Consider a linear system governed by the differential equation c x(t) + kx(t)=f(t) , for which the impulse response function was found in Example 5.1 as hx(t) = c^-1 e^-kt/c U(t) . Find the harmonic transfer function from the Fourier transform of hx(t) , as in Eq. 6.30, and compare with

Question

Consider a linear system governed by the differential equation [latex] c \dot {x}(t)+kx(t)=f(t) [/latex], for which the impulse response function was found in Example 5.1 as [latex] h_{x}(t)=c^{-1}e^{-kt/c}U(t) [/latex] . Find the harmonic transfer function from the Fourier transform of [latex] h_{x}(t) [/latex], as in Eq. 6.30

[latex]H_{x}(\omega )=2\pi h_{x}(\omega )=\int_{-\infty }^{\infty }{e^{-i\omega r}h_{x}(r)dr}[/latex]

and compare with the result from Eq. 6.33

[latex]H_{x}(\omega )=\left(\sum\limits_{j=0}^{n}{a_{j}(i\omega )^{j}} \right) ^{-1}[/latex]

Also find the autospectral density and the variance of the response [latex] \left\{X\left(t\right) \right\} [/latex] for the situation when [latex] f(t) [/latex] is replaced by a white noise (or deltacorrelated) process [latex] \left\{F\left(t\right) \right\} [/latex] with [latex] S_{FF}(\omega )=S_{0} [/latex] .

Accepted Answer

From Eq. 6.30

[latex]H_{x}(\omega )=2\pi h_{x}(\omega )=\int_{-\infty }^{\infty }{e^{-i\omega r}h_{x}(r)dr}[/latex]

we have

[latex] H_{x}(\omega )=\int_{-\infty }^{\infty }{e^{-i\omega r}} h_{x}(r)de=c^{-1}\int_{0}^{\infty }{e^{-i\omega r}e^{-kr/c}}dr=c^{-1}(i\omega +k/c)^{-1} [/latex]

[latex] =(k+i\omega c)^{-1} [/latex]

which is identical to what Eq. 6.33

[latex]H_{x}(\omega )=\left(\sum\limits_{j=0}^{n}{a_{j}(i\omega )^{j}} \right) ^{-1}[/latex]

gives.

The autospectral density of the response is now given by Eq. 6.31

[latex]S_{XX}(\omega )=H_{x}(\omega )H_{x}(-\omega )S_{FF}(\omega )=\left|H_{x}(\omega )\right| ^{2}S_{FF}(\omega )[/latex]

as

[latex] S_{XX}(\omega )=\left|H_{x}(\omega )\right| ^{2}S_{FF}(\omega )=\frac{S_{0}}{k^{2}+\omega ^{2}c^{2}} [/latex]

The most direct method to evaluate the response variance from this autospectral density is to note that [latex] S_{XX}(\omega ) [/latex] has poles at [latex] \omega =\pm ik/c [/latex] and use the calculus of residues.

Thus, if we evaluate the integral along the real axis of the complex [latex] \omega [/latex] space by closing the contour in the upper half-space, we get the integral as being equal to [latex] 2\pi i [/latex] times the residue at [latex] \omega =+ik/c [/latex]

[latex] \sigma ^{2}_{X}=\int_{-\infty }^{\infty }{\frac{S_{0}}{k^{2}+\omega ^{2}c^{2}} }d\omega =(2\pi i) \underset{\omega \rightarrow ik/c}{\lim } \frac{S_{0}(\omega -ik/c)}{c^{2}(\omega -ik/c)(\omega +ik/c)} =\frac{\pi S_{0}}{ck} [/latex]

Having a delta-correlated excitation also simplifies the time-domain evaluation of the response variance from Eq. 5.30

[latex]\sigma _{X}^{2}=\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }{G_{FF}(r_{2}-r_{1})h_{x}(r_{1})h_{x}(r_{2})dr_{1}dr_{2} }=\int_{-\infty }^{\infty }{G_{FF}(r_{e})h_{x,var}(r_{3})dr_{3}} [/latex]

or 5.40

[latex] \sigma _{X}^{2}=G_{XX}(0)= G_{0}\int_{-\infty }^{\infty }{h_{x}^{2}(r)dr\equiv G_{0}h_{x,var}(0)} [/latex]

. The result, of course, is identical to that obtained here.

Question 6.3: Consider a linear system governed by the differential equati...

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