Transcript Slide 1
Chapter 5 Sinusoidal Input 1 Chapter 5 Processes are also subject to periodic, or cyclic, disturbances. They can be approximated by a sinusoidal disturbance: 0 for t 0 U sin t A sin t for t 0 where: A = amplitude, ω = angular frequency A U sin ( s ) 2 s 2 Examples: 1. 24 hour variations in cooling water temperature. 2. 60-Hz electrical noise (in USA!) 2 For a sine input to the 1st order process: K U ( s) A 2 and G s 2 s s 1 Chapter 5 output is... 0 1s 2 K A Y ( s) 2 2 2 2 2 s 1 s s 1 s s 2 By partial fraction decomposition, AK 2 0 2 2 1 AK 1 2 2 1 AK 2 2 2 1 3 Inverting, this term dies out for large t AK t AK e sin(t ) 2 2 2 2 1 1 AK y sin(t ) B sin(t ) 2 2 1 where B K amplitude ratio 2 2 A 1 Chapter 5 y (t ) arctan( ) phase angle 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… 1 2 2 1 4 Figure 13.1 Attenuation and time shift between input and output sine waves (K= 1). The phase angle φ of the output signal is given φ t / P 360 , where t is the (period) shift and P by is the period of oscillation. Frequency Response Characteristics of a First-Order Process ˆ sin ωt φ as t where : For x(t ) A sin t , y t A Aˆ KA ω2 τ 2 1 and φ tan 1 ωτ 1. The output signal is a sinusoid that has the same frequency, , as the input signal, x t Asin t 2. The amplitude of the output signal, Aˆ , is a function of the frequency and the input amplitude, A: Aˆ KA ω2 τ 2 1 Dividing both sides by the input signal amplitude A yields the amplitude ratio (AR) Aˆ K AR A ω2 τ 2 1 which can, in turn, be divided by the process gain to yield the normalized amplitude ratio (ARN) (or magnitude ratio): AR N 1 ω2 τ 2 1 3. The output has a phase shift, φ, relative to the input. The amount of phase shift depends on frequency, i.e., t 2 tan 1 P Basic Theorem of Frequency Response The frequency response of a system can be found by substituting j for s in the system transfer function, i.e. Laplace domain G s s j ferquency domain G j 8 Properties G j AR amplitude ratio G j phase angle Note, G j R jI G j e jG j G j R I 2 G j tan 1 2 I R 9 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 G jω . 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: AR R 2 I 2 Memorize tan 1 ( R / I ) Example 13.1 Find the frequency response of a first-order system, with 1 G s τs 1 (13-16) Solution First, substitute s jω in the transfer function 1 1 G jω τjω 1 jωτ 1 (13-17) Then multiply both numerator and denominator by the complex conjugate of the denominator, that is, jωτ 1 jωτ 1 jωτ 1 G jω 2 2 jωτ 1 jωτ 1 ω τ 1 1 ω τ 1 2 2 j ωτ ω τ 1 2 2 R jI (13-18) R where: I 1 (13-19a) ω τ 1 ωτ 2 2 (13-19b) ω τ 1 2 2 From Step 3 of the Shortcut Method, 2 1 ωτ AR R 2 I 2 2 2 2 2 ω τ 1 ω τ 1 or 2 2 2 ω τ 1 1 ω2 τ 2 AR Also, φ tan 1 1 ω2 τ 2 1 2 (13-20a) I 1 1 tan ωτ tan ωτ (13-20b) R Complex Transfer Functions Consider a complex transfer G(s), Ga s Gb s Gc s G s G1 s G2 s G3 s Substitute s=j, Ga jω Gb jω Gc jω G jω G1 jω G2 jω G3 jω (13-22) (13-23) From complex variable theory, we can express the magnitude and angle of G jω as follows: G jω Ga jω Gb jω Gc jω (13-24a) G1 jω G2 jω G3 jω G jω Ga jω Gb jω Gc jω [G1 jω G2 jω G3 jω ] (13-24b) Transfer Functions in Series n Y s G s Gi s X s i 1 s j n G j Gi j i 1 n Gi j i 1 G j e Gi j e Gi j e i 1 i 1 n n ln G j jG j ln Gi j j Gi j i 1 i 1 G j n Gi j n n ln G j ln Gi j i 1 n G j Gi j i 1 14 Transfer Functions in Series Gi s Kp e s zi s 1 1 pi s 1 1 2 2 i s 2 i i s 1 0 i 1 15 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. 16 Bode Plot of A First-order System Recall: AR N 1 ω2 τ 2 1 and φ tan 1 ωτ At low frequencies ( ω 0 and ω AR N 1 and At high frequencies ( ω 0 and ω 1) : 1) : AR N 1/ ωτ and 17 Figure 13.2 Bode diagram for a first-order process. • Note that the asymptotes intersect at ω ωb 1/ τ, known as the break frequency or corner frequency. Here the value of ARN from (13-21) is: AR N ω ωb 1 0.707 11 (13-30) • Some books and software defined AR differently, in terms of decibels. The amplitude ratio in decibels ARd is defined as AR d 20 log AR (13-33) Negative Zeros Consider a process zero term, G s K ( sτ 1) Substituting s=j gives G jω K ( jωτ 1) Thus: AR G jω K ω2 τ 2 1 φ G jω tan 1 ωτ Note: In general, a multiplicative constant (e.g., K) changes the AR by a factor of K without affecting φ . 20 Negative Zeros 1 AR N ω τ 1 and φ tan ωτ At low frequencies ( ω 0 and ω 1) : 2 2 AR N 1 and At high frequencies ( ω 0 and ω 1) : AR N ωτ and 21 Positive Zeros G s 1 s AR N ω2 τ 2 1 and φ tan 1 ωτ At low frequencies ( ω 0 and ω AR N 1 and 1) : At high frequencies ( ω 0 and ω 1) : AR N ωτ and 22 Examples 1 2 1 1s 1 2 s 1 z1s 1 p1 s 1 p 2 s 1 23 n-th Power of s G s sn n 1, 2, G j j n AR n n Im G j tan Re G j 1 24 Integrating Elements The transfer function for an integrating element was given in Chapter 5: Y s 1 G s U s s 1 1 AR G j j φ G jω K 90 25 Differentiating Element G s s AR G j φ G jω tan 1 0 90 26 Second-Order System G s K τ 2 s 2 2ζτs 1 G j K 1 2 2 1 2 2 2 4 j 2 2 2 2 K 1 2 2 2 4 2 2 2 1 ARN 1 tan 2 2 2 2 2 2 1 2 2 1 27 Second-Order System Low-frequency asymptote (slope=0): 0 ARN 1 and 0 High-frequency asymptote (slope=-2): ARN 1 2 2 and Beaking frequency: 1 1 b 1 ARN b and tan 2 2 28 Resonant Peak d ARN 1 0 d 2 2 2 1 2 2 0 3 2 2 2 1 2 2 8 2 1 2 2 1 2 * 1 2 1 AR* 2 1 2 2 Notice that, since and must both be real, 1 ( 0.707) 2 29 Figure 13.3 Bode diagrams for second-order processes. Time Delay Its frequency response characteristics can be obtained according to G jω e jωθ which can be written in rational form by substitution of the Euler identity, G jω e jωθ cosωθ j sin ωθ Thus AR G jω cos 2 ωθ sin 2 ωθ 1 sin ωθ φ G jω tan 1 cos ωθ or φ ωθ Figure 13.6 Bode diagram for a time delay, e θs. Figure 13.7 Phase angle plots for e θs and for the 1/1 and 2/2 Padé approximations (G1 is 1/1; G2 is 2/2). Example e s G s s 1 Let 1 G1 s e and G2 s s 1 AR1 G1 j 1 s AR2 1 1 2 AR AR1 AR2 1 2 tan 1 34 Nyquist Diagrams Consider the transfer function 1 G s 2s 1 (13-76) with AR G jω 1 2ω 1 2 (13-77a) and φ G jω tan 1 2ω (13-77b) 35 Figure 13.12 The Nyquist diagram for G(s) = 1/(2s + 1) plotting Re G jω and Im G jω . 36 Figure 13.13 The Nyquist diagram for the transfer function in Example 13.5: 5(8s 1)e6s G( s) (20s 1)(4s 1) 37 Frequency Response Characteristics of Feedback Controllers Proportional Controller. Consider a proportional controller with positive gain Gc s Kc (13-57) In this case Gc jω K c , which is independent of . Therefore, AR c K c (13-58) and φc 0 (13-59) Proportional-Integral Controller. A proportional-integral (PI) controller has the transfer function (cf. Eq. 8-9), τI s 1 1 Gc s Kc 1 Kc τ s τ s I I Substitute s=j: (13-60) jτ I 1 1 1 Gc j Kc 1 Kc Kc 1 τ I j jτ I τI j Thus, the amplitude ratio and phase angle are: AR c Gc jω K c 1 1 ωτ I 2 Kc ωτ I 2 1 ωτ I φc Gc jω tan 1 1/ ωτ I tan 1 ωτ I 90 (13-62) (13-63) 10 s 1 G s 2 Figure 13.9 Bode plot of a PI controller, c 10 s Ideal Proportional-Derivative Controller. For the ideal proportional-derivative (PD) controller (cf. Eq. 8-11) Gc s Kc 1 τ D s (13-64) The frequency response characteristics are similar to those of a LHP zero: AR c K c ωτ D 2 1 φ tan 1 ωτ D (13-65) (13-66) Proportional-Derivative Controller with Filter. The PD controller is most often realized by the transfer function τDs 1 Gc s Kc ατ s 1 D (13-67) Figure 13.10 Bode plots of an ideal PD controller and a PD controller with derivative filter. Idea: Gc s 2 4s 1 With Derivative Filter: 4s 1 Gc s 2 0.4 s 1 PID Controller Forms • Parallel PID Controller. The simplest form in Ch. 8 is 1 Gc s Kc 1 τDs τ1s Series PID Controller. The simplest version of the series PID controller is τ1s 1 Gc s Kc (13-73) τ D s 1 τ1s Series PID Controller with a Derivative Filter. τ1s 1 τ D s 1 Gc s Kc τ1s τ D s 1 Figure 13.11 Bode plots of ideal parallel PID controller and series PID controller with derivative filter (α = 1). Idea parallel: 1 Gc s 2 1 4s 10 s Series with Derivative Filter: 10 s 1 4 s 1 Gc s 2 10 s 0.4 s 1