Question

a) the open - loop transfer function of a system is given: $g(s)h(s)=\frac{k(s + 10)}{s(s + 3)(s^{2}+4s + 8)}$ (i) determine the value of gain k when the system oscillates, (ii) calculate the roots of the closed - loop system for the k determined in part (i).

Explanation:

Step1: Write the characteristic equation

The closed - loop characteristic equation is $1 + G(s)H(s)=0$, so $s(s + 3)(s^{2}+4s + 8)+k(s + 10)=0$. Expand it:

$$
LATEXBLOCK0
$$

For the system to oscillate, the real - part of the roots of the characteristic equation is zero. We use the Routh - Hurwitz criterion. The Routh array is:

$$
LATEXBLOCK1
$$

For oscillation, the $s^{1}$ row must be zero. So $\frac{140+7k - 10k}{7}=0$.

Step2: Solve for k

$$
LATEXBLOCK2
$$

Step3: Substitute k into the characteristic equation and find the roots

Substitute $k = \frac{140}{3}$ into $s^{3}+7s^{2}+(20 + k)s+10k = 0$. The equation becomes $s^{3}+7s^{2}+(20+\frac{140}{3})s+10\times\frac{140}{3}=0$, or $s^{3}+7s^{2}+\frac{60 + 140}{3}s+\frac{1400}{3}=0$, or $3s^{3}+21s^{2}+200s + 1400=0$.
Since the system is oscillating, we know that there are pure imaginary roots. Let $s = j\omega$. Then $(j\omega)^{3}+7(j\omega)^{2}+(20+\frac{140}{3})(j\omega)+\frac{1400}{3}=0$.

$$
LATEXBLOCK3
$$

From the real - part $-7\omega^{2}+\frac{1400}{3}=0$, we get $\omega^{2}=\frac{200}{3}$, so $\omega=\pm\sqrt{\frac{200}{3}}$.
Substitute $\omega$ into the imaginary part to verify.
The roots of the closed - loop system are $s = j\sqrt{\frac{200}{3}},s=-j\sqrt{\frac{200}{3}}$ and the third root can be found by long - division or synthetic division. Divide $3s^{3}+21s^{2}+200s + 1400$ by $s^{2}+\frac{200}{3}$.

$$
LATEXBLOCK4
$$

The roots are $s = j\sqrt{\frac{200}{3}},s=-j\sqrt{\frac{200}{3}},s=- 7$.

Answer:

(i) $k=\frac{140}{3}$
(ii) $s = j\sqrt{\frac{200}{3}},s=-j\sqrt{\frac{200}{3}},s=-7$