Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2024 / Paper 2 / 5

Use the Fourier transform conventionThe convolution is $(f\ast g)(t)={\int }_{-\infty }^{\infty }f(\tau )g(t-\tau )d\tau$. For integrable $f,g$, Fubini's theorem and the substitution $s=t-\tau$ giveHence the convolution theorem is

Take the causal solution, extending the input and output by zero before the device starts at $t=0$. The two zero initial conditions ensure that this extension adds no impulsive terms to the first two distributional derivatives. The Fourier transform of a derivative is multiplication by $i\omega$, soFactoring the quadratic and taking partial fraction decomposition yieldsThe inverse Fourier transform follows directly by integrating decaying exponentials:where $H$ is the Heaviside step function. This damped-oscillator impulse response has $R(0^{+})=0$ and $R^{\prime }(0^{+})=1$, as required for a unit impulse.

The convolution theorem gives the causal responseDefine the integrated impulse response, for $s\ge 0$, byDirect differentiation gives $F^{\prime }(s)=e^{-ps}\sin (qs)/q$ and $F(0)=F^{\prime }(0)=0$. ThusIn particular, after switch-off the answer isThe response and its first derivative are continuous at both switching times, and the post-pulse motion is a freely decaying damped harmonic oscillator. The expression described as valid for $t<1$ in the paper is understood for $0<t<1$; before the pulse the causal output is zero.

Use the same Fourier transform convention, now in the sense of tempered distributions, since the homogeneous oscillations are not integrable. Transforming givesSince $q>0$, the two roots are distinct. A distribution annihilated by this polynomial is a linear combination of Dirac delta distributions at its roots:Fourier inversion therefore givesThe constants may be complex. Differentiating the factor $e^{-irt}$ also verifies that the original ordinary differential equation reduces to $\ddot{\psi }+q^2\psi =0$ for $\varphi =e^{-irt}\psi$.