Making it possible to do a Fourier transform on it: $\frac{1}{(k+w)^2(a^2 +w^2)}$

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Sorry for all the edits, I'm very stressed and not so used to Latex. Full question: consider a filter with impulse response

$$h(t)=e^{-bt} u(t)$$

where $u$ is the unit step function. The input signal is a wss with spectral density

$$S_{xx}(t)= \frac{a}{w^2 +1}$$

Find: A) the spectral density of the output B) the autocorrelation of the output C) the power of the output

My original question: How to break these up so it's possible to do a Fourier transform on it:

$$\frac{a}{(k+w)^2(1 +w^2)} -> \frac{a}{k^2 +1}(\frac{1}{w^2 +k^2}-\frac{1}{w^2 +1})$$

I don't know how to perform the partial fraction of a equation like this and it have taken me a long time to find anything..

My attempt at a similar question but with the nominator 1 instead of a My attempt at a similar question but with the nominator 1:

My exam on stochastic processes is tomorrow and this knowledge is needed for at least one question per exam if not more.

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$$\dfrac{a}{(k+w)^2(w^2+a^2)} = \dfrac{A}{k+w} + \dfrac{B}{(k+w)^2} + \dfrac{Cw+D}{w^2+a^2} \implies$$

$$\dfrac{a}{(k+w)^2(w^2+a^2)} = \dfrac{A(k+w)(w^2+a^2)+B(w^2+a^2)+Cw(k+w)^2+D(k+w)^2}{(k+w)^2(w^2+a^2)}$$

$$\implies a = Aw^3+Awa^2+Akw^2+Aka^2+Bw^2+Ba^2+Cw^3+2Ckw^2+Cwk^2+Dk^2+Dw^2+2Dkw$$

$$\implies 0w^3+0w^2+0w+a = w^3(A+C)+w^2(Ak+B+2Ck+D)+w(Aa^2+Ck^2+2Dk)+(Aka^2+Ba^2+Dk^2)$$

The system is

$$ \begin{align*} A+C = 0\\ Ak+B+2Ck+D = 0\\ Aa^2+Ck^2+2Dk=0\\ Aka^2+Ba^2+Dk^2 =a \end{align*}$$

From this you can solve the values and remeber, $a$ and $k$ are constanst

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I got help from a Brazilian friend, thanks so much for all the help. I feel a bit silly now but without asking I wouldn't have found it out. solution