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I’m in the final year of my education, and for the exam training I had to find the K (or max gain) value of the given system when the desired damping ration equals 0,7.

What i have done to addres the problem

I designed for my self a 6 step plan to calculate these kind of systems in general.

  1. Calculate the transfer function
  2. Draw the poles and zero's into a plot containing the unity circle
  3. Determin the 'z' point of the gain for a given overshoot. (or when the system wil get unstable)
  4. Take the denomerator of the transfer function and equal it to '0'
  5. enter the found 'z' into this formula
  6. calculate the gain (K)

Given system

To answer the question I first determined the transfer function

\$ \LARGE \frac{C(z)}{R(z)}=\frac{K(z+1)}{(z-1)(z-0.5)+K(z+1)} \$

after which I draw the polar plot with unity circle (in the picture "eenheidscirkel")

enter image description here

The question stated I had to find the damping ratio of 0,7 which was used to draw the following graph. enter image description here

From which the "z" value was 0,719+0,215j. This Z value i enterd into the denomerator of my transferfunction equal to zero. $$ (z-1)(z-0.5)+k(z+1)=0 \\ (0.719+0.215j-1) (0.719+0.215j-0.5)+k(0.719+0.215j+1)=0 $$

This function I rewrote to the final statement for K with the next steps

(-0.281+0.215j)(0.219+0.215j)+k(1.719+0.215j)=0

-0.062-0.060j+0.047j-0.046+k(1.719+0.215j)=0

-0.108-0.013j+k(1.719+0.215j)=0

k=(0.108+0.013j)/(1.719+0.215j)

And here my problem starts

My teacher states that have to devide the first part (ignoring the IM part) of the function to get my gain (K).

0.108/1.719 = 0.063

I could also use the second part for this.

0.013/0.215 = 0.061

Or the sum ignoring the J again

(0.108+0.013)/(1.719+0.215) = 0.063

resulting in K=0.063

This result is correct but I don't understand why.

Why can I ignore the imaginairy part of the final function.

With a complex devision I would expect atleast a K value of a+bj

I tryed this for the final step:

k=(0.108 +0.013j)/(1.719+0.215j)*(1.719-0.215j)/(1.719-0.215j)

k=(0.189-0.046j)/3

RE : k= 0.189/3 = 0.063

IM : k=0.046j/3 = 0.015j

what makes me think the result shoud be

k= 0.063-0.015j

I know this smells like a home work question and there systems that can calculate everything with the snap of a finger, im trying to understand why im allowed (if im allowed) to ignore the imaginary value of the gain of this system.

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migrated from mathematica.stackexchange.com Jan 16 '15 at 15:28

This question came from our site for users of Wolfram Mathematica.

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    \$\begingroup\$ This doesn't really look like a question about the software product Mathematica. Are you sure you're posting on the correct site? \$\endgroup\$ – Sjoerd C. de Vries Jan 16 '15 at 10:54
  • \$\begingroup\$ No im not sure that im on the correct site. But this was the only stackexchange site which contained control stystems and transferfunctions \$\endgroup\$ – Nick Jan 16 '15 at 10:55
  • \$\begingroup\$ And if there is a better site available please point me to it. \$\endgroup\$ – Nick Jan 16 '15 at 11:09
  • \$\begingroup\$ physics.stackexchange.com maybe? \$\endgroup\$ – Öskå Jan 16 '15 at 11:16
  • \$\begingroup\$ There is something like replace the question to the electronics site? I tought its called migrating but i dont know how to do that \$\endgroup\$ – Nick Jan 16 '15 at 12:08
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You can ignore it because it is small, and the error comes because the pole has not been that precisely determined.

With the pole value you give (0.719 + 0.215 j), the solution from the characteristic equation for k is 0.0626793 - 0.0000849631 j. As the precision in the pole value is increased the imaginary part will become even closer to zero.

[I should also add that you are not plotting the polar (Nyquist) plot but rather the root locus plot.]

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  • \$\begingroup\$ So if I understand correctly you say I got the Im part of the system because my pole is not exactly determined? If it was exactly determined the Im part would not exist? \$\endgroup\$ – Nick Jan 17 '15 at 18:10
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    \$\begingroup\$ @Yes, that's it. Normaly you dont choose a imaginary gain. So just use the real part. You'll see that your poles with the real gain aren't that different from the ones of your choosen intersection. (I think you're not able to see it in the plot) \$\endgroup\$ – Phab Jan 19 '15 at 14:31
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    \$\begingroup\$ Yes, and to elaborate on Phab's comment, if you use the gain 0.0626793 and compute the poles, you will get \$ 0.71866\pm 0.214957 i \$. If you use those poles, the gain will be \$0.062679 -7.95* 10^{-18}\$. This is more than enough accuracy. You can now double-check if you got the desired damping ratio. \$\endgroup\$ – Suba Thomas Jan 19 '15 at 14:59

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