Fourier Series and Transforms Questions and Answers

Answer: C

The fluctuations are the Gibbs phenomenon.

When a function with a jump discontinuity, such as a square wave, is represented by a truncated Fourier series, the partial sums overshoot and ring near the jump. Adding more terms squeezes the ripples closer to the discontinuity but does not reduce the peak overshoot, which stays near 9% of the jump. J. Willard Gibbs explained this in 1899, and it is named after him.

Fourier gave the series itself, not the overshoot. Cauchy and Laplace are associated with other results (Cauchy's integral theorem, the Laplace transform and equation) and not with this effect.

Answer: B

The answer is \(y(10) = \frac{2}{\sqrt{15}} e^{-5/4} \sin\left(\frac{5\sqrt{15}}{4}\right)\).

Take Laplace transforms with zero initial data:

\[ \begin{aligned} (2s^2 + s + 2) Y &= e^{-5s} \\ Y &= \frac{e^{-5s}}{2[(s + \frac{1}{4})^2 + \frac{15}{16}]} \end{aligned} \]

Since \(\frac{1}{(s+a)^2 + b^2}\) inverts to \(\frac{1}{b} e^{-at} \sin bt\), with \(a = \frac{1}{4}\) and \(b = \frac{\sqrt{15}}{4}\), and the factor \(e^{-5s}\) shifts by 5:

\[ \begin{aligned} y(x) = &\frac{2}{\sqrt{15}} e^{-(x-5)/4} \\ &\times \sin \frac{\sqrt{15}(x-5)}{4} \end{aligned} \]

for \(x \gt 5\). At \(x = 10\), \(x - 5 = 5\), which gives the keyed value. The options with \(e^{5/4}\) have the wrong sign in the decay, and those with \(\frac{10}{\sqrt{15}}\) and \(\sin \frac{\sqrt{15}}{4}\) put the 5 in front instead of inside the sine.

Answer: A

The series has only sine terms.

\(x^{3}\) is an odd function, and the interval \(-1 \le x \lt 1\) is symmetric about zero, so its periodic continuation is also odd. For an odd function the constant term \(a_{0}\) and every cosine coefficient \(a_{n} = \int_{-1}^{1} x^{3}\cos(n\pi x)\,dx\) vanish, because each integrand is odd. Only the sine coefficients survive.

Only cosine terms would need an even function, both kinds would need a function with no symmetry, and a non-zero constant would need a non-zero average over the period, which an odd function on a symmetric interval does not have.

Answer: C

The coefficient is 0.5.

Use the double-angle identity to write the function directly as a cosine series:

\[ \begin{aligned} \cos^{2}x &= \frac{1 + \cos 2x}{2} \\ &= \frac{1}{2} + \frac{1}{2}\cos 2x \end{aligned} \]

Comparing with \(a_{0} + \sum a_{n}\cos nx\) gives \(a_{0} = 0.5\), \(a_{2} = 0.5\) and every other coefficient zero. The same result comes from \(a_{2} = \frac{2}{\pi}\int_{0}^{\pi}\cos^{2}x\cos 2x\,dx = 0.5\).

-0.5 has the wrong sign, which comes from using \(\sin^{2}x = \frac{1}{2} - \frac{1}{2}\cos 2x\) by mistake. 1.0 drops the factor 1/2, and 0 would mean cos²x had no cos 2x component.

Answer: B

The constant term is 2.

The constant term of a Fourier series is the average value of the function over one period. Here the period is 2 seconds: the function is 4 for one second and 0 for the next.

\[ \begin{aligned} a_{0} &= \frac{1}{2}\int_{0}^{2} F(t)\,dt \\ &= \frac{4 \times 1 + 0 \times 1}{2} = 2 \end{aligned} \]

4 is the peak value, not the mean; it would be the answer only if the wave stayed at 4 all the time. 1 and 3 are not the mean of this waveform, since the wave spends equal time at 4 and at 0. A symmetric square wave between 0 and 4 always averages to the midpoint, 2.

Answer: C

The value is \(\pi\).

By Dirichlet's theorem the Fourier series equals \(f(x)\) where f is continuous and the average of the left and right limits at a jump.

At \(x = -\pi/2\), f is continuous and equal to \(\pi + x\), so \(F(-\pi/2) = \pi - \pi/2 = \pi/2\). At \(x = 0\) there is a jump from \(f(0^-) = \pi\) to \(f(0^+) = 0\), so

\[ \begin{aligned} F(0) &= \frac{\pi + 0}{2} = \frac{\pi}{2} \\ F(-\tfrac{\pi}{2}) + F(0) &= \frac{\pi}{2} + \frac{\pi}{2} = \pi \end{aligned} \]

Taking \(F(0) = f(0) = 0\) gives \(\pi/2\), the trap of ignoring the jump, and taking \(F(0) = \pi\) gives \(3\pi/2\).