Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2025 / Paper 2 / 5

The Fourier transform of a derivative is $i\omega$ times the transform under the stated convention. Transforming the linear ordinary differential equation givesThe applied force is $F(t)=H(t)e^{-t}$, with $H$ the Heaviside step function, so $\tilde{F}=(1+i\omega {)}^{-1}$. Partial fractions giveUsing the transform pair $H(t)e^{-ct}\leftrightarrow (c+i\omega {)}^{-1}$ yieldsIts value and first derivative both vanish at $t=0^{+}$, so the joining introduces no spurious Dirac delta terms. Equivalently, this is the convolution of the force with the causal impulse response $H(t)(e^{-2t}-e^{-3t})$.

Write $q=i\omega$. A constant-coefficient linear ordinary differential equation has transfer function $1/P(q)$, where $P(q)={\sum }_{n=1}^c{a}_n{q}^n$. The specified simple poles correspond to $q=0,-2,1$. A minimal-order choice is thereforeThus one possible system isup to a common nonzero multiplicative constant. The root $q=1$ permits an exponentially growing homogeneous solution; the pole data therefore describe an unstable system. This does not prevent the requested algebraic reconstruction of its transfer function.