Investigate the behavior of the first and second moments of stochastic response of the system c.X(t) + k X(t) = 0, for which the impulse response function was derived in Example 5.1 as hx(t)=c^-1 e^-kt/c U(t). In particular, investigate μ X(t), KXX(tl,t2), and ΦXX(tl,t2) for this situation with

Question

Investigate the behavior of the first and second moments of stochastic response of the system [latex] c\dot{X}(t)+kX(t)=0 [/latex] , for which the impulse response function was derived in Example 5.1 as [latex] h_{x}(t)=c^{-1}e^{-kt/c}U(t) [/latex] . In particular, investigate [latex] \mu _{X}(t), K_{XX}(t_{1},t_{2}) [/latex] , and [latex] \Phi _{XX}(t_{1},t_{2}) [/latex] for this situation with no excitation process, but with a random variable initial condition of [latex] X(t_{0})= Y [/latex] with [latex] \mu _{Y} [/latex] and [latex] \sigma _{Y} [/latex] known.

Accepted Answer

We can use Eqs. 5.22

[latex]\mu _{X}(t)=\sum\limits_{j}{\mu _{Y_{j}}} g_{j}(t-t_{0})+\int_{t_{0}}^{\infty }{\mu _{F}(S)h_{x}(t-s)ds}[/latex]

and 5.23

[latex]K_{XX}(t_{1},t_{2})=\sum\limits_{j_{1}}\sum\limits_{j_{2}}{K_{Y_{j_{1}} Y_{j_{2}}}g_{j_{1}}(t_{1}-t_{0})g_{j_{2}}(t_{2}-t_{0})} + \sum\limits_{j}{\int_{t_{0}}^{\infty }{K_{Y_{j}}F_{(s)}} g_{j} (t_{1}-t_{0})[h_{x}(t_{1}-s)+h_{x}(t_{2}-s)ds] + \int_{t_{0}}^{\infty }\int_{t_{0}}^{\infty }{K_{FF}(s_{1},s_{2})h_{x}(t_{1}-s_{1})h_{x}(t_{2}-s_{2})ds_{1}ds_{2}} } [/latex]

if we know the value of the [latex] g(t) [/latex] function. Note that one needs only one initial condition for this first-order system, so the given value of [latex] X(t_{0})[/latex] is sufficient to describe a unique solution. Furthermore, Example 5.1 shows that the impulse response function [latex] h_{x}(t)=c^{-1}e^{-kt/c}U(t) [/latex] is the response of the system to an initial condition of [latex] X(0)=c^{-1} [/latex] , so we have [latex] g(t)=ch_{x}(t-t_{0})=e^{-k(t-t_{0})/c}U(t-t_{0}) [/latex] . Thus, for [latex] t\gt t_{0} [/latex] to we have

[latex] \mu _{X}(t)=\mu _{Y} g(t-t_{0})=\mu _{Y} e^{-k(t-t_{0}/c)} [/latex]

And

[latex] K_{XX}(t_{1},t_{2})=\sigma ^{2}_{Y} g(t_{1}-t_{0})g(t_{2}-t_{0})=\sigma ^{2}_{Y} e^{-k(t_{1}+t_{2}-2t_{0})/c} [/latex]

Putting these two together gives the autocorrelation function of the response as

[latex] \phi _{XX}(t_{1},t_{2})=E(Y ^{2}) g(t_{1}-t_{0})g(t_{2}-t_{0})=E(Y ^{2}) e^{-k(t_{1}+t_{2}-2t_{0})/c} [/latex]

By letting [latex] t_{1}=t_{2}=t [/latex] , one can also find the response variance as [latex] \sigma^{2}_{X} (t)=\sigma^{2}_{Y}e^{-2k(t-t_{0})/c} [/latex] , and the mean-squared value is given by a similar expression.

Question 5.5: Investigate the behavior of the first and second moments of ...

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