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Ergodic process; WSS random signals through LTI

Short answer

An ergodic process has time averages equal to ensemble averages, so mean and autocorrelation can be measured from a single record. Passing WSS X(t) through LTI h(t) gives output Y with mean m_X H(0) and autocorrelation R_Y(τ)=h(τ)*h(-τ)*R_X(τ), keeping Y WSS; output PSD is |H(f)|²S_X(f).

What do you mean by Ergodic Stochastic Process? Explain with necessary derivation passage of wide-sense random signals through a LTI.

Ergodic stochastic process

A stationary random process is ergodic if its time averages, taken over one long sample function, equal its ensemble averages, taken across all sample functions at one instant:

lim⁡T→∞1T∫−T/2T/2x(t) dt=E[X(t)]=mX,lim⁡T→∞1T∫−T/2T/2x(t)x(t+τ) dt=RX(τ)\lim_{T\to\infty}\frac1T\int_{-T/2}^{T/2}x(t)\,dt=E[X(t)]=m_X,\qquad \lim_{T\to\infty}\frac1T\int_{-T/2}^{T/2}x(t)x(t+\tau)\,dt=R_X(\tau)

Ergodicity allows the mean (DC value), power and autocorrelation of noise to be measured from a single long record. An ergodic process must be stationary, but a stationary process need not be ergodic.

Passage of WSS random signals through an LTI system

Let a WSS process X(t)X(t) (mean mXm_X, autocorrelation RX(τ)R_X(\tau), PSD SX(f)S_X(f)) be applied to a stable LTI system with impulse response h(t)h(t) and H(f)H(f):

Y(t)=∫−∞∞h(α)X(t−α) dαY(t)=\int_{-\infty}^{\infty}h(\alpha)X(t-\alpha)\,d\alpha
  X(t) (WSS)   +--------------+   Y(t)
 ------------->| h(t) , H(f)  |------------->
  R_X, S_X     +--------------+  R_Y, S_Y

1. Mean of output. Expectation and integration can be interchanged:

E[Y(t)]=∫−∞∞h(α)E[X(t−α)] dα=mX∫−∞∞h(α) dα=mX H(0)(a constant, independent of t)\begin{aligned} E[Y(t)] &= \int_{-\infty}^{\infty}h(\alpha)E[X(t-\alpha)]\,d\alpha = m_X\int_{-\infty}^{\infty}h(\alpha)\,d\alpha \\ &= m_X\,H(0) \quad(\text{a constant, independent of } t) \end{aligned}

2. Autocorrelation of output.

RY(t,t+τ)=E[Y(t)Y(t+τ)]=E[∫h(α)X(t−α)dα∫h(β)X(t+τ−β)dβ]=∫ ⁣ ⁣∫h(α)h(β) E[X(t−α)X(t+τ−β)] dα dβ=∫ ⁣ ⁣∫h(α)h(β) RX(τ+α−β) dα dβ\begin{aligned} R_Y(t,t+\tau) &= E[Y(t)Y(t+\tau)] \\ &= E\left[\int h(\alpha)X(t-\alpha)d\alpha\int h(\beta)X(t+\tau-\beta)d\beta\right] \\ &= \int\!\!\int h(\alpha)h(\beta)\,E[X(t-\alpha)X(t+\tau-\beta)]\,d\alpha\,d\beta \\ &= \int\!\!\int h(\alpha)h(\beta)\,R_X(\tau+\alpha-\beta)\,d\alpha\,d\beta \end{aligned}

because XX is WSS, its autocorrelation depends only on the time difference (t+τ−β)−(t−α)=τ+α−β(t+\tau-\beta)-(t-\alpha)=\tau+\alpha-\beta. The result depends only on τ\tau, not on tt:

RY(τ)=h(τ)∗h(−τ)∗RX(τ)R_Y(\tau)=h(\tau)*h(-\tau)*R_X(\tau)

3. Mean-square value E[Y2]=RY(0)=∫ ⁣ ⁣∫h(α)h(β)RX(α−β)dαdβE[Y^2]=R_Y(0)=\int\!\!\int h(\alpha)h(\beta)R_X(\alpha-\beta)d\alpha d\beta is finite and constant.

Since the mean is constant and the autocorrelation depends only on τ\tau, Y(t)Y(t) is also wide-sense stationary.

4. Output PSD. Taking the Fourier transform of RY(τ)R_Y(\tau) (h(τ)→H(f)h(\tau)\to H(f), h(−τ)→H∗(f)h(-\tau)\to H^*(f)):

SY(f)=H(f)H∗(f)SX(f)=∣H(f)∣2SX(f)S_Y(f)=H(f)H^*(f)S_X(f)=\lvert H(f)\rvert^2S_X(f)

and the output power is PY=RY(0)=∫−∞∞∣H(f)∣2SX(f) dfP_Y=R_Y(0)=\int_{-\infty}^{\infty}\lvert H(f)\rvert^2S_X(f)\,df. The phase of H(f)H(f) has no effect on the output PSD.

Example (white noise through RC low-pass filter). SY(f)=N0/21+(2πfRC)2S_Y(f)=\frac{N_0/2}{1+(2\pi fRC)^2}, RY(τ)=N04RCe−∣τ∣/RCR_Y(\tau)=\frac{N_0}{4RC}e^{-\lvert\tau\rvert/RC}, output power N04RC\frac{N_0}{4RC}.

Cross-correlation: RXY(τ)=h(τ)∗RX(τ)R_{XY}(\tau)=h(\tau)*R_X(\tau); for white-noise input, RXY(τ)=N02h(τ)R_{XY}(\tau)=\frac{N_0}{2}h(\tau), so correlating output with input measures h(t)h(t).

✓ Answer checked by Claude Opus 5.5 (high effort)

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