Z-Domain Root Locus

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Z-Domain Root Locus

• M. Sami Fadali
• Professor of Electrical Engineering
• UNR

2
Outline

• Root locus plots in z-plane.
• Pole locations and time response.
• Z-plane contours.
• Proportional control.

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Closed-loop Characteristic Equation

• C(z) controller transfer function
• GZAS(z) transfer function of DAC, analog
  subsystem, and sampler
• L(z) loop gain
• K gain

4
Observations

• Identical equation to s-domain equation with s
  replaced by z .
• All the rules derived for s-domain are applicable
  and can be used to obtain z-domain root locus
  plots.
• The plots can also be obtained using the root
  locus plots of most CAD programs (MATLAB rlocus)

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Example 6.1

• Obtain the root locus plot and the critical gain
  for the first order type 1 system with loop gain

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Solution

• Root locus rules give plot (MATLAB rlocus).
• Root locus real axis locus between pole and
  zero.
• For a stable discrete system, real axis loci must
  lie between (1,0) and (?1,0) in the z-plane.
• Critical gain Kcr is at point (?1,0).
• Closed-loop characteristic equation
• z ? 1 K 0
• Substituting z ?1 gives Kcr 2

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Root Locus for Example 6.1
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Example 6.2

• Obtain the root locus plot and the critical gain
  for the second order type 1 system with loop gain

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Solution

• Use root locus rules.
• RL like Example 5.1(i) but in RHP.
• Breakaway point zb (1.5)/2 0.75
• Kcr (critical) intersection of RL unit circle.
• Closed loop characteristic equation

On the unit circle, z 1 (complex conjugate)
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Root Locus of Example 6.2
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z-Domain Pole Locations Associated Temporal
Sequences
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Time Functions Real Poles
Continuous Laplace Transform Sampled z-Transform

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Time Functions Complex Conjugate Poles
Continuous
Laplace Transform
Sampled
z-Transform
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Observation
If the Laplace transform F(s) of a
continuous-time function f(t) has a pole ps ,
then the z-transform F(z) of its sampled
counterpart f(kT), with sampling period T has a
pole at
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Primary Strip
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Proportional Controlof Digital Systems

• z-domain characteristic polynomial for a 2nd
  order underdamped system

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Pole Contours in the s-Domain and the z-Domain
Contour s-Domain Poles Contour z-Domain Poles
? constant vertical line z e ?T constant circle
?d constant horizontal line ? z constant radial line
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Constant ? Contours
Constant ? contours in s-plane.
Constant ? contours in z-plane.
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Constant ?d Contours

Constant ?d contours in the s-plane. Constant ?d
contours in the z-plane.

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Constant ? Contours
?

• Logarithmic spirals that get smaller for larger
  values of ?.
• The spirals are defined by the equation

z magnitude of pole ? angle of pole
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Constant ?n Contours

• z magnitude of pole
• ? angle of pole

To obtain the expression, eliminate ?
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Z-Domain Grid
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Characteristics of Log Spirals

1. Two spiral for each ? value, corresponding to ??.
   The ?? spiral is below the real axis mirror
   image of ? spiral.
2. For any spiral, z drops logarithmically with
   ? increase.
3. At the same angle , increasing ? gives smaller
   z , i.e. spirals are smaller for larger ?
   values.
4. All spirals start at ? 0, z 1 but end at
   different points.
5. When given ? and z , obtain ? by substituting
   in the equation

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MATLAB

• gtgt gtf(num, den, T) sampling period T
• gtgt rlocus(g) Root locus plot
• gtgt zgrid(zeta, wn) Plot contours
• zeta vector of damping ratios
• wn vector of undamped natural
• frequencies

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z-Domain Design Specifications

• Similar to those for s-domain design.
• Often approximate values based on continuous time
  definitions.
• Allow selection of pole locations for z-domain
  design.

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Time Constant

• Time constant of exponential decay for the
  continuous envelope of sampled waveform.
• Not necessarily equal to a specified percentage
  of the final value after one time constant

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Settling Time

• Period after which the envelope of the sampled
  waveform stays within a specified percentage
  (usually 2) of the final value.
• Multiple of time constant depending on the
  specified percentage.
• Settling time for a 2 specification

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Other Specification

• Frequency of Oscillations ?d angle of the
  dominant complex conjugate poles divided by the
  sampling period.
• Other design criteria such as the percentage
  overshoot,?, ?n, defined analogously to the
  continuous case.
• As in analog design, select a dominant
  closed-loop pair in the complex plane to obtain a
  satisfactory time response.
• Analytical design possible for low order systems
  but more difficult than analog design.

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Example 6.3

• Design a proportional controller for the digital
  system with sampling period T0.1s to obtain
• ?d 5 rad/s
• A time constant of 0.5 s
• ? 0.7

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Solution

• Obtain the results with calculator or MATLAB.
• MATLAB rlocus
• (a) angle of the pole ?d T 5?0.1 0.5rad
  28.65?
• (b) 1/( time constant) ? ?n 1/ 0.5 2 rad /s
• pole magnitude exp(???n ) 0.82
• (c) Use ? directly to get the results of Table
  6.1.
• Sampled step response MATLAB command step
• Higher gain designs have a low ? (oscillatory
  response).

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P-Control Design Results
Design Gain ? ?n rad/s
a 0.23 0.3 5.24
b 0.17 0.4 4.60
c 0.10 0.7 3.63
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Time Response Plots
Designs of Table 6.1 (a) ? , (b) ?, (c) .
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Analytical Design
Closed-loop characteristic equation

Equating coefficients

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Design (a) ?d 5
z1 equation
z0 equation
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Design (b) ? 0.5 s
Solve for ?
? 0.436 ?n 4.586 rad/s
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Design (c) ? 0.7
z0 equation
Solve numerically by trial and error with a
calculator
z0 equation
Graphically draw root locus and a segment of
the constant ? spiral and find their
intersection. (Rough results and the solution is
difficult without MATLAB for all but a few simple
root loci).
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Example 6.4

• Design a proportional controller for the unity
  feedback digital control system with a sampling
  period T0.1 s to obtain
• a) e(?) due to ramp 10
• b) ? 0.7

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Solution

• Closed-loop characteristic equation

• Equation involves three parameters ? , ?n K
• Equating coefficients yields two equations
• Evaluate two unknowns and obtain the third from
  the design specification.

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(a) Design for e(?) 10

• Velocity error constant (Type 1)
• Same as the velocity error constant for the
  analog proportional control system 10
  steady-state error due to ramp.

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Root locus for K50
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Time response for K 50
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b) Design for ? 0.7

• Difficult analytical solution for constant ?
• MATLAB move cursor to ? 0.7 ? K 10.
• Kcr109, stable system at K 50 and K 10.
• Design specs. are met for both (a) and (b).
• Must check other design criteria.
• For (a), K50, ? 0.18 (highly oscillatory)
• For (b), K 10, e(?) 50 due to unit ramp.
• P-control cannot provide good steady-state error
  together with good transient response.

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Root Locus for Constant ? Design