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 $k$?
To solve the given problem, we need to find the value of \(k\) in the waveform \(f(x)\), constructed from the given periodic waveform \(g(x)\). This involves analyzing the Fourier series of \(f(x)\) and particularly focusing on the coefficients of the third harmonics.
- The function \(g(x)\) is defined as: \(g(x) = \begin{cases} -k, & -\pi < x \le 0 \\ +k, & 0 < x \le \pi \\ 0, & \text{otherwise}. \end{cases}\)This means \(g(x)\) is a periodic waveform with one period spanning from \(-\pi\) to \(\pi\), having a negative amplitude in the first half and a positive amplitude in the second half.
- The function \(f(x)\) is constructed by replicating \(g(x)\) over all integer multiples of \(2\pi\): \(f(x) = \sum_{m=-\infty}^{\infty} g(x + 2\pi m)\)Hence, \(f(x)\) is \(2\pi\)-periodic and can be expressed as a Fourier series.
- We need to focus on the sum of the coefficients of the third harmonics (i.e., terms with frequency \(3\omega\)) in the Fourier series of \(f(x)\). For a \(2\pi\)-periodic function, the Fourier series is given by: \(f(x) = a_0 + \sum_{n=1}^{\infty} \left[ a_n \cos(nx) + b_n \sin(nx) \right]\)with the coefficients: \(a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx, \quad b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx\)where the integrals effectively focus on contributions from each harmonic component.
- For \(n = 3\) (third harmonics): \(a_3 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(3x) \, dx = \frac{1}{\pi} \left( \int_{-\pi}^{0} (-k) \cos(3x) \, dx + \int_{0}^{\pi} k \cos(3x) \, dx \right)\)Simplifying, we get: \(a_3 = \frac{2k}{3\pi} \sin(3\pi) = 0\)since \(\sin(3\pi) = 0\).
- Similarly, for \(b_3\): \(b_3 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(3x) \, dx = \frac{1}{\pi} \left( \int_{-\pi}^{0} -k \sin(3x) \, dx + \int_{0}^{\pi} k \sin(3x) \, dx \right)\)Evaluating these: \(b_3 = \frac{2k}{3\pi}\)
- The sum of the coefficients of the third harmonics is: \(a_3 + b_3 = 0 + \frac{2k}{3\pi} = \frac{2k}{3\pi}\) We know it equals \(\frac{2}{3\pi}\), so: \(\frac{2k}{3\pi} = \frac{2}{3\pi} \quad \Rightarrow \quad k = \frac{1}{2}\)
Therefore, the value of \(k\) is \(\frac{1}{2}\). This matches the correct answer option.
Important Questions from Fourier Series
If we use the Fourier transform ϕ(x, y) = \(\int {{{\rm{e}}^{{\rm{ikx}}}}} {ϕ _{\rm{k}}}\left( {\rm{y}} \right){\rm{dk}}\) to solve the partial differential equation \({\rm{ - }}\frac{{{\partial ^2}ϕ \left( {x,y} \right)}}{{\partial {y^2}}}\, - \,\frac{1}{{{y^2}}}\frac{{{\partial ^2}ϕ \left( {x,y} \right)}}{{\partial {x^2}}} + \frac{{{m^2}}}{{{y^2}}}ϕ \left( {x,y} \right) = 0\) in the half-plane {(x, y) : -∞ < x < ∞, 0 < y < ∞} the Fourier modes ϕ k(y) depend on y as y α and y β . The values of α and β are
When a time-domain signal is converted into its Fourier representation, which of the following is/are conserved?
I. Energy
II. Power
The trigonometric Fourier series of a periodic time function can have
The Fourier series expansion of x3 in the interval −1 ≤ x < 1 with periodic continuation has
The Fourier series to represent x-x2 for –π ≤ x ≤ π is given by \(x - {x^2} = \frac{{{a_0}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}cosnx + \mathop \sum \limits_{n = 1}^\infty {b_n}sinnx\)
The value of a0 (round off to two decimal places), is
