Suppose we choose to control the plant by scaling our error signal by $K$ and then put the controller on the feedback path like in Figure 7. It would be helpful for us to understand how the closed loop poles of the feedback system change as $K$ is varied over the range $[0, \infty)$. First, suppose

$G(s) = \frac{N_G}{D_G} \qquad H(s) = \frac{N_H}{D_H}.$

$\frac{Y(s)}{R(s)} = \frac{kG(s)}{1+kG(s)H(s)} = \frac{KN_GD_H}{D_GD_H+KN_GN_H}.$

$\Delta(s) = D_GD_H+KN_GN_H = 1 + K\frac{N_GN_H}{D_GD_H} = 0 \qquad (14)$

Clearly, no matter what $K$ is, the poles must satisfy two criteria.

$\left|k\frac{N_GN_H}{D_GD_H}\right| = 1 \qquad \angle K\frac{N_GN_H}{D_GD_H} = (2r+1)\pi,\quad r\in\mathbb{Z} \qquad (15)$

The root locus is the set of all $s\in\mathbb{C}$ such that $\exists K$ where $\Delta(s) = 0$.

First, notice that the roots of $\Delta(s)$ are the closed loop poles of the system.

The number of branches in the root locus is equal to the number of closed loop poles where a branch is the path traveled by a single pole as $K$is varied.

$\angle K \frac{\prod_{i=1}^m (s-z_i)}{\prod_{i=1}^n (s-p_i)} = \sum_{i=1}^m \angle (s-z_i) - \sum_{i=1}^n \angle (s-p_i)$

where $z_i$ are open loop zeros and $p_i$ are open loop poles. If we restrict ourselves to the real axis, then given a closed loop pole $s$, each $z_i > s$ will contribute $-180$˚and each $p_i > s$ will contribute $180$˚while the $z_i, p_i < s$ will contribute $0$˚.

When $K$ is small, then the poles look like the open loop poles. As $K$ grows very large, then the poles look like the open loop zeros.

If there are more poles ($n$) then zeros ($m$), then not all of the poles will end up at a zero in the limit. This means that $n-m$ poles must branch off to infinity.

$\lim_{|s|\to\infty} kH(s)G(s) \approx \lim_{|s|\to\infty}k\frac{s^m}{s^n} = (2l+1)\pi \implies s^{n-m} = re^{j\theta(n-m)} \implies \theta = \frac{(2l+1)\pi}{n-m}$

$\theta = \frac{-(2l+1)\pi}{n-m},$

$\sigma = \frac{\sum_{i=1}^m p_i - \sum_{i=1}^n z_i}{n-m}.$

$\sum_{i=1}^n \frac{1}{\sigma + p_i} = \sum_{i=1}^m \frac{1}{\sigma+z_i}.$

$\sum_{i=1}^m \angle (j\omega + z_i) - \sum_{i=1}^n \angle (j\omega + p_i) = (2l+1)\pi.$

Poles beginning at complex locations will depart at an angle $\theta$ where

$\sum_{i=1}^m (p + z_i) - \sum_{i=1}^n (p + p_i) = (2l+1)\pi$

Finally, since $|KG(s)H(s)|=1$, we can determine $K$ if we know a particular pole location.

Given a pole location $p$,

$K = \left|\frac{1}{G(p)H(p)}\right|.$

Because the Root Locus rules are derived from the characteristic polynomial $\Delta(s)$ of the closed-loop system, they can be used not just to find how the closed loop poles vary with a gain, but also to find how the closed loop poles vary with an open loop pole. Suppose that $G(s) = \frac{N_G}{(s+k)\prod_i(s+p_i)}$ and $H(s) = 1$. Then

$\Delta(s) = (s+k)\prod_i(s+p_i) + N_G = 1 + k\frac{\prod_i (s+p_i) }{N_G + s\prod_i(s+p_i)} = 0$

$Y(s) = \frac{\prod_i (s+p_i) }{N_G + s\prod_i(s+p_i)}$

$G(j\omega) = K \frac{(j\omega)^{N_{z0}}}{(j\omega)^{N_{p0}}}\frac{\prod_{i=0}^{n}{(1+\frac{j\omega}{\omega_{zi}})}}{\prod_{k=0}^{m}{(1+\frac{j\omega}{\omega_{pk}})}} = Ke^{j\frac{\pi}{2}(N_{z0}-N_{p0})} \frac{\prod_{i=0}^{n}{r_{zi}}}{\prod_{k=0}^{m}{r_{pk}}} e^{j(\sum_{i=0}^{n}{z_i} - \sum_{k=0}^{m}{p_k})}.$

Each $r$ is the magnitude of a factor $1 + \frac{j\omega}{\omega_n}$ where $\omega_n$ is either a zero or a pole, $z_i, p_k$ are the phases of each factor, and $N_{z0}, N_{p0}$ are the number of zeros and poles at 0. By writing $G(\omega)$ this way, it is clear that

$|G(\omega)| = K \frac{\prod_{i=0}^{n}{r_{zi}}}{\prod_{k=0}^{m}{r_{pk}}}.$

$20\log(|G(\omega)|) = 20\log(K) + 20\sum_{i=0}^{n}{\log(r_{zi})} - 20\sum_{k=0}^{m}{\log(r_{pk})}$

Likewise, the exponential form of $G(\omega)$ tells us that

$\angle G(\omega) = \frac{\pi}{2}(N_{z0}-N_{p0})+ (\sum_{i=0}^{n}{z_i} - \sum_{k=0}^{m}{p_k}).$

Each $p_k$ and $z_i$ are of the form $1 + \frac{j\omega}{\omega_n}$. If $\omega > 10\omega_n$, then $p_k, z_i \approx \omega_n$. Likewise, if $\omega < \frac{\omega_n}{10}$, $p_k, z_i \approx 1$. This means we can approximate bode plots using piece-wise linear segments using the following rules.

Each zero $\omega_z$ contributes $20\text{dB/dec}$ to magnitude starting at $\omega_z$.

Each pole $\omega_p$ contributes $-20\text{dB/dec}$ to magnitude starting at $\omega_p$.

Each zero $\omega_z$ contributes $45\text{˚/dec}$ to phase starting at $\frac{\omega_z}{10}$ and ending at $10\omega_z$.

Each pole $\omega_p$ contributes $-45\text{˚/dec}$ to phase starting at $\frac{\omega_p}{10}$ and ending at $10\omega_p$.

$G(s) = \frac{N_G}{D_G} \qquad H(s) = \frac{N_H}{D_H}.$

$\frac{Y(s)}{R(s)} = \frac{G}{1+G(s)H(s)} = \frac{N_GD_H}{D_GD_H + N_GN_H}.$

$1+GH = 1 + \frac{N_GN_H}{D_GD_H} = \frac{N_GN_H+D_GD_H}{D_GD_H}$

From here, we can see that the poles of $1+GH$ are the poles of the open loop system whereas the zeros of $1+GH$ are the poles of the closed loop system.

The Nyquist Criterion says that if $N$ is the number of counter-clockwise encirclements of zero of a contour mapped by a transfer function $F(s)$, $P$ is the number of poles in the contour, and $Z$ is the number of zeros in the contour, then

Thus, given an open loop transfer function $GH$, we can determine its stability. We already know $P$ from the poles of the open loop system, and we can find $N$ by defining a contour which encapsulates the right half plane and use Equation 16 to find $Z$. However, remember that we need to find the RHP poles of $1+GH$. This shifts our mapping to the right by $1$, so we can instead just let $N$ be the number of encirclements of $-1$. Once we have $Z$, we know how many RHP poles the closed loop transfer function will have because they are the same as the RHP zeros of the $1+GH$. We can extend the Nyquist Criterion to finding a range of gains that would make the open-loop system $kG(s)H(s)$ stable by looking for the encirclements of $\frac{-1}{K}$.