I have created the following system for practice purposes. From this system I want to determine the system function H(s).

enter image description here

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.


Your transfer function looks correct. Note that you can rewrite $H(s)$ as


or, equivalently,


Since multiplication by $s$ corresponds to differentiation in the time domain you obtain from $(2)$


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  • $\begingroup$ Hi, and thanks for confirming my calculation. I had two theories about this: 1. derive $-5y(t)-3\int_{0}^{\tau} x(\tau) d\tau = 3x(t)$ to get rid of the integral so that a differential equation is created, or 2. multiply H(s) by "s/s", then you can rewrite the term. I had this checked by realizing the TF with H(s) in Matlab (as a block), the result was the same. So that the calculation could also be checked "experimentally". Thank you again for the confirmation and your contribution! $\endgroup$ – P_Gate Feb 1 at 12:52

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