Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 1 / 14D / b

Taking Laplace transforms and using $y(0)=0$ givesHence the transfer function isIts pole is at $-1/\alpha$, and its causal impulse response isIt is absolutely integrable exactly when

For $\alpha >0$,Its only pole is $-(1+k)/\alpha$, so direct inspection gives stability exactly whenapart from the excluded boundary.

For the Nyquist stability criterion, let $L(s)=kG(s)$. The open loop has no right-half-plane poles, so $P=0$. On the clockwise right-half-plane contour, the winding number $N$ of $1+L$ about zero, equivalently of $L$ about $-1$, satisfieswhere $Z$ is the number of unstable closed-loop poles. The Nyquist locusis the circle with diameter joining $0$ and $k$ on the real axis, completed by its conjugate half.

If $k>-1$, this circle does not wind around $-1$, so $N=0$, $Z=0$, and the loop is stable. If $k<-1$, it winds once with $N=-1$, so $Z=1$, and the loop is unstable. Nyquist therefore gives precisely the same condition $k>-1$.