Mid1

DC Motor¶

\begin{matrix} τ = K_{t} i_{a} \\ e = K_{e} {\dot{θ}}_{m} \\ K_{t} = K_{e} (for same unit) \end{matrix}

\begin{matrix} τ = J_{m} \ddot{θ} + b \dot{θ} = K_{t} i_{a} \\ v_{a} = L_{a} i_{a}^{'} + r_{a} i_{a} + K_{e} {\dot{θ}}_{m} \end{matrix}

\begin{matrix} G (s) = \frac{V (s)}{U (s)} = \frac{\sum_{i} G_{i} Δ_{i}}{Δ} \end{matrix}

\begin{aligned} Δ = 1 & - sum of all individual loop gains \\ + sum of gain products of all possible two loops which do not touch \\ - sum of gain products of all possible three loops which do not touch \\ + \dots \end{aligned}

path gain of the $i$ th forward path $G_{i}$
value of $Δ$ for the part of the signal-flow graph that does not touch the $i$ th forward path.
see example.

\begin{matrix} H (s) = \frac{w_{n}^{2}}{s^{2} + 2 ζ ω_{n} s + ω_{n}^{2}} \end{matrix}

\begin{matrix} s = - ζ ω_{n} \pm j ω_{n} \sqrt{1 - ζ^{2}} = - σ \pm j ω_{d} \end{matrix}

\begin{matrix} Y (s) = \frac{1}{s} H (s) = \frac{1}{s} - \frac{s + 2 ζ ω_{n}}{s^{2} + 2 ζ ω_{n} s + ω_{n}^{2}} \end{matrix}

after some trivial calculation,

\begin{matrix} y (t) = 1 - K e^{- σ t} \sin (ω_{d} t + θ) \end{matrix}

note that $K$ is not important.

\begin{aligned} y^{'} & = 0 \\ y^{'} & = K^{'} e^{- σ t} \sin (ω_{d} t) \end{aligned}

max overshoot happens at $ω_{d} t_{p} = π$

\begin{matrix} t_{p} = \frac{π}{ω_{d}} \end{matrix}

\begin{aligned} y (t_{p}) & = 1 + M_{p} \\ = 1 + e^{- σ π / ω_{d}} \end{aligned}

\begin{matrix} ⟹ M_{p} = e^{- σ π / ω_{d}} = e^{- σ t_{p}} \end{matrix}

{\begin{matrix} e^{- t_{1} / τ} = 0.9 \\ e^{- t_{2} / τ} = 0.1 \end{matrix}

\begin{matrix} t_{r} = t_{2} - t_{1} = τ (- \ln 0.1 - (- \ln 0.9)) = τ \ln 9 \approx 2 τ \end{matrix}

\begin{aligned} t_{r} & \approx \frac{1.8}{ω_{n}} \\ \approx \frac{0.8 + 1.1 ζ + 1.4 ζ^{2}}{ω_{n}} \end{aligned}

\begin{matrix} e^{- t_{s} / τ} = 0.01 \\ ⟹ - t_{s} / τ = \ln 0.01 \\ t_{s} = - \ln 0.01 τ \approx 4.6 τ \end{matrix}

for characteristic equation: $s^{2} + 2 ζ ω_{n} + ω_{n}^{2} = (s - (σ + j ω_{d})) (s - (σ - j ω_{d}))$

\begin{matrix} t_{s} = - \frac{\ln 0.01}{σ} \end{matrix}

\begin{matrix} H (s) = K \frac{s + α σ}{(s + (σ + j ω_{d})) (s + (σ - j ω_{d}))} \end{matrix}

when $α$ is small ( $α < 4$ ), the extra zero would increase the overshoot $M_{p}$ .

when $α \to \infty$ , zero would be trivial.

\begin{matrix} H (s) = K \frac{1}{(s + α σ) (s + (σ + j ω_{d})) (s + (σ - j ω_{d}))} \end{matrix}

when $α$ is small ( $α < 4$ ), the extra pole decrease the rising time $t_{r}$ .

when $α \to \infty$ , the extra pole is trivial.

\begin{matrix} y (\infty) = {[s Y (s)]}_{s \to 0} \end{matrix}

proof:

\begin{matrix} L {y^{'}}_{s \to 0} = {[s Y (s) - y (0)]}_{s \to 0} = \int_{0}^{\infty} e^{- s t} y^{'} d t = \int_{0}^{\infty} y^{'} d t = y (\infty) - y (0) \\ ⟹ {[s Y (s)]}_{s \to 0} = y (\infty) \end{matrix}

\begin{matrix} {[s F (s)]}_{s \to \infty} = f (0^{+}) \end{matrix}

proof is similar to final value theorem.

\begin{matrix} s^{n} + a_{1} s^{n - 1} + a_{2} s^{n - 2} + \dots + a_{n - 1} s + a_{n} = 0 \end{matrix}

The system would be stable iff the elements of the first column ( $a_{1}, b_{1}, c_{1}, \dots$ ) are all positive.

In addition, for a stable system, the coefficients of polynomial are all positive (not even $0$ ).

A simple explanation is that for a stable system which have its all roots at L.H.P, we can write

\begin{matrix} (s + n_{1}) (s + n_{2}) (s + n_{3}) \dots \cdot (s + n_{n}) = 0 \end{matrix}

example 3.32

\begin{matrix} a (s) = s^{6} + 4 s^{5} + 3 s^{4} + 2 s^{3} + s^{2} + 4 s + 4 \end{matrix}

There are two sign changes, $s^{2}$ to $s^{1}$ , and $s^{1}$ to $s^{0}$ .

Thus, There are two roots in RHP