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Stochastic Process 5 - Random Process in LTI System

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Stochastic Process 5 - Random Process in LTI System

This note covers basic concept of random process in LTI system. Filters are discussed as an examples.

Recall what a linear time invariant system is. Given an imput \(x(t)\) and the impulse response function \(h(t)\), the output \(y(t)\) can be expressed as \[ y(t) = \int_{-\infty}^\infty h(t - s)x(s) \mathrm{d}s \]

which can be denoted as \(y(t) = h(t) * x(t).\) Here \(*\) means convolution.

Remark:

\[ \begin{aligned} (f*g)(t) &= \int_{-\infty}^\infty f(s)g(t-s) \mathrm{d}s = \int_{-\infty}^\infty f(t-s)g(s) \mathrm{d}s \\ (f*g)(a+b) &= \int_{-\infty}^\infty f(a+s)g(b-s) \mathrm{d}s = \int_{-\infty}^\infty f(b+s)g(a-s) \mathrm{d}s \end{aligned} \]

What if the input is a random process?

Let \(X(t)\) be a 2nd random process with autocorrelation \(R_X(s, t).\) Suppose that \[\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} h(t-s)h(t-u)R_X(s, u) \mathrm{d}s\mathrm{d}u\] exists and is finite for all \(t \in \mathbb{R}.\) Then we have \[ Y(t) = \int_{-\infty}^{\infty} h(t-s)X(s) \mathrm{d} s \qquad \text{mean square integral} \] denoted as \(Y(t) = h(t)*X(t).\)

Directly we can derive the following properties:

  • \(\mu_Y(t) = (h*\mu_X)(t) = \int_{-\infty}^{\infty} h(t-s)\mu_X(t) \mathrm{d} s = \mu_X(t) \int_{-\infty}^{\infty}h(s) \mathrm{d} s\) Here you may notice that the later part \(\int_{-\infty}^{\infty}h(s) \mathrm{d} s\) is independent of time \(t.\)

  • \(R_{YX}(s, t) = E[Y(s)X(t)] = E[\int_{-\infty}^{\infty} h(s -v)X(v)\mathrm{d} v \cdot X(t)] = \int_{-\infty}^{\infty} h(s-v) R_X(v, t) \mathrm{d} v =\int_{-\infty}^{\infty}h(s-v)R_X(v - t) \mathrm{d} v = (h*R_X)(s-t).\) Similarly, we derive that \(R_{XY}(s, t) = (h*R_X)(t-s)\) and \(R_{YX}(s, t) = R_{XY}(t, s).\)

  • Denote \(\hat{h}(t) = h(-t).\) Then we can derive that \(R_Y(s, t)\) only depends on (s-t). \[ \begin{aligned} R_Y(s, t) &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} h(t-u)h(s-v)R_x(v-u)\mathrm{d} v\mathrm{d} u \\ &= \int_{-\infty}^{\infty}h(t-u) \int_{-\infty}^{\infty}h(s-v)R_x(v-u)\mathrm{d} v\mathrm{d} u \\ &= \int_{-\infty}^{\infty}h(t-u) \cdot (h*R_X)(s-u) \mathrm{d} u \\ &= (\hat{h}*h*R_X)(s-t) \end{aligned} \]

Theorem:

Suppose \(X\) is a WSS process with mean \(\mu_X\) and autocorrelation \(R_x.\) X is input to a LTI system with impule response \(h(t).\) Then, (a) Output \(Y(t)\) is a a WSS with \[ \mu_Y = \mu_X(t) \int_{-\infty}^{\infty}h(s) \mathrm{d}\] \[ R_Y(\tau) = (\hat{h}*h*R_X)(\tau).\] (b) \(X(t)\) and \(Y(t)\) are jointly WSS. \[R_{YX}(s, t) = R_{XY}(t, s), R_{YX}(\tau) = R_{XY}(-\tau).\] (c) Autocovariance \[C_{YX}(\tau) = (h*C_X)(\tau), C_Y(\tau) = (\hat{h}*h*C_X)(\tau).\]

TBD