Lecture 9 Notes: Mapping the s-plane to the z-plane

Lecture 9: Mapping the s-plane to the z-plane

Overview

• This lecture focuses on mapping the s-plane to the z-plane.
• The aim is to establish how domains in the s-plane change under the z-operator.

Objectives

1. Revise the influence of s-plane poles on input responses.
2. Map the s-plane into the z-plane.
3. Review several examples.

Characteristic Roots and Stability

• The general form of a second-order system is given by:
  ÿ + 2ζω0ẏ + ω0^2y = u
  
• The characteristic roots (λ) are determined by:
  λ{1,2} = −ω0(ζ ± \sqrt{ζ^2 − 1})
  

Stability Regions

• Relationship between s-plane and z-plane:
  
  s = σ + jω \implies z = e^{sT} = e^{σT} ⋅ e^{jωT}
  

Mapping the s-plane to the z-plane

• The mapping $e^{sT}$ is periodic with a period of $jω_s = j\frac{2π}{T}$ because:
  
  e^{(s+ \frac{2πj}{T})T} = e^{sT}e^{2πj} = e^{sT}
  

Example 1: Pole at z = 1 - Critically Stable, No Oscillations

• Transfer function:
  
  \frac{Y(z)}{U(z)} = \frac{z}{z-1} = \frac{1}{1-z^{-1}}
  
• Difference equation:
  
  yn = y{n-1} + u_n
  
• Impulse response (where $u_n = 1$ for $n = 0$ and $0$ for $n ≥ 1$):
  • $y<em>0 = y</em>{-1} + u_0 = 0 + 1 = 1$
  • $y<em>1 = y</em>0 + u_1 = 1 + 0 = 1$
  • $y<em>2 = y</em>1 + u_2 = 1 + 0 = 1$
  • $y<em>3 = y</em>2 + u_3 = 1 + 0 = 1$
  • …

Example 2: Pole at z = 0.6 - Stable, No Oscillations

• Transfer function:
  
  \frac{Y(z)}{U(z)} = \frac{z}{z-0.6} = \frac{1}{1-0.6z^{-1}}
  
• Difference equation:
  
  yn = 0.6y{n-1} + u_n
  
• Impulse response (where $u_n = 1$ for $n = 0$ and $0$ for $n ≥ 1$):
  • $y<em>0 = 0.6y</em>{-1} + u_0 = 0 + 1 = 1$
  • $y<em>1 = 0.6y</em>0 + u_1 = 0.6 × 1 + 0 = 0.6$
  • $y<em>2 = 0.6y</em>1 + u_2 = 0.6 × 0.6 + 0 = 0.36$
  • $y<em>3 = 0.6y</em>2 + u_3 = 0.6 × 0.36 + 0 = 0.216$
  • …

Example 3: Pole at z = -1 - Critically Stable, With Oscillations

• Transfer function:
  
  \frac{Y(z)}{U(z)} = \frac{z}{z+1} = \frac{1}{1+z^{-1}}
  
• Difference equation:
  
  yn = -y{n-1} + u_n
  
• Impulse response (where $u_n = 1$ for $n = 0$ and $0$ for $n ≥ 1$):
  • $y<em>0 = -y</em>{-1} + u_0 = 0 + 1 = +1$
  • $y<em>1 = -y</em>0 + u_1 = -1 + 0 = -1$
  • $y<em>2 = -y</em>1 + u_2 = -(-1) + 0 = +1$
  • $y<em>3 = -y</em>2 + u_3 = -1 + 0 = -1$
  • …

Example 4: Pole at z = -0.6 - Stable, With Oscillations

• Transfer function:
  
  \frac{Y(z)}{U(z)} = \frac{z}{z+0.6} = \frac{1}{1+0.6z^{-1}}
  
• Difference equation:
  
  yn = -0.6y{n-1} + u_n
  
• Impulse response (where $u(nT) = 1$ for $n = 0$ and $0$ for $n ≥ 1$):
  • $y<em>0 = -0.6y</em>{-1} + u_0 = 0 + 1 = 1$
  • $y<em>1 = -0.6y</em>0 + u_1 = -0.6 × 1 + 0 = -0.6$
  • $y<em>2 = -0.6y</em>1 + u_2 = 0.6 × 0.6 + 0 = +0.36$
  • $y<em>3 = -0.6y</em>2 + u_3 = -0.6 × 0.36 + 0 = -0.216$
  • …

Summary: Mapping the s-plane to the z-plane

• The influence of s-plane poles on input responses.
• Mapped the s-plane into the z-plane.
• Reviewed several examples.

Reading

• Ogata’95, Sections 4.1 and 4.2.

Next Lecture

• Lecture 10: Digital control design in the w-domain.