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Provided I have a system of linear differential equations (in time domain) such as:

$$\begin{cases} \dot{x}(t)=Ax(t)+By(t)+Cz(t)\\ \dot{y}(t)=A'x(t)+B'y(t)+C'z(t)\\ \dot{r}(t)=B''y(t)\\ \end{cases}$$

I want to find the transfer function (in Laplace domain) such as:

$$H(s)=\dfrac{r(s)}{z(s)}$$

What should I do to obtain the desired transfer function? I am thinking of the removing derivatives by using the "multiply by s" property of Laplace transform. But after that, how should I continue to obtain the numerator and denominator of the transfer function?

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Assume all initial conditions are zero, input and output are scalar so $H(s) = r(s)/z(s)$ makes sense, and all matrices are constants i.e. $A' \neq \dot A$ and invertible/full rank

Then you can compute:

$$sx(s) = Ax(s) + By(s) + Cz(s)$$ $$sy(s) = A'x(s) + B'y(s) +C'z(s)$$ $$sr(s) = B''y(s)$$

(1) $$(sI - A)x(s) = By(s) + Cz(s) \Rightarrow x(s) = (sI-A)^{-1}(By(s)+Cz(s))$$

(2) $$sy(s) = A'x(s) + B'y(s) +C'z(s) = A'((sI-A)^{-1}(By(s)+Cz(s))) + B'y(s) + C'z(s) = (A(sI-A)^{-1}B+B')y(s) + (A(sI-A)^{-1}C+C')z(s)$$

$$\Rightarrow y(s) = (sI-(A(sI-A)^{-1}B+B'))^{-1} (A(sI-A)^{-1}C+C')z(s)$$

(3) $$sr(s) = B''y(s) =B''(sI-(A(sI-A)^{-1}B+B'))^{-1} (A(sI-A)^{-1}C+C')z(s) $$

$$\Rightarrow r(s) = s^{-1}B''(sI-(A(sI-A)^{-1}B+B'))^{-1} (A(sI-A)^{-1}C+C')z(s) $$

Then $H(s) = \dfrac{r(s)}{z(s)} = s^{-1}B''(sI-(A(sI-A)^{-1}B+B'))^{-1} (A(sI-A)^{-1}C+C')$

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  • $\begingroup$ Thank you very much for your help! $\endgroup$ – Airman01 Mar 6 '16 at 10:02

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