Reminder, class tomorrow at 6 pm. No class Oct. 1, 2015. Test on sept 29. Exam everything up to and including MIMO.
Consolidating example We will cover main points in class so far, from modeling, to finding T.F. to building block diagrams and MIMO. The example is to model Armature controlled DC motor. From physical system to differential equations.
Applying kirchoff voltage rule on the circuit gives
\begin{equation} V_{a}\left ( t\right ) =R_{a}i_{a}+L_{a}\frac{di_{a}}{dt}+k_{1}\omega \tag{1} \end{equation}
Where \(k_{1}\omega \) is the backemf voltage induced by the rotating arm and \(\omega \) is the angular velocity \(\frac{d\theta }{dt}\) of the motor arm. In addition, we have a mechanical relation between the applied torque \(t_{L}\) and \(i_{a}\)
\begin{equation} J\frac{d\omega }{dt}=k_{2}i_{a}-t_{L} \tag{2} \end{equation}
Finally
\begin{equation} t_{L}=k_{L}i_{a} \tag{3} \end{equation}
The above are the three equations needed. Let \(k_{1}=k_{2}=k_{m}\). Taking Laplace transform of each gives
\begin{align} V_{a}\left ( s\right ) & =R_{a}I_{a}\left ( s\right ) +sL_{a}I_{a}\left ( s\right ) +sk_{m}\theta \left ( s\right ) \tag{1A}\\ sJW\left ( s\right ) & =k_{2}I_{a}\left ( s\right ) -T_{L}\left ( s\right ) \tag{2A}\\ T_{L}\left ( s\right ) & =k_{L}I_{a}\left ( s\right ) \tag{3A} \end{align}
Notice that \(W\left ( s\right ) \) is the Laplace transform of \(\omega \) and that \(\omega =\frac{d\theta }{dt}\) or \(W\left ( s\right ) =s\theta \left ( s\right ) \), hence \(\frac{W\left ( s\right ) }{s}=\theta \left ( s\right ) \). Now we build the block diagram from the above three equations. The input is \(V_{a}\left ( s\right ) \) and the output is \(\theta \left ( s\right ) \). From the above we find\begin{align*} I_{a}\left ( s\right ) & =\frac{V_{a}\left ( s\right ) -k_{m}W\left ( s\right ) }{R_{a}+sL}\\ W\left ( s\right ) & =\frac{k_{m}I_{a}\left ( s\right ) -T_{L}\left ( s\right ) }{Js} \end{align*}
And the block diagram is
Reader: Find the transfer function \(\frac{\theta \left ( s\right ) }{V_{a}\left ( s\right ) }\). Consider step input. Find steady state \(\theta \left ( \infty \right ) \) for step input. Does it go to one? Note, every RLC circuit is stable circuit. Called passive circuit.
Reader answer
I get this \begin{align*} E & =R\left ( s\right ) -s\theta k_{m}\\ \theta & =E\left ( s\right ) \left ( \frac{k_{m}-k_{L}}{\left ( R_{a}+sL\right ) s^{2}J}\right ) \end{align*}
Hence
\begin{align*} \theta & =\left ( R\left ( s\right ) -s\theta k_{m}\right ) \left ( \frac{k_{m}-k_{L}}{\left ( R_{a}+sL\right ) s^{2}J}\right ) \\ \theta & =R\left ( s\right ) \left ( \frac{k_{m}-k_{L}}{\left ( R_{a}+sL\right ) s^{2}J}\right ) -s\theta k_{m}\left ( \frac{k_{m}-k_{L}}{\left ( R_{a}+sL\right ) s^{2}J}\right ) \\ \theta \left ( 1+sk_{m}\left ( \frac{k_{m}-k_{L}}{\left ( R_{a}+sL\right ) s^{2}J}\right ) \right ) & =R\left ( s\right ) \left ( \frac{k_{m}-k_{L}}{\left ( R_{a}+sL\right ) s^{2}J}\right ) \\ \frac{\theta \left ( s\right ) }{R\left ( s\right ) } & =\frac{\left ( \frac{k_{m}-k_{L}}{\left ( R_{a}+sL\right ) s^{2}J}\right ) }{1+sk_{m}\left ( \frac{k_{m}-k_{L}}{\left ( R_{a}+sL\right ) s^{2}J}\right ) }\\ & =\frac{k_{m}-k_{L}}{s^{2}J\left ( R_{a}+sL\right ) +sk_{m}\left ( k_{m}-k_{L}\right ) }\\ & =\frac{k_{m}-k_{L}}{s^{3}JL+s^{2}JR_{a}+sk_{m}^{2}-sk_{m}k_{L}}\\ & =\frac{\left ( k_{m}-k_{L}\right ) }{JL}\frac{1}{s^{3}+s^{2}\frac{R_{a}}{L^{2}}+s\left ( \frac{k_{m}^{2}-k_{m}k_{L}}{JL}\right ) } \end{align*}
What if we have another output of interest? say \(I_{a}\left ( s\right ) \) as output? And what if we have another input, a disturbance \(d\left ( t\right ) \) as shown
Reader Find the MIMO transfer function\[\begin{pmatrix} \theta \left ( s\right ) \\ I_{a}\left ( s\right ) \end{pmatrix} =\begin{pmatrix} T_{11} & T_{12}\\ T_{21} & T_{22}\end{pmatrix}\begin{pmatrix} V_{a}\left ( s\right ) \\ D\left ( s\right ) \end{pmatrix} \] Where \(T_{11}\) is the transfer function from \(V_{a}\left ( s\right ) \) to \(\theta \left ( s\right ) \), and \(T_{12}\) is the transfer function from \(\theta \left ( s\right ) \) to \(D\left ( s\right ) \) and \(T_{21}\) is the transfer function from \(V_{a}\left ( s\right ) \) to \(I_{a}\left ( s\right ) \) and \(T_{22}\) is the transfer function from \(D\left ( s\right ) \) to \(I_{a}\left ( s\right ) \).\begin{align*} \theta \left ( s\right ) & =T_{11}V_{a}\left ( s\right ) +T_{12}D\left ( s\right ) \\ I_{a}\left ( s\right ) & =T_{21}V_{a}\left ( s\right ) +T_{22}D\left ( s\right ) \end{align*}
Hence\begin{align*} T_{11} & =\frac{\theta \left ( s\right ) }{V_{a}\left ( s\right ) }\\ T_{12} & =\frac{\theta \left ( s\right ) }{D\left ( s\right ) }\\ T_{21} & =\frac{I_{a}\left ( s\right ) }{V_{a}\left ( s\right ) }\\ T_{22} & =\frac{I_{a}\left ( s\right ) }{D\left ( s\right ) } \end{align*}
We now start on signal graph. First we convert block diagram to signal graph. The block become a branch, and the variable become a node.
Then we will start on Mason rule, which uses the signal graph to obtain the transfer function.