Laplace Transforms Questions and Answers

Answer: A

The Laplace-domain solution is \(\dfrac{1}{s(ms^{2} + k)}\).

Take the Laplace transform with zero initial conditions. The second derivative transforms to \(s^{2}X(s)\) and the unit step to \(1/s\):

\[ \begin{aligned} ms^{2}X(s) + kX(s) &= \frac{1}{s} \\ X(s) &= \frac{1}{s(ms^{2} + k)} \end{aligned} \]

\(1/(ms^{2} + k)\) is the transfer function, which is the response to a unit impulse, not a step. \(s/(ms^{2} + k)\) and \(1/[s^{2}(ms^{2} + k)]\) correspond to the derivative of the impulse and to a ramp input respectively.

Answer: D

The values are A = 9, B = 12 and C = 2.

Write \(s^{2} = (s-3)^{2} + 6(s-3) + 9\) and split the fraction, then use \(\mathcal{L}^{-1}\{1/(s-a)^{k}\} = e^{at}t^{k-1}/(k-1)!\):

\[ \begin{aligned} \frac{s^{2}}{(s-3)^{3}} &= \frac{1}{s-3} + \frac{6}{(s-3)^{2}} + \frac{9}{(s-3)^{3}} \\ f(t) &= e^{3t}\left(1 + 6t + \tfrac{9}{2}t^{2}\right) \\ &= \frac{e^{3t}}{2}\left[9t^{2} + 12t + 2\right] \end{aligned} \]

Check at t = 0: the initial value theorem gives \(\lim s \cdot s^{2}/(s-3)^{3} = 1\), and \(C/2 = 1\) needs C = 2, which only the keyed option has.

Answer: B

The Laplace transform is \(24/s^{5}\).

The standard result for a power of t is

\[ \mathcal{L}\{t^{n}\} = \frac{n!}{s^{n+1}} \]

With \(n = 4\), \(n! = 24\) and the power of s is \(n + 1 = 5\), so \(\mathcal{L}\{t^{4}\} = 24/s^{5}\).

120 is \(5!\), the factorial for \(t^{5}\), so \(120/s^{5}\) and \(120/s^{4}\) use the wrong factorial. \(24/s^{4}\) has the right numerator but forgets that the power of s is one more than the power of t.

Answer: A

\(\frac{s}{s^{2} + a^{2}}\) is the Laplace transform of \(\cos at\).

From the definition, \(\mathcal{L}\{e^{iat}\} = \frac{1}{s - ia} = \frac{s + ia}{s^{2} + a^{2}}\). Its real part, \(\frac{s}{s^{2} + a^{2}}\), is the transform of \(\cos at\), and its imaginary part, \(\frac{a}{s^{2} + a^{2}}\), is the transform of \(\sin at\).

The hyperbolic functions have a minus sign in the denominator: \(\cosh at\) transforms to \(\frac{s}{s^{2} - a^{2}}\) and \(\sinh at\) to \(\frac{a}{s^{2} - a^{2}}\). So sin, sinh and cosh are all ruled out.

Answer: D

The correct equation is \(\dfrac{dY}{ds} + \dfrac{sY}{s^{2} + 1} = 0\).

\(J_{0}\) satisfies Bessel's equation \(xy'' + y' + xy = 0\) with \(y(0) = 1\), \(y'(0) = 0\). Using \(\mathcal{L}\{xf\} = -\frac{d}{ds}\mathcal{L}\{f\}\):

\[ \begin{aligned} \mathcal{L}\{xy''\} &= -\tfrac{d}{ds}(s^{2}Y - s) = -2sY - s^{2}Y' + 1 \\ \mathcal{L}\{y'\} &= sY - 1, \quad \mathcal{L}\{xy\} = -Y' \\ 0 &= -(s^{2} + 1)Y' - sY \end{aligned} \]

Dividing by \(-(s^{2} + 1)\) gives the equation. Its solution, \(Y = 1/\sqrt{s^{2} + 1}\), is the known transform of \(J_{0}\). The options with \(2sY\) come from dropping the \(sY\) of the \(y'\) term, and the minus signs reverse the result.

Answer: D

The inverse transform is \(f(t)=1-e^{-t}\).

Split into partial fractions and invert term by term:

\[ \begin{aligned} \frac{1}{s(s+1)} &= \frac{1}{s}-\frac{1}{s+1} \\ \mathcal{L}^{-1} &: 1 - e^{-t} \end{aligned} \]

A quick check: the final value theorem gives \(\lim_{s \to 0} sF(s) = 1\), and \(1-e^{-t}\) does tend to 1.

\(\sin t\) is the inverse of \(1/(s^{2}+1)\), \(e^{-t}\sin t\) the inverse of \(1/((s+1)^{2}+1)\), and \(e^{-t}\) the inverse of \(1/(s+1)\) alone, which is only one of the two partial fractions.

Answer: A

The inverse transform is \(\left(-1 + \frac{1}{2}e^{-t} + \frac{1}{2}e^{t}\right)u(t)\).

Factor the denominator and split into partial fractions:

\[ \begin{aligned} \frac{1}{s^{3} - s} &= \frac{1}{s(s - 1)(s + 1)} \\ &= -\frac{1}{s} + \frac{1/2}{s - 1} + \frac{1/2}{s + 1} \end{aligned} \]

The coefficients come from the cover-up rule: at \(s = 0\) the rest is \(1/((-1)(1)) = -1\); at \(s = 1\) it is \(1/(1 \times 2) = 1/2\); at \(s = -1\) it is \(1/((-1)(-2)) = 1/2\). Inverting term by term gives \(-1 + \frac{1}{2}e^{t} + \frac{1}{2}e^{-t}\) for \(t \ge 0\), that is \((\cosh t - 1)u(t)\).

The two options with \(u(t - 1)\) are delayed by one unit, which would need a factor \(e^{-s}\) in the transform; there is none. The option \(\left(\frac{1}{3}e^{-t} - e^{t}\right)u(t)\) fails a quick check: since the transform falls like \(1/s^{3}\), the function must start at 0, but that option starts at \(1/3 - 1 = -2/3\).

Answer: A

The transform is \(\frac{16s}{(s^{2}+4)(s^{2}+36)}\).

Use \(\sin A\sin B = \frac{1}{2}[\cos(A - B) - \cos(A + B)]\): \(\sin 2t\sin 4t = \frac{1}{2}(\cos 2t - \cos 6t)\). With \(L\{\cos at\} = s/(s^{2} + a^{2})\):

\[ \begin{aligned} F(s) &= \frac{1}{2}\left(\frac{s}{s^{2}+4} - \frac{s}{s^{2}+36}\right) \\ &= \frac{1}{2}\cdot\frac{32s}{(s^{2}+4)(s^{2}+36)} = \frac{16s}{(s^{2}+4)(s^{2}+36)} \end{aligned} \]

\(32s/\ldots\) forgets the factor \(\frac{1}{2}\). The options without \(s\) in the numerator would be transforms of sine terms; a product of two sines reduces to cosines, which bring the factor \(s\).