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We have a filter (or a system) that has two poles at $f_{p1} = f_{p2} = 20$ GHz and one zero at $f_{z1} = 15$ GHz. Let's assume the gain is $K = 1$. Then how do you mark these poles and zero in the s-plane?

Is it right that we consider it as three points at the $j \omega$ axis with real part equal to 0? i.e:

$s_{z_1} = 0 + j2 \pi f_{z_1} = j \omega_{z_1}$,

$s_{p_1} = 0 + j2 \pi f_{p_1} = j \omega_{p_1}$, and

$s_{p_2} = 0 + j2 \pi f_{p_2} = j \omega_{p_2}$

So in this case if I want to write the transfer function it would be

$$ H(s) = \frac{s - j \omega_{z_1}}{(s - j \omega_{p_1})(s - j \omega_{p_2})}$$

Then I can derive $H(j\omega)$ by substituting $s$ with $j \omega$

$$ H(j\omega) = \frac{j\omega - j \omega_{z_1}}{(j\omega - j \omega_{p_1})(j\omega - j \omega_{p_2})} = \frac{\omega - \omega_{z_1}}{(\omega - \omega_{p_1})(\omega - \omega_{p_2})}$$

Is that right so far?

Now if I want to convert the zero location into discrete time domain then which of the following should I use:

a) $\frac{1+\frac{\omega_{z_1}}{f_s}}{1-\frac{\omega_{z_1}}{f_s}}$

or

b) $\frac{1+\frac{s_{z_1}}{f_s}}{1-\frac{s_{z_1}}{f_s}}$ = $\frac{1+\frac{j\omega_{z_1}}{f_s}}{1-\frac{j\omega_{z_1}}{f_s}}$

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    $\begingroup$ "Is it right that we consider it as three points at the jω axis with real part equal to 0?" ---- i don't think it's a good idea to have that restriction. and especially with the poles, you don't want them sitting on the jω axis in the s-plane. $\endgroup$ – robert bristow-johnson Feb 28 at 23:54
  • $\begingroup$ Robert, could you please elaborate more why we don't want to have poles on the $j\omega$ axis? And what is the right or better way to model such a system? $\endgroup$ – shampar Mar 1 at 15:09
  • $\begingroup$ poles sitting on the axis are, at best, "critically stable" or "marginally stable". they really must be left of the axis(having a real part that is negative) for the system to be safely stable. unless you're trying to make an oscillator, i do not see why you would want this marginally stable poles literally ringing forever. $\endgroup$ – robert bristow-johnson Mar 1 at 20:16

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