Question:medium
A time-limited waveform \(g(x)\) is specified as follows:
\[
g(x)=\begin{cases}-k, & -\pi<x\le0\\ +k, & 0<x\le\pi\\ 0, & \text{otherwise}\end{cases}
\]
A new waveform \(f(x)\) is constructed from \(g(x)\) as follows:
\[
f(x)=\sum_{m=-\infty}^{\infty}g(x+2\pi m),\qquad \text{for all } x\in\mathbb{R}
\]
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 \(\dfrac{2}{3\pi}\).
What is the value of \(k\)?
A time-limited waveform \(g(x)\) is specified as follows:
\[ g(x)=\begin{cases}-k, & -\pi<x\le0\\ +k, & 0<x\le\pi\\ 0, & \text{otherwise}\end{cases} \]
A new waveform \(f(x)\) is constructed from \(g(x)\) as follows:
\[ f(x)=\sum_{m=-\infty}^{\infty}g(x+2\pi m),\qquad \text{for all } x\in\mathbb{R} \]
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 \(\dfrac{2}{3\pi}\).
What is the value of \(k\)?
\[ g(x)=\begin{cases}-k, & -\pi<x\le0\\ +k, & 0<x\le\pi\\ 0, & \text{otherwise}\end{cases} \]
A new waveform \(f(x)\) is constructed from \(g(x)\) as follows:
\[ f(x)=\sum_{m=-\infty}^{\infty}g(x+2\pi m),\qquad \text{for all } x\in\mathbb{R} \]
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 \(\dfrac{2}{3\pi}\).
What is the value of \(k\)?
Show Hint
f(x) is an odd square wave of period 2pi, so only sine terms exist; use the standard series 4k/(n*pi) for the nth odd harmonic.
Updated On: Jul 20, 2026
- \(1\)
- \(\dfrac{1}{2}\)
- \(\dfrac{1}{3}\)
- \(\dfrac{1}{4}\)
Show Solution
The Correct Option is B
Solution and Explanation
Step 1: Set up the coefficient formulas.
For a period $2\pi$ function, the Fourier cosine and sine coefficients are
\[ a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)\,dx,\qquad b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(nx)\,dx \]
Step 2: Compute $a_3$ directly.
Split the integral over the two pieces of $f(x)$:
\[ a_3=\frac{1}{\pi}\left[\int_{-\pi}^{0}(-k)\cos(3x)\,dx+\int_{0}^{\pi}k\cos(3x)\,dx\right] \]
Because $\cos(3x)$ is an even function, the two integrals, one negated and one not, taken over mirror-image ranges, are exact negatives of each other, so they cancel and $a_3=0$. This confirms there is no cosine content, as expected for an odd function.
Step 3: Compute $b_3$ directly.
\[ b_3=\frac{1}{\pi}\left[\int_{-\pi}^{0}(-k)\sin(3x)\,dx+\int_{0}^{\pi}k\sin(3x)\,dx\right] \]
Using $\int\sin(3x)\,dx=-\frac{\cos(3x)}{3}$, the second integral gives
\[ \int_{0}^{\pi}k\sin(3x)\,dx=k\left[-\frac{\cos(3x)}{3}\right]_0^{\pi}=k\left(-\frac{\cos3\pi}{3}+\frac{1}{3}\right)=k\left(\frac{1}{3}+\frac{1}{3}\right)=\frac{2k}{3} \]
since $\cos(3\pi)=-1$. By the odd symmetry of $\sin(3x)$, the first integral over $(-\pi,0)$ contributes the same amount again, giving a total inside the bracket of $\frac{4k}{3}$.
Step 4: Finish computing $b_3$.
\[ b_3=\frac{1}{\pi}\cdot\frac{4k}{3}=\frac{4k}{3\pi} \]
Step 5: Match to the given sum and solve.
The sum $a_3+b_3=\frac{4k}{3\pi}$ is given as $\frac{2}{3\pi}$, so $4k=2$, giving $k=\frac{1}{2}$.
\[ \boxed{k=\dfrac{1}{2}} \]
For a period $2\pi$ function, the Fourier cosine and sine coefficients are
\[ a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)\,dx,\qquad b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(nx)\,dx \]
Step 2: Compute $a_3$ directly.
Split the integral over the two pieces of $f(x)$:
\[ a_3=\frac{1}{\pi}\left[\int_{-\pi}^{0}(-k)\cos(3x)\,dx+\int_{0}^{\pi}k\cos(3x)\,dx\right] \]
Because $\cos(3x)$ is an even function, the two integrals, one negated and one not, taken over mirror-image ranges, are exact negatives of each other, so they cancel and $a_3=0$. This confirms there is no cosine content, as expected for an odd function.
Step 3: Compute $b_3$ directly.
\[ b_3=\frac{1}{\pi}\left[\int_{-\pi}^{0}(-k)\sin(3x)\,dx+\int_{0}^{\pi}k\sin(3x)\,dx\right] \]
Using $\int\sin(3x)\,dx=-\frac{\cos(3x)}{3}$, the second integral gives
\[ \int_{0}^{\pi}k\sin(3x)\,dx=k\left[-\frac{\cos(3x)}{3}\right]_0^{\pi}=k\left(-\frac{\cos3\pi}{3}+\frac{1}{3}\right)=k\left(\frac{1}{3}+\frac{1}{3}\right)=\frac{2k}{3} \]
since $\cos(3\pi)=-1$. By the odd symmetry of $\sin(3x)$, the first integral over $(-\pi,0)$ contributes the same amount again, giving a total inside the bracket of $\frac{4k}{3}$.
Step 4: Finish computing $b_3$.
\[ b_3=\frac{1}{\pi}\cdot\frac{4k}{3}=\frac{4k}{3\pi} \]
Step 5: Match to the given sum and solve.
The sum $a_3+b_3=\frac{4k}{3\pi}$ is given as $\frac{2}{3\pi}$, so $4k=2$, giving $k=\frac{1}{2}$.
\[ \boxed{k=\dfrac{1}{2}} \]
Was this answer helpful?
0
