2.4 Lecture 3, Thursday Sept. 10, 2015, Steady state and transient responses

Open loop vs. closed loop. In closed loop we add a controller to obtain desired response.

Another possibility is to put the controller in the feedback path

Now we will find the closed loop TF for the feed forward path configuration.

\begin{align*} Y\left ( s\right ) & =E\left ( s\right ) H\left ( s\right ) G\left ( s\right ) \\ E\left ( s\right ) & =R\left ( s\right ) -Y\left ( s\right ) \end{align*}

Substituting the second equation in the first gives\begin{align*} Y\left ( s\right ) & =\left ( R\left ( s\right ) -Y\left ( s\right ) \right ) H\left ( s\right ) G\left ( s\right ) \\ & =R\left ( s\right ) H\left ( s\right ) G\left ( s\right ) -Y\left ( s\right ) H\left ( s\right ) G\left ( s\right ) \end{align*}

Hence\begin{align*} Y\left ( s\right ) \left ( 1+H\left ( s\right ) G\left ( s\right ) \right ) & =R\left ( s\right ) H\left ( s\right ) G\left ( s\right ) \\ \frac{Y\left ( s\right ) }{R\left ( s\right ) } & =\frac{H\left ( s\right ) G\left ( s\right ) }{1+H\left ( s\right ) G\left ( s\right ) } \end{align*}

Reader: Do the above for the feedback path configuration. We should get \(\frac{Y\left ( s\right ) }{R\left ( s\right ) }=\frac{G\left ( s\right ) }{1+H\left ( s\right ) G\left ( s\right ) }\)

Reader: For the feedforward case, find the error transfer function \(\frac{E\left ( s\right ) }{R\left ( s\right ) }\)

Working with the feedforward case. \(E=R-Y\). We want \(E=0\) for tracking. Most common choices of \(H\left ( s\right ) \) are

1.

\(H=k\) a constant. Called proportional or pure gain.

2.

\(H=k_{1}+\frac{k_{2}}{s}\) The second term is an integrator. This is called PI controller.

3.

\(H=k_{1}+\frac{k_{2}}{s}+sk_{3}\). This is called PID. Note that \(s\) is derivative. So if there is lots of noise in the \(E\) signal, this will cause problems since derivative of noise generated large signal (large actuating signal). So it is safe to use an integrator, but not always save to use derivative, unless we know that the error signal will always be smooth.

For example, let \(G\left ( s\right ) =\frac{1}{s+2}\) with a unit step \(\left ( \frac{1}{s}\right ) \) input. Then in the openloop, \(Y\left ( s\right ) =\frac{1}{s+2}\frac{1}{s}\) or \(y\left ( t\right ) =\frac{1}{2}-\frac{1}{2}e^{-2t}\). Now we close the loop, using the feedforward configuration and add a pure gain controller \(k\). Hence\begin{align*} \frac{Y}{R} & =\frac{HG}{1+HG}\\ & =\frac{k\frac{1}{s+2}}{1+k\frac{1}{s+2}}\\ & =\frac{k}{s+\left ( 2+k\right ) } \end{align*}

Now \(R\left ( s\right ) =\frac{1}{s}\), hence\[ Y\left ( s\right ) =\frac{1}{s}\frac{k}{s+\left ( 2+k\right ) }\] And \[ y\left ( t\right ) =\overset{\text{steady state response}}{\overbrace{\frac{k}{k+2}}}-\overset{\text{transient response}}{\overbrace{\frac{k}{k+2}e^{-t\left ( k+2\right ) }}}\] By design, we want \(y\left ( t\right ) \) to track \(r\left ( t\right ) \) which is unit step in this case. Also we want the transient response to go away quickly. In this example, only when \(k\) very large do we approach a steady state close to one. But in practice having very large gain is not good due to sensitivity problems. (will talk about sensitivity later in the course). Also large \(k\) might lead to actuating signal that can not be satisfied. Note also, no matter how large \(k\) is, we can’t obtain perfect tracking. Some application might require perfect tracking.

Reader: Redo the analysis above using  integrator only controller. i.e. \(H\left ( s\right ) =\frac{k}{s}\) and see if \(y\left ( t\right ) \) will now track the input at steady state.

Final value theorem: Suppose \(F\left ( s\right ) =\frac{N\left ( s\right ) }{D\left ( s\right ) }\) is stable, then \[ \lim _{t\rightarrow \infty }f\left ( t\right ) =\lim _{s\rightarrow 0}sF\left ( s\right ) \]