Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2023 / Paper 2 / 5

The signal is continuous at zero with $f(0)=1$. Its envelope is $e^{-\lambda |t|}$, but its angular frequency falls from $2$ on the incoming side to $1$ on the outgoing side. Its one-sided slopes are $f^{\prime }(0^{-})=\lambda$ and $f^{\prime }(0^{+})=-\lambda$, producing a cusp at the peak. Both tails decay to zero.

Use $\tilde{f}(\omega )={\int }_{\mathbb{R}}{e}^{-i\omega t}f(t)dt$. Expanding each cosine into two exponentials and integrating its half-line givesEquivalently,The positive $\lambda$ guarantees convergence of the Fourier transform integrals.

The Fourier transform of a derivative gives $(-{\omega }^2-1)\tilde{y}=-\tilde{f}$. For $\lambda =1$,This selects the solution decaying at both time infinities. Indeed, the inverse transform of $1/(1+{\omega }^2)$ is the one-dimensional modified Helmholtz Green function $K(t)=e^{-|t|}/2$, and the convolution theorem gives $y=K\ast f$. The printed equation has homogeneous solutions $e^t,e^{-t}$, so these two decay conditions matter; no causal initial-value condition was specified.

For $t<0$, evaluate Fourier inversion by closing the contour in the lower half-plane, where $e^{i\omega t}$ decays. The contour is clockwise. The enclosed simple poles and residues of $\tilde{y}$ areThus inversion equals $-i$ times the sum of the residues of $e^{i\omega t}\tilde{y}$, givingAs a check, applying $d^2/d{t}^2-1$ gives $-e^t\cos 2t$. The convolution also gives $y(t)=e^{-t}[\frac{1}{8}+\frac{1}{5}\cos t+\frac{2}{5}\sin t]$ for $t\ge 0$; the two pieces have matching values $13/40$ and matching first derivatives $3/40$ at zero.