Periodogram

Periodogram

The periodogram of a length-$n$ time series $x_1,\dots ,x_n$ is the rescaled squared modulus of the DFT at every Fourier frequency ${\omega }_j=j/n$:
$P({\omega }_j)=\frac{4}{n}|d({\omega }_j){|}^2=\frac{4}{n^2}[({\sum }_t{x}_t\cos 2\pi t{\omega }_j{)}^2+({\sum }_t{x}_t\sin 2\pi t{\omega }_j{)}^2].$
It is the sample estimate of how power is distributed across frequencies, the frequency-domain analogue of the sample autocovariance in the time domain. By the symmetry of the DFT, $P({\omega }_j)=P(1-{\omega }_j)$ for real-valued data, so plots conventionally show only the half-band $\omega \in [0,1/2]$ and double the value at every interior frequency to preserve total power.

The periodogram estimates the spectral density $f(\omega )$: the population quantity that says how variance is distributed across frequencies in the data-generating process. The estimator is asymptotically unbiased (the average of many periodograms approaches $f(\omega )$) but is not consistent at any individual frequency: its variance does not shrink as the sample size grows. The raw periodogram therefore stays jagged regardless of how much data is collected, and consistent estimation requires either smoothing across nearby frequencies or averaging across non-overlapping windows.

Under Gaussian white noise with variance ${\sigma }^2$, the rescaled periodogram values $2P({\omega }_j)/f({\omega }_j)$ are approximately distributed as ${\chi }_2^2$ random variables, independently across distinct interior Fourier frequencies. Since ${\chi }_2^2$ has variance $4$, the standard deviation of $P({\omega }_j)$ is approximately $f({\omega }_j)$ at every frequency, a $100\%$ relative error that does not improve with $n$.

Inconsistency of the Raw Periodogram $n$. Consistent spectral-density estimation requires either smoothing (averaging $P({\omega }_j)$ across $L$ neighboring Fourier frequencies, which reduces variance by $1/L$ at the cost of resolving power $1/(nL)$) or Welch's method, which averages periodograms across overlapping windows of the record.

Adding more data does not make the periodogram smoother. Each Fourier frequency receives only two degrees of freedom (the sine and cosine coefficients), so each estimate sees a constant amount of information regardless of

Intuition

The periodogram is to the spectral density what the histogram is to a probability density: a sample-based decomposition of how the data’s variability is distributed, here across frequencies instead of across values. A spike in $P({\omega }_j)$ at some frequency ${\omega }^{\ast }$ says “a sinusoid at frequency ${\omega }^{\ast }$ explains a substantial part of the variation in the observed series.” More precisely, $P({\omega }_j)$ equals (up to scaling) the residual-sum-of-squares improvement when adding the $\cos (2\pi t{\omega }_j)$ and $\sin (2\pi t{\omega }_j)$ pair to a regression, so the periodogram is, literally, a regression diagnostic in disguise.

The roughness of the raw periodogram is not a bug to patch away with more data; it is the unavoidable price of resolving frequency at all. With $n$ samples there are $n/2$ frequency bins, each receiving exactly two degrees of freedom (the orthogonal sine-cosine pair). Increasing $n$ creates more bins rather than tightening the existing ones. To stabilize individual frequency estimates the analyst must pool information across nearby frequencies (smoothing) or across record segments (averaging), both of which trade frequency resolution for variance reduction.

Properties

(Periodogram-Autocovariance Identity) The periodogram is the discrete Fourier transform of the sample autocovariance.

For mean-centered data,
$P({\omega }_j)={\sum }_{h=-(n-1)}^{n-1}\hat{\gamma }(h)e^{-2\pi ih{\omega }_j},\hat{\gamma }(h)=\frac{1}{n}{\sum }_t(x_t-\bar{x})(x_{t+|h|}-\bar{x}).$
Up to normalization, the periodogram is the DFT of the sample autocovariance function. This is the sample-side reflection of the population identity $f(\omega )={\sum }_h\gamma (h)e^{-2\pi ih\omega }$, and is why the time-domain and frequency-domain analyses contain “the same information” rearranged.

(Symmetry) For real-valued data, $P({\omega }_j)=P(1-{\omega }_j)$.

The squared modulus inherits the conjugate symmetry of the DFT, so all the information is contained in the half-band $[0,1/2]$. Plots typically show this half and double the values at interior frequencies (so that integration over the visible band recovers the total variance). The Nyquist frequency $\omega =1/2$ is the highest oscillation that a discretely sampled series can express.

(Sampling Distribution) Under Gaussian white noise, $\{2P({\omega }_j)/f({\omega }_j)\}$ are approximately i.i.d. ${\chi }_2^2$ across distinct Fourier frequencies.

A ${\chi }_2^2$ has mean $2$ and variance $4$, so $\mathbb{E}[P({\omega }_j)]\approx f({\omega }_j)$ (asymptotically unbiased) but $\mathrm{SD}(P({\omega }_j))\approx f({\omega }_j)$ (relative error of $\approx 100\%$). Approximate $(1-\alpha )$ confidence bands at each frequency are
$[\frac{2P({\omega }_j)}{{\chi }_{2,1-\alpha /2}^2},\frac{2P({\omega }_j)}{{\chi }_{2,\alpha /2}^2}],$
which are wide because the ${\chi }_2^2$ distribution is heavy-tailed. A practical consequence: a single periodogram peak that fails to clear a $5\%$ confidence band is weak evidence of a true spectral feature.

(Smoothed Periodogram) Averaging over $L$ neighboring Fourier frequencies stabilizes the estimate.

The smoothed periodogram $\bar{P}({\omega }_j)=\frac{1}{L}{\sum }_{k=-m}^{m}P({\omega }_{j+k})$ with $L=2m+1$ has approximate distribution $\bar{P}({\omega }_j)\cdot 2L/f({\omega }_j)\sim {\chi }_{2L}^2/L$. The variance shrinks by a factor of $1/L$, but the resolving power degrades by $L$: too little smoothing leaves the estimate noisy, too much washes out genuine sharp peaks. Choosing $L$ is a bias-variance trade-off, typically guided by visual inspection or by the expected width of the spectral features being sought.

(Welch’s Method) Splitting the record into $K$ overlapping segments, computing periodograms separately, and averaging gives a consistent spectral estimator.

Welch’s method has effective degrees of freedom $\approx 2K$, so its variance is $K$ times smaller than the raw periodogram. Overlapping windows (typically $50\%$) recover information lost at segment boundaries by the windowing taper. This is the production-grade spectral estimator implemented in scipy.signal.welch and is what most signal-processing pipelines actually use.

(Bias from Trends and Means) Mean and trend dominate the low-frequency end.

If the series has a non-zero mean or a polynomial trend, $P({\omega }_j)$ near $\omega =0$ inflates dramatically and can mask higher-frequency structure. Standard pre-processing for spectral analysis is therefore to subtract the sample mean, optionally remove a low-order polynomial trend by regression, and sometimes apply a window taper (Hanning, Hamming) to reduce edge effects before computing the DFT. Skipping these steps produces a periodogram dominated by a giant spike at $\omega =0$ that has nothing to do with the spectral structure of interest.

Examples

Periodogram of white noise

For $x_1,\dots ,x_n\stackrel{\text{iid}}{\sim }\mathcal{N}(0,{\sigma }^2)$, the true spectral density is the constant $f(\omega )={\sigma }^2$. What does the sample periodogram look like, and why doesn’t it converge to a flat line?

Solution $P({\omega }_j)$ is approximately ${\sigma }^2\cdot {\chi }_2^2/2$, so the expected value at every Fourier frequency is ${\sigma }^2$, correct on average. But the standard deviation is also $\approx {\sigma }^2$, so individual periodogram values fluctuate by roughly $\pm 100\%$ around the true level no matter how large $n$ is. A plot looks like a noisy fence at level ${\sigma }^2$ rather than a flat line, with occasional spikes that can be mistaken for spectral features. Smoothing across $L=11$ neighboring frequencies reduces the standard deviation by $\sqrt{11}\approx 3.3\times$ and produces a visibly flat estimate.

Each

Detecting a hidden sinusoid

For $x_t=R\cos (2\pi f_{0}t+\varphi )+w_t$ with $w_t\sim \mathrm{WN}(0,{\sigma }^2)$ and $f_0=k/n$ on the Fourier grid, how strong does the sinusoid need to be to show up against the noise floor?

Solution $\approx n{R}^2/4$ at ${\omega }_j=f_0$ (and its mirror $1-f_0$). The noise contributes the white-noise periodogram described above, with mean ${\sigma }^2$ at every bin. The spike is visible against the noise floor when its height exceeds the typical noise excursion, roughly $n{R}^2/4\gg {\sigma }^2$, equivalently, when the per-sample signal-to-noise ratio satisfies $R^2/{\sigma }^2\gg 4/n$. So a long enough record can detect a sinusoid that is much smaller than the per-sample noise standard deviation.

The signal contributes a deterministic spike of height