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A cascade control system with proportional controller is shown below

schematic

Theroritically the largest values of the gain K1,and K2 that can be set withot causing instability of the closed loop sytem are

Given \$G1=\frac{1}{(s+1)(2s+1)}\$ and \$G_2=\frac{1}{(3s+1)}\$

\$(A)\$10 and 100 \$(B)\$100 AND 10 \$(C)\$10 AND 10\$(D)\$\$\infty\$ and \$\infty\$

The closed loop T.F will be

\$\frac{C(s)}{R(s)}=\frac{K_1K_2}{(s+1)(2s+1)(3s+1)+K_2(3s+1)+K_1K_2}\$

Now C.E will be $$6s^3+11s^2+(6+3K_2)s+1+K_2+K_1K_2=0$$

Now after applying R.H criteria I got two conditon for stability

that is $$20+9K_2-2K_1K_2>0$$ and $$1+K_2(1+K_1)>0$$

Now after satifying Options one by one all options becoming unstable for this equations.

Is the options are wrong or I am doing it wrong?

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  • \$\begingroup\$ You have a typo. C.E must be \$6 s^3+\ldots\$. \$\endgroup\$ – Suba Thomas Oct 30 '17 at 13:54
  • \$\begingroup\$ @Suba Thomas sorry \$\endgroup\$ – Rohit Oct 30 '17 at 13:56
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You seem to be doing things right. I verified them using Mathematica. So either you have \$G_1\$ and \$G_2\$ wrong, or the question is wrong.

enter image description here

There are two possibilities:

  1. The max value of \$K1\$ is \$\frac{9}{2}\$, and the max value of \$K2\$ is \$\infty\$.
  2. \$K1\$ is some finite value greater than \$\frac{9}{2}\$, and \$0<\text{K2}<\frac{20}{2 \ \text{K1}-9}\$.
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  • \$\begingroup\$ Thank you sir...Is there any alternate method to do this? and did you get the range with the software can't I do it manually? \$\endgroup\$ – Rohit Oct 30 '17 at 15:14
  • \$\begingroup\$ Yes you can. If \$K1>0\$ and \$K2>0\$, the second condition is trivially satisfied. The first one can be written as \$20+K2(9-2 K1)>0\$. Now you have to consider two conditions \$(9-2 K1)>0\$ and \$(9-2 K1)<0\$, from which the two conditions follow. (You can also stop addressing me sir. :)) \$\endgroup\$ – Suba Thomas Oct 30 '17 at 15:33

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