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.


  • $\begingroup$ 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.
    Commented Jan 30, 2015 at 10:17
  • $\begingroup$ Then this assingment is incomplete, because i cant find any values that are given to the sα1 and sα2 $\endgroup$ Commented Jan 30, 2015 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.
    Commented Jan 30, 2015 at 10:56

1 Answer 1


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.


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