= Solution
Center the five coefficients at lags $-2,-1,0,1,2$. Their transfer function is real:
$$
H(\omega)=\frac{-e^{-2i\omega}+2e^{-i\omega}+4+2e^{i\omega}-e^{2i\omega}}6
=\frac{4+4\cos\omega-2\cos(2\omega)}6.
$$
Choosing consecutive causal lags instead only adds a phase and leaves the squared modulus unchanged. After $k$ applications, the <white noise> <spectral density> has become
$$
\boxed{f_k(\omega)=\frac{\sigma^2}{2\pi}|H(\omega)|^{2k}.}
$$
Write $c=\cos\omega$. Completing the square gives
$$
H(\omega)=1+\frac23c-\frac23c^2
=\frac76-\frac23(c-\tfrac12)^2.
$$
For $-1\leq c\leq1$, its range is $[-1/3,7/6]$, and its absolute maximum is $7/6$, attained exactly when $c=1/2$. Consequently, on the conventional nonnegative frequency range $[0,\pi]$,
$$
\boxed{\frac{f_k(\omega)}{f_k(\pi/3)}
=\left(\frac{|H(\omega)|}{7/6}\right)^{2k}\longrightarrow0
\quad\text{for }\omega\ne\pi/3.}
$$
On the two-sided range $[-\pi,\pi]$ there is also a peak at $-\pi/3$; the ratio there is one for every $k$. This is a counterexample to the claim if read literally on a signed frequency domain. The correct two-sided exclusion is $\omega\ne\pm\pi/3$.
The result illustrates how <repeated symmetric filtering can amplify an oscillation>. Although $H(0)=1$ preserves a constant component, $H(\pi/3)=7/6>1$ amplifies period-six oscillations. Repetition selects an increasingly narrow frequency band around those peaks; it does not simply smooth all noise away. In fact the unnormalized output <variance> eventually grows without bound, since a fixed interval around either peak has $|H|>1$.
Back to article page
