GATE EE 2017 Set 1 — Question 16

Symmetry of $g(t)$For $t>0$ (non-integer), $\lceil -t \rceil = -\lfloor t \rfloor$, so$$g(-t) = -t - \lceil -t \rceil = -t + \lfloor t \rfloor = -\left(t - \lfloor t \rfloor\right) = -g(t).$$Hence $g$ is an odd function, and every cosine (even) coefficient of a harmonic expansion of $g$ must vanish:$$a_n = 0 \quad \text{for all } n, \text{ in particular } a_2 = 0.$$Is $g(t)$ periodic?On $t \ge 0$, $g(t) = t - \lfloor t \rfloor$ is the fractional part: a ramp rising from $0$ to $1$ on every interval $[k, k+1)$, so it repeats with period $1$. On $t < 0$, $g(t) = t - \lceil t \rceil$ is a ramp rising from $-1$ to $0$ on every interval $(k, k+1]$, again repeating with period $1$. However, the two halves do not join into one periodic function:$$g(-0.5) = -0.5 - 0 = -0.5, \qquad g(-0.5+1) = g(0.5) = 0.5 - 0 = +0.5 .$$So $g(t+1) \neq g(t)$, and since any period of the fractional part must be an integer, $g$ has no period at all on $\mathbb{R}$. A Fourier series therefore does not represent $g(t)$ on the whole real line.What the harmonics would beIf one works only with the branch for $t \ge 0$, i.e. the $1$-periodic sawtooth $g(t) = t$ on $(0,1)$ with $\omega_0 = 2\pi$, the standard expansion is$$t - \lfloor t \rfloor = \frac{1}{2} - \sum_{n=1}^{\infty} \frac{\sin(2\pi n t)}{n\pi},$$so the second harmonic has$$a_2 = 0, \qquad b_2 = -\frac{1}{2\pi} \approx -0.16 .$$Thus the "coefficient of the second harmonic" is $0$ if the cosine coefficient is meant (forced by the odd symmetry), and $-1/(2\pi) \approx -0.16$ if the sine coefficient of the $t \ge 0$ sawtooth is meant - and strictly speaking neither applies, because $g(t)$ as defined is not periodic.Answer: No single value can be assigned - $g(t)$ is not periodic on $\mathbb{R}$, and for the admissible readings the second-harmonic coefficient is either $a_2 = 0$ or $b_2 = -1/(2\pi) \approx -0.16$.