1
$\begingroup$

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.

| improve this question | |
$\endgroup$
3
$\begingroup$

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

$$H(s)=-\frac{3s}{5s+3}\tag{1}$$

or, equivalently,

$$5sY(s)+3Y(s)=-3sX(s)\tag{2}$$

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

$$5y'(t)+3y(t)=-3x'(t)\tag{3}$$

| improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.