Hi I asked a question earlier about this problem:
$$T(s) = \frac{G(s)}{1+G(s)H(s)}$$
Where
$$H(s) = \frac{s}{s+1}$$ $$G(s) = \frac{k(s+4)}{(s+2)(s^2+s+6)}$$ .
Poles resulting: $$-1, -2, -0.5 + 2.4j, -0.5 - 2.4j$$
Zeros resulting: $$0, -4$$
centroid: $$= (-1 + -2 + -0.5 + 2.4j + -0.5 - 2.4j + 4) /2 = 0$$
I am now calculating the breakaway point using the characteristic equation.
$$1 + G(s)H(s) = 0$$
After simplification and taking the derivative with respect to s
$$ 2s^5 + 16s^4+32s^3+12s^2+6s + 24 = 0$$
A 5th order polynomial?
This question is pulled of an old final. No computers aloud. We were showed a trick using division but the highest we did in class was 4th order. I feel like I'm doing something wrong any thoughts?
Note that for transfer functions we can always write $$H = \frac{P}{R}$$ where $P$ and $R$ are polynomials. You wrote $$1 + GH = 1+\frac{Q}{L}\frac{P}{R} = 0,$$ or $$LR + QP = 0.$$ For any two polynomials $P,R$, $\deg(PR) = \deg P + \deg R$ and $\deg (P + R) \leq \max(\deg P,\deg R)$, with $<$ only following if they are the same order have have some number of leading coefficients opposite--which is not the case for any polynomial here. Thus $$\deg(LR + QP) = \max(\deg L + \deg R, \deg Q + \deg P) = \max(1+3,1+1) = 4.$$ You must have made a mistake.