1 Chapter 5 Sinusoidal Input. 2 Chapter 5 Examples: 1.24 hour variations in cooling water temperature. 2.60-Hz electrical noise (in USA!) Processes are.

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Presentation transcript:

1 Chapter 5 Sinusoidal Input

2 Chapter 5 Examples: 1.24 hour variations in cooling water temperature Hz electrical noise (in USA!) Processes are also subject to periodic, or cyclic, disturbances. They can be approximated by a sinusoidal disturbance: where: A = amplitude, ω = angular frequency

3 For a sine input to the 1st order process: output is... By partial fraction decomposition, Chapter 5

4 Note that the amplitude ratio and phase angle is not a function of t but of τ and ω. For large t, y(t) is also sinusoidal, output sine is attenuated by… Inverting, this term dies out for large t Chapter 5

Figure 13.1 Attenuation and time shift between input and output sine waves (K= 1). The phase angle of the output signal is given by, where is the (period) shift and P is the period of oscillation.

1.The output signal is a sinusoid that has the same frequency,  as the input signal, 2.The amplitude of the output signal,, is a function of the frequency  and the input amplitude, A: Frequency Response Characteristics of a First-Order Process

which can, in turn, be divided by the process gain to yield the normalized amplitude ratio (AR N ) (or magnitude ratio): Dividing both sides by the input signal amplitude A yields the amplitude ratio (AR) 3.The output has a phase shift, φ, relative to the input. The amount of phase shift depends on frequency, i.e.,

8 Basic Theorem of Frequency Response

9 Properties

Shortcut Method for Finding the Frequency Response The shortcut method consists of the following steps: Step 1. Set s=j  in G(s) to obtain. Step 2. Rationalize G(j  ); We want to express it in the form. G(j  )=R + jI where R and I are functions of  Simplify G(j  ) by multiplying the numerator and denominator by the complex conjugate of the denominator. Step 3. The amplitude ratio and phase angle of G(s) are given by: Memorize

Example 13.1 Find the frequency response of a first-order system, with Solution First, substitute in the transfer function Then multiply both numerator and denominator by the complex conjugate of the denominator, that is,

where: From Step 3 of the Shortcut Method, or Also,

Consider a complex transfer G(s), From complex variable theory, we can express the magnitude and angle of as follows: Complex Transfer Functions Substitute s=j 

14 Transfer Functions in Series

15 Transfer Functions in Series

16 Bode Diagrams A special graph, called the Bode diagram or Bode plot, provides a convenient display of the frequency response characteristics of a transfer function model. It consists of plots of AR and phase angle as a function of frequency. Ordinarily, frequency is expressed in units of radians/time.

17 Bode Plot of A First-order System

Figure 13.2 Bode diagram for a first-order process.

Note that the asymptotes intersect at, known as the break frequency or corner frequency. Here the value of AR N from (13-21) is: Some books and software defined AR differently, in terms of decibels. The amplitude ratio in decibels AR d is defined as

20 Negative Zeros Substituting s=j  gives Consider a process zero term, Thus: Note: In general, a multiplicative constant (e.g., K) changes the AR by a factor of K without affecting.

21 Negative Zeros

22 Positive Zeros

23 Examples

24 n-th Power of s

25 Integrating Elements

26 Differentiating Element

27 Second-Order System

28 Second-Order System

29 Resonant Peak

Figure 13.3 Bode diagrams for second-order processes.

which can be written in rational form by substitution of the Euler identity, Thus or Time Delay Its frequency response characteristics can be obtained according to

Figure 13.6 Bode diagram for a time delay,.

Figure 13.7 Phase angle plots for and for the 1/1 and 2/2 Padé approximations (G 1 is 1/1; G 2 is 2/2).

34 Example

35 Nyquist Diagrams Consider the transfer function with and

36 Figure The Nyquist diagram for G(s) = 1/(2s + 1) plotting and

37 Figure The Nyquist diagram for the transfer function in Example 13.5:

Frequency Response Characteristics of Feedback Controllers Proportional Controller. Consider a proportional controller with positive gain In this case, which is independent of . Therefore, and

Proportional-Integral Controller. A proportional-integral (PI) controller has the transfer function (cf. Eq. 8-9), Thus, the amplitude ratio and phase angle are: Substitute s=j  :

Figure 13.9 Bode plot of a PI controller,

Ideal Proportional-Derivative Controller. For the ideal proportional-derivative (PD) controller (cf. Eq. 8-11) The frequency response characteristics are similar to those of a LHP zero: Proportional-Derivative Controller with Filter. The PD controller is most often realized by the transfer function

Figure Bode plots of an ideal PD controller and a PD controller with derivative filter. Idea: With Derivative Filter:

Series PID Controller. The simplest version of the series PID controller is Series PID Controller with a Derivative Filter. PID Controller Forms Parallel PID Controller. The simplest form in Ch. 8 is

Figure Bode plots of ideal parallel PID controller and series PID controller with derivative filter (α = 1). Idea parallel: Series with Derivative Filter: