Mid1
DC Motor
emf voltage: \(e\)
armature current: \(i_a\)
viscous friction coefficient: \(b\)
inertia: \(J_m\)
\[
\begin{gather}
\tau = K_t\ i_a
\\\\
e = K_e\ \dot\theta_m
\\\\
K_t = K_e \ (\text{for same unit})
\end{gather}
\]
Modeling
voltage across motor: \(v_a\)
\[
\begin{gather}
\tau = J_m\ddot\theta + b\dot \theta = K_t\ i_a
\\\\
v_a = L_a\ i_a' + r_a i_a + K_e\ \dot\theta_m
\end{gather}
\]
Mason's Rule
\[
\begin{gather}
G(s) = \frac{V(s)}{U(s)}= \frac{\sum_i{G_i\ \Delta_i}}{\Delta}
\end{gather}
\]
\[
\begin{align}
\Delta = 1 &- \text{sum of all individual loop gains}
\\\\
&+ \text{sum of gain products of all possible two loops which do not touch}
\\\\
&- \text{sum of gain products of all possible three loops which do not touch}
\\\\
&+ \dots
\end{align}
\]
Second System Response
Transfer function
\[
\begin{gather}
H(s) = \frac{w_{n}^{2}}{s^{2}+2\zeta \omega_n s + \omega_n^{2}}
\end{gather}
\]
Poles:
\(\sigma = \zeta \omega_n\) \(\omega_d = \omega_n\sqrt{1-\zeta^{2}}\)
\[
\begin{gather}
s = - \zeta \omega_n \pm j \omega_n\sqrt{1-\zeta^{2}} = -\sigma \pm j\omega_d
\end{gather}
\]
\[
\begin{gather}
Y(s) = \frac{1}{s}H(s) = \frac{1}{s} - \frac{s+2\zeta \omega_n}{s^{2}+2\zeta \omega_n s + \omega_n^{2}}
\end{gather}
\]
after some trivial calculation,
\[
\begin{gather}
y(t) = 1 - K\ e^{-\sigma t}\ \sin(\omega_dt + \theta)
\end{gather}
\]
note that \(K\) is not important.
Peak Time
\[
\begin{align}
y' &= 0
\\\\
y' &= K' e^{-\sigma t}\sin(\omega_d t)
\end{align}
\]
Max Overshoot
max overshoot happens at \(\omega_d \ t_p = \pi\)
\[
\begin{gather}
t_p = \frac{\pi}{\omega_d}
\end{gather}
\]
overshoot \(M_p\) (memorize directly)
\[
\begin{align}
y(t_p) &= 1 + M_p
\\\\
&= 1 + e^{-\sigma\pi/\omega_d}
\end{align}
\]
\[
\begin{gather}
\implies M_p = e^{-\sigma\pi/\omega_d} = e^{-\sigma\, t_p}
\end{gather}
\]
rise time
\(t_r \equiv t_2 - t1\) for characteristic equation: \(s+1/\tau\)
\[
\left\{
\begin{gather}
e^{-t_1/\tau}=0.9
\\\\
e^{-t_2/\tau}=0.1
\end{gather}\right.
\]
\[
\begin{gather}
t_r=t_2-t_1=\tau(-\ln0.1-(-\ln0.9))=\tau\ln9\approx2\tau
\end{gather}
\]
for characteristic equation: \(s^{2}+2\zeta \omega_n+\omega_n^{2}\)
\[
\begin{align}
t_r &\approx \frac{1.8}{\omega_n}
\\\\
&\approx \frac{0.8+1.1\zeta+1.4 \zeta^{2}}{\omega_n}
\end{align}
\]
settling time
\[
\begin{gather}
e^{-t_s/\tau} = 0.01
\\\\
\implies -t_s/\tau = \ln0.01
\\\\
t_s = -\ln 0.01 \tau \approx 4.6 \tau
\end{gather}
\]
for characteristic equation: \(s^{2}+2\zeta \omega_n+\omega_n^{2}=(s-(\sigma+j\omega_d))(s-(\sigma-j\omega_d))\)
\[
\begin{gather}
t_s = -\frac{\ln 0.01}{\sigma}
\end{gather}
\]
\[
\begin{gather}
H(s) = K\frac{s+\alpha\sigma}{(s+(\sigma+j\omega_d))(s+(\sigma-j\omega_d))}
\end{gather}
\]
when \(\alpha\) is small (\(\alpha <4\) ), the extra zero would increase the overshoot \(M_p\) .
when \(\alpha \to \infty\) , zero would be trivial.
\[
\begin{gather}
H(s) = K\frac{1}{(s+\alpha \sigma)(s+(\sigma+j\omega_d))(s+(\sigma-j\omega_d))}
\end{gather}
\]
when \(\alpha\) is small (\(\alpha < 4\) ), the extra pole decrease the rising time \(t_r\) .
when \(\alpha \to \infty\) , the extra pole is trivial.
Final Value Theorem
poles have to be on left-half plane (converge)
\[
\begin{gather}
y(\infty) = \left[sY(s)\right]_{s\to 0}
\end{gather}
\]
proof:
\[
\begin{gather}
L\left\{y'\right\}_{s\to0} =\left[sY(s)-y(0)\right]_{s\to 0}= \int_0^{\infty}{e^{-st}y' dt} = \int_0^{\infty}{y'dt} = y(\infty) - y(0)
\\\\
\implies \left[sY(s)\right]_{s\to0} = y(\infty)
\end{gather}
\]
\[
\begin{gather}
\left[sF(s)\right]_{s\to \infty} = f(0^+)
\\\\
\end{gather}
\]
proof is similar to final value theorem.
Routh-Hurwitz stability criterion
\[
\begin{gather}
s^{n} + a_1 s^{n-1} + a_2 s^{n-2} +\dots + a_{n-1}s +a_n=0
\end{gather}
\]
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
\[
\begin{gather}
(s+n_1)(s+n_2)(s+n_3)\dots \cdot (s+n_n)=0
\end{gather}
\]
# of roots in the RHP == # of sign changes in the first column.
example 3.32
\[
\begin{gather}
a(s) = s^{6} + 4s^{5}+3s^{4}+2s^{3}+s^{2}+4s+4
\end{gather}
\]
\(s^{6}\) 1
\(s^{5}\) 4
\(s^{4}\) 5/2
\(s^{3}\) 2
\(s^{2}\) 3
\(s^{1}\) -76/15
\(s^{0}\) 4
There are two sign changes, \(s^{2}\) to \(s^{1}\) , and \(s^{1}\) to \(s^{0}\) .
Thus, There are two roots in RHP
