Question 5 of 6: Gain for a specified damping ratio

Question 5: Gain for a specified damping ratio (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Unity negative feedback around $G(s)=K\left(s^{2}+6s+12\right)/\left[s^{2}(s+1)\right]$: a type-2 plant with a double pole at the origin, a pole at $-1$, and a complex zero pair at $-3\pm j1.732$. Target damping ratio $\zeta=0.707$.

Find. The gain $K$ that places the complex closed-loop pair on the $\zeta=0.707$ ray, and the resulting root locations.

Approach. Write the characteristic polynomial, force it to factor as a $\zeta=0.707$ quadratic times a real linear term, match coefficients to eliminate the unknown real root, and solve the single remaining equation for the pair location.

1. Form the characteristic equation. With unity feedback, $1+G(s)=0$ gives $s^{2}(s+1)+K\left(s^{2}+6s+12\right)=0$, i.e.$$s^{3}+(1+K)s^{2}+6Ks+12K=0 .$$It is cubic, so the closed loop has one real root and one complex pair for the gains of interest.
2. Impose the damping specification. A damping ratio $\zeta=1/\sqrt{2}=0.707$ puts the pair on a ray $45^{\circ}$ from the negative real axis, so the roots are $s=-a\pm ja$ with $a>0$, and the corresponding quadratic factor is $s^{2}+2as+2a^{2}$. Writing the third root as $-c$,$$\left(s^{2}+2as+2a^{2}\right)(s+c)=s^{3}+(2a+c)s^{2}+\left(2a^{2}+2ac\right)s+2a^{2}c .$$Comparing with the characteristic equation gives three conditions: $2a+c=1+K$, $2a^{2}+2ac=6K$ and $2a^{2}c=12K$.
3. Eliminate $K$ and $c$. Dividing the third condition by the second removes $K$: $2a^{2}c/\left(2a^{2}+2ac\right)=2$, which rearranges to $ac=2(a+c)$ and therefore $c=2a/(a-2)$. The third condition then gives $K=a^{2}c/6$, and substituting both into the first leaves one equation in $a$ alone:$$2a+\frac{2a}{a-2}=1+\frac{a^{2}}{6}\cdot\frac{2a}{a-2} .$$Solving numerically (bisection on $a>2$, which is required for $c>0$) gives a single positive root $a=4.1958$.
4. Back-substitute. With $a=4.1958$, $c=2(4.1958)/2.1958=3.8216$ and$$\boxed{\,K=\frac{a^{2}c}{6}=\frac{(4.1958)^{2}(3.8216)}{6}=11.21\,}$$The closed-loop roots at that gain are$$\boxed{\,s=-4.196\pm j4.196\ \ (\zeta=0.707,\ \omega_{n}=5.934),\qquad s=-3.822\,}$$Substituting $K=11.2133$ into $s^{3}+12.2133s^{2}+67.28s+134.56$ and factoring reproduces exactly these three roots, and $-4.196/\left|{-4.196+j4.196}\right|=0.7071$ confirms the damping ratio.
5. Examine the claim in the question. The paper asks the candidate to show that the complex roots are $-2.3\pm j2.3$. They are not. Substituting $s=-2.3+j2.3$ into the characteristic polynomial gives $\left(24.33-1.8K\right)+j\left(13.75+3.22K\right)$. The real part vanishes at $K=13.52$, but the imaginary part is then $57.3$, not zero — and no value of $K$ can null both simultaneously. At $K=13.52$ the actual complex pair is $-5.005\pm j3.306$ with $\zeta=0.835$. The stated roots therefore satisfy neither the characteristic equation nor the damping specification.
6. Comment on "dominant". At $K=11.21$ the real root $-3.822$ lies closer to the imaginary axis than the complex pair at $-4.196$, so strictly the real root is the dominant one and the $\zeta=0.707$ pair governs only the oscillatory content of the response. The question’s wording follows the usual textbook shorthand; the design is still well behaved, but a second-order approximation based on the complex pair alone would under-predict the settling time.