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I've the following image and I want to find the transfer function from input $x(t)$ to output $y(t)$. I know that I have to apply Laplace Transform, so the integrator becomes $\dfrac{1}{s}$, but I don't know what to do with the numbers $a$ and $b$. Should it still be the same?

If a and b still the same, then I found that $H(s) = \dfrac{Y(s)}{X(s)} = s^2-as-b$.

Block diagram

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In such cases it can be useful to introduce auxiliary variables at the input of the integrators. For the given diagram you could use a signal $u(t)$ at the input of the second integrator. The equation for its Laplace transform $U(s)$ becomes

$$U(s)=\frac{1}{s}\big[X(s)-bY(s)\big]-aY(s)\tag{1}$$

You need another equation relating $U(s)$ to $Y(s)$, but that one is trivial. From those two equations you can express $Y(s)$ in terms of $X(s)$, and in this way you can obtain the transfer function $H(s)=Y(s)/X(s)$. The solution in your question is not correct. The correct solution must be a rational function, not a polynomial.

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  • $\begingroup$ I got it. So the correct solution should be the inverse of the one I gave changing some signals, right? $H(s) = \dfrac{1}{s^2+as+b}$. Another question regards the numbers $a$ and $b$, when using Laplace Transform the numbers still the same or I should use Laplace on them? $\endgroup$ – FY Gamer Jul 10 '20 at 12:40
  • $\begingroup$ @FYGamer: Try to answer this yourself: what is the Laplace transform of $ax(t)$? $\endgroup$ – Matt L. Jul 10 '20 at 12:48
  • $\begingroup$ You're right. Thank you. $\endgroup$ – FY Gamer Jul 10 '20 at 13:15

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