We are given the following process:
- $X_t = A\cos(\frac{\pi}{3}t)+B\sin(\frac{\pi}{3}t)$, where $A, B$ are uncorrelated random variables with mean $0$ and variance $1$;
and the following time-invariant linear filter:
- $\sum_{j=-\infty}^{\infty}\psi_jX_{t-j}$, with $\psi_0=1, \ \psi_1 = -2\alpha, \ \psi_2 = 1, \ \psi_j = 0$ for $j\not = 0,1,2$.
The question is: if you wish to use the filter to suppress sinusoidal oscillations with period 6, which value of $\alpha$ would you select?
Applying the filter, we obtain: $Y_t = \sum_{j=-\infty}^{\infty}\psi_jX_{t-j} = A\cos(\frac{\pi}{3}t)+B\sin(\frac{\pi}{3}t) -2 \alpha (A\cos(\frac{\pi}{3}(t-1))+B\sin(\frac{\pi}{3}(t-1))) + A\cos(\frac{\pi}{3}(t-2))+B\sin(\frac{\pi}{3}(t-2))$
but I do not know what is the most efficient way to proceed to answer the question above. Any suggestions or help would be appreciated. Thanks in advance.
When looking at the filtered signal $Y_t$, it is clear that the coefficient $\alpha$ affect the amplitude of your signal only. In addition, since the Fourier transform of sinus and cosinus are Dirac delta function, in the frequency domain you've frequency components at the angular frequency $\omega=\pm \pi/3$, where the minus sign is for the image frequency. So if you want suppress frequency oscillation with period $T=6$, i.e. $\omega=2 \pi/T=\pi/3$ nothing is left.