How a Filter Changes Frequency Content
A linear filter combines present and past values of a sequence. At each frequency, that combination has a complex gain: it can change both amplitude and phase. The spectral measure records how much variance is present at each frequency, so the variance contribution is multiplied by the squared amplitude of that gain. This tutorial derives that rule and shows how it works for both continuous spectral densities and discrete spectral lines.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), and let \((X_n)_{n\in\mathbb{Z}}\) be a real-valued, weakly stationary process with state space \(\mathbb{R}\) and discrete time \(n\in\mathbb{Z}\). Weak stationarity means that \(\mathbb{E}[X_n]=m_X\) does not depend on \(n\), and that \(\operatorname{Cov}(X_{n+h},X_n)=\gamma_X(h)\) depends only on the lag \(h\). Use the Fourier convention established in the earlier material: the spectral measure \(F_X\) is a finite, nonnegative measure on \([-\pi,\pi)\) satisfying \(\gamma_X(h)=\int_{[-\pi,\pi)}e^{ih\lambda}F_X(d\lambda)\). In particular, \(F_X([-\pi,\pi))=\gamma_X(0)\).
Let \((a_j)_{j\geq0}\) be real filter coefficients satisfying \(\sum_{j\geq0}|a_j|<\infty\). Define the filtered output and its transfer function by
The absolute summability assumption ensures that the output is well-defined in mean square. Indeed, the centered terms have \(L^2\) norm \(\sqrt{\gamma_X(0)}\), so the \(L^2\) norm of a tail is at most \(\sqrt{\gamma_X(0)}\sum_{j>m}|a_j|\), which tends to zero. The mean also converges, since it is \(m_X\sum_j a_j\). Thus \(\mathbb{E}[Y_n]=m_X\sum_j a_j\). The transfer series converges uniformly, because its tail is bounded in absolute value by \(\sum_{j>m}|a_j|\).
Proof. First truncate the filter at \(m\), writing \(Y_n^{(m)}=\sum_{j=0}^{m}a_jX_{n-j}\). Its mean is independent of \(n\). For any integer \(h\), bilinearity of covariance gives
Each term depends on \(h\), not on \(n\), so the truncated output is weakly stationary. By the spectral representation of \(\gamma_X\), the right side equals
where \(H_m(\lambda)=\sum_{j=0}^{m}a_je^{-ij\lambda}\); the last equality uses real coefficients, so the second sum is the complex conjugate of the first. As \(m\) increases, \(Y_n^{(m)}\) converges in \(L^2\) to \(Y_n\), and hence its mean and covariance converge to those of \(Y_n\). Also \(H_m\) converges uniformly to \(H\). The functions \(H_m\) are uniformly bounded by \(\sum_j|a_j|\), so \(|H_m|^2\) converges uniformly to \(|H|^2\). Since \(F_X\) is finite, the displayed integrals converge to \(\int e^{ih\lambda}|H(\lambda)|^2F_X(d\lambda)\). This expression depends only on \(h\), proving weak stationarity and the asserted covariance representation. The measure \(|H|^2F_X\) is finite and nonnegative, and therefore is the spectral measure of the output. For a density, the measure identity gives the stated density formula. \(\square\)
The squared modulus is essential. A complex transfer value describes both amplitude and phase, but spectral mass is nonnegative variance. The phase disappears when a gain is multiplied by its complex conjugate. In particular, filtering can reweight existing frequencies; it does not introduce spectral mass at frequencies absent from the input measure.
Proof. For each Borel set \(C\), the filtering theorem gives \(F_Y(C)=\int_C|H(\lambda)|^2F_X(d\lambda)\). Substituting the given decomposition splits this integral into \(\int_C|H|^2f_X\,d\lambda\), \(\sum_{\lambda_r\in C}c_r|H(\lambda_r)|^2\), and \(\int_C|H|^2F_s(d\lambda)\). These are respectively the density, atomic, and singular contributions stated. Since a bounded continuous function times a measure supported on a set of Lebesgue measure zero is still supported on that set, the last contribution remains singular; since \(F_s\) has no atoms, weighting it by \(|H|^2\) creates none. \(\square\)
Worked Example: Filtering White Noise
Let \((X_n)\) be mean-zero white noise with variance \(\sigma^2\), so \(\gamma_X(0)=\sigma^2\) and \(\gamma_X(h)=0\) for \(h\ne0\). Its spectral density under the stated convention is \(f_X(\lambda)=\sigma^2/(2\pi)\). Apply the filter \(Y_n=X_n+\tfrac12X_{n-1}\).
The transfer function is \(H(\lambda)=1+\tfrac12e^{-i\lambda}\). Multiplying by its conjugate gives
Therefore \(f_Y(\lambda)=\tfrac{\sigma^2}{2\pi}(\tfrac54+\cos\lambda)\). Check this against the covariance calculation. Expanding the two terms in \(Y_n\), the variance is \(\operatorname{Var}(X_n)+\tfrac14\operatorname{Var}(X_{n-1})=\sigma^2+\tfrac14\sigma^2=\tfrac54\sigma^2\). At lag \(1\), the only shared white-noise term is \(X_n\), with coefficient \(1\) in \(Y_n\) and \(\tfrac12\) in \(Y_{n+1}\), so \(\gamma_Y(1)=\tfrac12\sigma^2\); all lags with absolute value greater than \(1\) have covariance zero. These are exactly the Fourier coefficients of \(\tfrac{\sigma^2}{2\pi}(\tfrac54+\cos\lambda)\).
Worked Example: A Filtered Periodic Component
Let \(\Theta\) be uniform on \([0,2\pi)\), and set \(X_n=c\cos(n\pi/3+\Theta)\), where \(c\) is a real constant. This process has mean zero. Averaging over the uniform phase gives \(\gamma_X(h)=\tfrac{c^2}{2}\cos(h\pi/3)\), so its spectral measure has mass \(c^2/4\) at each of \(\pi/3\) and \(-\pi/3\). Apply \(Y_n=X_n+X_{n-1}\).
Here \(H(\lambda)=1+e^{-i\lambda}\), so \(|H(\lambda)|^2=2+2\cos\lambda\). At either spectral line, \(\cos(\pi/3)=\tfrac12\), giving \(|H(\pi/3)|^2=|H(-\pi/3)|^2=2+2(\tfrac12)=3\). The output measure therefore has mass \(3c^2/4\) at each of the same two frequencies.
Check directly using the trigonometric sum: \(Y_n=2c\cos(\pi/6)\cos(n\pi/3+\Theta-\pi/6)=\sqrt{3}c\cos(n\pi/3+\Theta-\pi/6)\), since \(2\cos(\pi/6)=\sqrt{3}\). Its covariance is \(\tfrac{3c^2}{2}\cos(h\pi/3)\). The two output spectral masses sum to \(3c^2/2\), its variance, and their locations give the stated cosine covariance.
Worked Example: Filtering Geometric Covariance
Suppose a mean-zero weakly stationary input has covariance \(\gamma_X(h)=(\tfrac12)^{|h|}\). Its spectral density is \(f_X(\lambda)=\tfrac{3}{2\pi(5-4\cos\lambda)}\), as follows by summing the geometric Fourier series. Apply \(Y_n=X_n+\tfrac12X_{n-1}\). The transfer calculation from the first example gives
Verify the result from covariances. Expanding the filter gives, for every integer \(h\), \(\gamma_Y(h)=\tfrac54\gamma_X(h)+\tfrac12\gamma_X(h+1)+\tfrac12\gamma_X(h-1)\). At \(h=0\), this is \(\tfrac54+\tfrac12(\tfrac12)+\tfrac12(\tfrac12)=\tfrac74\). At \(h=1\), it is \(\tfrac54(\tfrac12)+\tfrac12(\tfrac14)+\tfrac12(1)=\tfrac58+\tfrac18+\tfrac12=\tfrac54\). For \(h\geq2\), \(\gamma_X(h+1)=\tfrac12\gamma_X(h)\) and \(\gamma_X(h-1)=2\gamma_X(h)\), so \(\gamma_Y(h)=(\tfrac54+\tfrac14+1)\gamma_X(h)=\tfrac52(\tfrac12)^h\). For negative \(h\), covariance symmetry gives \(\gamma_Y(h)=\gamma_Y(-h)\). These values are the covariance coefficients of the density displayed above, as guaranteed by the filtering formula.
Using the Formula Carefully
The formula is useful because it turns an operation on an entire covariance sequence into pointwise multiplication in frequency. If an input has a density, the output density follows by multiplying by \(|H|^2\). If the input has spectral lines, each line’s mass is multiplied by the squared gain at its own frequency. For an input with both kinds of components, the same measure formula handles them together.
A phase change alone does not change spectral mass. For example, a pure delay has transfer function \(H(\lambda)=e^{-ir\lambda}\) for an integer \(r\), and \(|H(\lambda)|^2=1\); it shifts the sequence in time without changing its spectrum. More generally, two filters with the same squared transfer-function modulus have the same effect on the spectral measure, even if their phases differ.
Common mistake. Replacing \(|H(\lambda)|^2\) by \(H(\lambda)^2\) is incorrect. For the filter \(H(\lambda)=1+\tfrac12e^{-i\lambda}\), at \(\lambda=\pi/2\) one has \(H(\pi/2)=1-\tfrac12i\). Squaring gives \(H(\pi/2)^2=\tfrac34-i\), which is complex and cannot be a nonnegative spectral weight. The correct calculation is \(H(\pi/2)\overline{H(\pi/2)}=(1-\tfrac12i)(1+\tfrac12i)=\tfrac54\).
The assumptions also matter. Absolute summability supplies a convenient sufficient condition for mean-square convergence and a uniformly convergent transfer series; it is not a claim that every sequence of coefficients defines a filter for every stationary input. When a different convergence condition is used, convergence of the filtered process and the corresponding transfer-function limit must be justified separately.
Check Your Understanding
Use the Fourier convention and definitions stated above, and give a brief justification for each answer.
- For \(Y_n=X_n-X_{n-1}\), calculate \(H(\lambda)\) and \(|H(\lambda)|^2\).
- If the input has an atom of mass \(c\) at frequency \(\lambda_0\), what is the output mass at that frequency?
- Why does a pure delay leave the spectral measure unchanged?
- For white noise of variance \(\sigma^2\) filtered by \(Y_n=X_n+\tfrac12X_{n-1}\), what is the output variance? Verify it from the filter coefficients.
- For the same two-term filter, explain why \(H(\lambda)^2\) cannot replace \(|H(\lambda)|^2\) in the spectral formula.