Part 8 · Chapter 29

Random Signals and Stochastic Processes

Every signal in this book so far has been a known function of time, yet the two signals that matter most in practice — the message, which is unpredictable or it would carry no information, and the noise, which is unpredictable by nature — are neither; describing them requires replacing the single waveform by an ensemble, the Fourier transform by the power spectral density, and convolution by a correlation identity that survives the change intact.

Signals and Systems Prof. Mithun Mondal Reading time ≈ 65 min
i What you'll learn
  • Why a noise waveform has no Fourier transform, and how Chapter 2's split between energy and power signals predicted this failure long before we met it.
  • The ensemble picture of a random process — a family of sample functions, one drawn at random — and the two functions that describe it: the mean \(\mu_x(t)\) and the autocorrelation \(R_{xx}(t_1,t_2)\).
  • Wide-sense stationarity, what it does and does not assume, and ergodicity — the property that lets one measured record stand in for an ensemble no laboratory can produce.
  • The autocorrelation function: why \(R_x(0)\) is average power, why \(|R_x(\tau)| \le R_x(0)\), and how to read a mean, a periodicity and a correlation time straight off its graph.
  • The Wiener–Khinchin theorem \(S_x(\omega) = \mathcal{F}\{R_x(\tau)\}\), and why the power spectral density is the correct frequency-domain object for a signal that has no spectrum.
  • The central engineering result \(S_{yy}(\omega) = |H(j\omega)|^2 S_{xx}(\omega)\), derived from convolution, together with \(\mu_y = \mu_x H(0)\).
  • White noise, why it is a fiction and a useful one, noise-equivalent bandwidth, and how cross-correlation with a noise input measures an unknown impulse response.
  • How the same results carry over unchanged to discrete-time random signals, where the spectral density is periodic and the integrals become sums.
Section 29-1

Where Determinism Runs Out

Every signal in the previous twenty-eight chapters has been a known function. We wrote \(x(t) = e^{-2t}u(t)\), or \(x[n] = \cos(0.2\pi n)\), transformed it, filtered it, and read off the answer. That habit conceals an assumption so complete it is easy to miss: to compute a Fourier transform you must already know the signal for all time, past and future.

Consider what that assumption means for the systems of Chapter 28. A radio link exists to convey a message the receiver does not already know — if it did, there would be no reason to transmit. Shannon's observation is exactly this: information and unpredictability are the same thing seen from two sides, and a perfectly predictable signal conveys nothing. The moment a signal is worth transmitting, it cannot be written down in advance.

The second failure is more concrete. Every resistor in a receiver generates a fluctuating voltage from the thermal agitation of its charge carriers. Every amplifier adds shot noise from the discreteness of charge. Neither can be written as a formula, and neither can be measured once and reused, because the next measurement gives a different waveform. Yet these signals decide whether a link works, so a theory that can say nothing about them is a theory that cannot design a receiver.

There is also a mathematical obstruction, and Chapter 2 anticipated it. A noise waveform runs forever with roughly constant strength, so its total energy \(\int|x(t)|^2dt\) diverges — it is a power signal, not an energy signal. Chapter 13 required absolute integrability, or at least finite energy, for the Fourier transform to exist. For a noise sample function no such integral converges, and \(X(j\omega)\) simply does not exist. Even if we forced one into existence, it would be different for every sample function and therefore useless as a description of the noise.

The change of question. The repair is not a better transform but a better question. We stop asking "what is this waveform?" and ask instead "what are the statistics of the family of waveforms this source produces?" — its average value, its average power, and how strongly the value at one instant predicts the value a little later. Those averages turn out to be perfectly well behaved, to have Fourier transforms, and to pass through LTI systems by a rule as clean as \(Y = HX\).
Section 29-2

Random Variables, Briefly

A random variable \(X\) is a rule assigning a number to each outcome of an experiment. It is described completely by its cumulative distribution \(F_X(x) = P\{X \le x\}\), or equivalently by the probability density \(f_X(x) = dF_X/dx\), whose integral over an interval gives the probability of landing in it and whose total integral is 1.

Three averages carry almost all the weight in what follows. The expectation is the probability-weighted mean; the mean square is the expectation of \(X^2\); and the variance measures spread about the mean.

Expectation, mean square, variance
\[ E[X] = \int_{-\infty}^{\infty}\! x f_X(x)\,dx, \qquad E[X^2] = \int_{-\infty}^{\infty}\! x^2 f_X(x)\,dx, \qquad \sigma_X^2 = E\big[(X-\mu_X)^2\big] = E[X^2] - \mu_X^2 \]

The last identity is worth verifying once, because the whole of Section 29-6 rests on it: expanding \((X-\mu)^2 = X^2 - 2\mu X + \mu^2\) and taking expectations term by term, using \(E[X] = \mu\), gives \(E[X^2] - 2\mu^2 + \mu^2 = E[X^2] - \mu^2\). Expectation is linear — that single fact is what makes every derivation in this chapter possible, and it is the direct analogue of the superposition that made Chapters 6 to 8 work.

Two densities appear repeatedly. The uniform density is constant over an interval and models a phase of which nothing is known. The Gaussian density is the model for thermal noise, and its pre-eminence is not a matter of convenience: the central limit theorem says that the sum of many small independent contributions tends to a Gaussian whatever the individual distributions are, and thermal noise is precisely the sum of an enormous number of independent electron motions.

The two densities used throughout
\[ f_X(x) = \frac{1}{b-a},\ \ a \le x \le b \qquad\qquad f_X(x) = \frac{1}{\sqrt{2\pi}\,\sigma}\exp\!\left(-\frac{(x-\mu)^2}{2\sigma^2}\right) \]

One property of the Gaussian will matter in Section 29-8: a Gaussian random variable driven through a linear operation stays Gaussian. Since an LTI system computes a convolution — a weighted sum of input values — Gaussian noise entering a filter emerges Gaussian, with only its mean and variance altered. No other common density has this stability, and it is the reason a receiver can be analysed with two numbers rather than a full distribution.

Section 29-3

The Ensemble

A random process \(X(t)\) assigns not a number but a whole function of time to each outcome. Perform the experiment — switch on one particular resistor, at one particular moment — and you obtain one waveform, called a sample function. Perform it again and you obtain a different one. The entire collection of possible waveforms, together with their probabilities, is the ensemble, and the ensemble is what the theory describes. The single waveform on your oscilloscope is one draw from it.

The picture below is worth holding onto, because it contains the two directions in which averages can be taken, and confusing them is the commonest error in this subject.

x⁽¹⁾(t) x⁽²⁾(t) x⁽³⁾(t) t₁ f(x; t₁) time average runs along a trace ⟶ ensemble average runs down the slice
One process, many sample functions — a vertical slice is a random variable

Freeze the time at \(t_1\). The value \(X(t_1)\) is not yet a number, because we have not said which sample function was drawn; it is a random variable, with a density of its own. Freeze at a second instant \(t_2\) and you get a second random variable, generally correlated with the first. A random process is therefore an indexed family of random variables, one for every instant, and a complete description would require the joint density of every finite collection of them — an object nobody can write down.

Engineering practice retreats to the first two moments, and finds that they are enough for everything linear.

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The two functions that describe a process
\[ \mu_x(t) = E\big[X(t)\big], \qquad R_{xx}(t_1,t_2) = E\big[X(t_1)X(t_2)\big] \]

The mean says where the process sits on average at each instant; the autocorrelation says how strongly its values at two instants are related. Both are ordinary deterministic functions — the randomness has been averaged out of them — and both are, in principle, measurable.

Note carefully what the expectation in these definitions means. It is an average across the ensemble, at fixed time: a vertical average in the figure. It is not an average along a single waveform. Whether the two coincide is the question of Section 29-5, and the answer is not automatic.

Section 29-4

Stationarity

A process whose statistics change with time is intractable. What is needed is the probabilistic version of the time-invariance that Chapter 5 imposed on systems: the assumption that the mechanism generating the signal does not care what time it is. A resistor at constant temperature satisfies it; a resistor being heated does not.

Strict-sense stationarity states the assumption in full: every joint density is unchanged by a shift of the time origin, so that \(X(t_1),\dots,X(t_n)\) and \(X(t_1+\tau),\dots,X(t_n+\tau)\) have the same joint density for every \(n\) and every \(\tau\). This is a very strong requirement and almost never verifiable in practice. The workable weakening asks only that the first two moments be shift-invariant.

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Wide-sense stationary (WSS)
\[ \text{(i)}\ \ \mu_x(t) = \mu_x = \text{constant} \qquad \text{(ii)}\ \ R_{xx}(t, t+\tau) = R_x(\tau) \]

The mean does not drift, and the correlation between two instants depends only on the gap between them, not on where the pair sits on the time axis. A third condition, \(E[X^2(t)] = R_x(0) \lt \infty\), is usually included to exclude processes of infinite power.

Strict-sense stationarity implies wide-sense, since if every density is shift-invariant then so is every moment computed from it. The converse fails in general — a process can have constant mean and shift-invariant correlation while its third moment drifts — with one important exception: for a Gaussian process the mean and autocorrelation determine every joint density completely, so wide-sense stationarity implies strict-sense. Since thermal noise is Gaussian, the distinction rarely troubles a communications engineer.

Two examples fix the idea. The random-phase sinusoid \(X(t) = A\cos(\omega_0 t + \Theta)\) with \(\Theta\) uniform on \([0,2\pi]\) is WSS: Worked Example 1 shows that its mean is zero and its autocorrelation is \((A^2/2)\cos\omega_0\tau\), with no residual dependence on \(t\). Remove the random phase, however, and \(X(t) = A\cos\omega_0 t\) is a deterministic signal whose "ensemble" has one member; its mean \(A\cos\omega_0 t\) varies with time, and it is not stationary. The random phase is what makes the process indifferent to the choice of time origin, which is a fair description of what stationarity means physically.

Why WSS is exactly the right amount of assumption. Everything this chapter needs — the autocorrelation properties, the Wiener–Khinchin theorem, the input–output relation for filters — follows from wide-sense stationarity alone. Nothing needs the full strict-sense condition. This is the same economy that made Chapter 8 possible: assume linearity and time invariance, and a very great deal follows; assume more, and you gain little but lose most of the systems you wanted to describe.
Section 29-5

Ergodicity

There is a practical gap in everything above. The definitions call for averages across an ensemble, but a laboratory has one resistor and one oscilloscope, and the ensemble is a mathematical fiction. What is actually measurable is a time average along one long record.

Time averages along a single sample function
\[ \langle x(t)\rangle = \lim_{T\to\infty}\frac{1}{2T}\int_{-T}^{T}\! x(t)\,dt, \qquad \langle x(t)x(t+\tau)\rangle = \lim_{T\to\infty}\frac{1}{2T}\int_{-T}^{T}\! x(t)x(t+\tau)\,dt \]

A process is ergodic when these time averages equal the corresponding ensemble averages, for almost every sample function. Ergodicity is the licence to substitute one long observation for the ensemble, and without it the theory would be unusable — every result in Sections 29-6 to 29-9 is stated in ensemble terms and applied to measured records.

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Ergodic in the mean and in autocorrelation
\[ \langle x(t)\rangle = \mu_x \qquad\text{and}\qquad \langle x(t)x(t+\tau)\rangle = R_x(\tau) \]

An ergodic process must be stationary — if the statistics drifted, no single long average could represent them — but a stationary process need not be ergodic. Ergodicity is the stronger and less checkable assumption, and in engineering it is almost always assumed rather than proved.

The standard counterexample makes the gap vivid. Let \(X(t) = A\) for all time, with \(A\) drawn once, uniformly on \([-1,1]\). The process is stationary: its mean is \(E[A] = 0\) at every instant and \(R_x(\tau) = E[A^2] = 1/3\) for every \(\tau\). But each sample function is a constant, so its time average is \(A\) itself — which is 0 only for the one realisation that happens to have drawn \(A=0\). Averaging along a record here reveals the realisation, not the process. The physical reading is that the randomness was exhausted at the start and never renewed; ergodicity requires a process that keeps producing fresh randomness as time passes, and a single frozen draw does not.

Practically, ergodicity is why measured spectra look the way they do. Chapter 26 gave the FFT of a finite record; applied to a noise waveform, that transform is itself random and jumps about wildly from one record to the next. Averaging many such records, or equivalently smoothing in frequency, converges to the power spectral density of the next section — and it does so because the process is assumed ergodic.

Section 29-6

The Autocorrelation Function

For a WSS process the autocorrelation \(R_x(\tau) = E[X(t)X(t+\tau)]\) carries an unexpected amount of information, and four properties let it be read almost by inspection.

Setting \(\tau = 0\) gives \(R_x(0) = E[X^2(t)]\), the mean-square value — which for a voltage across one ohm is the average power. That single evaluation is the reason the autocorrelation, rather than any other second moment, is the object of interest.

Symmetry follows from stationarity. Since the statistics do not depend on the time origin, \(R_x(-\tau) = E[X(t)X(t-\tau)] = E[X(t')X(t'+\tau)] = R_x(\tau)\) after writing \(t' = t-\tau\). The autocorrelation of a real WSS process is an even function, always.

The bound \(|R_x(\tau)| \le R_x(0)\) comes from a device worth remembering, because the same trick proves the Cauchy–Schwarz inequality. The quantity \(E\big[(X(t) \pm X(t+\tau))^2\big]\) is an expectation of a square and therefore cannot be negative. Expanding it gives \(2R_x(0) \pm 2R_x(\tau) \ge 0\), which is the bound with both signs at once.

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Properties of the autocorrelation
\[ R_x(0) = E[X^2] = P_{\text{avg}}, \qquad R_x(-\tau) = R_x(\tau), \qquad |R_x(\tau)| \le R_x(0), \qquad R_x(\pm\infty) = \mu_x^2 \]

The last holds whenever the process contains no periodic component: values far apart become independent, so the expectation of their product factors into \(\mu_x\cdot\mu_x\). The autocorrelation therefore displays the DC power as its floor and the total power as its peak, with the AC power \(\sigma_x^2\) the difference between them.

These four facts turn the graph of \(R_x(\tau)\) into a diagnostic chart. The peak at the origin is the total average power. The level the curve settles to far from the origin is the squared mean, the DC power. The width of the central peak is the correlation time — how long the process remembers, or how far ahead it can be usefully predicted. Any periodic component in the process survives in the autocorrelation as an undecaying periodic component of the same period, because a sinusoid stays correlated with itself forever; this is how a periodic signal buried in noise is detected.

τ R_x(τ) σ² 1/a R_x(τ) = σ² e^(−a|τ|)
Peak height is power; peak width is memory
ω S_x(ω) 2σ²/a ±a S_x(ω) = 2aσ²/(a² + ω²)
The Fourier transform of the same information

The two panels above are one Fourier pair, and the inverse relation between their widths is the same time–bandwidth reciprocity that Chapter 14 established for deterministic signals: a process with a short memory has a wide spectrum, and a process with a long memory is spectrally narrow. That observation is the content of the next section, stated precisely.

Section 29-7

Power Spectral Density

We still want a frequency-domain description, and Section 29-1 established that the transform of a sample function does not exist. The way out is to transform something that does exist. The autocorrelation of a WSS process is a deterministic, even, bounded function which decays for most processes of interest, so it has a perfectly ordinary Fourier transform. Defining the power spectral density as that transform is the content of the Wiener–Khinchin theorem.

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The Wiener–Khinchin theorem
\[ S_x(\omega) = \int_{-\infty}^{\infty}\! R_x(\tau)e^{-j\omega\tau}\,d\tau \qquad\qquad R_x(\tau) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\! S_x(\omega)e^{j\omega\tau}\,d\omega \]

The autocorrelation and the power spectral density are a Fourier transform pair. Everything Chapter 14 proved about such pairs — the shifting, scaling and convolution properties, the reciprocity of widths — applies unchanged.

The name is justified by setting \(\tau = 0\) in the inverse transform. The left side becomes \(R_x(0)\), the average power; the right side becomes the total area under \(S_x(\omega)/2\pi\). So \(S_x(\omega)\) really does distribute the average power over frequency, and the power in any band is obtained by integrating over that band alone. This is Parseval's relation of Chapter 14 in its correct form for power signals.

Total power, and power in a band
\[ P = R_x(0) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\! S_x(\omega)\,d\omega, \qquad P_{[\omega_1,\omega_2]} = \frac{1}{\pi}\int_{\omega_1}^{\omega_2}\! S_x(\omega)\,d\omega \]

Three properties of \(S_x(\omega)\) follow immediately and should be checked on every answer. It is real and even, because \(R_x(\tau)\) is real and even. It is non-negative for every \(\omega\) — a negative power density in some band would be meaningless, and the formal proof runs through the filtering result of the next section, by passing the process through a narrow bandpass filter and observing that the output power cannot be negative. And its area, divided by \(2\pi\), is the total power. A candidate spectral density that is odd, or dips below zero, is wrong.

An alternative and equally valid definition builds the density from finite records rather than from the autocorrelation. Truncate a sample function to \([-T,T]\), call the result \(x_T(t)\), which now has finite energy and therefore a transform \(X_T(j\omega)\), and form the average of \(|X_T|^2/2T\) over the ensemble as \(T\) grows.

The periodogram definition — the same object from the other side
\[ S_x(\omega) = \lim_{T\to\infty} \frac{1}{2T}\,E\Big[\big|X_T(j\omega)\big|^2\Big] \]

The expectation is essential. Without it the quantity \(|X_T|^2/2T\) — the periodogram — does not converge as \(T\) grows: its fluctuation stays as large as its mean no matter how long the record. Lengthening the record buys finer frequency resolution but no reduction in variance whatever, which is a genuinely counter-intuitive fact and the reason every practical spectrum estimator averages many short periodograms instead of computing one long one. Chapter 26's FFT supplies the periodograms; the averaging supplies the convergence.

Two definitions, one object. Wiener–Khinchin says the correlation route and the periodogram route give the same \(S_x(\omega)\). That equivalence is what makes the subject usable: theory is done with autocorrelations, where the algebra is easy, and measurement is done with FFTs, where the instrument is easy, and the two are guaranteed to agree.

The discrete-time statements are identical with sums in place of integrals. For a WSS sequence, \(R_x[m] = E\{x[n]x[n+m]\}\) and \(S_x(e^{j\Omega}) = \sum_m R_x[m]e^{-j\Omega m}\), which — like every discrete-time spectrum in Chapter 16 — is periodic in \(\Omega\) with period \(2\pi\). The average power is \(R_x[0] = (1/2\pi)\int_{-\pi}^{\pi}S_x(e^{j\Omega})\,d\Omega\).

Five pairs cover most of what is met in practice, and they are worth knowing by sight in the way the transform pairs of Chapter 13 are.

Process\(R_x(\tau)\)\(S_x(\omega)\)Reading
Constant level \(A\)\(A^2\)\(2\pi A^2\delta(\omega)\)All power at zero frequency
Random-phase sinusoid\(\tfrac{A^2}{2}\cos\omega_0\tau\)\(\tfrac{\pi A^2}{2}\big[\delta(\omega-\omega_0)+\delta(\omega+\omega_0)\big]\)Never decays; two lines
First-order (Markov) noise\(\sigma^2 e^{-a|\tau|}\)\(\dfrac{2a\sigma^2}{a^2+\omega^2}\)Correlation time \(1/a\)
White noise\(\tfrac{N_0}{2}\delta(\tau)\)\(\tfrac{N_0}{2}\)No memory; infinite power
Ideal lowpass noise, band \(W\) Hz\(N_0 W\operatorname{sinc}(2W\tau)\)\(\tfrac{N_0}{2}\) for \(|\omega| \le 2\pi W\)Correlation time \(1/2W\)
Section 29-8

Random Signals Through LTI Systems

This is the section the chapter exists for. A WSS process enters a stable LTI system with impulse response \(h(t)\); what comes out? Chapter 7 gave \(y(t) = \int h(\alpha)x(t-\alpha)\,d\alpha\) for each individual sample function, and since convolution is linear, expectations pass straight through it.

Take the mean first. Expectation is linear and \(h\) is deterministic, so it commutes with the integral:

Deriving the output mean
\[ \mu_y = E\!\left[\int h(\alpha)X(t-\alpha)\,d\alpha\right] = \int h(\alpha)\,E\big[X(t-\alpha)\big]\,d\alpha = \mu_x\!\int h(\alpha)\,d\alpha = \mu_x H(0) \]

The output mean is the input mean times the DC gain — which is exactly what it must be, since a constant is a signal at zero frequency and the system responds to it with \(H(0)\).

Now the autocorrelation. Write \(y(t)\) and \(y(t+\tau)\) as two convolution integrals, multiply, and take the expectation inside both:

Deriving the output autocorrelation
\[ R_y(\tau) = E\!\left[\iint h(\alpha)h(\beta)X(t-\alpha)X(t+\tau-\beta)\,d\alpha\,d\beta\right] = \iint h(\alpha)h(\beta)\,R_x(\tau + \alpha - \beta)\,d\alpha\,d\beta \]

The step that makes this work is the second equality: because the input is WSS, the inner expectation depends only on the difference of the two time arguments, \((t+\tau-\beta)-(t-\alpha) = \tau+\alpha-\beta\), and the absolute time \(t\) has vanished. The output is therefore WSS as well — a conclusion worth stating on its own, since it means the machinery is closed under filtering.

The double integral is a pair of convolutions in disguise, and it is far easier to read after transforming. Take the Fourier transform with respect to \(\tau\), and exchange the order of integration:

Transforming — the exponentials separate
\[ S_y(\omega) = \iint h(\alpha)h(\beta)\left[\int R_x(\tau+\alpha-\beta)e^{-j\omega\tau}d\tau\right]d\alpha\,d\beta = \iint h(\alpha)h(\beta)\,S_x(\omega)e^{j\omega(\alpha-\beta)}\,d\alpha\,d\beta \]

using the shifting property on the inner integral. The remaining double integral now factors completely: the \(\alpha\) part is \(\int h(\alpha)e^{j\omega\alpha}d\alpha = H(-j\omega) = H^*(j\omega)\) for real \(h\), and the \(\beta\) part is \(\int h(\beta)e^{-j\omega\beta}d\beta = H(j\omega)\). Their product is \(|H(j\omega)|^2\).

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The fundamental input–output relation for random signals
\[ S_y(\omega) = |H(j\omega)|^2\,S_x(\omega) \qquad\Longleftrightarrow\qquad R_y(\tau) = h(\tau) * h(-\tau) * R_x(\tau) \]

Power spectral densities multiply by the squared magnitude of the frequency response. The phase of \(H\) disappears entirely — it rearranges a waveform in time but cannot change how much power sits at each frequency.

S_x(ω) × |H(jω)|² S_y(ω) FREQUENCY DOMAIN R_x(τ) ∗ h(τ)∗h(−τ) R_y(τ) TIME DOMAIN the phase of H(jω) affects neither row
Filtering a random signal — the same statement in both domains

One more correlation is worth defining, because it turns the result into a measurement technique. The cross-correlation between output and input is \(R_{yx}(\tau) = E[Y(t+\tau)X(t)]\), and running the same argument with only one convolution to expand gives \(R_{yx}(\tau) = h(\tau) * R_x(\tau)\).

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System identification by cross-correlation
\[ R_{yx}(\tau) = h(\tau) * R_x(\tau) \;\xrightarrow{\ R_x(\tau) = \tfrac{N_0}{2}\delta(\tau)\ }\; R_{yx}(\tau) = \frac{N_0}{2}h(\tau) \]

Drive an unknown system with white noise and cross-correlate the output against the input: the answer is the impulse response, scaled. No impulse need ever be applied — which matters, since a real impulse large enough to measure would drive most systems out of their linear range.

Section 29-9

White Noise, Bandwidth and Signal-to-Noise Ratio

The idealisation that makes the previous section practical is white noise: a process whose power spectral density is flat, \(S_x(\omega) = N_0/2\) for all \(\omega\). The name borrows from white light, which contains all visible frequencies in equal measure. Transforming back gives an autocorrelation that is an impulse.

White noise in both domains
\[ S_x(\omega) = \frac{N_0}{2} \ \ \text{for all } \omega \qquad\Longleftrightarrow\qquad R_x(\tau) = \frac{N_0}{2}\,\delta(\tau) \]

Read the second statement literally and its two faces appear at once. Any two distinct instants are uncorrelated, however close together — the process has no memory at all. And \(R_x(0)\) is infinite, so white noise has infinite average power and cannot exist. It is a fiction, and a legitimate one, for the same reason the impulse of Chapter 3 was: what is actually required is that the density be flat over the band the system responds to, and every real system has a finite bandwidth. Thermal noise is flat to roughly \(10^{12}\) Hz, which is white enough for any circuit anyone is building.

The physics fixes the constant. A resistor at absolute temperature \(T\) delivers an available noise power spectral density of \(kT\) watts per hertz, one-sided, with \(k = 1.38\times10^{-23}\) J/K; equivalently \(N_0 = kT\) and the two-sided density is \(N_0/2 = kT/2\). At room temperature \(N_0 \approx 4\times10^{-21}\) W/Hz, a number worth remembering, since it sets the floor beneath which no receiver can be designed.

Because white noise is flat, the output power of a filter driven by it depends only on the area under \(|H|^2\). That observation is packaged as the noise-equivalent bandwidth: the width of the ideal brick-wall filter, of the same peak gain, that would pass the same noise power.

Defining noise-equivalent bandwidth
\[ P_{\text{out}} = \frac{N_0}{2}\cdot\frac{1}{2\pi}\int_{-\infty}^{\infty}\!\big|H(j\omega)\big|^2 d\omega \;\triangleq\; N_0 B_N \big|H\big|_{\max}^2 \quad\Longrightarrow\quad B_N = \frac{1}{|H|_{\max}^2}\int_{0}^{\infty}\!\big|H(j2\pi f)\big|^2 df \]
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Noise bandwidth of a first-order lowpass
\[ B_N = \frac{1}{4RC} = \frac{\pi}{2}\,B_{3\text{dB}} \]

A single-pole filter passes \(\pi/2 \approx 1.57\) times the noise that its \(-3\) dB bandwidth would suggest, because its gentle rolloff keeps admitting noise well past the corner. Sharper filters have \(B_N\) closer to \(B_{3\text{dB}}\), which is one quantitative reason to spend order on a receiver's filters.

With noise power in hand, the figure of merit that decides whether a link works is the signal-to-noise ratio, \(\text{SNR} = P_{\text{signal}}/P_{\text{noise}}\), usually quoted in decibels as \(10\log_{10}\) of that ratio. Every design decision in Chapter 28 can now be re-read as an SNR decision: AM's carrier wastes power that would otherwise raise the SNR; FM's wide bandwidth admits more noise but the discriminator's nonlinearity converts the extra bandwidth into an SNR gain proportional to \(\beta^2\); PCM's regenerators refuse to accumulate noise at all, which is why a digital link either works or does not, with very little in between.

One final result closes the circle with Chapter 15. Suppose a known pulse \(s(t)\) of duration \(T\) arrives buried in white noise, and we may choose any filter we like to maximise the output signal-to-noise ratio at the sampling instant. The answer is not a lowpass filter, and it does not depend on the noise beyond its density: the optimum is the matched filter \(h(t) = s(T-t)\), the time-reverse of the pulse itself.

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The matched filter
\[ h(t) = s(T-t) \qquad\Longrightarrow\qquad \text{SNR}_{\max} = \frac{2E_s}{N_0}, \qquad E_s = \int_0^T s^2(t)\,dt \]

The best achievable ratio depends on the pulse's energy and on nothing else — not its shape, not its duration, not its bandwidth. Convolving with \(s(T-t)\) is correlating with \(s(t)\), so the optimum receiver is the one that asks how much of the expected pulse is present, which is exactly what Section 29-8's cross-correlation does.

Where this book ends, and what begins. Twenty-nine chapters have gone from "what is a signal" to a receiver that extracts a known pulse from thermal noise as well as any receiver ever can. Every tool in that last sentence was built here: the impulse and convolution of Part 2, the transforms of Parts 3 and 5, the sampling theorem of Part 6, and the statistics of this chapter. What follows — digital communications, statistical signal processing, estimation and detection, adaptive filtering, and the learned filters of machine learning — takes these same objects and asks harder questions of them. None of it replaces what is here. Convolution stays convolution; the transform stays the transform; and the phrase "linear, time-invariant" keeps doing the same work it has done since Chapter 6.
Section 29-10

Worked Examples

1 The random-phase sinusoid

Problem. Let \(X(t) = A\cos(\omega_0 t + \Theta)\) with \(A\) and \(\omega_0\) fixed and \(\Theta\) uniform on \([0,2\pi]\). Find \(\mu_x(t)\) and \(R_x(t,t+\tau)\), decide whether the process is WSS, and find its power spectral density and average power.

Solution. The density of \(\Theta\) is \(1/2\pi\) on \([0,2\pi]\), so the mean is an average of a cosine over one full period:

Working — the mean
\[ \mu_x(t) = \frac{A}{2\pi}\int_0^{2\pi}\cos(\omega_0 t + \theta)\,d\theta = \frac{A}{2\pi}\Big[\sin(\omega_0 t+\theta)\Big]_0^{2\pi} = 0 \]

Zero, and independent of \(t\) — condition (i) holds. For the autocorrelation, use \(\cos P\cos Q = \tfrac12\cos(P-Q) + \tfrac12\cos(P+Q)\):

Working — the autocorrelation
\[ R_x = A^2E\big[\cos(\omega_0 t + \Theta)\cos(\omega_0 t + \omega_0\tau + \Theta)\big] = \frac{A^2}{2}\cos\omega_0\tau + \frac{A^2}{2}E\big[\cos(2\omega_0 t + \omega_0\tau + 2\Theta)\big] \]

The second expectation is again a cosine averaged over a full period of \(\Theta\) — note that \(2\Theta\) sweeps two full periods, so the average is still zero — leaving \(R_x(\tau) = (A^2/2)\cos\omega_0\tau\). It depends on \(\tau\) alone, so condition (ii) holds and the process is WSS.

The power spectral density is the transform of a cosine, which Chapter 13 gives as a pair of impulses:

Working — the density
\[ S_x(\omega) = \frac{\pi A^2}{2}\Big[\delta(\omega-\omega_0) + \delta(\omega+\omega_0)\Big] \]

Average power is \(R_x(0) = A^2/2\), the familiar mean-square value of a sinusoid. Checking against the density: \((1/2\pi)(\pi A^2/2)(1+1) = A^2/2\) — agreement. Notice that the autocorrelation does not decay, exactly as Section 29-6 predicted for a process containing a periodic component.

2 Reading a process off its autocorrelation

Problem. A WSS process has \(R_x(\tau) = 4 + 25e^{-3|\tau|}\). Find the mean, the mean-square value, the variance, the DC power, the AC power and the power spectral density. Verify the total power two ways.

Solution. Apply the four properties in order. As \(|\tau|\to\infty\) the exponential dies and \(R_x \to 4\), which must be \(\mu_x^2\), so \(\mu_x = \pm 2\) — the autocorrelation cannot resolve the sign, and no second-order statistic can. At \(\tau = 0\),

Working — the moments
\[ E[X^2] = R_x(0) = 4 + 25 = 29\ \text{W}, \qquad \sigma_x^2 = 29 - 4 = 25\ \text{W} \]

So 4 W sits in the DC component and 25 W in the fluctuation. The correlation time is \(1/3\) s: beyond about a third of a second the process has largely forgotten itself.

For the density, transform term by term. A constant transforms to an impulse, \(4 \leftrightarrow 8\pi\delta(\omega)\), and the two-sided exponential is the standard pair \(e^{-a|\tau|} \leftrightarrow 2a/(a^2+\omega^2)\) from Chapter 13, here with \(a = 3\):

Working — the spectral density
\[ S_x(\omega) = 8\pi\,\delta(\omega) + \frac{25\cdot 6}{9+\omega^2} = 8\pi\,\delta(\omega) + \frac{150}{9+\omega^2} \]

Real, even and non-negative, as required. Verifying the power from the density, using \(\int_{-\infty}^{\infty}d\omega/(9+\omega^2) = \pi/3\):

Working — the check
\[ P = \frac{1}{2\pi}\left[8\pi + 150\cdot\frac{\pi}{3}\right] = 4 + 25 = 29\ \text{W} \]

which matches \(R_x(0)\). The impulse at the origin carries exactly the DC power and the Lorentzian carries exactly the AC power — the split visible in the autocorrelation survives the transform.

3 Thermal noise into a capacitor

Problem. A resistor \(R\) at temperature \(T\) drives a capacitor \(C\), forming a first-order lowpass. The resistor's thermal noise has two-sided density \(S_x(\omega) = 2kTR\) V²/Hz. Find the output power spectral density, the output autocorrelation, the noise-equivalent bandwidth and the mean-square output voltage. Evaluate for \(C = 1\) nF at 290 K.

Solution. The transfer function is that of Chapter 19's \(RC\) lowpass:

Working — the output density
\[ H(j\omega) = \frac{1}{1+j\omega RC} \;\Longrightarrow\; S_y(\omega) = |H|^2 S_x(\omega) = \frac{2kTR}{1+(\omega RC)^2} \]

Inverting with the same pair as in Example 2, written as \(2kTR = (kT/C)\cdot 2RC\cdot\frac{1}{RC}\cdot\frac{1}{2}\cdot 2\), or more simply by matching \(2a\sigma^2/(a^2+\omega^2)\) with \(a = 1/RC\):

Working — the output autocorrelation
\[ R_y(\tau) = \frac{kT}{C}\,e^{-|\tau|/RC} \]

The noise bandwidth follows from the definition, using \(\int_0^\infty df/[1+(2\pi f RC)^2] = 1/(4RC)\):

Working — noise bandwidth and output power
\[ B_N = \frac{1}{4RC}\ \text{Hz}, \qquad P_{\text{out}} = 4kTR\cdot B_N = \frac{4kTR}{4RC} = \frac{kT}{C} \]

This is the celebrated \(kT/C\) noise, and its most striking feature is that \(R\) has cancelled. A larger resistor generates more noise but also narrows the filter, and the two effects cancel exactly. Numerically, with \(kT = 1.38\times10^{-23}\times 290 = 4.0\times10^{-21}\) J,

Working — the number
\[ \overline{v^2} = \frac{4.0\times10^{-21}}{1\times10^{-9}} = 4.0\times10^{-12}\ \text{V}^2 \quad\Longrightarrow\quad v_{\text{rms}} = 2.0\ \mu\text{V} \]

Two microvolts, set entirely by the capacitor and the temperature. This is why sampling capacitors in data converters are made large: the only way to lower the noise floor of a sample-and-hold is to add capacitance or cool the circuit.

4 Stationary but not ergodic

Problem. Let \(X(t) = A\) for all \(t\), with \(A\) drawn once, uniformly on \([-1,1]\). Show the process is WSS, compute the time averages of a single sample function, and decide whether it is ergodic.

Solution. The ensemble averages are ordinary expectations of \(A\):

Working — ensemble averages
\[ \mu_x = E[A] = \frac{1}{2}\int_{-1}^{1}a\,da = 0, \qquad R_x(\tau) = E[A\cdot A] = \frac{1}{2}\int_{-1}^{1}a^2\,da = \frac{1}{3} \]

Both are constants, independent of \(t\) and of \(\tau\), so the process is WSS with mean 0 and average power \(1/3\) W.

Now take one sample function. Once \(A\) has been drawn, the waveform is the constant \(a\), so its time averages are trivial:

Working — time averages of one record
\[ \langle x(t)\rangle = \lim_{T\to\infty}\frac{1}{2T}\int_{-T}^{T}a\,dt = a, \qquad \langle x(t)x(t+\tau)\rangle = a^2 \]

The time-averaged mean is \(a\), not 0, for every realisation except the single one with \(a=0\); the time-averaged power is \(a^2\), not \(1/3\). Time and ensemble averages disagree, so the process is stationary but not ergodic. The moral is practical: no matter how long you observe this source, you learn the value of one draw and nothing about the distribution it came from. Ergodicity fails whenever the randomness is settled once and never refreshed — a manufacturing tolerance, a fixed but unknown offset, a channel that does not change during the call.

5 Identifying a system with noise

Problem. An unknown stable LTI system is driven by zero-mean white noise of two-sided density \(0.5\) W/Hz. The measured cross-correlation of output against input is \(R_{yx}(\tau) = 2e^{-4\tau}u(\tau)\). Find \(h(t)\), \(H(s)\), the output power spectral density and the output power. Verify the power a second way.

Solution. With \(S_x(\omega) = 0.5\) the input autocorrelation is \(R_x(\tau) = 0.5\,\delta(\tau)\), so the convolution in Section 29-8 collapses:

Working — recovering the impulse response
\[ R_{yx}(\tau) = h(\tau) * 0.5\,\delta(\tau) = 0.5\,h(\tau) \;\Longrightarrow\; h(t) = 4e^{-4t}u(t) \]

The system is causal, as the \(u(\tau)\) in the measurement already told us, and Chapter 17 gives its transform at once: \(H(s) = 4/(s+4)\), a first-order lowpass with DC gain 1 and a pole at \(-4\). The output density follows from the fundamental relation:

Working — output density
\[ S_y(\omega) = |H(j\omega)|^2 S_x(\omega) = \frac{16}{16+\omega^2}\cdot 0.5 = \frac{8}{16+\omega^2} \]
Working — output power, via the spectrum
\[ P_y = \frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{8\,d\omega}{16+\omega^2} = \frac{1}{2\pi}\cdot 8\cdot\frac{\pi}{4} = 1\ \text{W} \]

For an independent check, use the time-domain route. For white noise in, \(R_y(0) = (N_0/2)\int h^2(t)\,dt\), so

Working — output power, via the impulse response
\[ P_y = 0.5\int_0^{\infty}16e^{-8t}\,dt = 0.5\cdot\frac{16}{8} = 1\ \text{W} \]

The two agree, as Parseval requires. The wider lesson is that the entire measurement was made without ever applying an impulse: a noise generator and a correlator recover \(h(t)\) while the system stays comfortably inside its linear range.

6 A discrete-time noise filter

Problem. White noise \(x[n]\) with zero mean and variance \(\sigma^2 = 4\) drives the FIR filter \(y[n] = x[n] + 0.5\,x[n-1]\). Find \(R_y[m]\), the output variance and \(S_y(e^{j\Omega})\), and check that the two agree.

Solution. White in discrete time means \(R_x[m] = \sigma^2\delta[m]\) — successive samples are uncorrelated, and unlike the continuous-time case there is nothing pathological about it, since \(R_x[0] = \sigma^2\) is finite. The discrete form of Section 29-8's result is \(R_y[m] = \sigma^2\sum_k h[k]h[k+m]\), with \(h[0]=1\), \(h[1]=0.5\):

Working — the output autocorrelation
\[ R_y[0] = 4(1^2+0.5^2) = 5, \qquad R_y[\pm 1] = 4(1)(0.5) = 2, \qquad R_y[m] = 0 \ \ \text{for } |m|\ge 2 \]

The output variance is \(R_y[0] = 5\), up from 4 at the input: the filter has DC gain 1.5 and passes low frequencies with gain, so it raises the power even though it is a smoother. The output is correlated at lag 1 and uncorrelated beyond — a two-tap filter can only create one lag of memory.

In the frequency domain, \(H(e^{j\Omega}) = 1 + 0.5e^{-j\Omega}\), so

Working — the output density
\[ S_y(e^{j\Omega}) = \sigma^2\big|1+0.5e^{-j\Omega}\big|^2 = 4\big(1 + 0.25 + \cos\Omega\big) = 5 + 4\cos\Omega \]

Reading the last expression as a Fourier series in \(\Omega\), and writing \(4\cos\Omega = 2e^{j\Omega} + 2e^{-j\Omega}\), the coefficients are exactly \(R_y[0]=5\) and \(R_y[\pm1]=2\) — the discrete Wiener–Khinchin relation, confirmed. Checking the power: \((1/2\pi)\int_{-\pi}^{\pi}(5+4\cos\Omega)\,d\Omega = 5\), since the cosine integrates to zero over a full period. The density is non-negative everywhere, reaching \(9\) at \(\Omega = 0\) and \(1\) at \(\Omega = \pi\) — a lowpass shaping of what was a flat input, which is what "colouring" white noise means.

Review

Chapter Summary

Why the transform fails

A noise sample function is a power signal of infinite energy and unknown form, so \(X(j\omega)\) does not exist. Statistics of the ensemble replace it.

Mean and autocorrelation

\(\mu_x(t)\) and \(R_{xx}(t_1,t_2)\) describe a process to second order. Both are deterministic functions and both are measurable.

WSS and ergodic

WSS: constant mean, \(R_x\) depending only on \(\tau\). Ergodic: time averages equal ensemble averages, so one record suffices.

Reading \(R_x(\tau)\)

Peak \(R_x(0)\) is total power, floor \(R_x(\infty)\) is \(\mu_x^2\), width is correlation time, and periodic components never decay.

Wiener–Khinchin

\(S_x(\omega) = \mathcal{F}\{R_x(\tau)\}\), real, even and non-negative, with area \(2\pi R_x(0)\). Equivalently, the averaged periodogram.

Through a filter

\(S_y = |H(j\omega)|^2 S_x\) and \(\mu_y = \mu_x H(0)\). Phase does not matter; the output of a filter driven by a WSS process is WSS.

White noise

Flat density \(N_0/2\), impulsive autocorrelation, infinite power — a fiction that is exact over any real system's band. \(N_0 = kT\).

Correlation as a tool

\(R_{yx} = h * R_x\) identifies a system from noise; the matched filter \(s(T-t)\) maximises SNR at \(2E_s/N_0\).

Practice

Problems

Problems 1 to 3 test stationarity and the reading of autocorrelations; 4 to 6 push signals through filters; 7 and 8 are design calculations of the kind a receiver requires. Where a spectral density is asked for, check that your answer is real, even and non-negative before going on.

  1. Let \(X(t) = A\cos(\omega_0 t) + B\sin(\omega_0 t)\) where \(A\) and \(B\) are zero-mean, uncorrelated random variables with equal variance \(\sigma^2\). Show that \(X(t)\) is WSS with \(R_x(\tau) = \sigma^2\cos\omega_0\tau\). Then show that if \(A\) and \(B\) have different variances the process is no longer WSS, and identify which condition fails.
  2. A WSS process has \(R_x(\tau) = 9e^{-2|\tau|} + 16\cos(10\tau)\). Find the mean, the total average power, the DC power, and the power in the periodic component. Sketch \(S_x(\omega)\), labelling the impulses and the continuous part.
  3. Decide, with a reason, which of the following can be a valid autocorrelation of a real WSS process: (a) \(e^{-|\tau|}\); (b) \(\tau e^{-|\tau|}\); (c) \(1/(1+\tau^2)\); (d) a rectangular pulse of unit height for \(|\tau| \lt 1\). For those that fail, name the property violated.
  4. White noise of two-sided density \(N_0/2\) drives an ideal bandpass filter of unit gain passing \(f_1 \le |f| \le f_2\) and zero elsewhere. Find \(S_y(\omega)\), the output power, and \(R_y(\tau)\). Show that as \(f_1 \to 0\) your answer reduces to the ideal-lowpass result \(R_y(\tau) = N_0 f_2\operatorname{sinc}(2f_2\tau)\).
  5. A zero-mean WSS process with \(R_x(\tau) = 5e^{-2|\tau|}\) is applied to a system with \(h(t) = e^{-t}u(t)\). Find \(S_x(\omega)\), \(S_y(\omega)\) and the output average power. Find the output mean, and explain why it is what it is without computing an integral.
  6. The discrete-time process \(y[n] = x[n] - 0.8\,y[n-1]\) is driven by white noise of variance \(\sigma^2\). Find \(H(e^{j\Omega})\), the output power spectral density, and the output variance. Compare the variance with the input's and account for the difference physically.
  7. An amplifier has a single-pole response with \(-3\) dB bandwidth 1 MHz and midband gain 40 dB, and its input noise density is \(N_0 = 4\times10^{-21}\) W/Hz. Find the noise-equivalent bandwidth and the output noise power. If the wanted signal delivers 1 pW at the input, what is the output SNR in decibels?
  8. A rectangular pulse of amplitude \(A\) and duration \(T\) arrives in white noise of density \(N_0/2\). Write the matched filter's impulse response, sketch its output when the pulse is present, and show that the peak signal-to-noise ratio is \(2A^2T/N_0\). Explain why doubling \(T\) while halving \(A^2\) leaves the answer unchanged.
Tip: almost every problem in this chapter is solved by deciding, first, which domain the question lives in. Powers, variances and correlation times are read from \(R_x(\tau)\); filtering, band-limiting and noise bandwidths are read from \(S_x(\omega)\). Move between them with Wiener–Khinchin, and check the answer in the other domain — as the worked examples do, every time.