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I am really struggling to see how the constants Kp and Ks can be solved for the closed loop unity feedback system shown in the image below.

Unity feedback control system with PD compensator

This compensator in pink is being designed such that the complex conjugate poles of the system are shifted so that s1 = -4+4j. I understand that the CLCE equation of this system configuration is as follows. $$(s+1)(s+2)+3(K_P+sK_D)=0$$

But then according to my lecture notes the Kp and Kd can be solved directly from this equation but does not elaborate as to how (except with a subscript saying that the complex pole should be substituted in place of s). $$[(s+1)(s+2)+3(K_P+sK_D)=0] (s=-4+4j)$$

I have tried substituting s=1 and getting Kp in terms of Kd and then substituting this value into the compensated CLCE with the pole substituted in place of s but this is to no avail because every time I get a quadratic in terms of Kd with no solution (after getting rid of the imaginary terms via multiplying by the complex conjugate). Can somebody please show and or tell me how to calculate the proportional and derivative gains in a circumstance like this.

I need to know how this works because I have a homework problem where I have to design a PD compensator for a third order system where the compensated transfer function is as follows: $$\frac{C(s)}{R(s)} = \frac{[K_P+sK_D]21}{s(s+1)(s+3)+[K_P+sK_D]21}$$ and so the CLCE is the denominator of this transfer function set equal to 0. This system has to have a pole at -1+j.

As usual any help is appreciated,

Simon.

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  • \$\begingroup\$ What are the implications if you are given one of a "complex conjugate" pair? \$\endgroup\$ – pgvoorhees May 16 '17 at 11:27
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In the equation \$(s+1)(s+2)+3(K_P+sK_D)=0\$, substitute \$s=-4+4 i\$ and \$s=-4-4 i\$.

This will give you two equations in the two unknowns \$K_P\$ and \$K_D\$.

The solution I am getting after this is \$K_P=10\$ and \$K_D=\frac{5}{3}\$.

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  • \$\begingroup\$ Of course, simultaneous equations. How obvious! I will try and apply this for the third order case and get this done. Thanks! \$\endgroup\$ – burton01 May 17 '17 at 1:48

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