Fourier and Laplace Methods Questions and Answers

Q: Laplace transform F(s) of a function f(x) = 1, for \(0 \le x \le \infty\) is equal to :

The Laplace transform of f(x) = 1 is \(1/s\).\[ \begin{aligned} F(s) &= \int_{0}^{\infty} e^{-sx}\cdot 1\,dx \\ &= \left[-\frac{e^{-sx}}{s}\right]_{0}^{\infty} = \frac{1}{s} \end{aligned} \]valid for s > 0, where the exponential vanishes at the upper limit. \(1/s^{2}\) is the transform of f(x) = x, not of a constant. Neither s nor 2s can be a Laplace transform of an ordinary function at all, since a transform must tend to zero as s grows.

Q: Consider a differential equation\[ \begin{aligned} &\left(\frac{d^{4}}{dx^{4}} + 2\frac{d^{2}}{dx^{2}}\right)f(x) \\ &= g(x) \end{aligned} \]Let F(k) and G(k) be the Fourier transforms of f(x) and g(x) respectively. Then F(k) is :

\(F(k) = \frac{G(k)}{k^{4} - 2k^{2}}\).Under the Fourier transform, each derivative \(\frac{d}{dx}\) becomes multiplication by \(ik\) (with the usual \(e^{-ikx}\) convention):\[ \begin{aligned} \frac{d^{4}}{dx^{4}} &\to (ik)^{4} = k^{4} \\ 2\frac{d^{2}}{dx^{2}} &\to 2(ik)^{2} = -2k^{2} \\ (k^{4} - 2k^{2})F(k) &= G(k) \end{aligned} \]Dividing gives the second option. The forms with \(G\) multiplied, rather than divided, invert the relation; \(+2k^{2}\) forgets that \(i^{2} = -1\); and \(-2ik^{2}\) keeps a stray \(i\) from a single derivative.

Q: Laplace transform of (1 + sin (2t)) is :

The transform is \(\frac{1}{s} + \frac{2}{s^{2} + 4}\).The Laplace transform is linear, so transform each term:\[ \begin{aligned} \mathcal{L}\{1\} &= \frac{1}{s} \\ \mathcal{L}\{\sin at\} &= \frac{a}{s^{2} + a^{2}}, \quad a \\ &= 2 \\ \mathcal{L}\{1 + \sin 2t\} &= \frac{1}{s} + \frac{2}{s^{2} + 4} \end{aligned} \]\(\frac{1}{s^{2} + 1}\) uses \(a = 1\); \(\frac{4s}{s^{2} + 4}\) confuses the sine with the cosine transform \(\frac{s}{s^{2} + a^{2}}\); and the option starting with 1 forgets that \(\mathcal{L}\{1\} = \frac{1}{s}\).

Q: The inverse Laplace transform of \(\frac{2s^{2}}{s^{4} + 5s^{2} + 4}\) is :

The inverse transform is \(\frac{4}{3}\sin(2x) - \frac{2}{3}\sin(x)\), \(x \gt 0\).The denominator factors as \((s^{2}+1)(s^{2}+4)\). Write the fraction in partial fractions:\[ \begin{aligned} \frac{2s^{2}}{(s^{2}+1)(s^{2}+4)} &= \frac{A}{s^{2}+1} + \frac{B}{s^{2}+4} \\ 2s^{2} &= A(s^{2}+4) + B(s^{2}+1) \end{aligned} \]Putting \(s^{2} = -1\) gives \(-2 = 3A\), so \(A = -2/3\); putting \(s^{2} = -4\) gives \(-8 = -3B\), so \(B = 8/3\). Using \(\mathcal{L}^{-1}[1/(s^{2}+a^{2})] = \sin(ax)/a\):\[ \begin{aligned} f(x) &= -\frac{2}{3}\sin x + \frac{8}{3}\cdot\frac{\sin 2x}{2} \\ &= \frac{4}{3}\sin 2x - \frac{2}{3}\sin x \end{aligned} \]Check at small \(x\): the transform behaves like \(2/s^{2}\) for large \(s\), so \(f(x) \approx 2x\); indeed \(\frac{8}{3}x - \frac{2}{3}x = 2x\). The other sign and coefficient combinations fail this check.

11.

Laplace transform F(s) of a function f(x) = 1, for \(0 \le x \le \infty\) is equal to :

A.\(1/s\)
B.\(1/s^{2}\)
C.\(s\)
D.\(2s\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

The Laplace transform of f(x) = 1 is \(1/s\).

\[ \begin{aligned} F(s) &= \int_{0}^{\infty} e^{-sx}\cdot 1\,dx \\ &= \left[-\frac{e^{-sx}}{s}\right]_{0}^{\infty} = \frac{1}{s} \end{aligned} \]

valid for s > 0, where the exponential vanishes at the upper limit. \(1/s^{2}\) is the transform of f(x) = x, not of a constant. Neither s nor 2s can be a Laplace transform of an ordinary function at all, since a transform must tend to zero as s grows.

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12.

A function f(x) satisfies the differential equation \(\frac{d^{2}f}{dx^{2}} - \omega^{2}f(x) = -\delta(x - a)\), where ω is positive. The Fourier transform \(F(k) = \int_{-\infty}^{\infty} f(x)e^{ikx}dx\) and solution f(x) of the differential equations are respectively :

A.\(\frac{e^{ika}}{k^{2} - \omega^{2}}\) and \(\frac{1}{2\omega}\left(e^{-i\omega|x-a|} + e^{i\omega|x-a|}\right)\)
B.\(\frac{e^{ika}}{k^{2} - \omega^{2}}\) and \(\frac{1}{2i\omega}\left(e^{-i\omega|x-a|} - e^{i\omega|x-a|}\right)\)
C.\(\frac{e^{ika}}{k^{2} + \omega^{2}}\) and \(\frac{1}{2\omega}\left(e^{-\omega|x-a|} + e^{\omega|x-a|}\right)\)
D.\(\frac{e^{ika}}{k^{2} + \omega^{2}}\) and \(\frac{1}{2\omega}e^{-\omega|x-a|}\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: D

The transform and solution are \(\frac{e^{ika}}{k^{2} + \omega^{2}}\) and \(\frac{1}{2\omega}e^{-\omega|x-a|}\).

Multiply the equation by \(e^{ikx}\) and integrate. Each derivative brings down a factor \(-ik\), so \(f''\) becomes \(-k^{2}F\), and the delta function picks out \(e^{ika}\):

\[ \begin{aligned} -k^{2}F - \omega^{2}F &= -e^{ika} \\ F(k) &= \frac{e^{ika}}{k^{2} + \omega^{2}} \end{aligned} \]

Away from x = a the equation \(f'' = \omega^{2}f\) has solutions \(e^{\pm\omega x}\); keeping the decaying one on each side gives \(Ce^{-\omega|x-a|}\). The jump in slope at x = a must equal -1, which fixes \(C = 1/2\omega\). The options with \(k^{2} - \omega^{2}\) belong to the opposite sign of the \(\omega^{2}\) term, and the option with \(e^{+\omega|x-a|}\) grows without limit, so it is not an acceptable solution.

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13.

Consider a differential equation

\[ \begin{aligned} &\left(\frac{d^{4}}{dx^{4}} + 2\frac{d^{2}}{dx^{2}}\right)f(x) \\ &= g(x) \end{aligned} \]

Let F(k) and G(k) be the Fourier transforms of f(x) and g(x) respectively. Then F(k) is :

A.\(F(k) = (k^{4} - 2k^{2})G(k)\)
B.\(F(k) = \frac{G(k)}{(k^{4} - 2k^{2})}\)
C.\(F(k) = (k^{4} + 2k^{2})G(k)\)
D.\(F(k) = \frac{G(k)}{(k^{4} - 2ik^{2})}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

\(F(k) = \frac{G(k)}{k^{4} - 2k^{2}}\).

Under the Fourier transform, each derivative \(\frac{d}{dx}\) becomes multiplication by \(ik\) (with the usual \(e^{-ikx}\) convention):

\[ \begin{aligned} \frac{d^{4}}{dx^{4}} &\to (ik)^{4} = k^{4} \\ 2\frac{d^{2}}{dx^{2}} &\to 2(ik)^{2} = -2k^{2} \\ (k^{4} - 2k^{2})F(k) &= G(k) \end{aligned} \]

Dividing gives the second option. The forms with \(G\) multiplied, rather than divided, invert the relation; \(+2k^{2}\) forgets that \(i^{2} = -1\); and \(-2ik^{2}\) keeps a stray \(i\) from a single derivative.

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14.

Laplace transform of (1 + sin (2t)) is :

A.\(\frac{1}{s} + \frac{1}{s^{2} + 1}\)
B.\(\frac{1}{s} + \frac{2}{s^{2} + 4}\)
C.\(\frac{1}{s} + \frac{4s}{s^{2} + 4}\)
D.\(1 + \frac{1}{s^{2} + 4}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

The transform is \(\frac{1}{s} + \frac{2}{s^{2} + 4}\).

The Laplace transform is linear, so transform each term:

\[ \begin{aligned} \mathcal{L}\{1\} &= \frac{1}{s} \\ \mathcal{L}\{\sin at\} &= \frac{a}{s^{2} + a^{2}}, \quad a \\ &= 2 \\ \mathcal{L}\{1 + \sin 2t\} &= \frac{1}{s} + \frac{2}{s^{2} + 4} \end{aligned} \]

\(\frac{1}{s^{2} + 1}\) uses \(a = 1\); \(\frac{4s}{s^{2} + 4}\) confuses the sine with the cosine transform \(\frac{s}{s^{2} + a^{2}}\); and the option starting with 1 forgets that \(\mathcal{L}\{1\} = \frac{1}{s}\).

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15.

The inverse Laplace transform of \(\frac{2s^{2}}{s^{4} + 5s^{2} + 4}\) is :

A.\(\frac{4}{3}\sin(2x) - \frac{2}{3}\sin(x), x \gt 0\)
B.\(\frac{2}{3}\sin(2x) - \frac{4}{3}\sin(x), x \gt 0\)
C.\(\frac{2}{3}\sin(2x) + \frac{4}{3}\sin(x), x \gt 0\)
D.\(\frac{4}{3}\sin(2x) + \frac{2}{3}\sin(x), x \gt 0\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The inverse transform is \(\frac{4}{3}\sin(2x) - \frac{2}{3}\sin(x)\), \(x \gt 0\).

The denominator factors as \((s^{2}+1)(s^{2}+4)\). Write the fraction in partial fractions:

\[ \begin{aligned} \frac{2s^{2}}{(s^{2}+1)(s^{2}+4)} &= \frac{A}{s^{2}+1} + \frac{B}{s^{2}+4} \\ 2s^{2} &= A(s^{2}+4) + B(s^{2}+1) \end{aligned} \]

Putting \(s^{2} = -1\) gives \(-2 = 3A\), so \(A = -2/3\); putting \(s^{2} = -4\) gives \(-8 = -3B\), so \(B = 8/3\). Using \(\mathcal{L}^{-1}[1/(s^{2}+a^{2})] = \sin(ax)/a\):

\[ \begin{aligned} f(x) &= -\frac{2}{3}\sin x + \frac{8}{3}\cdot\frac{\sin 2x}{2} \\ &= \frac{4}{3}\sin 2x - \frac{2}{3}\sin x \end{aligned} \]

Check at small \(x\): the transform behaves like \(2/s^{2}\) for large \(s\), so \(f(x) \approx 2x\); indeed \(\frac{8}{3}x - \frac{2}{3}x = 2x\). The other sign and coefficient combinations fail this check.

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