0
1.6kviews
Find the range of K for the system to be stable $s^4 + 7s^3 + 10s^2 + 2ks + k = 0$.

Mumbai University > Electronics Engineering > Sem 4 > Linear Control Systems

Topic: Stability Analysis in Time Domain

Difficulty : High

Marks : 5M

1 Answer
0
30views

Given, $s^{4}+7s^{3}+10s^{2}+2ks+k=0$

$s^{4}|1\quad 10\quad k$

$s^{3}|7\quad 2k\quad 0$

$s^{2}|\cfrac{70-2k}{7}\quad k\quad 0$

$s^{1}|\cfrac{\left(\cfrac{70-2k}{7}\right)2k-7k}{\left(\cfrac{70-2k}{7}\right)} \quad 0\quad 0$

$s^{0}|k \quad 0\quad 0$

At $s^0 \Longrightarrow k \gt 0$

For $s^{1}=\cfrac{\left(\cfrac{70-2k}{7}\right)2k-7k}{\left(\cfrac{70-2k}{7}\right)} \gt 0$

$k \left[ 2\left(\cfrac{70-2k}{7}\right)-7 \right] \gt 0$

$ \left[ \cfrac{140-4k-49}{7} \right] \gt 0$

$140-4k-49=0$

$91=4k$

$\cfrac{91}{4}=k$

$22.75 \gt k$

For $s^{2}=\cfrac{70-2k}{7} \gt 0$

$70-2k \gt 0$

$k \lt 35$

Combining the condition, k should be in the range of $0 \lt k \lt 35$.

Please log in to add an answer.