Question:
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
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The Correct Option is B
Solution and Explanation
Step 1: Recognize the periodic waveform.
The construction \(f(x)=\sum_m g(x+2\pi m)\) simply repeats \(g(x)\) every \(2\pi\), turning it into a periodic square wave of period \(2\pi\), equal to \(-k\) on \((-\pi,0]\) and \(+k\) on \((0,\pi]\) in every period.
Step 2: Check the symmetry of \(f(x)\).
Since \(f(-x)=-f(x)\) (a negative x maps into the \(-k\) region exactly when the corresponding positive x maps into the \(+k\) region), \(f(x)\) is an odd function. An odd periodic function has only sine terms in its Fourier series; every cosine coefficient \(a_n\) is zero.
Step 3: Recall the Fourier series of this standard square wave.
This is the classic odd square wave of amplitude \(k\) and period \(2\pi\). Its trigonometric Fourier series is
\[ f(x)=\frac{4k}{\pi}\sum_{n\ \text{odd}}\frac{\sin(nx)}{n} \]
so the sine coefficient of the \(n^{th}\) harmonic, for odd \(n\), is
\[ b_n=\frac{4k}{n\pi} \]
Step 4: Pick out the third harmonic.
For \(n=3\),
\[ b_3=\frac{4k}{3\pi} \]
and since \(a_3=0\) (no cosine terms at all, as shown in Step 2), the sum of the third harmonic's sine and cosine coefficients is just
\[ a_3+b_3=0+\frac{4k}{3\pi}=\frac{4k}{3\pi} \]
Step 5: Use the given value to solve for \(k\).
We are told this sum equals \(\dfrac{2}{3\pi}\):
\[ \frac{4k}{3\pi}=\frac{2}{3\pi} \]
\[ 4k=2\ \Rightarrow\ k=\frac{1}{2} \]
Step 6: Analyze the options.
(A) 1: Would give a third harmonic sum of \(4/(3\pi)\), too large. Incorrect.
(B) 1/2: Gives exactly \(2/(3\pi)\), matching the given value. Correct.
(C) 1/3: Would give \(4/(9\pi)\), too small. Incorrect.
(D) 1/4: Would give \(1/(3\pi)\), too small. Incorrect.
Final Answer:
\[ \boxed{k=\frac{1}{2}} \]
The construction \(f(x)=\sum_m g(x+2\pi m)\) simply repeats \(g(x)\) every \(2\pi\), turning it into a periodic square wave of period \(2\pi\), equal to \(-k\) on \((-\pi,0]\) and \(+k\) on \((0,\pi]\) in every period.
Step 2: Check the symmetry of \(f(x)\).
Since \(f(-x)=-f(x)\) (a negative x maps into the \(-k\) region exactly when the corresponding positive x maps into the \(+k\) region), \(f(x)\) is an odd function. An odd periodic function has only sine terms in its Fourier series; every cosine coefficient \(a_n\) is zero.
Step 3: Recall the Fourier series of this standard square wave.
This is the classic odd square wave of amplitude \(k\) and period \(2\pi\). Its trigonometric Fourier series is
\[ f(x)=\frac{4k}{\pi}\sum_{n\ \text{odd}}\frac{\sin(nx)}{n} \]
so the sine coefficient of the \(n^{th}\) harmonic, for odd \(n\), is
\[ b_n=\frac{4k}{n\pi} \]
Step 4: Pick out the third harmonic.
For \(n=3\),
\[ b_3=\frac{4k}{3\pi} \]
and since \(a_3=0\) (no cosine terms at all, as shown in Step 2), the sum of the third harmonic's sine and cosine coefficients is just
\[ a_3+b_3=0+\frac{4k}{3\pi}=\frac{4k}{3\pi} \]
Step 5: Use the given value to solve for \(k\).
We are told this sum equals \(\dfrac{2}{3\pi}\):
\[ \frac{4k}{3\pi}=\frac{2}{3\pi} \]
\[ 4k=2\ \Rightarrow\ k=\frac{1}{2} \]
Step 6: Analyze the options.
(A) 1: Would give a third harmonic sum of \(4/(3\pi)\), too large. Incorrect.
(B) 1/2: Gives exactly \(2/(3\pi)\), matching the given value. Correct.
(C) 1/3: Would give \(4/(9\pi)\), too small. Incorrect.
(D) 1/4: Would give \(1/(3\pi)\), too small. Incorrect.
Final Answer:
\[ \boxed{k=\frac{1}{2}} \]
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