I have the following question

enter image description here

I need to Draw the Bode Plot,so i need to find Magnitude and Phase Angles for a Frquency.I understand that i need to susbstitude S=jW(W>Omega) as per this video


But in my question that denominator is not of the from (1+xxS)(1+XXS) so how should i proceed.Please advice.


"denominator is not of the from (1+xxS)(1+XXS)"

If you like such a form (and I would agree with you for solving the present task), you have nothing to do than to find the complex roots (s1 and s2) of the denominator equation D(s)=0. Then you can write D(s)=(1-s1)(1-s2).

EDIT: Regarding the three other questions - I have some problems to understand the meaning. Stability parameters are defined for systems with feedback only. For all stability calculations we need the loop gain. Here, we have a given transfer function - and that´s all we have. Hence, these three questions make no sense, unless a feedback loop is shown.

  • \$\begingroup\$ Do you mean by using the equation img.sparknotes.com/figures/A/a28868a2f5deab01f6e8c7727c2ee7a4/… \$\endgroup\$ – jkooper Apr 30 '16 at 11:27
  • \$\begingroup\$ Yes, of course. D(s) is a simple quadratic equation to be solved applying the classical methods. \$\endgroup\$ – LvW Apr 30 '16 at 11:32
  • \$\begingroup\$ I got (1+.000246)(1+.000246).. please let me know if im right. \$\endgroup\$ – jkooper Apr 30 '16 at 11:37
  • \$\begingroup\$ No - that´s not looking good. A double real pole? It´s a simple quadratic equation to be solved! \$\endgroup\$ – LvW Apr 30 '16 at 12:00
  • \$\begingroup\$ Will solve it again..then.. \$\endgroup\$ – jkooper Apr 30 '16 at 12:19

Look at it like this: -

\$s^2+ s + 1\$ becomes \$1-\omega^2 +j\omega\$

In other words you have a real and imaginery part. I've just substituted s with jw remembering that \$j^2\$ = -1.

Multiply numerator and denominator by the complex conjugate (\$1-\omega^2 -j\omega\$) to bring all the terms of j to the numerator. Denominator becomes \$(1-\omega^2)^2 +\omega^2\$ and as you can see there are no "j" terms.

Is this enough of a hint to get you to a solution?

  • \$\begingroup\$ Thanks for your answer... but it seems the way in which the video proceeds need the numerator is specific form (1+xxS)(1+XXS).. ie:1+something.. Im a computer science guy.. dont know much about Bode Plot.. \$\endgroup\$ – jkooper Apr 30 '16 at 11:39
  • \$\begingroup\$ Do you know how to solve the roots of a quadratic equation? \$\endgroup\$ – Andy aka Apr 30 '16 at 17:44

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