<h4>Example</h4>

Consider an example model of a DC motor. The system is based on simple harmonic motion; there is inertia $M$, and damping $D$. The system function is given by 

\begin{align*}
\hat{G}(s) &= \frac{1}{Ms^2 + Ds} = \frac{1/M}{s(s + D/M)} \\
\end{align*}

For this example, consider $M = 1$, and $D=16$. 


(a) is this system BIBO?


The $i\omega$-axis is not in the region of convergence. The right-most pole of the system happens to be on the $i\omega$-axis. 


(b) what is the LCCDE of the system?


\begin{align*}
\hat{G}(s) &= \frac{ \hat{Y}(s) }{ \hat{Q}(s) } \\
\frac{ \hat{Y}(s) }{ \hat{Q}(s) } &= \frac{1}{s^2 + 16} \\
\lr{s^2 + 16} \hat{Y}(s)  &= \hat{Q}(s) \\
\end{align*}

Recall the differentiation property of the laplace transform:

\begin{nicebox}
\begin{align*}
\frac{d}{dt} x(t) \ltp s\hat{X}(s) \\
\end{align*}
$RoC \supseteq RoC(x)$
\end{nicebox}

We can use this to solve for the time domain expression:

\begin{align*}
\lr{s^2 + 16} \hat{Y}(s)  &= \hat{Q}(s) \\
s^2\hat{Y}(s) + 16\hat{Y}(s)  &= \hat{Q}(s) \\
\ddot{y}(t) + 16\dot{y}(t) &= q(t) \\
\end{align*}

(c) Stabilizing the system via the feedback loop in the image below, determine the overall system function $\hat{H}(s)$


% image of feedback
% \img{images/feedback/motor/motor.ps}

\begin{center}
\begin{pspicture}(0,-4)(12,4)

\rput(0.25,0){$x$}
\rput(2.3,0.35){$\epsilon$}
\rput(8,0){$y$}
\rput(5.75,0){$\hat{G}(s)$}
\rput(3.1,0){$K$}
\rput(1.5,-1.75){$H$}

% plus or minus for adder
\rput(2.1,-0.5){$-$}
\rput(1.45,0.35){$+$}

\psframe(0.75,-2)(7.25,1)

\pspolygon(2.8,0.3)(2.8,-0.3)(4,0)

\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(0.5,0)(1.5,0)
\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(1.75,-1.25)(1.75,-0.25)
%\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(5.5,-2)(4,-2)
\psline[linewidth=1.25 pt](6.75,-1.25)(1.75,-1.25)
\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(2,0)(2.8,0) % out to y
\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(4,0)(5,0) % out to y
\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(6.5,0)(7.85,0) % out to y
\psline[linewidth=1.25 pt](6.75,0)(6.75,-1.25)

\psframe(5, -.5)(6.5,0.5)

\pscircle(1.75,0){0.25}
\psline(1.5,0)(2,0)
\psline(1.75,0.25)(1.75,-0.25)

\end{pspicture}
\end{center}

$x$ is the desired position of the system, $y$ is the actual position of the system, and $\epsilon$ is the error signal, given by $x - y$. This system is called a proportional feedback controller.

\begin{align*}
\ltrans{\epsilon(t)} \hat{K}(s) \hat{G}(s) &= \hat{Y}(s) \\
\ltrans{ x(t) - y(t) } \hat{K}(s) \hat{G}(s) &= \hat{Y}(s) \\
\hat{X}(s) \hat{K}(s) \hat{G}(s) - \hat{Y}(s) \hat{K}(s) \hat{G}(s) &= \hat{Y}(s) \\
\hat{X}(s) \hat{K}(s) \hat{G}(s) &= \hat{Y}(s) ( 1 +  \hat{K}(s) \hat{G}(s) ) \\
\hat{H}(s) &= \frac{ \hat{K}(s) \hat{G}(s)  }{  1 +  \hat{K}(s) \hat{G}(s)  } \\ 
\end{align*}

In this case, we have $\hat{K}(s) = K \gt 0$. 

\begin{align*}
\hat{H}(s) &= \frac{ K \hat{G}(s)  }{  1 +  K \hat{G}(s)  } \\ 
\hat{H}(s) &= \frac{ K \frac{1}{s^2 + 16} }{ ( 1 +  K \frac{1}{s^2 + 16} ) } \\ 
\hat{H}(s) &= \frac{K}{s^2 + 16s + K} \\
\end{align*}

Another way to solve for this is using Blacks Formula, $\frac{\mbox{Forward Gain}}{1 - \mbox{Loop Gain}}$:

\begin{align*}
\frac{\mbox{Forward Gain}}{1 - \mbox{Loop Gain}} \\
\frac{K \hat{G}(s) }{1 + K\hat{G}(s) } \\
\end{align*}



(c) Is $H$ guaranteed to be BIBO for $K \gt 0$?



We need to look at the poles of the system. We can determine the poles by finding the roots of $s^2 + 16s + K$, using the quadratic formula:

\begin{align*}
\frac{-16 \pm \sqrt{ 16^2 - 4K} } { 2} \\
-8 \pm \sqrt{ 64 - K }
\end{align*}

We have two cases that we need to analyze:


I $64 - K \geq 0$



Here we have $64 - K \in \R$ and $\sqrt{ 64 - K} \lt 8$. This means that $-8 \pm \sqrt{ 64 - K} $ is always less than 0, or to the left of the $i\omega$-axis.


II  $64 - K \lt 0$.



In this case we have $\sqrt{ 64 - K} \in \C$, and we have $-8 \pm i\sqrt{K - 64}$. This is to the left of the $i\omega$-axis. 





In all cases, we have a causal, stable signal whose poles are always to the left of the $i\omega$-axis.



(e) if $x(t) = u(t)$, what is $\hat{Y}(s)$?



\begin{align*}
\hat{Y}{s} &= \hat{H}(s)\hat{X}(s) \\
\hat{Y}(s) &= \frac{ K } { (s^2 + 16s + K)s } \\
\end{align*}




(f) Let $K = 63$, what is the coefficient of the steady state response?




\begin{align*}
\hat{Y}(s) &= \frac{63}{s(s+7)(s+9)} \\
\hat{Y}(s) &= \frac{A}{s} + \frac{B}{(s+7)} + \frac{C}{(s+9)} \\
A &= \left. s \hat{Y}(s) \right|_{s = 0} = \hat{H}(0) = 1 \\
\end{align*}



Example DC Motor