Mid2

Sensitivity¶

sensitivity of transfer function $T$ w.r.t. its parameter $K$ : $S_{K}^{T}$

\begin{matrix} S_{K}^{T} = \frac{\frac{d T}{T}}{\frac{d K}{K}} = \frac{d \ln T}{d \ln K} = \frac{K}{T} \frac{d T}{d K} \end{matrix}

System Type for Tracking¶

Reference Tracking¶

Tracking definition
the output $Y (s)$ follow any reference input $R (s)$ as closely as possible.

\begin{matrix} E = R - Y \to 0 \\ E (s) = R (1 - \frac{D G}{1 + D G}) = \frac{1}{1 + D G} R \end{matrix}

System Classification
for type $k$ , we have the input $R$ as

\begin{matrix} r (t) = \frac{1}{k!} t^{k} 1 (t) ⟺ R (s) = \frac{1}{s^{k + 1}} \end{matrix}

such that

\begin{matrix} e (\infty) = const \neq 0 \end{matrix}

type $0$

\begin{aligned} e (\infty) & = [s E (s)]_{s \to 0} \\ = [s \frac{1}{1 + D G} R (s)]_{s \to 0} \\ = [s \frac{1}{1 + D G} \frac{1}{s}]_{s \to 0} \\ = \frac{1}{1 + D (0) G (0)} = \frac{1}{1 + K_{p}} \end{aligned}

$K_{p}$ : position error constant

type $1$

\begin{aligned} e (\infty) & = [s E (s)]_{s \to 0} \\ = [s \frac{1}{1 + D G} R (s)]_{s \to 0} \\ = [s \frac{1}{1 + D G} \frac{1}{s^{2}}]_{s \to 0} \\ = [\frac{1}{s + s D G}]_{s \to 0} = \frac{1}{K_{v}} \end{aligned}

\begin{matrix} ⟹ D (s) G (s) = \frac{1}{s} J (s) \\ ⟹ K_{v} = J (0) \end{matrix}

$K_{v}$ : velocity constant

type $2$

\begin{aligned} e (\infty) & = [s E (s)]_{s \to 0} \\ = [s \frac{1}{1 + D G} R (s)]_{s \to 0} \\ = [s \frac{1}{1 + D G} \frac{1}{s^{3}}]_{s \to 0} \\ = [\frac{1}{s^{2} + s^{2} D G}]_{s \to 0} = \frac{1}{K_{a}} \end{aligned}

\begin{matrix} ⟹ D (s) G (s) = \frac{1}{s^{2}} J (s) \\ ⟹ K_{a} = J (0) \end{matrix}

- $K_{a}$ : acceleration constant¶

Disturbance Rejection¶

We hope to reject the disturbance $W (s)$ . Thus, to determine the system type w.r.t. $W$ , first ignoring the effect of $R$ by giving $R (s) = 0$ .

\begin{matrix} E (s) = R - Y = - Y \\ Y = \frac{G}{1 + G D} W \end{matrix}

then

\begin{matrix} E (s) = \frac{- G}{1 + G D} W \end{matrix}

PID Controller¶

P (Proportional)

\begin{matrix} D_{c l} (s) = \frac{U (s)}{E (s)} = k_{p} \end{matrix}

I (Integral)¶

\begin{matrix} D_{c l} (s) = \frac{U (s)}{E (s)} = \frac{k_{I}}{s} \end{matrix}

fix steady-state error (with $K_{I} \neq 0$ , $e_{s s} = 0$ )

D (Derivative)¶

\begin{matrix} D_{c l} (s) = \frac{U (s)}{E (s)} = s k_{D} \end{matrix}

make the damping coefficient $ζ ↑$ , then the stability $↑$ , overshoot $↓$
has no effect on stead-state error

PI (P + I)

\begin{matrix} U (s) = E (s) k_{P} + \frac{k_{I}}{s} ⟺ u = k_{P} e + k_{I} \int e (τ) d τ \\ D_{c l} (s) = \frac{U (s)}{E (s)} = s k_{D} \end{matrix}

Root Locus¶

for the characteristic equation, all of these equations are equivalent.

\begin{matrix} 1 + K L (s) = 0 \\ 1 + K \frac{(s - z_{1}) (s - z_{2}) (s - z_{3}) \dots \cdot}{(s - p_{1}) (s - p_{2}) (s - p_{3}) \dots \cdot} = 0 \\ ((s - p_{1}) (s - p_{2}) \dots \cdot) + K ((s - z_{1}) (s - z_{2}) \dots \cdot) = 0 \\ \frac{1}{K} ((s - p_{1}) (s - p_{2}) \dots \cdot) + ((s - z_{1}) (s - z_{2}) \dots \cdot) = 0 \end{matrix} \begin{array}{r} (1) \\ (2) \\ (3) \\ (4) \end{array}

Rule 1: Find zeros and poles.
$K = 0$ , i.e. start of the root locus is located at the poles (observing (3))
$K = \infty$ , i.e. end of the root locus is located at the zeros (observing (4))
Rule 2: Find the real axis positions of the locus.
Let $s = s_{0}$ , for $s_{0} \in ℝ$

\begin{matrix} ∠ L (s_{0}) = (ψ_{z 1} + ψ_{z 2} + \dots +) - (ϕ_{p 1} + ϕ_{p 2} + \dots +) = π, - π, 3 π, - 3 π, \dots \end{matrix}

if none of $s_{0}$ satisfy the equation above then the root locus is not on the real axis.
Otherwise, we can solve the range of $s_{0}$

Rule 3: Find the asymptotes for large $K$
When $K \to \infty$ , there is not only the result show in Rule 1 ( $s$ is finite)
Consider (2), for $K, s$ is large we can modify (2) as

\begin{matrix} {[1 + K \frac{(s - z_{1}) (s - z_{2}) (s - z_{3}) \dots \cdot}{(s - p_{1}) (s - p_{2}) (s - p_{3}) \dots \cdot}]}_{s \to \infty} = 0 \\ ⟹ 1 + K \frac{s^{(\# of zeros)}}{s^{(\# of poles)}} = 0 \\ ⟹ 1 + K s^{m - n} = 0 \\ ⟹ s^{n - m} + K = 0 \end{matrix}

for $n - m = 0$ ,

\begin{matrix} K \neq 0 \end{matrix}

In this case there is no asymptote. As Rule 1 says, all root go from poles to finite zeros.

- for $n - m = 1$

\begin{matrix} s = - K = - \infty \end{matrix}

this indicates that there is an imaginary zero at $- \infty$ .

for $n - m > 1$

\begin{matrix} s^{n - m} = - K = K ∠ π \\ ⟹ s = k ∠ (\frac{π + ℓ \cdot 2 π}{n - m}) \\ k = K^{1 / (n - m)} = \infty \end{matrix}

e.g. $n - m = 3$

s = {\begin{matrix} k ∠ \frac{π}{3} \\ k ∠ \frac{3 π}{3} = & k ∠ π \\ k ∠ \frac{- π}{3} \end{matrix}

note that this result only tell us there are three imaginary zeros at different directions of $\infty$ .
However, the source (intersection) is no need to be at the origin ( $s = 0$ ). This is because, for a finite $s_{0} < 0$ .

s_{0} = m ∠ π ⟹ {\begin{matrix} (k ∠ \frac{π}{3} - s_{0}) = k ∠ \frac{π}{3} \\ (k ∠ π - s_{0}) = k ∠ π \\ (k ∠ \frac{- π}{3} - s_{0}) = k ∠ \frac{- π}{3} \end{matrix}

Thus we can write the asymptotes as

\begin{matrix} ((s - p_{1}) (s - p_{2}) \dots \cdot) + K ((s - z_{1}) (s - z_{2}) \dots \cdot) = 0 \\ ⟺ (s^{n} + a s^{n - 1} + \dots +) + K (s^{m} + b s^{m - 1} + \dots +) = 0 \\ ⟹ \sum p_{i} = - a \end{matrix}

\begin{matrix} (s^{n} + a s^{n - 1} + \dots +) + K (s^{m} + b s^{m - 1} + \dots +) = 0 \\ ⟺ s^{n} + a s^{n - 1} + \dots + K (s^{m} + b s^{m - 1} + \dots +) = 0 \\ ⟺ (s - r_{1}) (s - r_{2}) \dots \cdot (s - r_{n}) = 0 \\ ⟹ \sum r_{i} = - a \\ ⟹ \sum r_{i} = - a = \sum p_{i} \end{matrix}

now rewrite the characteristic equation

\begin{matrix} [1 + K \frac{(s - z_{1}) (s - z_{2}) (s - z_{3}) \dots \cdot}{(s - p_{1}) (s - p_{2}) (s - p_{3}) \dots \cdot}]_{s \to \infty} \approx 1 + K \frac{1}{(s - α)^{n - m}} = 0 \\ ⟹ (s - α)^{n - m} + K = 0 \\ ⟺ (s - r_{1}) (s - r_{2}) \dots \cdot (s - r_{n - m}) = 0 \\ ⟹ \sum_{i = 1}^{n - m} r_{i} = - (- (n - m) α) = (n - m) α \end{matrix}

notice that all of the roots go from poles to zeros, therefore for $n > m$ case, there are some poles go to infinite imaginary zeros.
Thus we have

\begin{matrix} \sum (all roots) = \sum (roots go to finite zeros) + \sum (roots go to infinte img. zeros) \end{matrix}

\begin{aligned} ⟹ \sum_{i = 1}^{n} r_{i} & = \sum_{i = 1}^{m} z_{i} + \sum_{i = m + 1}^{n} z_{i m g, i} \\ = \sum z_{i} + (n - m) α \end{aligned}

\begin{matrix} ⟹ α = \frac{\sum r_{i} - \sum z_{i}}{n - m} = \frac{\sum p_{i} - \sum z_{i}}{n - m} \end{matrix}

Rule 4: departure and arrival angle
- for departure angle take $s_{0} \to p_{i}$ and solve $∠ L (s_{0}) = n π$
- for departure angle take $s_{0} \to z_{i}$ and solve $∠ L (s_{0}) = n π$
Rule 5: points on Image( $j ω$ ) axis
similar to rule 2, just let $s_{0} = j ω_{0}$ and try to solve the characteristic equation.
Rule 6: find breakaway points (location of multiple roots)
consider there is a multiple roots $r_{1}$ then we can write

\begin{matrix} 1 + K L (s) = 0 \\ ⟹ a (s) + K b (s) \equiv (s - r_{1})^{p} (s - r_{2}) \cdot \dots \cdot = 0 \\ \frac{d}{d s} [a (s) + K b (s)]_{s = r_{1}} = p (s - r_{1})^{p - 1} (s - r_{2}) \cdot \dots \cdot = 0 \end{matrix}

thus, by solving the equation

\begin{matrix} \frac{d}{d s} [a (s) + K b (s)] = 0 \end{matrix}

we can probably find the breakaway point $r_{1}$ .