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Lecture Outline Regeltechniek Previous lecture: Root locus, - - PowerPoint PPT Presentation

May 03, 2023 264 likes •334 views

Lecture Outline Regeltechniek Previous lecture: Root locus, frequency response derivation. Lecture 7 Frequency Response, Bode Plots Robert Babu ska Today: Handout for the remaining computer sessions. Delft Center for Systems and

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SLIDE 1

Regeltechniek

Lecture 7 – Frequency Response, Bode Plots

Robert Babuˇ ska Delft Center for Systems and Control Faculty of Mechanical Engineering Delft University of Technology The Netherlands e-mail: r.babuska@dcsc.tudelft.nl www.dcsc.tudelft.nl/˜babuska tel: 015-27 85117

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 1

Lecture Outline

Previous lecture: Root locus, frequency response derivation. Today:

  • Handout for the remaining computer sessions.
  • Bode plots.
  • Non-minimum-phase systems.
  • System type in Bode plots.

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 2

Frequency Response

Periodic input: u(t) = M sin ωt Steady-state output: y(t) = |G(jω)| · M sin

  • ωt + ∠G(jω)
  • |G(jω)| . . . magnitude (gain)

∠G(jω) . . . phase

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Magnitude and Phase

Magnitude: |G(jω)| =

  • {Re[G(jω)]}2 + {Im[G(jω)]}2

Phase: ∠G(jω) = tan−1 Im[G(jω)] Re[G(jω)]

  • Both the magnitude and phase are generally functions of ω!

Fully describe G(s), can also can be measured experimentally.

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 4

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Magnitude and Phase: Example

G(s) G(s)

U(s) Y(s)

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Magnitude and Phase Plot

5 10 15 20 25 30 0.5 1 1.5 2

ω [rad/s] Magnitude

5 10 15 20 25 30 −100 −80 −60 −40 −20

ω [rad/s] Phase

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Bode Plot

Plotting on a linear scale is not so useful – plots are hard to inter- pret and cannot be easily drawn by hand. If logarithmic scales are introduced, drawing becomes easier. Such a logarithmic plot is called the Bode plot: – frequency is plotted on a logarithmic scale (log10) – amplitude is plotted using logarithmic units (decibels) – phase is plotted on a linear scale (degrees)

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 7

Logarithmic Scale: Decibels

The 10-base logarithm of a power gain is called a Bell (B): x B = log10 Pout Pin This unit appeared too large (x was usually small) the decibel (dB) was introduced: x dB = 10 log10 Pout Pin

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SLIDE 3

Logarithmic Scale: Decibels

Furthermore, power is proportional to the square of voltage: x dB = 10 log10 αV 2

  • ut

βV 2

in

= 20 log10 αVout βVin Therefore for a gain K the corresponding value in dB is: x = 20 log10(K) That is: x is the value of K expressed in dB.

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 9

Decomposing Transfer Functions

Decompose a transfer function into: G(s) = G1(s)G2(s) · · · Gn(s) Letting s = jω we have: G(jω) = |G(jω)| ej∠G(jω) with |G(jω)| = |G1(jω)| · · · |Gn(jω)| ∠G(jω) = ∠G1(jω) + · · · + ∠Gn(jω) In words: magnitudes multiply, phases add.

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Expressing Magnitude in Decibels

dB(|G(jω)|) = 20 log10 |G(jω)| which implies: dB(|G(jω)|) = dB(|G1(jω)|) + · · · + dB(|Gn(jω)|) In words: in dB, we can add magnitudes too.

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 11

Bode Form of Transfer Function

G(s) = K · sk ·

  • i(τis + 1) ·

i[( s ωn,i)2 + 2ζ( s ωn,i) + 1]

  • j(τjs + 1) ·

j[( s ωn,j)2 + 2ζ( s ωn,j) + 1]

Example: G(s) = 2000(s + 0.5) s(s + 10)(s + 50) = 2(s/0.5 + 1) s(s/10 + 1)(s/50 + 1) G(jω) = 2(jω/0.5 + 1) jω(jω/10 + 1)(jω/50 + 1)

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Bode Plots: Basic Transfer Functions

Any transfer function G(s) can be represented as a product of (some of) the following terms:

  • K
  • (s)±1
  • (τs + 1)±1
  • [( s

ωn)2 + 2ζ( s ωn) + 1]±1

We can draw the magnitudes and phases of these basic terms and add them up graphically.

Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 13

Bode Plots: G(jω) = K, K > 0

Frequency(rad/sec) Frequency(rad/sec)

20 log10 K ∠K

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Bode Plots: G(jω) = jω

20

  • 20

90

20 log10 |jω| ∠(jω) log(ω) log(ω) 1 1 10 10 0.1 0.1

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Bode Plots: G(jω) = 1

jω

20

  • 20
  • 90

20 log10 |1/(jω)| ∠1/(jω) log(ω) log(ω) 0.1 0.1 1 1 10 10

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Bode Plots: G(jω) = jωτ + 1

20 30 10 60 30 90

20 log10 |(jωτ + 1)| ∠(jωτ + 1) log(ω) log(ω) 0.1/τ 0.1/τ 1/τ 1/τ 10/τ 10/τ

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Bode Plots: G(jω) =

1 jωτ+1

  • 20
  • 10
  • 30
  • 60
  • 90

20 log10 |1/(jωτ + 1)| ∠1/(jωτ + 1) log(ω) log(ω) 0.1/τ 0.1/τ 1/τ 1/τ 10/τ 10/τ

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Bode Plots: G(jω) = [( s

ωn)2 + 2ζ( s ωn) + 1]−1

  • 40
  • 20

20

  • 180
  • 90

20 log10 |G(jω)| ∠G(jω) log(ω) log(ω) 0.1ωn 0.1ωn ωn ωn 10ωn 10ωn ζ = 0.1..1 steps of 0.1 ζ = 0.1..1 steps of 0.1

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Non-Minimum-Phase Systems

A system with zeros zi such that Re{zi} > 0 is called non-minimum phase system. Example: G1(s) = s + 1 s + 10 G2(s) = −s + 1 s + 10 System G2(s) undergoes a larger net change in phase than G1(s), i.e., G2(s) is called non-minimum phase.

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SLIDE 6

MP vs. NMP System: Bode Plots

10

−1

10 10

1

10

2

10

3

−20 −15 −10 −5

Frequency [rad/s] Magnitude [dB]

10

−1

10 10

1

10

2

10

3

−200 −150 −100 −50 50 100

Frequency [rad/s] Phase [deg]

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Type of System

From the bode diagram of the open loop system G(s) it is possi- ble to see of what type the loop transfer L(s) is, if proportional controller is used.

  • If the slope of the magnitude on the extreme left of the Bode

plot is 0 ⇒ no pure integrator ⇒ Type 0.

  • If the slope of the magnitude on the extreme left of the Bode

plot is −20n dB/decade ⇒ n integrators ⇒ Type n.

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Type 0 System: Example

10

−1

10 10

1

−40 −20 20

Frequency [rad/s] Magnitude [dB]

10

−1

10 10

1

10

−2

10

−1

10 10

1

Magnitude

10

−1

10 10

1

−200 −150 −100 −50

Frequency [rad/s] Phase [deg]

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Type 1 System: Example

10

−1

10 10

1

−60 −40 −20 20 40

Frequency [rad/s] Magnitude [dB]

10

−1

10 10

1

10

−3

10

−2

10

−1

10 10

1

10

2

Magnitude

10

−1

10 10

1

−300 −250 −200 −150 −100 −50

Frequency [rad/s] Phase [deg]

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