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For an absolutely summable linear filter of a stationary time series, the filter gain is the modulus of its frequency response $a(\lambda )={\sum }_r{a}_r{e}^{ir\lambda }$. It multiplies a sinusoidal amplitude, while its square multiplies the spectral density of a stationary process. A zero removes the corresponding frequency; the filter gain of a composition is the product of the two filter gains.

The transfer function is the filter generating function evaluated on the unit circle, with a sign convention matching the Fourier representation. Filtering a sinusoid of angular frequency $\lambda$ multiplies its complex amplitude by $H_A(\lambda )$. Its modulus is the filter gain; filtering a weakly stationary process multiplies its time-series spectral density by $|H_A(\lambda ){|}^2$. Using the opposite Fourier sign conjugates the response for real coefficients and leaves that squared modulus unchanged.

For differencing, the only nonzero filter coefficients are $a_0=1$ and $a_1=-1$. Therefore $a(\lambda )=1-e^{i\lambda }$ andThe left sketch shows the filter gain increasing from zero at frequency zero to two at $\pi$. This differencing filter removes a constant component and attenuates slowly varying components while relatively emphasizing higher frequencies. It is a high-pass filter, rather than an averaging or smoothing filter.

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.

The composition of absolutely summable time-series filters is $(1-B^{12})(1-B)$, with $B$ the backshift operator. Thusand its only nonzero coefficients are $c_0=1$, $c_1=-1$, $c_{12}=-1$, $c_{13}=1$. Multiplication of the two filter gains givesEquivalently, its spectral density transformation under a linear filter is $f_Z=16{\sin }^2(\lambda /2){\sin }^2(6\lambda )f_X$.

Use the two-sided spectral density of a stationary process conventionThus white noise of variance ${\sigma }^2$ has density ${\sigma }^2/(2\pi )$.

Absolute summability makes the linear filter of a stationary time series well defined in $L^2$: ${\sum }_r\Vert a_r{X}_{t-r}{\Vert }_2\le \Vert X_0{\Vert }_2{\sum }_r|a_r|<\infty$. Its mean is $m_Y=m_X{\sum }_r{a}_r$. For its autocovariance, continuity of the $L^2$ inner product givesThe sum is absolutely convergent because $|{\gamma }_X(j)|\le {\gamma }_X(0)$. It depends only on $k$, proving weak stationarity. Inserting the spectral inversion formula and exchanging the absolutely dominated sum with the integral yieldsHere the coefficients are real, so the two factors are complex conjugates. This is the spectral density transformation under a linear filter.

Applying a second linear filter of a stationary time series givesThe composition of absolutely summable time-series filters can also be calculated directly:Indeed ${\sum }_{\ell }|c_{\ell }|\le \Vert a{\Vert }_1\Vert b{\Vert }_1$; this bound also justifies the interchange of the two $L^2$ sums. The filter gain multiplies under this composition. The sketches for the two individual filters are: