Exam 1 on Tuesday Sept. 29, 2015. Closed book, closed notes. There will be office hours Monday 1-3 pm.
Keywords for exam:
Now for one more Mason problem. We use Mason any time we want to find a transfer function. Find \(\frac{Y\left ( s\right ) }{U\left ( s\right ) }\) for this signal graph
There are 4 forward paths from \(U\) to \(Y\), here they are, with the associated Mason \(\Delta \)
\begin{align*} M_{1} & =abcde,\Delta _{1}=1\\ M_{2} & =afhe,\Delta _{2}=1-i\\ M_{3} & =afcde,\Delta _{3}=1\\ M_{4} & =abhe,\Delta _{4}=1-i \end{align*}
We need \(\frac{Y}{U}=\frac{\sum _{i=1}^{4}M_{i}\Delta _{i}}{\Delta }\). The loops are \(\left \{ k,j,i,bcg,dl,fhl\ g,bhl\ g,fcg\right \} \), hence\[ \Delta =1-\left ( k+i+j+dk+fhl\ g+bhl\ g+fcg\right ) +\left ( ki+kj+ij+kdl+jbcg+jfcg\right ) -\left ( kij\right ) \] Hence\begin{align*} \frac{Y}{U} & =\frac{\sum _{i=1}^{4}M_{i}\Delta _{i}}{\Delta }\\ & =\frac{abcde\left ( 1\right ) +afhe\left ( 1-i\right ) +afcde\left ( 1\right ) +abhe\left ( 1-i\right ) }{1-\left ( k+i+j+dk+fhl\ g+bhl\ g+fcg\right ) +\left ( ki+kj+ij+kdl+jbcg+jfcg\right ) -\left ( kij\right ) } \end{align*}
Next topic we will start on is the benefits of feedback. So far we talked about tracking only. Other benefits are
We can use feedback to pre compensate a nonlinear system to make it approximately linear. Given a non-linear device, say diode, with input \(U=X\) which represent voltage and output \(Y\) which is nonlinear function of the input such as \(Y=NX\) can we use feedback to make the output closed to linear?
Warning: When the system is non-linear, we can not use transfer functions and can not use Laplace. These are only for linear systems. Transfer functions and Laplace transforms are used only when the system is linear. So how do we analyze non-linear system? We use time domain. For example, if \(Y=X^{2}\)
Closed loop is nonlinear. We need relation between \(R\left ( s\right ) \) and \(Y\left ( s\right ) \)
There are two type of nonlinearity, saturations and dead-zone. Dead zone is an area where the input is not yet sufficient to cause any output to be generated, it might be a threshold for the device to start operating. Here is a typical output from a non linear device
Reader For this open loop, if \(x=2\sin t\), sketch \(y\left ( t\right ) \). Another example
We know \begin{equation} Y=N\left ( x\right ) \tag{1} \end{equation} and \[ x=2\left ( R-Y\right ) \] hence \(2Y=2R-x\) and \begin{equation} Y=R-\frac{x}{2} \tag{2} \end{equation}
(1) and (2) must both hold. For each \(R\) input, we solve (1,2) for \(Y,x\) and plot them. For example, for \(R=\left \{ 0,0.1,0.2,\cdots \right \} \) for each \(R\left ( i\right ) \) we solve for \(Y\left ( i\right ) \)
| \(R\) | \(Y=R-\frac{x}{2}\) |
| \(0\) | \(0-\frac{x}{2}\) |
| \(0.1\) | \(0.1-\frac{x}{2}\) |
| \(0.2\) | \(0.2-\frac{x}{2}\) |
| \(\vdots \) | \(\vdots \) |
For each line in the above, such as \(-\frac{x}{2},0.1-\frac{x}{2},\cdots \), we now draw this line on top of the original \(Y\left ( x\right ) \) plot, and see where this line intersect with the original \(Y\left ( x\right ) \). The point of intersection is the new value of \(Y\). This is done for each entry of \(R\), so we obtain
So we obtain this table
| \(R\) | \(Y=R-\frac{x}{2}\) |
| \(0\) | \(0\) |
| \(0.1\) | \(0\) |
| \(0.2\) | \(0\) |
| \(\vdots \) | \(\vdots \) |
| big | \(.3\) |
| \(\vdots \) | \(1\) |
Reader: calculate and obtain closed loop. We might obtain this
We see that dead zone has shrunk (good) and linear region increased (good).
Reader: Redo with \(k=10\). For large \(K\) we should obtain
