Lecture 8B Frequency Response

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# Lecture 8B Frequency Response - PowerPoint PPT Presentation

Lecture 8B Frequency Response. Frequency Response. The most interesting type of system response Uses a sine wave forcing function input Interest is in the steady-state response instead of the transient response

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### Lecture 8BFrequency Response

Frequency Response
• The most interesting type of system response
• Uses a sine wave forcing function input
• Interest is in the steady-state response instead of the transient response
• Steady-state response to a step change only provides information about process gain
• Transient response to sine wave change involves a fast change (gone within 4 Tp values) to consistent response oscillating at the input frequency
Attributes of Frequency Response
• Two major characteristics of frequency response:
• Output amplitude with respect to the input
• Output phase lag or lead with respect to the input
• Amplitude is attenuated or amplified depending on input frequency
• Output oscillating out of phase withinput also depends on input frequency
First Order Frequency Response

For a sine wave input to a first order process

the output equals

By partial fraction decomposition, numerators are:

First Order Frequency Response

Now, by inverting back to the time domain:

dies out as t increases

Note: φis a function of Tpandω, but not of t.

So for values of t/Tp≥ 4.0 (e-4 = 0.018), the process output , is a sine wave attenuated by

whose value depends on ω being fast or slow.

First Order Frequency Response

Attenuation and time shift between input and output sine waves (Kp=1). The phase angle φ of the output signal is calculated from φ = – 180*Δt/P, where Δt is the time shift and P is the period of oscillation.

Bodé Diagrams

- for a First Order System

Note that ωn = 1/Tp, the natural frequency of the system, so the X-axis is the relative frequency ω/ωn

Why is Frequency Response Analysis so Useful?
• The amplitude ratio and phase lag provide info about system stability
• These can be determined using the Laplace Transform directly to generate a Bodé Diagram
• Can be used to study:

• System identification (create a model)

• Controller tuning

• Stability of the control system

• Robustness of the control system

• Designing noise filters

Laplace and Complex Numbers

Consider a complex number W defined as

a + bj

wherea represents the real part of W or Re(W)

b represents the imaginary part of W or Im(W).

Define the following terms to characterize this number in the complex domain (polar coordinates)

The Modulus The Phase Angle

Laplace and Complex Numbers

These two terms can be used to plot the complex number in a polar coordinate plane

a = |W|cosθand b = |W|sinθ

and W = |W|cosθ + j|W|sinθ

Laplace and Complex Numbers

Since cosθ = (e+jθ + e-jθ)/2 and sinθ = (e+jθ – e-jθ)/2j

Then W = |W|[(e+jθ + e-jθ)/2]+ j|W|[(e+jθ – e-jθ)/2j]

W = |W|e+jθ

Apply this complex knowledge to a First Order System:

Substitute jω for s:

Laplace and Complex Numbers

In polar coordinate form, this becomes:

So by substituting jω for s and rearranging the transform into polar coordinate form, the amplitude ratio and phase angle for the system are obtained directly.

Second Order Frequency Response

The original second order transform:

Substitute jω for s:

Second Order Frequency Response

In polar coordinate form, this becomes:

So, steady state response is obtained directly for the second order system in a few steps and the Bodé Diagram can be plotted for different values of ωTp and ζ.

Bode Diagram

- for Second Order System

Note that ωn = 1/Tp, the natural frequency of the system, so the X-axis is the relative frequency ω/ωn

Notes on Second Order Frequency Response

2nd order response has maximum lag of 180° - twice a 1st order system. If phase angle exceeds 90°, then it is a higher order.

The real problem faced by 2nd and higher order systems shows up at the natural frequency ωn . The amplitude ratio can be >>>> 1.0 and if ζapproaches0, the response is amplified to  . With a 3rd order system, two critical frequencies may exist while for a 4th order system there can be three. Clearly such system responses are not desired or tolerable and must be avoided.

As well the phase lag of higher order systems for high frequency inputs is equal to -90n where n is the system order. With such high lags, these systems are more difficult to control and design.

Bode Diagram for PI Controller
• PI Controller with Kc = 2.0 and TI = 10
Bode Diagram for PD Controller
• PD Controller with Kc = 2.0 and TD = 10
Bode Diagram for PD Controller
• PID Controller with Kc = 2.0, TI = 10 and TD = 4

Notch filters are often used in control systems to eliminate the effects of mechanical resonance.

The general transfer function for many types of filters, including notch filters, is equivalent to the transfer function of a PID controller.

Controller Design using Frequency Response
• Applicable to all orders & types of dynamic models
• Designer can specify desired closed-loop response
• Information on stability and sensitivity is obtained
• Approach is iterative and hence, is time-consuming–interactive computer graphics are necessary
• Results often taken to be "gospel" (far from reality of most processes and plants)
• Diagrams represent expected results for a model – not necessarily the real process
Frequency Response Stability Criteria

Two principal methods to assess stability:

- Bode Stability Criterion

- Nyquist Stability Criterion

The Bode Stability Criterion

The Bode stability criterion states that a closed-loop system is unstable if the Frequency Response of the open-loop Transfer Function has an amplitude ratio greater than 1.0 at the critical frequency ωc. Otherwise the closed-loop system is stable. For the analysis, ωc is the value of ω where open-loop phase angle is -180°

Bode Stability Criteria
• The Bode stability criterion
• a closed-loop system is unstable if the Frequency Response of the open-loop Transfer Function has an amplitude ratio > 1.0 at the critical frequency Cω.
• Otherwise the closed-loop system is stable.
• Cωis where the open-loop phase angle = 180°
Bode Stability Criteria

Example:

• A process has a Transfer Function:
• For proportional control, determine the closed-loop stability for three values of Kc = 1, 4, and 20.
Bode Stability Criteria

Solution:

• The Open Loop Transfer Function
• The Bode plots for the three values of Kc are shown on the next slide. Note that the phase angle curves are identical for all three cases. (Explain?)
Bode Stability Criteria

AROL

φOL

ωc

ω/ωn

Ultimate Gain and Ultimate Period

The Ultimate Controller Gain, KcU, is the maximum value of Kcthat results in stable closed loop system with proportional-only control (continuous oscillation).

The Ultimate Period is the reciprocal of ωc as follows:

KcUis determined from the Open Loop Frequency Response with P-only control and Kc =1. So:

Gain Margin

GM is determined from the Amplitude Ratio Bodé Diagram.

• Determine critical frequency (ωc) where phase lag = 180°.
• Determine the value of the Amplitude Ratio (Ac) at ωc.
• The Gain Margin is then defined as:

GM = 1/Ac

4. The Bode Stability Criterion states:

If the value of GM < 1.0, the system is stable for all input sine wave frequencies.

Phase Margin

PM is determined from the Phase Lag Bodé Diagram.

• Find the frequency ωgwhere the Amplitude Ratio= 1.0.
• Determine the value of the phase lag φgat this ωg value
• The phase margin is defined as

PM = 180° + φg

The Bode Stability Criterion states

If PM is less than 0 , the system is stable for all sine wave input frequencies

Gain and Phase margin

AROL

ωg

ωc

ωc

ω/ωn

φg

φOL

ωg

ωc

ω/ωn

Gain Margin and Phase Margin

1.7 or 2.0 ≤ GM

30° or 45°≤ PM

So, one can adjust Kc such that the GM is at least 1.7 or the PM must be more than 30°

Comparison of Transient Response
• System response to a step change in set point for a PI controller tuned by frequency response ( ) and by Zeigler-Nichols rules ( ).