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the transfert function is attached. he was close to find the answers but apparently did something wrong along the way

$H(s)=(s+25)_{2}+40000s_{2}+40000 $
(b) (15\%) The circuit in (a) has poles that are too close to the imaginary axis. Proportional feedback control $(G(s)=K)$ is used to move the poles farther to the left in the $s$ plane. The closed-loop poles must have a real part equal to -50 . Find the appropriate values for the coefficient $K$. (Note: the imaginary part of the poles may change too, and this is acceptable.)
Proportional (P) Feedback: we choose $G(s)=K$ $□$ For systems with a single pole, it is sufficient afor systems with $≥2$ poles, it may be insufficient $□$ We need more flexibility (degrees of freedom) Control Theory Proportional +Derivative (PD) Feedback: $G(s)=K_{1}+K_{2}s$ Proportional +Integral (PI) Feedback: $G(s)=K_{1}+sK_{2} $ Prop. + Integral +Derivative (PID) Feedback: $G(s)=K_{1}+K_{2}s+sK_{3} $
$(s+2s)_{2}+40000x=2a−b±b_{2}−4ac a(s)=s_{2}+50s+4062ss_{2}+40000 ⋅s_{2}+50s+406asx=2(0)−50±(s0)_{2}−4(1)(40625) x=s_{2}+40000−sc±−160000 Q(s)=[s_{2}+50s+40625]+ks_{2}+40000ks_{2}+40000 Q(s)=1+k((s+2s)+40000s+40000 )(s+2s)+40000s+40000 =(s+2s)_{2}+40000+k(s_{2}+40000)s_{2}+40000 fors=−50X(s+50) $

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