Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / 4

Use the Fourier transform convention $\tilde{a}(\omega )={\int }_{-\infty }^{\infty }a(t)e^{-i\omega t}dt$, with inverse factor $1/(2\pi )$. Direct integration over the pulse givesThe apparent singularity at zero is removable. Changing the values of the pulse at its two endpoints does not affect the transform.

The filter is a real even triangular function. Its Fourier transform iswith $\tilde{f}(0)=1$. Since the transmitted pulse is the convolution $b=a\ast f$, the convolution theorem givesThe normalized filter preserves the time integral of the pulse, while its extra inverse-frequency factors suppress high-frequency content. It is a smoothing filter rather than an exact compact-bandwidth cutoff.

By the Fourier modulation and translation identity, delaying the pulse by $\tau$ multiplies its transform by $e^{-i\omega \tau }$. HenceThe removable value at zero is $2(1+\alpha )$. For real $\alpha$, the squared spectral magnitude is multiplied by $1+{\alpha }^2+2\alpha \cos (\omega \tau )$, showing the interference between the direct and reflected signals.

The Fourier transform of a Gaussian givesThe Fourier transform of a derivative obeys $\mathcal{F}[q^{\prime \prime }]=-{\omega }^2\tilde{q}$, so multiplication of the spectrum by $1+ϵ{\omega }^2$ corresponds to $q-ϵq^{\prime \prime }$. Since $q^{\prime \prime }=(2/\sqrt{\pi })(4t^2-2)e^{-t^2}$,This is real and even, as required by the given spectrum. Its time integral is two for every $ϵ$, consistently with $\tilde{s}(0)=2$.