I have created the following system for practice purposes. From this system I want to determine the system function H(s).
In the picture I have worked with auxiliary (dummy) variables, which should simplify the whole calculation (see blue marking).
How did I approach the matter now? Well, first I have drawn in my auxiliary variables, then I have looked at the summing elements.
I know the relation $H(s)=\frac{Y(s)}{X(s)}$, so I have to transfer my equations into this form.
So these are my equations:
$$\text{Eq 1:}\quad X(s) + G(s) = P(s)$$ $$\text{Eq 2:}\quad 3P(s)=Y(s),$$ $$\text{Eq 3:}\quad G(s) = s^{-1}Y(s)+2Y(s)$$ Eq 3 in 1 results in: $$X(s) + s^{-1}Y(s)+2Y(s) = P(s)$$ Using Eq. 2 results in: $$Y(s) = 3X(s) + 3s^{-1}Y(s) + 6Y(s)$$ Therefor: $$Y(s)(-5-3s^{-1})= 3X(s)$$ $$H(s)=\frac{Y(s)}{X(s)}=\frac{3}{-5-3s^{-1}}$$
My first question is, if you see the system function $H(s)$, is it correct so far?
Then I would be interested in the conversion into a differential equation, here I have already prepared something:
$$Y(s)(-5-3s^{-1})= 3X(s)$$
This should now be handled with the inverse Laplace transformation:
$$-5y(t)+\mathcal{L}^{-1}(-3s^{-1})= 3x(t) \\-5y(t)-3\int_{0}^{\tau} x(\tau) d\tau = 3x(t)$$
My problem at this point is that when I look into my Laplace transform table, I find that $s^{-1}Y(s)$ transformed back results in an integral. Now it is said that these systems can also be given in a differential equation (in terms ot the z-trafo this would be comparable to the difference equation there). How would one come to this differential equation? I have a hunch but I would be interested in your answers :)
I hope my question is so far understandable. Thank you very much.
