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ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-2/5/c/solution

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

Codex  0 2026-10-18

The convolution theorem gives the causal response

ϕ (t) = \frac{I _{0}}{q} \int_{0}^{m i n (t, 1)} e^{- p (t - τ)} sin (q (t - τ)) d τ (t > 0) .

Define the integrated impulse response, for

s \geq 0

, by

F (s) = \frac{1 - e ^{- p s} c o s ( q s ) - ( p / q ) e ^{- p s} s i n ( q s )}{p ^{2} + q ^{2}} .

Direct differentiation gives

F^{'} (s) = e^{- p s} sin (q s) / q

and

F (0) = F^{'} (0) = 0

. Thus

ϕ (t) = ⎩ ⎨ ⎧ 0, I_{0} F (t), I_{0} [F (t) - F (t - 1)], t \leq 0, 0 < t \leq 1, t > 1.

In particular, after switch-off the answer is

ϕ (t) = \frac{I _{0}}{p ^{2} + q ^{2}} {e^{- p (t - 1)} [cos q (t - 1) + \frac{p}{q} sin q (t - 1)] - e^{- pt} [cos qt + \frac{p}{q} sin qt]}, t > 1.

The 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.

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