Steady-State Analysis of Control Systems

 

Steady-State Analysis of Control Systems

A system’s steady-state error (SSE) measures the difference between the desired input and the actual output as $t\to \mathrm{\infty }$. It depends on the system type (number of pure integrators in the open-loop transfer function) and the error constants.

1. System Types and Steady-State Error

• Type 0 system (no integrator in open-loop transfer function):

  • Step input → finite error

  • Ramp input → infinite error

  • Parabolic input → infinite error

• Type 1 system (one integrator):

  • Step input → zero error

  • Ramp input → finite error

  • Parabolic input → infinite error

• Type 2 system (two integrators):

  • Step input → zero error

  • Ramp input → zero error

  • Parabolic input → finite error

2. Steady-State Error Constants

These constants quantify system accuracy for different inputs:

• Position Constant ($K_p$)

$$K_p=\underset{s\to 0}{\lim }G(s)$$

Steady-state error for step input:

$$e_{ss}=\frac{1}{1+K_p}$$

• Velocity Constant ($K_v$)

$$K_v=\underset{s\to 0}{\lim }sG(s)$$

Steady-state error for ramp input:

$$e_{ss}=\frac{1}{K_v}$$

• Acceleration Constant ($K_a$)

$$K_a=\underset{s\to 0}{\lim }{s}^{2}G(s)$$

Steady-state error for parabolic input:

$$e_{ss}=\frac{1}{K_a}$$

Simple Numerical Example

Given: Open-loop transfer function:

$$G(s)=\frac{10}{s(s+2)}$$

• System type: One integrator → Type 1 system.

• Position constant:

$$K_p=\underset{s\to 0}{\lim }G(s)=\mathrm{\infty }\Rightarrow e_{ss}=0\text{(step input)}$$

• Velocity constant:

$$K_v=\underset{s\to 0}{\lim }sG(s)=\underset{s\to 0}{\lim }\frac{10}{s+2}=5\mathrm{<br>}\mathrm{<br>}\mathrm{<br>}$$

$$e_{ss}=\frac{1}{K_v}=\frac{1}{5}=0.2\text{(ramp input)}$$

• • Acceleration constant:

  $$K_a=\underset{s\to 0}{\lim }{s}^{2}G(s)=0\Rightarrow e_{ss}=\mathrm{\infty }\text{(parabolic input)}$$

Error Constants Recap

• Position constant ($K_p$) → step input accuracy

• Velocity constant ($K_v$) → ramp input accuracy

• Acceleration constant ($K_a$) → parabolic input accuracy

Simple Numerical Illustration

Given:

$$G(s)=\frac{20}{s(s+5)}$$

• System type → Type 1 (one integrator).

• Step input:

$$K_p=\underset{s\to 0}{\lim }G(s)=\mathrm{\infty }\Rightarrow e_{ss}=0$$

• Ramp input:

$$K_v=\underset{s\to 0}{\lim }sG(s)=\frac{20}{5}=4\Rightarrow e_{ss}=\frac{1}{4}=0.25$$

• Parabolic input:

$$K_a=\underset{s\to 0}{\lim }{s}^{2}G(s)=0\Rightarrow e_{ss}=\mathrm{\infty }$$

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