0
$\begingroup$

Hi so we have a sinus signal that is sent to a feedback loop system and I'm trying to understand signals and stability better. I let $\arg(G_o(iw_o))=0$ and get that $y$ has no bound although $r=0$.

Suppose that we have a stable system by the nyquist criteria. Let r=0, then $-e(t)=y(t)$ but we also have that $y(t)_{\text{new}}=G_oe(t)=-G_oy(t)$. Further let $\arg(G_o(iw_o)=0$ and $|G(iw_o)|=k>1$. Then $G(iw_o)=k$ thus $y(t)_{\text{new}}=-ky(t)$. So $y$ has no bound. The feedback inverts y and going through $G_o$ the absolute value increases with a factor of $k$.

Since the system is stable there must be a damping effect on $y$ I can't see right?

enter image description here

$\endgroup$
3
  • $\begingroup$ Please explain the switch and what it is doing. Also, please state which criterion is being used to determine stability. $\endgroup$ Commented Mar 12 at 18:41
  • $\begingroup$ If switch 2 is on then the system is open if switch 1 is on then it is closed feedback system. Signal is only sent when swtich 2 is on and then switch 1 is put on and y and e is observed. The system is stable only under my assumption we could say Z=P+N where P=N=0, so G_o has no unstable poles. $\endgroup$ Commented Mar 12 at 19:37
  • $\begingroup$ All the stuff in the above comment needs to be incorporated into the text of the question. You can edit the text by clicking on the Edit link below the text. you also need to specify all the time instants when the switch is in position 2 and position 1, e.g. "Switch is in position $2$ for $t$ ranging from $-\infty$ to $0^-$. At $t=0$ it switches instantly to position $1$ and remains there from $t=0^+$ onwards." or whatever you want to tell us about what the switch does. Also, $G_0(\omega)$ is a constant $k > 1$? and there is no filtering going on, just gains of $k$ and $-1$? $\endgroup$ Commented Mar 13 at 2:57

0

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.