Waveform detection and estimation

Waveform detection and estimation

• Use the Karhunen–Loève expansion to reduce continuous signal detection in white noise to a finite-dimensional Gaussian problem, and derive the correlation/matched filter receiver.
• Derive the general binary test with two signals correlated by \rho, and recognise antipodal signals (\rho=-1) as optimal.
• Extend to M-ary detection, build the "largest of" receiver for equal-energy orthogonal signals, and compute the error probability by integration.
• Maximum-likelihood-estimate a linear signal parameter using a correlator, and confirm it attains the Cramér–Rao bound.
• Derive the likelihood ratio test for a random-phase signal, recognise the Bessel function I_0, and build the incoherent matched filter (quadrature receiver).

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From discrete parameters to a continuous waveform

In chapters 5, 6, 9 (modules 21, 22, 24), the observation was a finite number of random variables. This chapter extends to a continuous waveform: under H_1, the transmitter sends a known deterministic signal s(t); under H_0, nothing is sent. At the receiver, H_1{:}\,Y(t)=s(t)+W(t), H_0{:}\,Y(t)=W(t), 0\le t\le T, with W(t) zero-mean white Gaussian noise of two-sided power spectral density N_0/2 (Barkat, pp. 533 to 535).

Karhunen–Loève: turning the continuous signal into discrete coefficients

To reuse the chapter-5 framework, a countable set of random variables is needed. The Karhunen–Loève expansion represents Y(t) through coefficients Y_k=\int_0^TY(t)\varphi_k(t)\,dt over a complete orthonormal basis \{\varphi_k(t)\}. Choosing \varphi_1(t)=s(t)/\sqrt E (with E=\int_0^Ts^2(t)\,dt the signal energy) as the first basis function; the remaining \varphi_k(t), k\ge2, need only be orthogonal to \varphi_1(t) (Barkat, pp. 535 to 536).

Only Y_1 carries information about the hypothesis

With this choice, Y_1=\sqrt E+W_1 under H_1 and Y_1=W_1 under H_0; every coefficient Y_k, k\ge2, equals just W_k (the Karhunen–Loève noise coefficient) regardless of the hypothesis — carrying no decision information. Since the W_j,W_k are uncorrelated and jointly Gaussian they are statistically independent; hence Y_k, k\ge2, is independent of Y_1 and does not affect the decision. The sufficient statistic is T(Y)=Y_1 — the infinite-dimensional observation space reduces to just one dimension (Barkat, pp. 536 to 537).

Exactly Example 5.1, only the data-generation route differs

Since W_1 is zero-mean Gaussian with variance E[W_1^2]=(N_0/2)\int_0^T\varphi_1^2(t)\,dt=N_0/2 (since \int\varphi_1^2=1), the problem reduces to exactly Y_1=\sqrt E+W_1 or Y_1=W_1 with m=\sqrt E, \sigma^2=N_0/2 — exactly Example 5.1 (module 21) (Barkat, p. 537). Applying that result directly: threshold \gamma=(N_0/2\sqrt E)\ln\eta+\sqrt E/2, and detection parameter d^2=2E/N_0.

The correlation receiver and the matched filter

There are two ways to realize computing Y_1=\int_0^Ty(t)\varphi_1(t)\,dt: a correlation receiver (multiply y(t) by \varphi_1(t) then integrate) or a matched filter with impulse response h(t)=\varphi_1(T-t), 0\le t\le T, sampling the output at t=T (Barkat, pp. 537 to 538). Both give the same Y_1; the matched filter is so named because its impulse response is matched to the time-reversed shape of the signal being sought.

Measured: d^2, threshold, P_D,P_F match simulation

Measured: with E=4, N_0=1, \eta=1: d^2=2E/N_0 = 8, threshold \gamma = 1. A Monte Carlo simulation of 300,000 trials: the theoretical P_D=Q((\gamma-\sqrt E)/\sqrt{N_0/2}) = 0.9214 matches the simulated 0.9216; the theoretical P_F=Q(\gamma/\sqrt{N_0/2}) = 0.0786 matches the simulated 0.0785.

Self-check, part 1

• Why does only Y_1 carry decision information, while Y_k, k\\ge2, does not?
• What form does the matched filter's impulse response take?
• What formula gives the detection parameter d^2 of simple binary detection?

Hints: because \varphi_1 is chosen matching the signal shape, while the other \varphi_k contain only noise regardless of hypothesis; h(t)=\varphi_1(T-t); d^2=2E/N_0.

Two arbitrary signals: two basis functions are needed

Generalising: H_1{:}\,Y(t)=s_1(t)+W(t), H_0{:}\,Y(t)=s_0(t)+W(t), with energies E_1,E_0 and correlation coefficient \rho=\frac{1}{\sqrt{E_0E_1}}\int_0^Ts_0(t)s_1(t)\,dt, -1\le\rho\le1. Using Gram–Schmidt: \varphi_1(t)=s_1(t)/\sqrt{E_1}, and \varphi_2(t) is the part of s_0(t) orthogonal to \varphi_1(t), renormalised (Barkat, pp. 544 to 546).

The problem reduces to two dimensions: exactly module 24, part 2

In the new coordinate system, Y_1,Y_2 are independent Gaussian, with the same covariance C=(N_0/2)I under both hypotheses, and mean vectors m_1=[\sqrt{E_1},0]^T, m_0=[\rho\sqrt{E_0},\sqrt{E_0(1-\rho^2)}]^T (Barkat, pp. 546 to 548). This is exactly module 24's equal-covariance problem: the test linearises to T(y)=y^T(m_1-m_0)\gtrless\gamma_1, and can be rewritten with a single correlator, T(y)=\int_0^Ty(t)[s_1(t)-s_0(t)]\,dt.

The detection parameter and the error probability

Computing d^2=\Delta m^TC^{-1}\Delta m with C=(N_0/2)I: d^2=\dfrac{2}{N_0}(E_1+E_0-2\rho\sqrt{E_1E_0}) (Barkat, pp. 550 to 551). With equal priors, the minimum-error-probability criterion, module 21's result applies directly: P(\varepsilon)=Q(d/2)=Q\!\left(\sqrt{2\alpha/N_0}\right), with \alpha\triangleq E_1+E_0-2\rho\sqrt{E_1E_0} (Barkat, pp. 551 to 552).

Why antipodal signals (\rho=-1) are always optimal

P(\varepsilon) decreases as \alpha increases (with N_0 fixed); from \alpha=E_1+E_0-2\rho\sqrt{E_1E_0}, with fixed energies E_1,E_0, \alpha is maximised when \rho is minimised, i.e. \rho=-1 (Barkat, p. 552). Then s_0(t)=-s_1(t) — antipodal signals. If in addition E_0=E_1=E, the test reduces to T(y)=y^T(m_1-m_0)\gtrless0 — the threshold sits exactly midway between the two signals, favouring neither.

Measured: P(\varepsilon) against \rho, confirming antipodal is optimal

Measured: with E_1=4, E_0=3, N_0=1, \rho=-0.5: \alpha = 10.4641, d^2 = 20.9282, theoretical P(\varepsilon) = 0.0111, a Monte Carlo simulation of 300,000 trials matches 0.0111. Comparing three cases of the same energies: \rho=-1 (antipodal) gives P(\varepsilon) = 0.0042, markedly smaller than \rho=0 (orthogonal) giving 0.0307, and both smaller than \rho=-0.5 above — confirming antipodal is always best.

Self-check, part 2

• What does the correlation coefficient \\rho between the two signals measure?
• Why does the general binary detection problem reduce exactly to module 24?
• What value of \\rho gives the smallest error probability, for fixed signal energy?

Hints: how similar the waveforms s_0(t) and s_1(t) are; because after Gram–Schmidt, the covariance is equal under both hypotheses (C=(N_0/2)I); \rho=-1 (antipodal signals).

From two hypotheses to M hypotheses

Generalising: H_k{:}\,Y(t)=s_k(t)+W(t), k=1,\ldots,M, each signal of energy E_k, possibly correlated through a coefficient \rho_{jk} (Barkat, pp. 557 to 558). Using Gram–Schmidt on the M signals, only K\le M basis functions are needed if just K of the M signals are linearly independent. Every Y_k, k>K, contains only noise, not affecting the decision — the infinite-dimensional observation space reduces to K dimensions.

MAP reduces to ML when the priors are equal

The optimal decision chooses the hypothesis maximising the posterior P(H_j|Y) (module 21). If P_j=1/M for every j, maximising the posterior is equivalent to maximising f_{Y|H_j}(y|H_j) — the MAP criterion reduces to ML. Because the density is Gaussian, maximising f_{Y|H_j} is equivalent to minimising \sum_{k=1}^K(y_k-s_{kj})^2=\|y-s_j\|^2 — the squared Euclidean distance between the observation and each signal (Barkat, pp. 559 to 560).

Orthogonal, equal-energy signals: the "largest of" receiver

If every signal has the same energy E, expanding \|y-s_j\|^2=\|y\|^2-2y^Ts_j+E; since \|y\|^2 and E are the same for every j, minimising the distance is equivalent to maximising y^Ts_j=\int_0^Ty(t)s_j(t)\,dt (Barkat, p. 560). The optimal receiver computes M correlation values T_j(y)=\int_0^Ty(t)s_j(t)\,dt and chooses the hypothesis with the largest value — the \"largest of\" receiver.

The error probability for M orthogonal signals

With orthogonal signals (\rho_{jk}=0), the T_k are independent Gaussian: under H_1, T_1\sim\mathcal N(E,N_0/2) while T_k\sim\mathcal N(0,N_0/2), k\ge2. The probability of a correct decision is P_c=P(T_2<T_1,\ldots,T_M<T_1|H_1), computed by a nested integral: P_c=\int_{-\infty}^\infty f_{T_1|H_1}(t_1|H_1)\left[\int_{-\infty}^{t_1}f_{T_k|H_1}(t_k|H_1)\,dt_k\right]^{M-1}dt_1, and P(\varepsilon)=1-P_c (Barkat, pp. 560 to 562).

Measured: the closed integral matches a direct simulation

Measured: with M=4, E=2, N_0=1: the error probability computed by numerical integration of the formula above is 0.1772, and a Monte Carlo simulation of 300,000 trials (directly generating independent Gaussian T_1,\ldots,T_M and counting how often some T_k, k\ge2, exceeds T_1) gives 0.1766 — agreeing, confirming the integral formula is correct without needing a lookup table or approximation.

Self-check, part 3

• When the priors are equal, what does the MAP criterion reduce to?
• When can the \"largest of\" receiver be applied, in terms of the signal set?
• How is the error probability for M orthogonal signals computed?

Hints: the maximum likelihood (ML) criterion; when every signal is orthogonal and of equal energy; by a nested integral of the independent Gaussian densities.

The model: a parameter multiplying a known waveform linearly

Consider Y(t)=\theta s(t)+W(t), 0\le t\le T, with s(t) known (energy E) and \theta a scalar parameter to estimate — for example an unknown signal amplitude. The mapping \theta\mapsto\theta s(t) is linear (s(t,\theta_1+\theta_2)=s(t,\theta_1)+s(t,\theta_2)), so this is called a linear estimation problem (Barkat, pp. 572 to 573).

Reducing again to Y_1: exactly part 1's Karhunen–Loève technique

Using the same \varphi_1(t)=s(t)/\sqrt E: Y_1=\int_0^TY(t)\varphi_1(t)\,dt=\theta\sqrt E+W_1, while Y_k, k>1, equals just W_k, independent of \theta. Hence Y_1 is a sufficient statistic for \theta, Gaussian of mean \theta\sqrt E, variance N_0/2 (Barkat, p. 573).

The ML estimate: solving the likelihood equation for Gaussian Y_1

The likelihood function L(\theta)=f_{Y_1|\Theta}(y_1|\theta) is Gaussian in \theta; solving \partial\ln L/\partial\theta=0 gives (\sqrt E/N_0)(y_1-\theta\sqrt E)\cdot2=0, i.e. \hat\theta_{ml}=Y_1/\sqrt E — a correlator between Y(t) and s(t), normalised by \sqrt E (Barkat, pp. 573 to 574). Physically, this is exactly part 1's correlation receiver, only reading out an estimated amplitude instead of issuing a binary decision.

Unbiased and attaining the Cramér–Rao bound

E[\hat\theta_{ml}]=E[Y_1]/\sqrt E=\theta\sqrt E/\sqrt E=\theta — unbiased. From differentiating the likelihood function, \partial\ln L/\partial\theta=(2\sqrt E/N_0)(y_1-\theta\sqrt E) has exactly the form c(\theta)[\hat\theta-\theta] with c=2E/N_0 — module 22's efficiency condition (Barkat, pp. 574 to 575). The Cramér–Rao bound is attained exactly, equal to N_0/(2E).

Measured: mean, variance, and the Cramér–Rao bound match simulation

Measured: with E=4, N_0=1, true \theta = 2.3, a simulation of 300,000 trials: the measured mean of \hat\theta_{ml} is 2.2989 (matching \theta=2.3); the measured variance is 0.1249, matching the Cramér–Rao bound N_0/(2E) = 0.125 — confirming \hat\theta_{ml} is both unbiased and efficient.

Self-check, part 4

• Why is the model Y(t)=\\theta s(t)+W(t) called a \"linear\" estimation problem?
• What physical device from part 1 does the ML estimate \\hat\\theta_{ml} correspond to?
• What quantity does the variance of \\hat\\theta_{ml} attain exactly?

Hints: because the mapping \theta\mapsto\theta s(t) satisfies the linear superposition principle; the correlation receiver (correlator); the Cramér–Rao bound N_0/(2E).

When the signal is not fully known in advance: an unwanted parameter

In many applications (a radar echo, a fading channel), the received signal has one or more unknown parameters — called unwanted parameters, unlike the earlier parts where s_1(t),s_0(t) were fully specified in advance. If the parameter is random with a known prior density, average the likelihood ratio over that density as K\to\infty in the Karhunen–Loève expansion (Barkat, pp. 580 to 582).

The standard example: a phase uniform on [-\pi,\pi]

Consider H_1{:}\,Y(t)=A\cos(\omega_ct+\Theta)+W(t), H_0{:}\,Y(t)=W(t), known amplitude A and frequency \omega_c, with the phase \Theta uniform on [-\pi,\pi] (Barkat, pp. 583 to 584). Define the two quadrature components y_c=\int_0^Ty(t)\cos\omega_ct\,dt, y_s=\int_0^Ty(t)\sin\omega_ct\,dt; expand \cos(\omega_ct+\theta)=\cos\omega_ct\cos\theta-\sin\omega_ct\sin\theta to separate the \theta-dependent part.

Averaging over phase: the Bessel function I_0 appears

Averaging the likelihood ratio over \theta uniform on [-\pi,\pi] gives \frac{1}{2\pi}\int_{-\pi}^\pi\exp\left[\frac{2A}{N_0}(y_c\cos\theta-y_s\sin\theta)\right]d\theta — exactly the integral definition of the modified Bessel function of order 0: \frac1{2\pi}\int_{-\pi}^\pi e^{a\cos x+b\sin x}\,dx=I_0(\sqrt{a^2+b^2}) (Barkat, pp. 584 to 585). The result: \Lambda=e^{-A^2T/2N_0}I_0\!\left(\frac{2A}{N_0}\sqrt{y_c^2+y_s^2}\right).

The test needs only the envelope r: the incoherent matched filter

Since I_0(\cdot) is monotonically increasing, the test \Lambda\gtrless\eta is equivalent to r=\sqrt{y_c^2+y_s^2}\gtrless\gamma_2 — needing only the envelope of the two quadrature components, without knowing the true phase \Theta (Barkat, pp. 585 to 587). Realized with matched filters (or correlators) against \cos\omega_ct and \sin\omega_ct, squaring each branch, summing, then comparing to a threshold — called the incoherent matched filter or the quadrature receiver.

Measured: the phase-averaged integral matches the closed Bessel function

Measured: with A=1.5, N_0=1, true signal phase \theta=0.7 rad, computing y_c,y_s from a sample signal segment, then r=\sqrt{y_c^2+y_s^2}. The direct numerical integral \frac1{2\pi}\int_{-\pi}^\pi\exp\left[\frac{2A}{N_0}(y_c\cos\theta-y_s\sin\theta)\right]d\theta gives 2.7172, matching the closed Bessel function value I_0(2Ar/N_0) = 2.7172 — confirming the I_0 formula is correct, needing no further assumption beyond its integral definition.

Self-check, part 5

• How does an \"unwanted\" parameter differ from a fully known-in-advance parameter?
• Averaging the likelihood ratio over a uniformly random phase leads to what special function?
• What quantity of the received signal does the incoherent matched filter need, without knowing the true phase?

Hints: its value is not known in advance, only its probability density (random); the modified Bessel function of order 0, I_0; the envelope r=\sqrt{y_c^2+y_s^2} of the two quadrature components.

Takeaways

• The Karhunen–Loève expansion reduces continuous waveform detection in white noise to a finite-dimensional Gaussian problem; for simple binary detection, only one dimension (Y_1) carries information — exactly Example 5.1.
• General binary detection reduces exactly to module 24's equal-covariance problem: d^2=2(E_1+E_0-2\rho\sqrt{E_1E_0})/N_0; antipodal signals (\rho=-1) are always optimal.
• M-ary detection with equal priors reduces MAP to ML (minimum distance); equal-energy orthogonal signals give the "largest of" receiver, the error probability computed by a nested integral of independent Gaussian densities.
• Linear estimation of \theta s(t) reuses exactly the Karhunen–Loève technique, giving \hat\theta_{ml}=Y_1/\sqrt E — a correlator, unbiased, attaining the Cramér–Rao bound N_0/(2E).
• With a random phase (an unwanted parameter), averaging the likelihood ratio gives the Bessel function I_0; the optimal test needs only the envelope r=\sqrt{y_c^2+y_s^2} — the incoherent matched filter.