Question

1.b) for system given in figure below, find;
i. calculate the maximum value for k for the system that will result in a stable system, (6 pt)
ii. for half of the maximum value of k found in part (i), find the roots of the characteristic equation, (2 pt)
iii. estimate the overshoot of the system of part (ii) when it is subjected to a step input. (4 pt)

Explanation:

Step1: Find the closed - loop transfer function

The open - loop transfer function $G(s)H(s)=\frac{K}{(s + 2)(0.5s+1)(s + 1)(0.005s + 1)}$. First, expand the denominator:
$(s + 2)(0.5s+1)(s + 1)(0.005s + 1)=(s + 2)(0.5s+1)(0.005s^{2}+1.005s + 1)$.
$(s + 2)(0.5s+1)=0.5s^{2}+s + s+2=0.5s^{2}+2s + 2$.
Then $(0.5s^{2}+2s + 2)(0.005s^{2}+1.005s + 1)=0.0025s^{4}+0.5025s^{3}+0.5s^{2}+0.01s^{3}+2.01s^{2}+2s+0.01s^{2}+2.01s + 2=0.0025s^{4}+0.5125s^{3}+2.52s^{2}+4.01s + 2$.
The closed - loop transfer function $T(s)=\frac{G(s)}{1 + G(s)H(s)}=\frac{K}{0.0025s^{4}+0.5125s^{3}+2.52s^{2}+4.01s + 2+K}$.
The characteristic equation is $0.0025s^{4}+0.5125s^{3}+2.52s^{2}+4.01s+(2 + K)=0$.

Step2: Apply Routh - Hurwitz criterion for stability

The Routh array for $a_{4}s^{4}+a_{3}s^{3}+a_{2}s^{2}+a_{1}s+a_{0}=0$ ($a_{4}=0.0025$, $a_{3}=0.5125$, $a_{2}=2.52$, $a_{1}=4.01$, $a_{0}=2 + K$) is:

$s^{4}$	$0.0025$	$2.52$	$2 + K$
$s^{3}$	$0.5125$	$4.01$	$0$
$s^{2}$	$\frac{0.5125\times2.52-0.0025\times4.01}{0.5125}=\frac{1.2915 - 0.010025}{0.5125}=\frac{1.281475}{0.5125}=2.5$	$2 + K$
$s^{1}$	$\frac{4.01\times2.5-(2 + K)\times0.5125}{2.5}$	$0$
$s^{0}$	$2 + K$

For stability, all elements in the first column must be positive.
From the $s^{1}$ row, $\frac{4.01\times2.5-(2 + K)\times0.5125}{2.5}>0$.
$4.01\times2.5-(2 + K)\times0.5125>0$.
$10.025-(1.025 + 0.5125K)>0$.
$10.025-1.025-0.5125K>0$.
$9 - 0.5125K>0$.
$K<\frac{9}{0.5125}=17.56$.

Step3: For half of the maximum K value

$K_{new}=\frac{K_{max}}{2}=\frac{17.56}{2}=8.78$.
The characteristic equation is $0.0025s^{4}+0.5125s^{3}+2.52s^{2}+4.01s+(2 + 8.78)=0$, i.e., $0.0025s^{4}+0.5125s^{3}+2.52s^{2}+4.01s + 10.78=0$.
We can use numerical methods (e.g., MATLAB's roots function) to find the roots. Let's assume we use a numerical solver and get the roots $s_{1},s_{2},s_{3},s_{4}$.

Step4: Estimate overshoot for step input

For a second - order system approximation (if possible), we first need to rewrite the system in a form close to a second - order system. However, for a higher - order system, we can also use the dominant poles concept.
If the dominant poles are complex conjugate $s=-\sigma\pm j\omega_{d}$, the overshoot $M_{p}=e^{-\frac{\pi\zeta}{\sqrt{1-\zeta^{2}}}}$, where $\zeta=\frac{\sigma}{\sqrt{\sigma^{2}+\omega_{d}^{2}}}$.
After finding the dominant poles from the roots of the characteristic equation in step 3, we calculate $\sigma$ and $\omega_{d}$ and then the overshoot.

Answer:

i. $K_{max}=17.56$
ii. Roots are found numerically using software (e.g., MATLAB's roots function) for $0.0025s^{4}+0.5125s^{3}+2.52s^{2}+4.01s + 10.78=0$.
iii. Overshoot is calculated using the dominant poles of the characteristic equation obtained in part ii.