The root locus of the feedback control system having the real characteristic equation s ²+ 6Ks + 2s + 5 = 0 where K > 0, enters into the real axis at

Understanding the Characteristic Equation

The problem asks us to find the point where the root locus of a feedback control system enters the real axis. The system's characteristic equation is given as:

$s^2+6Ks+2s+5=0$

where K is a positive gain (K > 0).

Rewriting the Characteristic Equation

First, let's combine the terms involving s and rearrange the equation to identify the standard root locus form, which is $1+KG(s)H(s)=0$.

Combining the terms with s:

$s^2+(6K+2)s+5=0$

Now, we isolate the term containing K:

$s^2+2s+5=-6Ks$

To fit the $1+KG(s)H(s)=0$ format, we divide both sides by $-6s$ (assuming $s\ne 0$):

$-\frac{s^2+2s+5}{6s}=K$

This gives us the transfer function related part:

$1+K\left(\frac{-s}{s^2+2s+5}\right)=0$

Or, more commonly written as:

$1+K\frac{6s}{s^2+2s+5}=0$

Here, $G(s)H(s)=\frac{s}{s^2+2s+5}$ and the gain is effectively $6K$.

Finding Breakaway/Break-in Points

The points where the root locus enters or exits the real axis are found by calculating the value of K as a function of s and finding the points where K has local maximum or minimum values. This is done by taking the derivative of K with respect to s and setting it to zero.

We have:

$K(s)=-\frac{s^2+2s+5}{6s}$

Now, let's find the derivative $\frac{dK}{ds}$ using the quotient rule:

$\frac{dK}{ds}=-\frac{1}{6}\frac{d}{ds}\left(\frac{s^2+2s+5}{s}\right)$

$\frac{dK}{ds}=-\frac{1}{6}\left(\frac{(2s+2)s-(s^2+2s+5)(1)}{s^2}\right)$

Set the derivative to zero:

$\frac{(2s^2+2s)-(s^2+2s+5)}{s^2}=0$

This requires the numerator to be zero:

$2s^2+2s-s^2-2s-5=0$

Simplify the equation:

$s^2-5=0$

Solving for s:

$s^2=5$

$s=\pm \sqrt{5}$

Verifying the Root Locus Entry Point

We found two potential points on the real axis: $s=\sqrt{5}$ and $s=-\sqrt{5}$. We need to determine which one is relevant for the root locus with K > 0.

The original poles of the system (from $s^2+2s+5=0$) are complex: $s=\frac{-2\pm \sqrt{2^2-4(1)(5)}}{2}=\frac{-2\pm \sqrt{4-20}}{2}=\frac{-2\pm \sqrt{-16}}{2}=-1\pm 2j$.

The root locus starts from these poles. The zero is at $s=0$. The locus must connect the poles to the zeros or to infinity.

Let's calculate the value of K at each potential point:

- For $s=\sqrt{5}$: $K=-\frac{(\sqrt{5}{)}^2+2(\sqrt{5})+5}{6(\sqrt{5})}=-\frac{5+2\sqrt{5}+5}{6\sqrt{5}}=-\frac{10+2\sqrt{5}}{6\sqrt{5}}$ Since this value of K is negative, and we are given K > 0, the point $s=\sqrt{5}$ is not on the root locus for positive K.
- For $s=-\sqrt{5}$: $K=-\frac{(-\sqrt{5}{)}^2+2(-\sqrt{5})+5}{6(-\sqrt{5})}=-\frac{5-2\sqrt{5}+5}{-6\sqrt{5}}=\frac{10-2\sqrt{5}}{6\sqrt{5}}$ Simplify K: $K=\frac{5-\sqrt{5}}{3\sqrt{5}}=\frac{5\sqrt{5}-5}{15}=\frac{\sqrt{5}-1}{3}$ Since $\sqrt{5}\approx 2.236$, this value of K is positive ($K=\frac{2.236-1}{3}\approx \frac{1.236}{3}\approx 0.412$).

The root locus starts from the complex poles $-1\pm 2j$. As K increases from 0, the locus moves towards the real axis. The value $s=-\sqrt{5}$ is the point where the locus branches first meet the real axis (a breakaway point from the complex plane into the real axis). This occurs at the calculated positive gain $K=\frac{\sqrt{5}-1}{3}$.

Conclusion

The root locus enters the real axis at $s=-\sqrt{5}$.