}\sin(3\omega_o t)+ \ldots \right]\)
- Option 4: \(f(t)=A/2+\frac{2A}{\pi}\left[\sin(\omega_o t)+\frac{1}{3}\sin(2\omega_o t)+ \ldots \right]\)
Upon inspection, Option 1 is incorrect due to the erroneous structure \(\sin(\sin 3\omega_o t)\). Option 2 does not have the correct DC term. Option 3 scales the DC term incorrectly, and Option 4 includes even harmonics, which are not present in the square wave Fourier series.
Therefore, the correct expression is:
\(f(t) = \frac{A}{2} + \frac{2A}{\pi} \left[\sin(\omega_o t) + \frac{1}{3}\sin(3\omega_o t) + \ldots \right]\)
Important Questions from Fourier Series
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