Sinusoidal Forcing of a FirstOrder Process. / τ


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1 Frequency Response Analysis Chapter 3 Sinusoidal Forcing of a FirstOrder Process For a firstorder transfer function with gain K and time constant τ, the response to a general sinusoidal input, xt = A tis: sin ω KA () ( t / τ yt = ωτe ωτ cos ωt+ sin ω t) (525) ωτ + Note that y(t) and x(t) are in deviation form. The longtime response, y l (t), can be written as: KA y () t = sin( ωt+ φ) for t (3) ωτ + where: φ = tan ωτ
2 Figure 3. Attenuation and time shift between input and output sine waves (K= ). The phase angle φ of the output signal is given by φ = Time shift / P 360, where t is the (period) shift and P is the period of oscillation. 2
3 Frequency Response Characteristics of a FirstOrder Process ˆ For xt = Asin ωt, y t = Asin ωt+ φ as t where : ˆ KA A = and φ = tan ωτ ωτ +. The output signal is a sinusoid that has the same frequency, ω, as the input.signal, x(t) =Asinωt. 2. The amplitude of the output signal, Â, is a function of the frequency ω and the input amplitude, A: ˆ KA A = (32) ωτ + 3. The output has a phase shift, φ, relative to the input. The amount of phase shift depends on ω. 3
4 Dividing both sides of (32) by the input signal amplitude A yields the amplitude ratio (AR) Chapter 3 AR which can, in turn, be divided by the process gain to yield the normalized amplitude ratio (AR N ) AR N Aˆ = = A = K ωτ + ωτ + (33a) (33b) 4
5 Shortcut Method for Finding the Frequency Response The shortcut method consists of the following steps: Step. Set s=jω in G(s) to obtain G jω. Step 2. Rationalize G(jω); We want to express it in the form. G(jω)=R + ji where R and I are functions of ω. Simplify G(jω) 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 + I Memorize ϕ = tan ( I/ R) 5
6 Example 3. Find the frequency response of a firstorder system, with Solution G s = τs + First, substitute s = jω in the transfer function (36) G( jω ) = = (37) τω j + jωτ + Then multiply both numerator and denominator by the complex conjugate of the denominator, that is, jωτ + ( ω) G j jωτ + jωτ + = = ωτ + ωτ + ωτ + ( j )( j ) ( ωτ) = + j = R+ ji ωτ + ωτ + (38) 6
7 where: I R = = ωτ + ωτ ωτ + (39a) (39b) Chapter 3 From Step 3 of the Shortcut Method, ( ωτ + ) ωτ AR = R + I = + ωτ + ωτ + or ( + ωτ ) AR = = (320a) 2 ωτ + Also, I φ = tan = tan ( ωτ) = tan ( ωτ ) (320b) R 7
8 Consider a complex transfer G(s), Ga s Gb s Gc s G( s) = G s G s G s Substitute s=jω, Ga j Gb j Gc j G( jω ) = (323) G j G j G j ( ω) ( ω) ( ω) ( ω) ( ω) ( ω) 2 3 From complex variable theory, we can express the magnitude and angle of G jω as follows: Complex Transfer Functions ( ω ) G j 2 3 = ( ω) ( ω) ( ω) ( ω) ( ω) ( ω) G j G j G j a b c G j G j G j 2 3 ( ω) a( ω) b( ω) c( ω) G ( j ) G ( j ) G ( j ) G j = G j + G j + G j (322) (324a) [ ω + ω + ω + ] (324b) 8
9 Bode Diagrams Chapter 3 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 φ as a function of ω. Ordinarily, ω is expressed in units of radians/time. Bode Plot of A Firstorder System Recall: AR N = and φ = tan ωτ ωτ + At low frequencies ( ω 0andωτ ) : AR N = and ϕ=0 At high frequencies ( ω and ωτ ) : AR = / ωτ and ϕ= 90 N 9
10 Figure 3.2 Bode diagram for a firstorder process. 0
11 Note that the asymptotes intersect at ω= ωb = / τ, known as the break frequency or corner frequency. Here the value of AR N from (32) is: Chapter 3 AR N ( ω= ωb ) = = (330) + Some books and software defined AR differently, in terms of decibels. The amplitude ratio in decibels AR d is defined as AR d = 20 log AR (333)
12 Integrating Elements The transfer function for an integrating element was given in Chapter 5: Y( s) K G( s) = = (534) U s s K K AR = G( jω ) = = (334) jω ω K φ= G jω = = 90 (335) SecondOrder Process A general transfer function that describes any underdamped, critically damped, or overdamped secondorder system is K G( s) = (340) τ s + 2ζτs+ 2
13 Substituting s = jω and rearranging yields: AR = K ( ωτ) + ( 2ωτζ) (34a) Chapter 3 2ωτζ φ = tan (34b) ωτ Figure 3.3 Bode diagrams for secondorder processes. 3
14 which can be written in rational form by substitution of the Euler identity, From (354) or jωθ G jω = e = cos ωθ jsin ωθ (354) Time Delay Its frequency response characteristics can be obtained by substituting s = jω, ωθ = G jω e j (353) AR = G jω = cos ωθ + sin ωθ = (355) φ= G( jω) = tan sin ωθ cosωθ φ= ωθ (356) 4
15 Figure 3.6 Bode diagram for a time delay, e θs. 5
16 Figure 3.7 Phase angle plots for e θs and for the / and 2/2 Padé approximations (G is /; G 2 is 2/2). 6
17 Process Zeros Chapter 3 Consider a process zero term, Substituting s=jω gives Thus: = K( sτ + ) G s G jω = K( jωτ + ) AR = G jω = K ω τ + φ= G jω =+ tan ωτ Note: In general, a multiplicative constant (e.g., K) changes the AR by a factor of K without affecting φ. 7
18 Frequency Response Characteristics of Feedback Controllers Chapter 3 Proportional Controller. Consider a proportional controller with positive gain c = (357) G s K In this case Gc jω = Kc, which is independent of ω. Therefore, and c AR = K (358) c c φc = 0 (359) 8
19 ProportionalIntegral Controller. A proportionalintegral (PI) controller has the transfer function (cf. Eq. 89), Chapter 3 τ I s + Gc s = Kc + = Kc (360) τis τis Substitute s=jω: jω τ + G ( ω ) = + = I c j Kc Kc = Kc j τi jω jωτi τiω Thus, the amplitude ratio and phase angle are: c c c 2 c ( ωτi ) 2 ωτi + AR = G ( jω) = K + = K (362) ωτ ( j ) φ = G ω = tan / ωτ = tan ωτ 90 (363) c c I I I 9
20 Figure 3.9 Bode plot of a PI controller, G c ( s) 0s + = 2 0s 20
21 Ideal ProportionalDerivative Controller. For the ideal proportionalderivative (PD) controller (cf. Eq. 8) = ( + ) G s K τ s (364) c c D Chapter 3 The frequency response characteristics are similar to those of a LHP zero: 2 AR = K ωτ + (365) c c D φ = tan ωτ (366) ProportionalDerivative Controller with Filter. The PD controller is most often realized by the transfer function D τds + Gc( s) = Kc ατds + (367) 2
22 Figure 3.0 Bode plots of an ideal PD controller and a PD controller with derivative filter. Idea: = 24 ( + ) G s s c With Derivative Filter: 4s + Gc ( s) = s + 22
23 PID Controller Forms Parallel PID Controller. The simplest form in Ch. 8 is Chapter 3 G c s = K c + + τ τ D s s Series PID Controller. The simplest version of the series PID controller is τs + Gc( s) = Kc ( τ Ds+ ) (373) τs Series PID Controller with a Derivative Filter. τs+ τ s+ Gc( s) = K D c τs ατds+ 23
24 Figure 3. Bode plots of ideal parallel PID controller and series PID controller with derivative filter (α = ). Idea parallel: Gc ( s) = s 0s G c ( s) Series with Derivative Filter: 0s+ 4s+ = 2 0s 0.4s+ 24
25 Consider the transfer function Nyquist Diagrams Chapter 3 with and G s = (376) 2s + AR = G( jω ) = (377a) 2ω + 2 φ= G jω = tan 2ω (377b) 25
26 Figure 3.2 The Nyquist diagram for G(s) = /(2s + ) plotting Re ω and Im G jω. G( j ) 26
27 Figure 3.3 The Nyquist diagram for the transfer function in Example 3.5: Gs () = 6s 5(8s+ ) e (20s+ )(4s+ ) 27
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