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I am trying to solve the following question.

Consider the transfer function $$G(s) = \frac{1.247}{s^2+9.76s+23.8}$$ is in the forward path of a unity feedback control loop. Assume that it is compensated using a static gain K in the forward path. Now I have to plot the root locus of closed-loop system and determine if it is possible to find a value for K such that the settling time is 0.7963 s and the rise time is 0.4445 s.

I know how to do it on paper. The overall closed loop transfer function become $$G(s) = \frac{K*1.247}{s^2+9.76s+23.8+(K*1.247)}$$

If I compare this with standard form of second order system, I can see that $$2*\zeta*\omega_n = 9.76$$ and $$\zeta\omega_n = 4.88$$ From the formula for settling time, $$t_s = \frac{4}{\zeta\omega_n}$$ $$ \zeta\omega_n = \frac{4}{0.7963} = 5.02$$

Therefore, it is not possible to design the system by just varying the value of K. But my doubt is, is there any way to directly find it from the root-locus plot in Matlab. How do I plot the root locus in the first place without knowing the value of K?

Michio Kaku
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  • The formula for settling time is approximate. In this case, \$t_s \approx \frac{4}{4.88}=0.82\$, and you cannot change this value. You can, however, select K to achieve the required 10%-90% rise time (once again, the formula for this is approximate). So, yes, this can all be done via the root locus, but don't expect the system to produce the exact values given in the design specification. – Chu Apr 12 '21 at 12:02
  • Please add a plot of the root locus. I think the answer will be obvious from the root locus plot itself. Mark the area on the complex s-plane where settling time is 0.79 and rise time is 0.44 (formulae for both exist for second order systems). See if the root locus passes through that area. – AJN Apr 12 '21 at 12:21
  • "*How do I plot the root locus in the first place without knowing the value of K?*" Isn't root locus plotted by varying K from zero to infinity ? So, there is no need to know the "*final*" value of K while plotting the locus. – AJN Apr 12 '21 at 12:22
  • Why do you evaluate the closed loop, since you said it is in feed forward path, therefore open loop? make a depiction of the block diagram of the entire loop. – Marko Buršič Apr 12 '21 at 12:48

1 Answers1

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Therefore, it is not possible to design the system by just varying the value of K

I think that the mistake you are making is assuming \$\omega_n\$ is unaffected by \$K\$. In your closed-loop \$G(s)\$ formula, \$\omega_n\$ is in fact this: -

$$\sqrt{23.8 + 1.247\cdot K}$$

Hence, \$\zeta = \dfrac{5.02}{\sqrt{23.8 + 1.247\cdot K}}\$

Andy aka
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