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I have trouble understanding how you can from a pole/zero-plot say something about the impulse response. For instance I have 2 poles in

(-5 +/- 8 jw)

and a zero in 0. How would I know, from just that, how the impulse response would look like?

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    \$\begingroup\$ From the transfer function. You know the poles and zero. \$\endgroup\$ – Chu Sep 30 '16 at 12:38
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The impulse response is the inverse Laplace transform of the transfer function. Hence if you know this latter you can get the former.

If you have a poles-zeros diagram you effectively know the transfer function (neglecting a scale factor). Therefore you can get the general form of the impulse response.

For example, in your case, the transfer function is:

$$ H(s) = K \cdot \frac{s}{(s-s_p)(s-s_p^{*})} = K \cdot \frac{s}{s^2 - (s_p + s_p^{*}) s + |s_p|^2} $$

where \$s_p = -5 + 8 j\,\$ and \$K\$ is an unspecified scale factor. Inserting the value of the pole, you get:

$$ H(s) = K \cdot \frac{s}{s^2 + 10 s + 89} $$

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Find the expression for the unit impulse response of \$\dfrac{1}{s^2+10s+89}\$, then differentiate it (that takes into account the zero at \$s=0\$), and then multiply by whatever constant gain term is appropriate.

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If it were a low pass filter (forget the zero for now) then the poles would make a TF of the form: -

TF = \$\dfrac{\omega_n^2}{s^2 + 2\zeta \omega_ns+ \omega_n^2}\$

You can use the solution to the quadratic to calculate the poles: -

s = \$\dfrac{-2\zeta \omega_n+/-\sqrt{4\zeta^2\omega_n^2 - 4\omega_n^2} }{2}\$

s = \$\zeta \omega_n+/-\sqrt{\zeta^2\omega_n^2-\omega_n^2}\$

= \$\omega_n(\zeta +/-j\sqrt{1-\zeta^2})\$

Now that you have your poles indentified in terms of zeta and \$\omega_n\$ all that remains to do is investigate the standard responses of a 2nd order low pass filter for varying values of zeta and you can see what the impulse response will be.

Don't forget to bring the s (the zero) back to the top line - that makes it a band pass filter but the dominant part is the denominator.

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