Chapter 14 Frequency Response Analysis and Control System Design CHAPTER CONTENTS 14 1 Sinusoidal Forcing of a First Order Process 14 2 Sinusoidal Forcing of an nth Order Process 14 3 Bode Diagrams 14[.]
Chapter 14 Frequency Response Analysis and Control System Design CHAPTER CONTENTS 14.1 Sinusoidal Forcing of a First-Order Process 14.2 Sinusoidal Forcing of an nth-Order Process 14.3 Bode Diagrams 14.3.1 14.3.2 14.3.3 First-Order Process Integrating Process Second-Order Process 14.3.4 Process Zero 14.3.5 Time Delay 14.4 Frequency Response Characteristics of Feedback Controllers 14.5 Nyquist Diagrams 14.6 Bode Stability Criterion 14.7 Gain and Phase Margins Summary In previous chapters, Laplace transform techniques were used to calculate transient responses from transfer functions This chapter focuses on an alternative way to analyze dynamic systems by using frequency response analysis Frequency response concepts and techniques play an important role in stability analysis, control system design, and robustness assessment Historically, frequency response techniques provided the conceptual framework for early control theory and important applications in the field of communications (MacFarlane, 1979) We introduce a simplified procedure to calculate the frequency response characteristics from the transfer function model of any linear process Two concepts, the Bode and Nyquist stability criteria, are generally applicable for feedback control systems and stability 244 analysis Next we introduce two useful metrics for relative stability, namely gain and phase margins These metrics indicate how close a control system is to instability A related issue is robustness, that is, the sensitivity of control system performance to process variations and to uncertainty in the process model The design of robust feedback control systems is considered in Appendix J 14.1 SINUSOIDAL FORCING OF A FIRST-ORDER PROCESS We start with the response properties of a first-order process when forced by a sinusoidal input and show how the output response characteristics depend on the frequency of the input signal This is the origin of 14.1 the term frequency response The responses for firstand second-order processes forced by a sinusoidal input were presented in Chapter Recall that these responses consisted of sine, cosine, and exponential terms Specifically, for a first-order transfer function with gain K and time constant τ, the response to a general sinusoidal input, x(t) = A sin ωt, is KA (ωτe−t∕τ − ωτ cos ωt + sin ωt) (5-23) ω2 τ2 + where y is in deviation form If the sinusoidal input is continued for a long time, the exponential term (ωτe−t/τ ) becomes negligible The remaining sine and cosine terms can be combined via a trigonometric identity to yield y(t) = y𝓁 (t) = √ KA ω2 τ2 + sin (ωt + ϕ) (14-1) where ϕ = −tan−1 (ωτ) The long-time response y𝓁 (t) is called the frequency response of the first-order system and has two distinctive features (see Fig 14.1) The output signal is a sine wave that has the same frequency, but its phase is shifted relative to the input sine wave by the angle ϕ (referred to as the phase shift or the phase angle); the amount of phase shift depends on the forcing frequency ω The sine wave has an amplitude  that is a function of the forcing frequency: KA  = √ ω2 τ2 + (14-2) Dividing both sides of Eq 14-2 by the input signal amplitude A yields the amplitude ratio (AR) AR =  K =√ A ω2 τ2 + (14-3a) which can, in turn, be divided by the process gain to yield the normalized amplitude ratio (ARN ): ARN = AR =√ K ω τ2 + (14-3b) Next we examine the physical significance of the preceding equations, with specific reference to the blending P A A Output, y Time shift, Δt Input, x Time, t Figure 14.1 Attenuation and time shift between input and output sine waves The phase angle ϕ of the output signal is given by ϕ = Δt/P × 360∘, where Δt is the time shift and P is the period of oscillation Sinusoidal Forcing of a First-Order Process 245 process example discussed earlier In Chapter 4, the transfer function model for the stirred-tank blending system was derived as X ′ (s) = K3 K1 K2 X ′ (s) + W ′ (s) + W ′ (s) (4-84) τs + 1 τs + τs + 1 Suppose flow rate w2 is varied sinusoidally about a constant value, while the other inlet conditions are kept constant at their nominal values; that is, w′1 (t) = x′1 (t) = Because w2 (t) is sinusoidal, the output composition deviation x′ (t) eventually becomes sinusoidal according to Eq 5-24 However, there is a phase shift in the output relative to the input, as shown in Fig 14.1, owing to the material holdup of the tank If the flow rate w2 oscillates very slowly relative to the residence time τ(ω ≪ 1/τ), the phase shift is very small, approaching 0∘ , whereas the normalized amplitude ̂ ratio (A/KA) is very nearly unity For the case of a low-frequency input, the output is in phase with the input, tracking the sinusoidal input as if the process model were G(s) = K On the other hand, suppose that the flow rate is varied rapidly by increasing the input signal frequency For ω ≫ 1/τ, Eq 14-1 indicates that the phase shift approaches a value of −π/2 radians (−90∘ ) The presence of the negative sign indicates that the output lags behind the input by 90∘ ; in other words, the phase lag is 90∘ The amplitude ratio approaches zero as the frequency becomes large, indicating that the input signal is almost completely attenuated; namely, the sinusoidal deviation in the output signal is very small These results indicate that positive and negative deviations in w2 are essentially canceled by the capacitance of the liquid in the blending tank if the frequency is high enough High frequency implies ω ≫ 1/τ Most processes behave qualitatively similar to the stirred-tank blending system, when subjected to a sinusoidal input For high-frequency input changes, the process output deviations are so completely attenuated that the corresponding periodic variation in the output is difficult (perhaps impossible) to detect or measure Input–output phase shift and attenuation (or amplification) occur for any stable transfer function, regardless of its complexity In all cases, the phase shift and amplitude ratio are related to the frequency ω of the sinusoidal input signal In developments up to this point, the expressions for the amplitude ratio and phase shift were derived using the process transfer function However, the frequency response of a process can also be obtained experimentally By performing a series of tests in which a sinusoidal input is applied to the process, the resulting amplitude ratio and phase shift can be measured for different frequencies In this case, the frequency response is expressed as a table of measured amplitude ratios and phase shifts for selected values of ω However, the method is very time-consuming 246 Chapter 14 Frequency Response Analysis and Control System Design because of the repeated experiments for different values of ω Thus other methods, such as pulse testing (Ogunnaike and Ray, 1994), are utilized, because only a single test is required In this chapter, the focus is on developing a powerful analytical method to calculate the frequency response for any stable process transfer function Later in this chapter, we show how this information can be used to design controllers and analyze the properties of the closed loop system responses 14.2 The shortcut method can be summarized as follows: Step Substitute s = jω in G(s) to obtain G(jω) Step Rationalize G(jω), i.e., express G(jω) as the sum of real (R) and imaginary (I) parts R + jI, where R and I are functions of ω, using complex conjugate multiplication Step The √ output sine wave has amplitude −1 ̂ A = A R2 + I and phase angle √ϕ = tan (I/R) 2 The amplitude ratio is AR = R + I and is independent of the value of A SINUSOIDAL FORCING OF AN nTH-ORDER PROCESS This section presents a general approach for deriving the frequency response of any stable transfer function The physical interpretation of frequency response is not valid for unstable systems, because a sinusoidal input produces an unbounded output response instead of a sinusoidal response A rather simple procedure can be employed to find the sinusoidal response After setting s = jω in G(s), by algebraic manipulation we can separate the expression into real (R) and imaginary (I) terms (j indicates an imaginary component): G(jω) = R(ω) + jI(ω) ϕ = tan (I∕R) G(s) = (14-6b) Both  and ϕ are functions of frequency ω A simple but elegant relation for the frequency response can be derived, where the amplitude ratio is given by √  AR = (14-7) = |G| = R2 + I A The absolute value denotes the magnitude of G, and the phase shift (also called the phase angle or argument of G, ∠G) between the sinusoidal output and input is given by (14-8) ϕ = ∠G = tan−1 (I∕R) Because R(ω) and I(ω) (and hence AR and ϕ) can be obtained without calculating the complete transient response y(t), these characteristics provide a convenient shortcut method to determine the frequency response of transfer functions Equations 14-7 and 14-8 can calculate the frequency response characteristics of any stable G(s), including those with time-delay terms τs + (14-9) SOLUTION First substitute s = jω in the transfer function G(jω) = 1 = τjω + jωτ + (14-10) Then multiply both numerator and denominator by the complex conjugate of the denominator, that is, −jωτ + −jωτ + −jωτ + = 2 (jωτ + 1)(−jωτ + 1) ω τ +1 G(jω) = (14-5)  and ϕ are related to I(ω) and R(ω) by the following relations (Seborg et al., 2004): √ (14-6a)  = A R2 + I −1 Find the frequency response of a first-order system, with (14-4) Similar to Eq 14-1, we can express the long time response for a linear system (cf Eq 14-1) as y𝓁 (t) =  sin(ωt + ϕ) EXAMPLE 14.1 (−ωτ) +j 2 = R + jI ω2 τ2 + ω τ +1 = where (14-11) R= ω2 τ2 + (14-12a) I= −ωτ ω2 τ2 + (14-12b) and From Eq 14-7, √ AR = |G(jω|) = ( )2 ω2 τ2 +1 ( + −ωτ +1 )2 ω2 τ2 Simplifying, √ AR = (1 + ω2 τ2 ) = √ (ω2 τ2 + 1)2 ω2 τ2 + ϕ = ∠G(jω) = tan−1 (−ωτ) = −tan−1 (ωτ) (14-13a) (14-13b) If the process gain had been a positive value K instead of 1, AR = √ K ω2 τ2 + (14-14) and the phase angle would be unchanged (Eq 14-13b) Both the amplitude ratio and phase angle are identical to those values calculated in Section 14.1 using the time-domain derivation 14.3 From this example, we conclude that direct analysis of the complex transfer function G(jω) is computationally easier than solving for the actual long-time output response j𝓁 (t) The computational advantages are even greater when dealing with more complicated processes, as shown in the following Start with a general transfer function in factored form G (s)Gb (s)Gc (s) · · · (14-15) G(s) = a G1 (s)G2 (s)G3 (s) · · · G(s) is converted to the complex form G(jω) by the substitution s = jω: G (jω)Gb (jω)Gc (jω) · · · G(jω) = a (14-16) G1 (jω)G2 (jω)G3 (jω) · · · The magnitude and phase angle of G(jω) are as follows: |Ga (jω)‖Gb (jω)‖Gc (jω)| · · · |G(jω)| = (14-17a) |G1 (jω)‖G2 (jω)‖G3 (jω)| · · · ∠G(jω) = ∠Ga (jω) + ∠Gb (jω) + ∠Gc (jω) + · · · − [∠G1 (jω) + ∠G2 (jω) + ∠G3 (jω) + · · ·] (14-17b) Equations 14-17a and 14-17b greatly simplify the computation of |G(jω)| and ∠G(jω) and, consequently, AR and ϕ, for factored transfer functions These expressions eliminate much of the complex algebra associated with the rationalization of complicated transfer functions Hence, the factored form (Eq 14-15) may be preferred for frequency response analysis On the other hand, if the frequency response curves are generated using software such as MATLAB, there is no need to factor the numerator or denominator, as discussed in Section 14.3 Calculate the amplitude ratio and phase angle for the overdamped second-order transfer function G(s) = K (τ1 s + 1)(τ2 s + 1) SOLUTION Using Eq 14-15, let Ga = K G1 = τ1 s + G2 = τ2 s + Substituting s = jω Ga (jω) = K G1 (jω) = jωτ1 + G2 (jω) = jωτ2 + The magnitudes and angles of each component of the complex transfer function are |Ga | = K √ |G1 | = √ω2 τ21 + |G2 | = ω2 τ22 + ∠Ga = ∠G1 = tan−1 (ωτ1 ) ∠G2 = tan−1 (ωτ2 ) 247 Combining these expressions via Eqs 14-17a and 14-17b yields |Ga (jω)| |G1 (jω)‖G2 (jω)| K √ = √ ω2 τ21 + ω2 τ22 + AR = |G(jω)| = (14-18a) ϕ = ∠G(jω) = ∠Ga (jω) − (∠G1 (jω) + ∠G2 (jω)) = −tan−1 (ωτ1 ) − tan−1 (ωτ2 ) (14-18b) 14.3 BODE DIAGRAMS The Bode diagram (or Bode plot) provides a convenient display of the frequency response characteristics in which AR and ϕ are each plotted as a function of ω Ordinarily, ω is expressed in units of radians/time to simplify inverse tangent calculations (e.g., Eq 14-18b) where the arguments must be dimensionless, that is, in radians Occasionally, a cyclic frequency, ω/2π, with units of cycles/time, is used Phase angle ϕ is normally expressed in degrees rather than radians For reasons that will become apparent in the following development, the Bode diagram consists of: (1) a log–log plot of AR versus ω and (2) a semilog plot of ϕ versus ω These plots are particularly useful for rapid analysis of the response characteristics and stability of closed-loop systems 14.3.1 EXAMPLE 14.2 Bode Diagrams First-Order Process In the past, when frequency response plots had to be generated by hand, they were of limited utility A much more practical approach now utilizes spreadsheets or control-oriented software such as MATLAB to simplify calculations and generate Bode plots Although spreadsheet software can be used to generate Bode plots, it is much more convenient to use software designed specifically for control system analysis Thus, after describing the qualitative features of Bode plots of simple transfer functions, we illustrate how the AR and ϕ components of such a plot are generated by a MATLAB program in Example 14.3 For a first-order model, K/(τs + 1), Fig 14.2 shows a general log–log plot of the normalized amplitude ratio versus ωτ, for positive K For a negative valve of K, the phase angle is decreased by −180∘ A semilog plot of ϕ versus ωτ is also shown In Fig 14.2, the abscissa ωτ has units of radians If K and τ are known, ARN (or AR) and ϕ can be plotted as a function of ω Note that, at high frequencies, the amplitude ratio drops to an infinitesimal level, and the phase lag (the phase angle expressed as a positive value) approaches a maximum value of 90∘ Some books and software define AR differently, in terms of decibels The amplitude ratio in decibels 248 Chapter 14 Frequency Response Analysis and Control System Design ωb = 1/τ ϕ = ∠G(jω) = ∠K − tan−1 (j∞) = −90∘ (14-21) 14.3.3 Normalized amplitude 0.1 ratio, ARN Second-Order Process A general transfer function for a second-order system without numerator dynamics is G(s) = 0.01 0.01 0.1 ωτ 10 100 –30 ωb = 1/τ Phase angle –60 ϕ (deg) –90 –120 0.01 0.1 ωτ 10 100 Figure 14.2 Bode diagram for a first-order process ARdb is defined as ARdb = 20 log AR (14-19) The use of decibels merely results in a rescaling of the Bode plot AR axis The decibel unit is employed in electrical communication and acoustic theory and is seldom used today in the process control field Note that the MATLAB bode routine uses decibels as the default option; however, it can be modified to plot AR results, as shown in Fig 14.2 In the rest of this chapter, we only derive frequency responses for simple transfer functions (integrator, first-order, second-order, zeros, time delay) Software should be used for calculating frequency responses of more complicated transfer functions 14.3.2 Integrating Process The transfer function for an integrating process was given in Chapter Y(s) K G(s) = = (5-32) U(s) s Because of the single pole located at the origin, this transfer function represents a marginally stable process The shortcut method of determining frequency response outlined in the preceding section was developed for stable processes, that is, those that converge to a bounded oscillatory response for a sinusoidal input Because the output of an integrating process is bounded when forced by a sinusoidal input, the shortcut method does apply for this marginally stable process: |K| K (14-20) AR = |G(jω)| = || || = | jω | ω K τ2 s2 + 2ζτs + (14-22) Substituting s = jω and rearranging into real and imaginary parts (see Example 14.1) yields K AR = √ (14-23a) 2 (1 − ω τ )2 + (2ζωτ)2 [ ] −2ζωτ (14-23b) ϕ = tan−1 − ω2 τ2 Note that, in evaluating ϕ, multiple results are obtained because Eq 14-23b has infinitely many solutions, each differing by n180∘ , where n is a positive integer The appropriate solution of Eq 14-23b for the second-order system yields −180∘ < ϕ < Figure 14.3 shows the Bode plots for overdamped (ξ > 1), critically damped (ξ = 1), and underdamped (0 < ξ < 1) processes as a function of ωτ The lowfrequency limits of the second-order system are identical to those of the first-order system However, the limits are different at high frequencies, ωτ ≫ ARN ≈ 1∕(ωτ)2 ϕ ≈ −180o (14-24a) (14-24b) For overdamped systems, the normalized amplitude ̂ ratio is attenuated (A/KA < 1) for all ω For underdamped systems, the amplitude√ratio plot exhibits a maximum (for values of < ζ < 2∕2) at the resonant frequency √ − 2ζ2 (14-25) ωr = τ (ARN )max = √ 2ζ − ζ2 (14-26) These expressions can be derived by the interested reader The resonant frequency ωr is that frequency for which the sinusoidal output response has the maximum amplitude for a given sinusoidal input Equations 14-25 and 14-26 indicate how ωr and (ARN )max depend on ξ This behavior is used in designing organ pipes to create sounds at specific frequencies However, excessive resonance is undesirable, for example, in automobiles, where a particular vibration is noticeable only at a certain speed For industrial processes operated without feedback control, resonance is seldom encountered, although some measurement devices are designed to exhibit a limited amount of resonant behavior On the other hand, feedback controllers can be tuned to give the controlled process a slight amount of oscillatory 14.3 249 10 ζ=1 ζ = 0.2 0.1 ARN 0.01 0.001 0.0001 0.01 Bode Diagrams ARN 0.01 Slope = –2 0.1 ωτ 10 0.4 0.8 0.1 Slope = –2 0.001 0.01 100 0.1 ωτ 10 100 –45 ϕ (deg) –90 –45 –135 –180 0.01 ϕ (deg) ωτ 0.4 –135 ζ=1 0.1 –90 ζ = 0.2 10 100 –180 0.01 0.1 ωτ 0.8 10 100 Figure 14.3 Bode diagrams for second-order processes Right: underdamped Left: overdamped and critically damped or underdamped behavior in order to speed up the controlled system response (see Chapter 12) 14.3.4 Process Zero A term of the form τs + in the denominator of a transfer function is sometimes referred to as a process lag, because it causes the process output to lag the input (the phase angle is negative) Similarly, a process zero of the form τs + (τ > 0) in the numerator (see Section 6.1) causes the sinusoidal output of the process to lead the input (ϕ > 0); hence, a left-half plane (LHP) zero often is referred to as a process lead Next we consider the amplitude ratio and phase angle for this term Substituting s = jω into G(s) = τs + gives G(jω) = jωτ + from which AR = |G(jω)| = √ ω2 τ2 + ϕ = ∠G(jω) = tan−1 (ωτ) (14-27) (14-28a) (14-28b) Therefore, a process zero contributes a positive phase angle that varies between and +90∘ The output signal amplitude becomes very large at high frequencies (i.e., AR → ∞ as ω → ∞), which is a physical impossibility Consequently, in practice a process zero is always found in combination with one or more poles The order of the numerator of the process transfer function must be less than or equal to the order of the denominator, as noted in Section 6.1 Suppose that the numerator of a transfer function contains the term − τs, with τ > As shown in Section 6.1, a right-half plane (RHP) zero is associated with an inverse step response The frequency response characteristics of G(s) = − τs are √ (14-29a) AR = ω2 τ2 + ϕ = −tan−1 (ωτ) (14-29b) Hence, the amplitude ratios of LHP and RHP zeros are identical However, an RHP zero contributes phase lag to the overall frequency response because of the negative sign Processes that contain an RHP zero or time delay are sometimes referred to as nonminimum phase systems because they exhibit more phase lag than another transfer function that has the same AR characteristics (Franklin et al., 2014) Exercise 14.11 illustrates the importance of zero location on the phase angle 14.3.5 Time Delay The time delay e−θs is the remaining important process element to be analyzed Its frequency response characteristics can be obtained by substituting s = jω: G(jω) = e−jωθ (14-30) which can be written in rational form by substitution of the Euler identity G(jω) = cos ωθ − j sin ωθ (14-31) 250 Chapter 14 Frequency Response Analysis and Control System Design 10 AR AR 0.1 0.01 0.1 ωθ 10 ω (rad/min) –180 ϕ (deg) ϕ (deg) –360 –540 0.01 0.1 10 ωθ −θs Figure 14.4 Bode diagram for a time delay, e Figure 14.5 Bode plot of the transfer function in Example 14.3 From Eq 14.6, √ AR = |G(jω)| = cos2 ωθ + sin2 ωθ = ( ) sin ωθ ϕ = ∠G(jω) = tan−1 − cos ωθ (14-32) or ϕ = −ωθ (14-33) Because ω is expressed in radians/time, the phase angle in degrees is −180ωθ/π Figure 14.4 illustrates the Bode plot for a time delay The phase angle is unbounded, that is, it approaches −∞ as ω becomes large By contrast, the phase angles of all other process elements are smaller in magnitude than some multiples of 90∘ This unbounded phase lag is an important attribute of a time delay and is detrimental to closed-loop system stability, as is discussed in Section 14.6 EXAMPLE 14.3 Generate the Bode plot for the transfer function G(s) = ω (rad/min) 5(0.5s + 1)e−0.5s (20s + 1)(4s + 1) where the time constants and time delay have units of minutes SOLUTION The Bode plot is shown in Fig 14.5 The steady-state gain (K = 5) is the value of AR when ω → The phase angle at high frequencies is dominated by the time delay The MATLAB code for generating a Bode plot of the transfer function is shown in Table 14.1 In this code the normalized AR is used (ARN ) Table 14.1 MATLAB Program to Calculate and Plot the Frequency Response in Example 14.3 %Make a Bode plot for G = (0.5s + 1)e^–0.5s/(20s + 1) %(4s + 1) close all gain = 5; tdead = 0.5; num = [0.5 1]; den = [80 24 1]; G = tf (gain∗ num, den) %Define the system as a transfer %function points = 500; %Define the number of points ww = logspace (−2, 2, points); %Frequencies to be evaluated [mag, phase, ww] = bode (G,ww); % Generate numerical %values for Bode plot AR = zeros (points, 1); % Preallocate vectors for Amplitude %Ratio and Phase Angle PA = zeros (points, 1); for i = : points AR(i) = mag (1,1,i)/gain; %Normalized AR PA(i) = phase (1,1,i) – ((180/pi) ∗ tdead∗ ww(i)); end figure subplot (2,1,1) loglog(ww, AR) axis ([0.01 100 0.001 1]) title (‘Frequency Response of a SOPTD with Zero’) ylabel(‘AR/K’) subplot (2,1,2) semilogx(ww,PA) axis ([0.01 100 −270 0]) ylabel(‘Phase Angle (degrees)’) xlabel(‘Frequency (rad/time)’) 14.4 14.4 Frequency Response Characteristics of Feedback Controllers FREQUENCY RESPONSE CHARACTERISTICS OF FEEDBACK CONTROLLERS In order to use frequency response analysis to design control systems, the frequency-related characteristics of feedback controllers must be known for the most widely used forms of the PID controller discussed in Chapter In the following derivations, we generally assume that the controller is reverse-acting (Kc > 0) If a controller is direct-acting (Kc < 0), the AR plot does not change, because |Kc | is used in calculating the magnitude However, the phase angle is shifted by −180∘ when Kc is negative For example, a direct-acting proportional controller (Kc < 0) has a constant phase angle of −180∘ As a practical matter, it is possible to use the absolute value of Kc to calculate ϕ when designing closed-loop control systems, because stability considerations (see Chapter 11) require that Kc < only when Kv Kp Km < This choice guarantees that the open-loop gain (KOL = Kc Kv Kp Km ) will always be positive Use of this convention conveniently yields ϕ = 0∘ for any proportional controller and, in general, eliminates the need to consider the −180∘ phase shift contribution of the negative controller gain Proportional Controller Consider a proportional controller with positive gain Gc (s) = Kc (14-34) In this case, |Gc (jω)| = Kc , which is independent of ω Therefore, (14-35) AR = Kc and ϕ = 0∘ (14-36) Proportional-Integral Controller A proportionalintegral (PI) controller has the transfer function, ( ( ) ) τI s + 1 = Kc (14-37) Gc (s) = Kc + τI s τI s Substituting s = jω gives ( ( ) ) j = Kc − Gc (jω) = Kc + τI jω ωτI (14-38) Thus, the amplitude ratio and phase angle are √ √ (ωτI )2 + 1 AR = |Gc (jω)| = Kc + = K c (ωτI )2 ωτI (14-39) ϕ = ∠Gc (jω) = tan−1 (−1∕ωτI ) = tan−1 (ωτI ) − 90∘ (14-40) Based on Eqs 14-39 and 14-40, at low frequencies, the integral action dominates As ω → 0, AR → ∞, and ϕ → −90∘ At high frequencies, AR = Kc and ϕ = 0∘ ; neither is a function of ω in this region (cf the proportional controller) 251 Ideal Proportional-Derivative Controller The ideal proportional-derivative (PD) controller (cf Eq 8-11) is rarely implemented in actual control systems but is a component of PID control and influences PID control at high frequency Its transfer function is Gc (s) = Kc (1 + τD s) (14-41) The frequency response characteristics are similar to those of an LHP zero: √ AR = Kc (ωτD )2 + (14-42) ϕ = tan−1 (ωτD ) (14-43) Proportional-Derivative Controller with Filter As indicated in Chapter 8, the PD controller is most often realized by the transfer function ( ) τD s + Gc (s) = Kc (14-44) ατD s + where α has a value in the range 0.05–0.2 The frequency response for this controller is given by √ (ωτD )2 + AR = Kc (14-45) (αωτD )2 + ϕ = tan−1 (ωτD ) − tan−1 (αωτD ) (14-46) The pole in Eq 14-44 bounds the high-frequency asymptote of the AR lim AR = lim |Gc (jω)| = Kc ∕α = 2∕0.1 = 20 (14-47) ω→∞ ω→∞ Note that this form actually is an advantage, because the ideal derivative action in Eq 14-41 would amplify high-frequency input noise, due to its large value of AR in that region In contrast, the PD controller with derivative filter exhibits a bounded AR in the high-frequency region Because its numerator and denominator orders are both one, the high-frequency phase angle returns to zero Parallel PID Controller The PID controller can be developed in both parallel and series forms, as discussed in Chapter Either version exhibits features of both the PI and PD controllers The simpler version is the following parallel form (cf Eq 8-14): ) ( ( ) + τI s + τI τD s2 Gc (s) = Kc + + τD s = Kc τI s τI s (14-48) Substituting s = jω and rearranging gives ( ) [ ( )] 1 Gc (jω) = Kc + + jωτD = Kc + j ωτD − jωτI ωτI (14-49) 252 Chapter 14 Frequency Response Analysis and Control System Design 102 moves the amplitude ratio curve up or down, without affecting the width of the notch Generally, the integral time τI is larger than τD , typically τI ≈ 4τD Ideal With derivative filter AR 101 100 10–3 10–2 10–1 100 101 ω (rad/min) 100 50 ϕ (deg) –50 –100 10–3 10–2 10–1 100 101 ω (rad/min) Parallel PID Controller with a Derivative Filter The parallel controller with a derivative filter was described in Chapter and Table 8.1 (14-50) Figure 14.6 shows a Bode plot for an ideal PID controller, with and without a derivative filter (see Table 8.1) The controller settings are Kc = 2, τI = 10 min, τD = min, and α = 0.1 The phase angle varies from −90∘ (ω → 0) to +90∘ (ω → ∞) A comparison of the amplitude ratios in Fig 14.6 indicates that the AR for the controller without the derivative filter in Eq 14-48 is unbounded at high frequencies, in contrast to the controller with the derivative filter (Eq 14-50), which has a bounded AR at all frequencies Consequently, the addition of the derivative filter makes the series PID controller less sensitive to high-frequency noise For the typical value of α = 0.10, Eq 14-50 yields at high frequencies: ARω→∞ = lim |Gc (jω)| = Kc ∕α = 20Kc ω→∞ This controller transfer function can be interpreted as the product of the transfer functions for PI and PD controllers Because the transfer function in Eq 14-52 is physically unrealizable and amplifies high-frequency noise, a more practical version includes a derivative filter 14.5 Figure 14.6 Bode plots of ideal parallel PID controller and ideal parallel PID controller filter (α = 0.1) ( with derivative ) Ideal parallel: Gc (s) = + + 4s 10s ( ) 4s Parallel with derivative filter: Gc (s) = + + 10s 0.4s + ) ( τD s Gc (s) = Kc + + τI s ατD s + Series PID Controller The simplest version of the series PID controller is ( ) τ1 s + (14-52) Gc (s) = Kc (τD s + 1) τ1 s (14-51) When τD = 0, the parallel PID controller with filter is the same as the PI controller of Eq 14-37 By adjusting the values of τI and τD , one can prescribe the shape and location of the notch in the AR curve Decreasing τI and increasing τD narrows the notch, whereas the opposite changes broaden it Figure 14.6 indicates that the center of the notch is located at √ ω = 1∕ τI τD where ϕ = 0∘ and AR = Kc Varying Kc NYQUIST DIAGRAMS The Nyquist diagram is an alternative representation of frequency response information, a polar plot of G(jω) in which frequency ω appears as an implicit parameter The Nyquist diagram for a transfer function G(s) can be constructed directly from |G(jω)| and ∠G(jω) for different values of ω Alternatively, the Nyquist diagram can be constructed from the Bode diagram, because AR = |G(jω)| and ϕ = ∠G(jω) The advantages of Bode plots are that frequency is plotted explicitly as the abscissa, and the log–log and semilog coordinate systems facilitate block multiplication The Nyquist diagram, on the other hand, is more compact and is sufficient for many important analyses, for example, determining system stability (see Appendix J) Most of the recent interest in Nyquist diagrams has been in connection with designing multiloop controllers and for robustness (sensitivity) studies (Maciejowski, 1989; Skogestad and Postlethwaite, 2005) For single-loop controllers, Bode plots are used more often 14.6 BODE STABILITY CRITERION The Bode stability criterion has an important advantage in comparison with the alternative of calculating the roots of the characteristic equation in Chapter 11 It provides a measure of the relative stability rather than merely a yes or no answer to the question “Is the closed-loop system stable?” Before considering the basis for the Bode stability criterion, it is useful to review the General Stability Criterion of Section 11.1: A feedback control system is stable if and only if all roots of the characteristic equation lie to the left of the imaginary axis in the complex plane Thus, the imaginary axis divides the complex plane into stable and unstable regions Recall that the characteristic equation was defined in Chapter 11 as + GOL (s) = (14-53) where the open-loop transfer function in Eq 14-53 is GOL (s) = Gc (s)Gv (s)Gp (s)Gm (s) 14.6 Before stating the Bode stability criterion, we introduce two important definitions: A critical frequency ωc is a value of ω for which ϕOL (ω) = −180∘ This frequency is also referred to as a phase crossover frequency A gain crossover frequency ωg is a value of ω for which AROL (ω) = The Bode stability criterion allows the stability of a closed-loop system to be determined from the open-loop transfer function Bode Stability Criterion Consider an open-loop transfer function GOL = Gc Gv Gp Gm that is strictly proper (more poles than zeros) and has no poles located on or to the right of the imaginary axis, with the possible exception of a single pole at the origin Assume that the open-loop frequency response has only a single critical frequency ωc and a single gain crossover frequency ωg Then the closed-loop system is stable if the open-loop amplitude ratio AROL (ωc ) < Otherwise, it is unstable The root locus diagrams of Section 11.5 (e.g., Fig 11.27) show how the roots of the characteristic equation change as controller gain Kc changes By definition, the roots of the characteristic equation are the numerical values of the complex variable, s, that satisfy Eq 14-53 Thus, each point on the root locus also satisfies Eq 14-54, which is a rearrangement of Eq 14-53: (14-54) GOL (s) = −1 The corresponding magnitude and argument are (14-55) |G (jω)| = and ∠G (jω) = −180∘ OL OL For a marginally stable system, ωc = ωg and the frequency of the sustained oscillation, ωc , is caused by a pair of roots on the imaginary axis at s = ±ωc j Substituting this expression for s into Eq 14-55 gives the following expressions for a conditionally stable system: AROL (ωc ) = |GOL (jωc )| = ϕOL (ωc ) = ∠GOL (jωc ) = −180∘ (14-56) (14-57) for some specific value of ωc > Equations 14-56 and 14-57 provide the basis for the Bode stability criterion Some of the important properties of the Bode stability criterion are It provides a necessary and sufficient condition for closed-loop stability, based on the properties of the open-loop transfer function The Bode stability criterion is applicable to systems that contain time delays The Bode stability criterion is very useful for a wide variety of process control problems However, for any GOL (s) that does not satisfy the required conditions, the Nyquist stability criterion discussed in Appendix J can be applied Bode Stability Criterion 253 10000 100 AROL 0.01 –90 ϕOL –180 (deg) –270 –360 0.001 0.01 0.1 ω (radians/time) 10 100 Figure 14.7 Bode plot exhibiting multiple critical frequencies For many control problems, there is only a single ωc and a single ωg But multiple values for ωc can occur, as shown in Fig 14.7 In this somewhat unusual situation, the closed-loop system is stable for two different ranges of the controller gain (Luyben and Luyben, 1997) Consequently, increasing the absolute value of Kc can actually improve the stability of the closed-loop system for certain ranges of Kc For systems with multiple ωc or ωg , the Bode stability criterion has been modified by Hahn et al (2001) to provide a sufficient condition for stability As indicated in Chapter 11, when the closed-loop system is marginally stable, the closed-loop response exhibits a sustained oscillation after a set-point change or a disturbance Thus, the amplitude neither increases nor decreases In order to gain physical insight into why a sustained oscillation occurs at the stability limit, consider the analogy of an adult pushing a child on a swing The child swings in the same arc as long as the adult pushes at the right time and with the right amount of force Thus the desired sustained oscillation places requirements on both timing (i.e., phase) and applied force (i.e., amplitude) By contrast, if either the force or the timing is not correct, the desired swinging motion ceases, as the child will quickly protest A similar requirement occurs when a person bounces a ball To further illustrate why feedback control can produce sustained oscillations, consider the following thought experiment for the feedback control system shown in Fig 14.8 Assume that the open-loop system is stable and that no disturbances occur (D = 0) Suppose that the set-point is varied sinusoidally at the critical frequency, ysp (t) = A sin (ωc t), for a long period of time Assume that during this period, the measured output, ym , is disconnected, so that the feedback loop is broken before the comparator After the initial transient dies out, ym will oscillate at the excitation frequency ωc , because the response of a linear system to a sinusoidal input is a sinusoidal output at the same 254 Chapter 14 Frequency Response Analysis and Control System Design Figure 14.8 Sustained oscillation in a feedback control system Gd D=0 Yd Ysp Km Ysp + E – P Gc Gv Ym frequency (see Section 14.2) Suppose the two events occur simultaneously: (i) the set-point is set to zero, and (ii) ym is reconnected If the feedback control system is marginally stable, the controlled variable y will then exhibit a sustained sinusoidal oscillation with amplitude A and frequency ωc To analyze why this special type of oscillation occurs only when ω = ωc , note that the sinusoidal signal E in Fig 14.8 passes through transfer functions Gc , Gv , Gp , and Gm before returning to the comparator In order to have a sustained oscillation after the feedback loop is reconnected, signal Ym must have the same amplitude as E and a 180∘ phase shift relative to E Note that the comparator also provides a −180∘ phase shift because of its negative sign Consequently, after Ym passes through the comparator, it is in phase with E and has the same amplitude, A Thus, the closed-loop system oscillates indefinitely after the feedback loop is closed because the conditions in Eqs 14-56 and 14-57 are both satisfied But what happens if Kc is increased by a small amount? Then, AROL (ωc ) is greater than one, the oscillations grow, and the closed-loop system becomes unstable In contrast, if Kc is reduced by a small amount, the oscillation is damped and eventually dies out EXAMPLE 14.4 A process has the third-order transfer function (time constant in minutes), Gp (s) = (0.5s + 1)3 Also, Gv = 0.1 and Gm = 10 For a proportional controller, evaluate the stability of the closed-loop control system using the Bode stability criterion and three values of Kc : 1, 4, and 20 SOLUTION For this example, GOL = Gc Gv Gp Gm = (Kc )(0.1) 2Kc (10) = (0.5s + 1)3 (0.5s + 1)3 U Yu Gp + + Y Gm 100 Kc = 20 10 AROL Kc = Kc = 0.1 0.01 ϕOL –90 (deg) –180 –270 0.01 0.1 ωc ω (rad/min) 10 100 Figure 14.9 Bode plots for GOL = 2Kc /(0.5s + 1)3 Figure 14.9 shows a Bode plot of GOL for three values of Kc Note that all three cases have the same phase angle plot, because the phase lag of a proportional controller is zero for Kc > From the phase angle plot, we observe that ωc = 3.46 rad/min This is the frequency of the sustained oscillation that occurs at the stability limit, as discussed previously Next, we consider the amplitude ratio AROL for each value of Kc Based on Fig 14.9, we make the following classifications: Kc AROL (for ω = ωc ) 20 0.25 Classification Stable Marginally stable Unstable In Section 12.5.1, the concept of the ultimate gain was introduced For proportional-only control, the ultimate gain Kcu was defined to be the largest value of Kc that results in a stable closed-loop system The value of Kcu can be determined graphically from a Bode plot for transfer function G = Gv Gp Gm For proportional-only control, GOL = Kc G Because a proportional controller has zero phase lag, ωc is determined solely by G Also, AROL (ω) = Kc ARG (ω) (14-58) 14.6 where ARG denotes the amplitude ratio of G At the stability limit, ω = ωc , AROL (ωc ) = and Kc = Kcu Substituting these expressions into Eq 14-58 and solving for Kcu gives an important result: Kcu = ARG (ωc ) Bode Stability Criterion 100 b 10 AR (14-59) 255 a c 0.1 0.01 90 The stability limit for Kc can also be calculated for PI and PID controllers and is denoted by Kcm , as demonstrated by Example 14.5 ϕ (deg) a –90 c b –180 EXAMPLE 14.5 Consider PI control of an overdamped second-order process (time constants in minutes), Gp (s) = (s + 1)(0.5s + 1) Gm = Gv = (a) Determine the value of Kcu (b) Use a Bode plot to show that controller settings of Kc = 0.4 and τI = 0.2 produce an unstable closed-loop system (c) Find Kcm , the maximum value of Kc that can be used with τI = 0.2 and still have closed-loop stability (d) Show that τI = results in a stable closed-loop system for all positive values of Kc SOLUTION (a) In order to determine Kcu , we set Gc = Kc The open-loop transfer function is GOL = Kc G where G = Gv Gp Gm Because a proportional controller does not introduce any phase lag, G and GOL have identical phase angles (b) Consequently, the critical frequency can be determined graphically from the phase angle plot for G However, curve a in Fig 14.10 indicates that ωc does not exist for proportional control, because ϕOL is always greater than −180∘ As a result, Kcu does not exist, and thus Kc does not have a stability limit Conversely, the addition of integral control action can produce closed-loop instability Curve b in Fig 14.10 indicates that an unstable closed-loop system occurs for Gc (s) = 0.4(1 + 1/0.2s), because AROL > when ϕOL = −180∘ (c) To find Kcm for τI = 0.2 min, we note that ωc depends on τI but not on Kc , because Kc has no effect on ϕOL For curve b in Fig 14.10, ωc = 2.2 rad/min, and the corresponding amplitude ratio is AROL = 1.38 To find Kcm , multiply the current value of Kc by a factor, 1/1.38 Thus, Kcm = 0.4/1.38 = 0.29 (d) When τI is increased to min, curve c in Fig 14.10 results Because curve c does not have a critical frequency, the closed-loop system is stable for all positive values of Kc –270 0.01 ωc 0.1 10 100 ω (rad/min) Figure 14.10 Bode plots for Example 14.5 Curve a: G(s) ( ) Curve b: GOL (s): Gc (s) = 0.4 + 0.2s ( ) Curve c: GOL (s): Gc (s) = 0.4 + s EXAMPLE 14.6 Find the critical frequency for the following process and PID controller, assuming Gv = Gm = ( ) e−0.3s Gp (s) = Gc (s) = 20 + +s (9s + 1)(11s + 1) 2.5s SOLUTION Figure 14.7 shows the open-loop amplitude ratio and phase angle plots for GOL Note that the phase angle crosses −180∘ at three points Because there is more than one value of ωc , the Bode stability criterion cannot be applied EXAMPLE 14.7 Evaluate the stability of the closed-loop system for: 4e−s 5s + The time constant and time delay have units of minutes and, Gv = 2, Gm = 0.25, Gc = Kc Gp (s) = Obtain ωc and Kcu from a Bode plot Use an initial value of Kc = SOLUTION The Bode plot for GOL and Kc = is shown in Fig 14.11 For ωc = 1.69 rad/min, ϕOL = −180∘ , and AROL = 0.235 For Kc = 1, AROL = ARG and Kcu can be calculated from 256 Chapter 14 Frequency Response Analysis and Control System Design Eq 14-59 Thus, Kcu = 1/0.235 = 4.25 Setting Kc = 1.5 Kcu gives Kc = 6.38 A larger value of Kc causes the closed-loop system to become unstable Only values of Kc less than Kcu result in a stable closed-loop system AROL ARc = 1 GM ωg 100 AROL ω 0.235 0.1 ϕOL 0.01 90 ϕOL ωc ϕg Phase margin (deg) –180 –90 ωg (deg) –180 ω ωc –270 –360 0.01 Figure 14.12 Gain and phase margins on a Bode plot 0.1 10 100 ωc = 1.69 rad/min ω (rad/min) Figure 14.11 Bode plot for Example 14.7, Kc = 14.7 GAIN AND PHASE MARGINS Rarely does the model of a chemical process stay unchanged for a variety of operating conditions and disturbances When the process changes or the controller is poorly tuned, the closed-loop system can become unstable Thus, it is useful to have quantitative measures of relative stability that indicate how close the system is to becoming unstable The concepts of gain margin (GM) and phase margin (PM) provide useful metrics for relative stability Let ARc be the value of the open-loop amplitude ratio at the critical frequency ωc Gain margin GM is defined as: (14-60) GM ≜ ARc According to the Bode stability criterion, ARc must be less than one for closed-loop stability An equivalent stability requirement is that GM > The gain margin provides a measure of relative stability, because it indicates how much any gain in the feedback loop component can increase before instability occurs For example, if GM = 2.1, either process gain Kp or controller gain Kc could be doubled, and the closed-loop system would still be stable, although probably very oscillatory Next, we consider the phase margin In Fig 14.12, ϕg denotes the phase angle at the gain-crossover frequency ωg where AROL = Phase margin PM is defined as PM ≜ 180 + ϕg (14-61) The phase margin also provides a measure of relative stability In particular, it indicates how much additional time delay can be included in the feedback loop before instability will occur Denote the additional time delay as Δθmax For a time delay of Δθmax , the phase angle is −Δθmax ω (see Section 14.3.5) Thus, Δθmax can be calculated from the following expression, ( ) 180∘ PM = Δθmax ωg (14-62) π or ( )( ) PM π (14-63) Δθmax = ∘ ωg 180 where the (π/180∘ ) factor converts PM from degrees to radians Graphical representations of the gain and phase margins in a Bode plot are shown in Fig 14.12 The specification of phase and gain margins requires a compromise between performance and robustness In general, large values of GM and PM correspond to sluggish closed-loop responses, whereas smaller values result in less sluggish, more oscillatory responses The choices for GM and PM should also reflect model accuracy and the expected process variability Guideline In general, a well-tuned controller should have a gain margin between 1.7 and 4.0 and a phase margin between 30∘ and 45∘ Recognize that these ranges are approximate and that it may not be possible to choose PI or PID controller settings that result in specified GM and PM values Tan et al (1999) have developed graphical procedures for designing PI and PID controllers that satisfy GM and PM specifications The GM and PM concepts are easily evaluated when the open-loop system does not have multiple values of ωc or ωg However, for systems with multiple ωg , gain margins can be determined from Nyquist plots (Doyle et al., 2009) 14.7 Gain and Phase Margins 257 EXAMPLE 14.8 For the FOPTD model of Example 14.7, calculate the PID controller settings for the following approaches: (a) IMC (Table 12.1 with τc = 1) (b) Continuous Cycling: Use the Tyreus–Luyben tuning relations (Luyben and Luyben, 1997), which are Kc = 0.45 Kcu ; τI = 2.2 Pu ; τD = Pu /6.3 Assume that the two PID controllers are implemented in the parallel form with a derivative filter (α = 0.1) in Table 8.1 Plot the open-loop Bode diagram and determine the gain and phase margins for each controller For the Tyreus–Luyben settings, determine the maximum increase in the time delay Δθmax that can occur while still maintaining closed-loop stability SOLUTION GOL = Gc Gv Gp Gm = Gc Controller settings Kc τI (min) τD (min) IMC Tyreus–Luyben 1.83 1.91 5.5 8.2 0.45 0.59 The open-loop transfer function are ( ) ( −s ) 0.45s 2e GOL, IMC = 1.83 + + 5.5s 0.045s + 5s + ( ) ( −s ) 0.59s 2e GOL, T−L = 1.91 + + 8.18s 0.059s + 5s + Figure 14.13 shows the frequency response of GOL for the two controllers The gain and phase margins can be determined by inspection of the Bode diagram or by using the MATLAB command margin 2e−s 5s + (a) IMC tuning: Based on Table 12.1, (line H) for τc = 1, we have θ τ+ + 0.5 = 1.83; Kc = ( ) = θ 2(1 + 0.5) K τc + θ τI = τ + = 5.5 min; τθ τD = = = 0.45 2τ + θ 10 + (b) Tyreus–Luyben: From Example 14.7, the ultimate gain is Kcu = 4.25, and 2π = 3.72 Therefore, the ultimate period is Pu = 1.69 the PID controller settings are Controller IMC Tyreus–Luyben GM PM ωc (rad/min) 2.2 1.8 68.5∘ 76∘ 2.38 2.51 Based on Fig 14.13, the Tyreus–Luyben controller settings are close to the IMC tuning result The value of Δθmax is calculated from Eq 14-63, and the information in the preceding table: (76∘ )(π rad) = 1.7 Δθmax = (0.79 rad∕min)(180∘ ) Thus, time delay θ can increase by as much as 70% and still maintain closed-loop stability IMC Tyreus–Luyben 102 AR 100 10–2 10–1 100 101 102 101 102 ω (rad/min) (a) –200 ϕ –400 (deg) –600 –800 –2 10 IMC Tyreus–Luyben 10–1 100 ω (rad/min) (b) Figure 14.13 Comparison of GOL Bode plots for Example 14.8 258 Chapter 14 Frequency Response Analysis and Control System Design SUMMARY Frequency response techniques are powerful tools for the design and analysis of feedback control systems The frequency response characteristics of a process, its amplitude ratio AR and phase angle, characterize the dynamic behavior of the process and can be plotted as functions of frequency in Bode diagrams The Bode stability criterion provides exact stability results for a wide variety of control problems, including processes with time delays It also provides a convenient measure of relative stability, such as gain and phase margins Control system design involves trade-offs between control system performance and robustness (see Appendix J) Modern control systems are typically designed using a model-based technique, such as those described in Chapter 12 REFERENCES Doyle, J C., B A Francis, and A R Tannenbaum, Feedback Control Theory, Macmillan, New York, 2009 Franklin, G F., J D Powell, and A Emami-Naeini, Feedback Control of Dynamic Systems, 7th ed., Prentice Hall, Upper Saddle River, NJ, 2014 Hahn, J., T Edison, and T F Edgar, A Note on Stability Analysis Using Bode Plots, Chem Eng Educ 35(3), 208 (2001) Luyben, W L., and M L Luyben, Essentials of Process Control, McGraw-Hill, New York, 1997, Chapter 11 MacFarlane, A G J., The Development of Frequency Response Methods in Automatic Control, IEEE Trans Auto Control, AC-24, 250 (1979) Maciejowski, J M., Multivariable Feedback Design, Addison-Wesley, New York, 1989 Ogunnaike, B A., and W H Ray, Process Dynamics, Modeling, and Control, Oxford University Press, New York, 1993 Seborg, D E., T F Edgar, and D A., Mellichamp, Process Dynamics and Cantrol, 2nd ed., John Wiley and Sons, Hoboken, NJ, 2004 Skogestad, S., and I Postlethwaite, Multivariable Feedback Design: Analysis and Design, 2d ed., John Wiley and Sons, Hoboken, NJ, 2005 Tan, K K., Q.-G Wang, C C Hang, and T Hägglund, Advances in PID Control, Springer, New York, 1999 EXERCISES 14.1 A heat transfer process has the following transfer function between a temperature T (in ∘ C) and an inlet flow rate q where the time constants have units of minutes: 3(1 − s) T ′ (s) = Q′ (s) s(2s + 1) If the flow rate varies sinusoidally with an amplitude of L/min and a period of 0.5 min, what is the amplitude of the temperature signal after the transients have died out? 14.2 Using frequency response arguments, discuss how well e−θs can be approximated by a two-term Taylor series expansion, − θs Compare your results with those given in Section 6.2.1 for a 1/1 Padé approximation 14.3 A data acquisition system for environmental monitoring is used to record the temperature of an air stream as measured by a thermocouple It shows an essentially sinusoidal variation after about 15 s The maximum recorded temperature is 128 ∘ F, and the minimum is 120 ∘ F at 1.8 cycles per It is estimated that the thermocouple has a time constant of s Estimate the actual maximum and minimum air temperatures 14.4 A perfectly stirred tank is used to heat a flowing liquid The dynamic model is shown in Fig E14.4 0.1s Figure E14.4 where: P is the power applied to the heater Q is the heating rate of the system T is the actual temperature in the tank Tm is the measured temperature time constants have units of A test has been made with P′ varied sinusoidally as P′ = 0.5 sin 0.2t For these conditions, the measured temperature is Tm′ = 3.464 sin(0.2t + ϕ) Find a value for the maximum error bound between T ′ and Tm′ if the sinusoidal input has been applied for a long time 14.5 Determine if the following processes can be made unstable by increasing the gain of a proportional controller Kc to a sufficiently large value using frequency response arguments: (a) Gv Gp Gm = s+1 (b) Gv Gp Gm = (s + 1)(2s + 1) (c) Gv Gp Gm = (s + 1)(2s + 1)(3s + 1) 5e−s (d) Gv Gp Gm = 2s + 14.6 Two engineers are analyzing step-test data from a bioreactor Engineer A says that the data indicate a second-order overdamped process, with time constants of and but no time delay Engineer B insists that the best fit is a FOPTD model, with τ = and θ = Both engineers claim a proportional controller can be set at a large value for Kc to Exercises control the process and that stability is no problem Based on their models, who is right, who is wrong, and why? Use a frequency-response argument 14.7 Plot the Bode diagram (0.1 ≤ ω ≤ 100) of the third-order transfer function, (10s + 1)(2s + 1)(s + 1) Find both the value of ω that yields a −180∘ phase angle and the value of AR at that frequency G(s) = Using MATLAB, plot the Bode diagram of the following transfer function: 6(s + 1)e−2s G(s) = (4s + 1)(2s + 1) Repeat for the situation where the time-delay term is replaced by a 1/1 Padé approximation Discuss how the accuracy of the Padé approximation varies with frequency 14.8 14.9 Two thermocouples, one of them a known standard, are placed in an air stream whose temperature is varying sinusoidally The temperature responses of the two thermocouples are recorded at a number of frequencies, with the phase angle between the two measured temperatures as shown below The standard is known to have first-order dynamics and a time constant of 0.15 when operating in the air stream From the data, show that the unknown thermocouple also is a first order and determine its time constant 259 14.12 Develop expressions for the amplitude ratio as a function of ω of each of the two forms of the PID controller: (a) The parallel controller of Eq 8-13 (b) The series controller of Eq 8-15 Plot AR/Kc vs ωτD for each AR curve Assume τ1 = 4τD and α = 0.1 For what region(s) of ω are the differences significant? 14.13 You are using proportional control (Gc = Kc ) for a pro4 0.6 and Gp = (time constants in s) cess with Gv = 2s + 50s + You have a choice of two measurements, both of which exhibit 2 or Gm2 = first-order dynamic behavior, Gm1 = s+1 0.4s + Can Gc be made unstable for either process? Which measurement is preferred for the best stability and performance properties? Why? Frequency (cycles/min) Phase Difference (deg) 14.14 For the following statements, discuss whether they are always true, sometimes true, always false, or sometimes false Cite evidence from this chapter (a) Increasing the controller gain speeds up the response for a set-point change (b) Increasing the controller gain always causes oscillation in the response to a setpoint change (c) Increasing the controller gain too much can cause instability in the control system (d) Selecting a large controller gain is a good idea in order to minimize offset 0.05 0.1 0.2 0.4 0.8 1.0 2.0 4.0 4.5 8.7 16.0 24.5 26.5 25.0 16.7 9.2 14.15 Use arguments based on the phase angle in frequency response to determine if the following combinations of G = Gv Gp Gm and Gc become unstable for some value of Kc Gc = Kc (a) G = (4s + 1)(2s + 1) ) ( 1 (b) G = Gc = Kc + (4s + 1)(2s + 1) 5s 14.10 Exercise 5.19 considered whether a two-tank liquid surge system provided better damping of step disturbances than a single-tank system with the same total volume Reconsider this situation, this time with respect to sinusoidal disturbances; that is, determine which system better damps sinusoidal inputs of frequency ω Does your answer depend on the value of ω? 14.11 A process has the transfer function of Eq 6-14 with K = 2, τ1 = 10, τ2 = If τa has the following values, Case i: τa = 20 Case ii: τa = Case iii: τa = Case iv: τa = −2 Plot the composite amplitude ratio and phase angle curves on a single Bode plot for each of the four cases of numerator dynamics What can you conclude concerning the importance of the zero location for the amplitude and phase characteristics of this second-order system? (c) G = s+1 (4s + 1)(2s + 1) Gc = Kc (d) G = 1−s (4s + 1)(2s + 1) Gc = Kc (e) G = e−s 4s + (2s + 1) s Gc = Kc 14.16 Plot the Bode diagram for a composite transfer function consisting of G(s) in Exercise 14.8 multiplied by that of a parallel-form PID controller with Kc = 0.21, τI = 5, and τD = 0.42 Repeat for a series PID controller with filter that employs the same settings How different are these two diagrams? In particular, by how much the two amplitude ratios differ when ω = ωc ? 14.17 For the process described by the transfer function 12 G(s) = (8s + 1)(2s + 1)(0.4s + 1)(0.1s + 1) 260 Chapter 14 Frequency Response Analysis and Control System Design (a) Find second-order-plus-time-delay models that approximate G(s) and are of the form ̂ G(s) = Ke−θs (τ1 s + 1)(τ2 s + 1) One of the approximate models can be found by using the method discussed in Section 6.3; the other, by using a method from Chapter (b) Compare all three models (exact and approximate) in the frequency domain and a FOPTD model 14.18 Obtain Bode plots for both the transfer function: G(s) = 10(2s + 1)e−2s (20s + 1)(4s + 1)(s + 1) and a FOPTD approximation obtained using the method discussed in Section 6.3 What you conclude about the accuracy of the approximation relative to the original transfer function? (a) Using the process, sensor, and valve transfer functions in Exercise 11.21, find the ultimate controller gain Kcu using a Bode plot Using simulation, verify that values of Kc > Kcu cause instability (b) Next fit a FOPTD model to G and tune a PI controller for a set-point change What is the gain margin for the controller? 14.19 14.20 A process that can be modeled as a time delay (gain = 1) is controlled using a proportional feedback controller The control valve and measurement device have negligible dynamics and steady-state gains of Kv = 0.5 and Km = 1, respectively After a small set-point change is made, a sustained oscillation occurs, which has a period of 10 (a) What controller gain is being used? Explain (b) How large is the time delay? 14.21 The block diagram of a conventional feedback control system contains the following transfer functions: ) ( Gv = Gc = Kc + 5s Gm = s+1 5e−2s Gp = Gd = 10s + (a) (b) (c) (d) Plot the Bode diagram for the open-loop transfer function For what values of Kc is the system stable? If Kc = 0.2, what is the phase margin? What value of Kc will result in a gain margin of 1.7? 14.22 Consider the storage tank with sightglass in Fig E14.22 The parameter values are R1 = 0.5 min/ft2 , R2 = min/ft2 , A1 = 10 ft2 , Kv = 2.5 cfm/mA, A2 = 0.8 ft2 , Km = 1.5 mA/ft, and τm = 0.5 (a) Suppose that R2 is decreased to 0.5 min/ft2 Compare the old and new values of the ultimate gain and the critical frequency Would you expect the control system performance to become better or worse? Justify your answer (b) If PI controller settings are calculated using the ZieglerNichols rules, what are the gain and phase margins? Assume R2 = min/ft Figure E14.22 14.23 A process (including valve and sensor-transmitter) has the approximate transfer function, G(s) = 2e−0.2s /(s + 1) with time constant and time delay in minutes Determine PI controller settings and the corresponding gain margins by two methods: (a) Direct synthesis (τc = 0.3 min) (b) Phase margin = 40∘ (assume τI = 0.5 min) (c) Simulate these two control systems for a unit step change in set point Which controller provides the better performance? 14.24 Consider the feedback control system in Fig 14.8, and the following transfer functions: ) ( 2s + Gv = Gc = Kc 0.1s + 0.5s + 0.4 Gp = Gd = s(5s + 1) 5s + Gm = (a) Plot a Bode diagram for the open-loop transfer function (b) Calculate the value of Kc that provides a phase margin of 30∘ (c) What is the gain margin when Kc = 10? 14.25 Hot and cold liquids are mixed at the junction of two pipes The temperature of the resulting mixture is to be controlled using a control valve on the hot stream The dynamics of the mixing process, control valve, and temperature sensor/transmitter are negligible and the sensortransmitter gain is mA/mA Because the temperature sensor is located well downstream of the junction, an s time delay occurs There are no heat losses/gains for the downstream pipe (a) Draw a block diagram for the closed-loop system (b) Determine the Ziegler–Nichols settings (continuous cycling method) for both PI and PID controllers (c) For each controller, simulate the closed-loop responses for a unit step change in set point (d) Does the addition of derivative control action provide a significant improvement? Justify your answer 14.26 For the process in Exercise 14.23, the measurement is to be filtered using a noise filter with transfer function GF (s) = 1/(0.1s + 1) Would you expect this change to result in better or worse control system performance? Compare the ultimate gains and critical frequencies with and without the filter Justify your answer Exercises 14.27 The dynamic behavior of the heat exchanger shown in Fig E14.27 can be described by the following transfer functions (H S Wilson and L M Zoss, ISA J., 9, 59 (1962)): Process: 261 ∘ F∕lb∕min T′ = ′ Ws (0.432s + 1)(0.017s + 1) Control valve: 0.047 in∕psi X′ = P′ 0.083s + Ws′ lb = 112 X′ in Temperature sensor-transmitter: 0.12 psi∕∘ F P′ = ′ T 0.024s + The valve lift x is measured in inches Other symbols are defined in Fig E14.27 (a) Find the Ziegler–Nichols settings for a PI controller (b) Calculate the corresponding gain and phase margins 14.28 Consider the control problem of Exercise 14.28 and a PI controller with Kc = and τI = 0.3 (a) Plot the Bode diagram for the open-loop system (b) Determine the gain margin from the Bode plot Figure E14.27 Chapter 15 Feedforward and Ratio Control CHAPTER CONTENTS 15.1 Introduction to Feedforward Control 15.2 Ratio Control 15.3 Feedforward Controller Design Based on Steady-State Models 15.3.1 Blending System 15.4 Feedforward Controller Design Based on Dynamic Models 15.5 The Relationship Between the Steady-State and Dynamic Design Methods 15.5.1 Steady-State Controller Design Based on Transfer Function Models 15.6 Configurations for Feedforward–Feedback Control 15.7 Tuning Feedforward Controllers Summary In Chapter it was emphasized that feedback control is an important technique that is widely used in the process industries Its main advantages are It does not provide predictive control action to compensate for the effects of known or measurable disturbances Corrective action occurs as soon as the controlled variable deviates from the set point, regardless of the source and type of disturbance It may not be satisfactory for processes with large time constants and/or long time delays If large and frequent disturbances occur, the process may operate continuously in a transient state and never attain the desired steady state In some situations, the controlled variable cannot be measured on-line, so feedback control is not feasible Feedback control requires minimal knowledge about the process to be controlled; in particular, a mathematical model of the process is not required, although it can be very useful for control system design The ubiquitous PID controller is both versatile and robust If process conditions change, re-tuning the controller usually produces satisfactory control However, feedback control also has certain inherent disadvantages: No corrective action is taken until after a deviation in the controlled variable occurs Thus, perfect control, where the controlled variable does not deviate from the set point during disturbance or set-point changes, is theoretically impossible 262 For situations in which feedback control by itself is not satisfactory, significant improvement can be achieved by adding feedforward control But feedforward control requires that the disturbances be measured (or estimated) on-line In this chapter, we consider the design and analysis of feedforward control systems We begin with an overview of feedforward control Then ratio control, a special type of feedforward control, is introduced Next, design techniques for feedforward controllers are developed based on either steady-state or dynamic 15.1 models Then alternative configurations for combined feedforward–feedback control systems are considered This chapter concludes with a section on tuning feedforward controllers 15.1 Introduction to Feedforward Control 263 LC LT INTRODUCTION TO FEEDFORWARD CONTROL The basic concept of feedforward control is to measure important disturbance variables and take corrective action before they upset the process In contrast, a feedback controller does not take corrective action until after the disturbance has upset the process and generated a nonzero error signal Simplified block diagrams for feedforward and feedback control are shown in Fig 15.1 Feedforward control has several disadvantages: The disturbance variables must be measured on-line In many applications, this is not feasible To make effective use of feedforward control, at least an approximate process model should be available In particular, we need to know how the controlled variable responds to changes in both the disturbance variable and the manipulated variable The quality of feedforward control depends on the accuracy of the process model Ideal feedforward controllers that are theoretically capable of achieving perfect control may not be physically realizable Fortunately, practical approximations of these ideal controllers can provide very effective control Feedforward control was not widely used in the process industries until the 1960s (Shinskey, 1996) Since D Ysp Feedforward controller U Process Y D Ysp Feedback controller U Process Y Figure 15.1 Simplified block diagrams for feedforward and feedback control Steam Boiler drum Feedwater Hot gas Figure 15.2 Feedback control of the liquid level in a boiler drum then, it has been applied to a wide variety of industrial processes However, the basic concept is much older and was applied as early as 1925 in the three-element level control system for boiler drums We will use this control application to illustrate the use of feedforward control A boiler drum with a conventional feedback control system is shown in Fig 15.2 The level of the boiling liquid is measured and used to adjust the feedwater flow rate This control system tends to be quite sensitive to rapid changes in the disturbance variable, steam flow rate, as a result of the small liquid capacity of the boiler drum Rapid disturbance changes are produced by steam demands made by downstream processing units Another difficulty is that large controller gains cannot be used because level measurements exhibit rapid fluctuations for boiling liquids Thus a high controller gain would tend to amplify the measurement noise and produce unacceptable variations in the feedwater flow rate The feedforward control scheme in Fig 15.3 can provide better control of the liquid level The steam flow rate is measured, and the feedforward controller adjusts the feedwater flow rate so as to balance the steam demand Note that the controlled variable, liquid level, is not measured As an alternative, steam pressure could be measured instead of steam flow rate Feedforward control can also be used advantageously for level control problems where the objective is surge control (or averaging control), rather than tight level control For example, the input streams to a surge tank will be intermittent if they are effluent streams from batch operations, but the tank exit stream can be continuous Special feedforward control methods have been developed for these batch-to-continuous transitions to balance the surge capacity requirement ... +j 2 = R + jI ? ?2 ? ?2 + ω τ +1 = where (14-11) R= ? ?2 ? ?2 + (14-12a) I= −ωτ ? ?2 ? ?2 + (14-12b) and From Eq 14-7, √ AR = |G(jω|) = ( )2 ? ?2 ? ?2 +1 ( + −ωτ +1 )2 ? ?2 ? ?2 Simplifying, √ AR = (1 + ? ?2 ? ?2 )... G2 (jω) = jω? ?2 + The magnitudes and angles of each component of the complex transfer function are |Ga | = K √ |G1 | = √? ?2 ? ?21 + |G2 | = ? ?2 ? ?22 + ∠Ga = ∠G1 = tan−1 (ωτ1 ) ∠G2 = tan−1 (ω? ?2 ) 24 7... (14 -22 ) Substituting s = jω and rearranging into real and imaginary parts (see Example 14.1) yields K AR = √ (14 -23 a) 2 (1 − ω τ )2 + (2? ?ωτ )2 [ ] ? ?2? ?ωτ (14 -23 b) ϕ = tan−1 − ? ?2 ? ?2 Note that, in evaluating