A time-limited waveform g(x) is specified as follows: g(x) = −k,−π < x ≤ 0; g(x) = +k,0 < x ≤ π; 0, otherwise. A new waveform f(x) is constructed from g(x) as follows: f(x) = Σ^∞_m=−∞ g(x+2πm). The sum of the coefficients of the third har monics of the sine and cosine terms in the trigonometric Fourier series expansion of f(x) is 2/3π. What is the value of k?

The correct answer is 1/2.

Step 1: Understand the constructed periodic waveform
The given time-limited waveform is defined as:
g(x) = -k for -π < x ≤ 0
g(x) = +k for 0 < x ≤ π
g(x) = 0, otherwise.
A periodic waveform f(x) is constructed by repeating g(x) with a period of T = 2π:
$$f(x)=\sum _{m=-\infty }^{\infty }g(x+2\pi m)$$
Thus, over one period from -π to π, f(x) is given by:
f(x) = -k for -π < x ≤ 0
f(x) = +k for 0 < x ≤ π

Step 2: Identify the symmetry of the waveform
Since f(-x) = -f(x), the periodic function f(x) is an odd function.
For any odd periodic function, the trigonometric Fourier series consists only of sine terms. Therefore:
• The DC component, a₀ = 0
• All cosine coefficients, a_n = 0 for all n. Specifically, the third harmonic cosine coefficient is:
$$a_3=0$$

Step 3: Calculate the third harmonic sine coefficient (b₃)
The general formula for the sine coefficients b_n of a periodic function with period T = 2π is:
$$b_n=\frac{1}{\pi }{\int }_{-\pi }^{\pi }f(x)\sin (nx)dx$$
Since f(x) and sin(nx) are both odd, their product is an even function. We can simplify the integration range to [0, π]:
$$b_n=\frac{2}{\pi }{\int }_0^{\pi }k\sin (nx)dx$$
Integrating the function:
$$b_n=\frac{2k}{\pi }{\left[\frac{-\cos (nx)}{n}\right]}_0^{\pi }=\frac{2k}{n\pi }$$
For the third harmonic (n = 3):
$$b_3=\frac{2k}{3\pi }(1-\cos (3\pi ))$$
Since cos(3π) = -1:
$$b_3=\frac{2k}{3\pi }(1-(-1))=\frac{4k}{3\pi }$$

Step 4: Solve for k
We are given that the sum of the coefficients of the third harmonics of the sine and cosine terms is 2/3π:
$$a_3+b_3=\frac{2}{3\pi }$$
Substituting the values of a₃ and b₃:
$$0+\frac{4k}{3\pi }=\frac{2}{3\pi }$$
Multiplying both sides by 3π:
4k = 2 ⇒ k = 1/2