Watch today for email on HW1 solution and HW2. See this you tube on steady state error ??
Today we will spend more time on final value theorem. Then talk about typical specs for control system.
F.V.T. is important for tracking. Given \(F\left ( s\right ) \), does \(f\left ( t\right ) \) have a final value? ie. does \(\lim _{t\rightarrow \infty }f\left ( t\right ) \) exist? Sometimes infinity is allowed as final value. But many times we do not have a final value, such as for period signals such as \(\cos \left ( t\right ) \).
When can we apply F.V.T. ? In practice, \(F\left ( s\right ) \) should be stable. This means the poles should be in the left half plane. But we allow one pole to be at the origin. Example \(\frac{1}{s+1}\Longleftrightarrow e^{-t}\). How about \(\frac{1}{s+1}\frac{1}{s}\)? Final value still exist.
Reader Consider \(F\left ( s\right ) =\frac{1}{s^{2}}\frac{1}{s+1}\)
When \(F\left ( s\right ) \) is stable, then \(\lim _{t\rightarrow \infty }f\left ( t\right ) =\lim _{s\rightarrow 0}sF\left ( s\right ) .\)
Specification of control system: Consider classical response
Some specifications are settling time, and rise time (time to go from 10% to 90% of the step input) and amount of overshoot. Other specifications are given in terms of damping and sensitivity and steady state error. Steady state error is important and we will talk more about it today.
How to measure quality of tracking? Study \(E\left ( s\right ) \). What steady state error we accept depends on the application. The error transfer function is \(\frac{E\left ( s\right ) }{R\left ( s\right ) }=\frac{1}{1+H\left ( s\right ) G\left ( s\right ) }\). Now consider a step input, hence \(R\left ( s\right ) =\frac{1}{s}\) and the error becomes\[ E\left ( s\right ) =\frac{1}{s}\frac{1}{1+H\left ( s\right ) G\left ( s\right ) }\] Assuming F.V.T. applies (i.e. \(E\left ( s\right ) \) is stable) then\begin{align*} \lim _{t\rightarrow \infty }e\left ( t\right ) & =\lim _{s\rightarrow 0}sE\left ( s\right ) \\ & =\lim _{s\rightarrow 0}\frac{1}{1+H\left ( s\right ) G\left ( s\right ) }\\ & =\frac{1}{1+H\left ( 0\right ) G\left ( 0\right ) } \end{align*}
Example, if \(G\left ( s\right ) =\frac{1}{s^{2}+2s+4}\) then \(\lim _{t\rightarrow \infty }e\left ( t\right ) =\frac{4}{5}\) which is not good. We want this to be zero. So for \(\frac{1}{1+H\left ( 0\right ) G\left ( 0\right ) }\) to be zero, we want \(H\left ( 0\right ) G\left ( 0\right ) \) to be very large. This means \(H\left ( s\right ) \) should have \(\frac{1}{s}\) in it as a factor. This means an integrator. Hence an integrator in \(H\left ( s\right ) \) guarantees that error goes to zero when the input is step.
Reader For the mass spring damper, \(G=\frac{1}{ms^{2}+cs+k}\) design \(H\left ( s\right ) \) leading to zero steady state error for step command. Use \(\frac{k}{s}\) in \(H\left ( s\right ) \).
What if the input is ramp? which is \(\frac{1}{s^{2}}\)? Then \begin{align*} E\left ( s\right ) & =\frac{1}{s^{2}}\frac{1}{1+H\left ( s\right ) G\left ( s\right ) }\\ \lim _{t\rightarrow \infty }e\left ( t\right ) & =\lim _{s\rightarrow 0}sE\left ( s\right ) \\ & =\lim _{s\rightarrow 0}\frac{1}{s}\frac{1}{1+H\left ( s\right ) G\left ( s\right ) }\\ & =\lim _{s\rightarrow 0}\frac{1}{sH\left ( s\right ) G\left ( s\right ) } \end{align*}
So now we need \(H\left ( s\right ) \) to have \(\frac{1}{s^{2}}\) factor in it, so that it becomes very large at \(s=0\) and cause the error to go to zero. This means 2 integrator in series. This means to track \(r\left ( t\right ) =t^{k}\) we need \(k+1\) integrators in \(H\left ( s\right ) \) to get zero error at steady state.
Finally for any signal \(r\left ( t\right ) =\sum _{i=0}^{k}a_{i}t^{i}\), we need \(k+1\) integrators, since the largest term is the only term that needs to be satisfied, due to linearity of the system. So given a complicated polynomial, we look at the largest power and this tells us how many integrators we need for zero steady state error.