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 closed - loop 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 the left - hand side:

$$s(s^{2}+4s + 8+3s + 12)+k(s + 10)=0$$

$$s(s^{2}+7s + 20)+k(s + 10)=0$$

$$s^{3}+7s^{2}+20s+ks + 10k=0$$

$$s^{3}+7s^{2}+(20 + k)s+10k=0$$

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

$$
LATEXBLOCK0
$$

Set the \(s^{1}\) row element equal to zero for marginal stability (oscillation).

Step2: Solve for \(k\)

$$7(20 + k)-10k=0$$

$$140+7k-10k=0$$

$$140 - 3k=0$$

$$k=\frac{140}{3}$$

Step3: Substitute \(k\) into the characteristic equation and solve for roots

Substitute \(k = \frac{140}{3}\) into \(s^{3}+7s^{2}+(20 + k)s+10k=0\), we get \(s^{3}+7s^{2}+(20+\frac{140}{3})s + 10\times\frac{140}{3}=0\), or \(s^{3}+7s^{2}+\frac{60 + 140}{3}s+\frac{1400}{3}=0\), \(s^{3}+7s^{2}+\frac{200}{3}s+\frac{1400}{3}=0\), multiply through by 3 to get \(3s^{3}+21s^{2}+200s + 1400=0\).
Since the system is marginally stable, we know that the form of the roots is \(s=\pm j\omega\). Substitute \(s = j\omega\) into \(3s^{3}+21s^{2}+200s + 1400=0\):

$$3(j\omega)^{3}+21(j\omega)^{2}+200(j\omega)+1400=0$$

$$-3j\omega^{3}-21\omega^{2}+200j\omega + 1400=0$$

Separate real and imaginary parts:
Real part: \(-21\omega^{2}+1400 = 0\), \(\omega^{2}=\frac{1400}{21}=\frac{200}{3}\), \(\omega=\pm\sqrt{\frac{200}{3}}\)
Imaginary part: \(-3\omega^{3}+200\omega = 0\), \(\omega(-3\omega^{2}+200)=0\), which is consistent with \(\omega=\pm\sqrt{\frac{200}{3}}\)
The roots of the closed - loop system are \(s=\pm j\sqrt{\frac{200}{3}}\) and the third root can be found by long - division or synthetic division. Dividing \(3s^{3}+21s^{2}+200s + 1400\) by \(s^{2}+\frac{200}{3}\) gives \(s + 7\), so the third root is \(s=-7\)

Answer:

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