# Control Systems - Root Locus

In the root locus diagram, we can observe the path of the closed loop poles. Hence, we can identify the nature of the control system. In this technique, we will use an open loop transfer function to know the stability of the closed loop control system.

## Basics of Root Locus

The Root locus is the locus of the roots of the characteristic equation by varying system gain K from zero to infinity.

We know that, the characteristic equation of the closed loop control system is

$$1+G(s)H(s)=0$$

We can represent $G(s)H(s)$ as

$$G(s)H(s)=K\frac{N(s)}{D(s)}$$

Where,

• K represents the multiplying factor

• N(s) represents the numerator term having (factored) nth order polynomial of ‘s’.

• D(s) represents the denominator term having (factored) mth order polynomial of ‘s’.

Substitute, $G(s)H(s)$ value in the characteristic equation.

$$1+k\frac{N(s)}{D(s)}=0$$

$$\Rightarrow D(s)+KN(s)=0$$

Case 1 − K = 0

If $K=0$, then $D(s)=0$.

That means, the closed loop poles are equal to open loop poles when K is zero.

Case 2 − K = ∞

Re-write the above characteristic equation as

$$K\left(\frac{1}{K}+\frac{N(s)}{D(s)} \right )=0 \Rightarrow \frac{1}{K}+\frac{N(s)}{D(s)}=0$$

Substitute, $K = \infty$ in the above equation.

$$\frac{1}{\infty}+\frac{N(s)}{D(s)}=0 \Rightarrow \frac{N(s)}{D(s)}=0 \Rightarrow N(s)=0$$

If $K=\infty$, then $N(s)=0$. It means the closed loop poles are equal to the open loop zeros when K is infinity.

From above two cases, we can conclude that the root locus branches start at open loop poles and end at open loop zeros.

### Angle Condition and Magnitude Condition

The points on the root locus branches satisfy the angle condition. So, the angle condition is used to know whether the point exist on root locus branch or not. We can find the value of K for the points on the root locus branches by using magnitude condition. So, we can use the magnitude condition for the points, and this satisfies the angle condition.

Characteristic equation of closed loop control system is

$$1+G(s)H(s)=0$$

$$\Rightarrow G(s)H(s)=-1+j0$$

The phase angle of $G(s)H(s)$ is

$$\angle G(s)H(s)=\tan^{-1}\left ( \frac{0}{-1} \right )=(2n+1)\pi$$

The angle condition is the point at which the angle of the open loop transfer function is an odd multiple of 1800.

Magnitude of $G(s)H(s)$ is -

$$|G(s)H(s)|=\sqrt {(-1)^2+0^2}=1$$

The magnitude condition is that the point (which satisfied the angle condition) at which the magnitude of the open loop transfer function is one.