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
The Fourier series representation of a square wave is shown in the figure below. The fluctuations seen near $x = \pm 1$ are named after which one of the following scientists?
The Fourier series expansion of the function $f(x) = |\cos(x)|$ in the interval $(-\pi, \pi)$ is $\alpha + \beta \left[ \frac{1}{3}\cos(2x) - \frac{1}{15}\cos(4x) + \dots \right]$.
The values of $\alpha$ and $\beta$ respectively are ______.
Periodic function $f(x)$ is given below. $$f(x) = \begin{cases} -1, & \text{when } -\pi < x < 0 \\ 1, & \text{when } 0 < x < \pi \end{cases} ; f(x + 2\pi) = f(x)$$ The CORRECT option representing the Fourier series expansion of $f(x)$ is:
A time-limited waveform $g(x)$ is specified as follows: $$g(x) = \begin{cases} -k, & -\pi < x \le 0 \\ +k, & 0 < x \le \pi \\ 0, & \text{otherwise} \end{cases}$$ A new waveform $f(x)$ is constructed from $g(x)$ as follows: $$f(x) = \sum_{m=-\infty}^{\infty} g(x + 2\pi m), \quad \text{for all } x \in \mathbb{R}$$ The sum of the coefficients of the third harmonics of the sine and cosine terms in the trigonometric Fourier series expansion of $f(x)$ is $\frac{2}{3\pi}$.
What is the value of $k$?
The effective value of the current for the Sawtooth waveform as shown is:
