AME-558

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Topic 1: Review of Classical Control Theory

• 1.1 Open and closed loop poles
• 1.2 Stability analysis
• 1.3 Steady state errors
• 1.4 Transient response
• 1.5 Frequency response

Steady state errors

Consider the typical control system shown below.

Error can be defined in different ways. In unity-feedback systems (H(s)=1), the most natural definition of error is the difference of the reference input and the controlled system output, $R\left(t\right)-C\left(t\right)$. In general, it is usual to consider the error signal to be the difference between the input R(t) and the feedback signal F(t), represented in the s-domain by the $E\left(s\right)$ in the block diagram above.

"Steady-state error" is best defined for stable linear systems subject to piecewise constant, step inputs, because all signals of such systems go to steady-state values asymptotically. The final value theorem can be used to easily find the steady-state value of any signal, including error.

Considering a signal $E\left(t\right)$ with associated Laplace transform $E\left(s\right)$, the final value theorem says that if a steady-state value is reached (which requires stability), that value is

$E_{\text{ss}}=\underset{t\to \infty }{lim}E\left(t\right)=\underset{s\to 0}{lim}sE\left(s\right)$

Example

For example, consider a simple proportional controller of a third-order system.

The open-loop transfer function of this system is

$\frac{C\left(s\right)}{E\left(s\right)}=\frac{36K}{s(s^2+4s+36)}$

The closed-loop transfer function of this system is

$\frac{C\left(s\right)}{R\left(s\right)}=\frac{36K}{s^3+4s^2+36s+36K}$

Consider a step input of the reference $R\left(t\right)=\overline{R}$. The corresponding Laplace transform is $R\left(s\right)=\frac{\overline{R}}{s}$.

The output is thus

$C\left(s\right)=\frac{36K}{s^4+4s^3+36s^2+36Ks}\overline{R}$

The steady-state output is then

$C_{\text{ss}}=\underset{s\to 0}{\lim }sC\left(s\right)=\overline{R}$

Finally, the steady-state error (since H=1) is

$E_{\text{ss}}=R_{\text{ss}}-C_{\text{ss}}=\overline{R}-\overline{R}=0$

The steady-state error is zero, independently of $K$. This is because of the $\frac{1}{s}$term in the open-loop transfer function. The steady-state error would have been nonzero had the open-loop transfer function been $\frac{C\left(s\right)}{E\left(s\right)}=\frac{36K}{s^2+4s+36}$, for example.

It is also possible to look at the steady-state bounds of error assuming more complex inputs. This is typically done using frequency response methods.

What's next?

Go on to page 4, where we'll look at transient response.