0
$\begingroup$

I was hoping that you can help me understand how to find poles and zeros in this transfer function.

$$ H(s)=\frac{1}{(s-s_{\alpha1})(s-s_{\alpha2})} $$

I know that this function does not have any zeros, because of the 1.

From the solutions I can see that the poles are $s_{\alpha1}=-1+j$ and $s_{\alpha2}=-1-j$ and to tell you the truth, I have no idea how to get solution for the poles.

I tried to solve it using quadratic equation, but the only thing I get is a big mess.

Thanks!!

$\endgroup$
  • $\begingroup$ The poles simply are $s_{\alpha1}$ and $s_{\alpha2}$. That's it, you obviously can't get numerical values from this general transfer function. The transfer function also has a double zero at infinity. $\endgroup$ – Matt L. Jan 30 '15 at 10:17
  • $\begingroup$ Then this assingment is incomplete, because i cant find any values that are given to the sα1 and sα2 $\endgroup$ – depecheSoul Jan 30 '15 at 10:21
  • $\begingroup$ If the only thing you have is $H(s)$ then you can't do more than I told you. No figure, etc.? $\endgroup$ – Matt L. Jan 30 '15 at 10:56
1
$\begingroup$

Because $s-s_{\alpha1} = 0 $ and $s-s_{\alpha2} = 0 $ so poles: $s_1,s_2 = s_{\alpha1},s_{\alpha2} $ What numerical values you have for $s_{\alpha1},s_{\alpha2} $ are the What numerical values you are looking for.

$\endgroup$

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.