Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 80 / 4 / b / Solution

For Fourier stability analysis, take spatial modes $e^{im\theta }$ and putA bounded implicit step requires $D$ to stay away from zero, and von Neumann stability analysis requires $|G|\le 1$ for every frequency. Expanding the squared moduli and using $\cos 2\theta =2{\cos }^2\theta -1$ givesConsequently, apart from denominator invertibility, the necessary and sufficient sign condition isIts real solution set is $(-\infty ,-1]\cup [0,1]\cup [2,\infty )$.

To check the denominator rather than assume it, noteIf $\mu \ne 1/2$, a zero can occur only at zero or the Nyquist frequency. The zero-frequency value is nonzero, and the Nyquist zeros occur only at $\mu =(1\pm \sqrt{3})/2$, both outside the proposed stable intervals. At $\mu =1/2$, $D=9/4+(3/4)\cos \theta \ge 3/2$, also nonzero. Continuity on the frequency circle now gives a positive lower bound on $|D|$ for each allowed $\mu$.

The implicit convolution operator is therefore boundedly invertible on the square-summable grid sequences. By the Plancherel theorem, $|G|\le 1$ givesOutside these intervals the modulus difference is negative at every nonzero frequency where $D$ is nonzero. There is an interval on which $|G|>1$, giving arbitrarily large growth of powers of the step operator, so the method is unstable; a zero of $D$ additionally makes the step ill posed.

For the physical choice $k,h>0$, the stable range is $0<\mu \le 1$ or $\mu \ge 2$; zero is the limiting zero-step case. The exact-shift values have unit gain and are included.