Lecture notes. Written from the material of Berkeley Stat 153, licensed CC BY 4.0. These are notes, not a transcript: the material has been reorganised and rewritten. This adaptation carries the same licence, and the original is linked above.

10. Periodic Signals and the Periodogram¶

What this covers¶

This chapter builds the model for a deterministic wave hidden inside noisy data — a sinusoid with a given frequency, amplitude and phase — and then asks the natural next question: if a signal is a mixture of such waves at frequencies you don't know in advance, how do you find them from data alone? The answer is the periodogram, a device for reading amplitude and dominant frequency straight off a decomposition of the observed series. It assumes familiarity with (weak) stationarity, autocovariance, and ordinary least-squares regression.

Building a periodic signal¶

A pure periodic ("harmonic") component is $$y(t) := \beta_0 + R\cos(2\pi f t + \phi),$$ a wave of frequency $f$, amplitude $R$ and phase $\phi$, riding on a constant $\beta_0$. Expanding the cosine of a sum, $$R\cos(2\pi f t + \phi) = R\cos\phi\,\cos(2\pi f t) - R\sin\phi\,\sin(2\pi f t),$$ so the same wave can be written as a linear combination of a cosine and a sine at the same frequency: $$y(t) = \beta_0 + U_1\cos(2\pi f t) + U_2 \sin(2\pi f t), \qquad U_1 = R\cos\phi,\ \ U_2=-R\sin\phi.$$ This reparameterization is exact — no information is lost — and it trades the pair $(R,\phi)$ for the pair $(U_1,U_2)$: amplitude and phase are recovered from them as $A=\sqrt{U_1^2+U_2^2}$ (equal to $R$) and $\phi=\arctan(U_2/U_1)$.

A random-amplitude, random-phase model, and why it is stationary¶

Now let $U_1$ and $U_2$ be random — uncorrelated, mean zero, each with variance $\sigma^2$ — so that $y_t$ has the same frequency every time but a random amplitude and phase. With $\mathbb{E}(y_t)=0$ and writing $\lambda = 2\pi f$, the autocovariance is

\[ \begin{aligned} \gamma_y(t,s) &= \operatorname{cov}(y_t, y_s)\\ &= \operatorname{cov}\big(U_1\cos(\lambda t) + U_2\sin(\lambda t),\ U_1\cos(\lambda s)+U_2\sin(\lambda s)\big)\\ &= \operatorname{cov}(U_1\cos(\lambda t), U_1\cos(\lambda s)) + \operatorname{cov}(U_1\cos(\lambda t), U_2\sin(\lambda s))\\ &\quad + \operatorname{cov}(U_2\sin(\lambda t), U_1\cos(\lambda s)) + \operatorname{cov}(U_2\sin(\lambda t), U_2\sin(\lambda s))\\ &= \sigma^2\cos(\lambda t)\cos(\lambda s) + 0 + 0 + \sigma^2\sin(\lambda t)\sin(\lambda s)\\ &= \sigma^2\big[\cos(\lambda t)\cos(\lambda s) + \sin(\lambda t)\sin(\lambda s)\big]\\ &= \sigma^2\cos(\lambda(t-s)). \end{aligned} \]

The two cross terms vanish because $U_1$ and $U_2$ are uncorrelated; the two surviving terms share the same variance $\sigma^2$ and combine, by the cosine-difference identity, into a function of the lag $t-s$ alone. Mean constant, autocovariance a function of lag only: $y_t$ is (weakly) stationary, with $\gamma_y(h) = \sigma^2\cos(\lambda h)$ at lag $h$.

Mixtures of frequencies¶

Real signals rarely consist of one frequency, so generalize to a sum of $q$ harmonic components: $$y_t = \sum_{k=1}^q \big[U_{k1}\cos(2\pi f_k t) + U_{k2}\sin(2\pi f_k t)\big],$$ with $f_1,\dots,f_q$ distinct frequencies and all the $U_{k1},U_{k2}$ ($k=1,\dots,q$) uncorrelated with mean zero, $U_{k1}$ and $U_{k2}$ sharing variance $\sigma_k^2$. The covariance calculation above goes through term by term — cross terms between different components $k$ vanish for the same reason as before — so the mixture is stationary too.

The lecture's notebook illustrates this with three sinusoids at frequencies $f=6,10,39$ (chosen with no particular relationship to each other) and amplitude pairs $(U_1,U_2)=(2,3),(4,5),(6,7)$. Two things about summing sinusoids are worth holding onto, because they resurface once the frequencies are unknown and have to be estimated:

The sum's amplitude is bounded by the sum of the amplitudes. If component $k$ has amplitude $A_k=\sqrt{U_{k1}^2+U_{k2}^2}$, the combined series reaches at most $\sum_k A_k$ in absolute value.
That ceiling is only reached if every component's peak lines up at the same instant, which happens reliably only when the frequencies are harmonically related — integer multiples of a common base frequency. Frequencies picked with no such relationship, like $6, 10, 39$, essentially never line up, so the combined wave's envelope stays visibly below the sum-of-amplitudes bound even though each component individually reaches its own peak somewhere in the window.

The notebook repeats the same construction with three audio-frequency sinusoids added together and played as sound, to make audible what a sum of several periodic components looks/sounds like before any attempt is made to recover the frequencies from it.

From a model to data: the periodogram¶

Suppose instead you are handed $n$ observations $y_1,\dots,y_n$ and do not know what mixture of frequencies, if any, produced them. Rather than assuming some particular number of frequencies $q$ up front, write the data exactly as a sum over every frequency the sample can resolve: $$y_t = a_0 + \sum_{j=1}^{\lfloor n/2\rfloor}\big[a_j\cos(2\pi tj/n) + b_j \sin(2\pi tj/n)\big], \qquad t=1,\dots,n,$$ where $\lfloor\cdot\rfloor$ rounds down to the nearest integer. If $n$ is even the top term is special: $a_{n/2}\cos(2\pi t\cdot\tfrac12)=a_{n/2}(-1)^t$ and $b_{n/2}=0$.

This is not a modeling assumption — it is a change of basis, an exact representation of any $n$ numbers as a combination of cosines and sines at the frequencies $j/n$, $j=1,\dots,\lfloor n/2\rfloor$, called the Fourier frequencies. Here $j/n$ is a frequency in cycles per sample, and as $j$ runs from $1$ to $\lfloor n/2\rfloor$ it sweeps through every frequency distinguishable at this sampling rate. With $n=100$: $j=1$ completes exactly one full cycle over the whole window, the slowest oscillation that fits; $j=2$ completes two cycles, $j=3$ three, and so on; $j=50=n/2$ completes fifty cycles, the fastest oscillation resolvable — alternating up and down every single sample — the Nyquist frequency.

The coefficients come from regression: projecting the data onto the cosine and sine at each Fourier frequency, $$a_j = \frac{2}{n}\sum_{t=1}^n x_t \cos(2\pi tj/n), \qquad b_j = \frac{2}{n}\sum_{t=1}^n x_t \sin(2\pi tj/n).$$ $a_j$ and $b_j$ jointly set the amplitude and phase at frequency $j/n$ — the data-driven analogue of $U_1,U_2$ in the model above — and are estimated independently at each $j$.

This gives the (scaled) periodogram: $$P(j/n) = a_j^2 + b_j^2, \qquad j/n \neq 0, \tfrac12.$$ $P(j/n)$ is the sample variance carried by frequency $j/n$ — an estimate of $\sigma_j^2$, the variance a sinusoid at that frequency would need in the mixture model to explain what is seen in the data. A large value of $P(j/n)$ says frequency $j/n$ is doing real work in the series; a small value looks like noise.

Three genuine frequency components stand out as tall spikes against a noise floor of small values elsewhere — the periodogram's job is exactly to make that contrast visible.

Reading the periodogram: examples from the lecture¶

The lecture's notebook works through several data sets to show what the periodogram does and does not tell you.

Recovering known frequencies exactly. Three sinusoids at $f=6,10,39$ (amplitude pairs $(U_1,U_2)=(12,13),(4,5),(6,7)$) are summed and sampled at $100$ Hz over one second, so $n=100$. scipy.signal.periodogram computes $P(j/n)$ against frequency in Hz, and the peaks land exactly at $6$, $10$ and $39$ Hz — because with $n=100$ samples over a $1$-second window, the Fourier frequencies $j/n$ (cycles per sample) coincide exactly with $j$ Hz, and $6,10,39$ are all integers.

Frequency resolution is set by duration, not by the sampling rate. The notes are explicit about this: "our frequency resolution is $1/T$ where $T$ is the duration of our signal (in seconds! not samples!)." Sampling faster adds more Fourier frequencies up to a higher Nyquist frequency, but it does not pack them closer together — recording for longer, not sampling faster, is what lets two nearby frequencies be told apart.

The sunspot cycle. Sunspot counts sampled twice a year from 1749 to 1978 ($f_s=2$ cycles/year) are run through the same periodogram. Plotted against frequency in cycles/year, it shows a peak near $1/11$ — the roughly 11-year sunspot cycle — and, once the low-frequency end is stretched out by plotting against period ($1/\text{frequency}$) on a log axis rather than against frequency directly, a second, much slower peak near $1/110$, the Gleissberg cycle. The two would be hard to compare on a single linear frequency axis, since one sits close to zero and the other much further along; a log-period axis brings both into view together.

Detecting a nuisance frequency. A recorded vowel sound has a synthetic $60$ Hz tone (a cosine plus a sine at $f=60$, standing in for electrical mains interference) added onto it. The periodogram of the contaminated recording, zoomed into $0$–$600$ Hz, shows the injected interference as a spike at exactly $60$ Hz standing out from the speech spectrum around it — the periodogram used diagnostically, to locate a known unwanted periodic component rather than to discover a new one.

Sources¶

Slides, 01-periodic-signals.md — the periodic signal model, its reparameterization, the stationarity derivation, and the generalization to a sum of $q$ frequencies.
Slides, 02-periodogram.md — the exact Fourier-frequency expansion of an observed series, the regression formulas for $a_j,b_j$, and the definition of the scaled periodogram.
Notebook, Lecture12.md — the worked examples: three synthetic sinusoids and the amplitude-bound comment, the musical-tone audio demo, exact recovery of $6,10,39$ Hz via scipy.signal.periodogram, the frequency-resolution note, the sunspot-cycle periodogram (via astsa.load_sunspotz), and the line-noise detection example. Several figures and one audio widget referenced in the notebook were not reproduced in the conversion (marked "figure omitted" in the source).
No transcript, written notes, or exercise set was supplied for this lecture.
The slides assign Chapter 4 of Shumway and Stoffer as the reading for this material; that text was referred to but not supplied. The periodogram notes also point forward to relating $P(j/n)$ to the Discrete Fourier Transform, described as material for "next time" and likewise not covered here.

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