 Solution

The seasonal difference operator has $b_0=1$, $b_{12}=-1$ and all other coefficients zero. Its frequency response is $1-e^{12i\lambda }$, soThe right sketch shows zeros at $\lambda =j\pi /6$ for $j=0,\dots ,6$, and maxima of height two at $(2j+1)\pi /12$ for $j=0,\dots ,5$. The seasonal difference operator removes every exactly twelve-periodic component, including a constant. Its filter gain has a comb pattern: it selectively suppresses seasonal frequencies, rather than increasing monotonically with frequency.