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 14.1 First-Order Process 14.3 Second-Order Process 14.4 Frequency Response Characteristics of Feedback Controllers 14.6 Bode Stability Criterion 14.7 Gain and Phase Margins Summary In previous chapters, Laplace transform techniques analysis. Next we introduce two useful metrics for rel- were used to calculate transient responses from trans- ative stability, namely gain and phase margins. These fer functions. This chapter focuses on an alternative metrics indicate how close a control system is to insta- way to analyze dynamic systems by using frequency bility.
A related issue is robustness, that is, the sensitivity response analysis. Frequency response concepts and of control system performance to process variations and techniques play an important role in stability analy- to uncertainty in the process model. sis, control system design, and robustness assessment. The design of robust feedback control systems is con- Historically, frequency response techniques provided sidered in Appendix J.
the conceptual framework for early control theory and important applications in the field of communications 14.1 SINUSOIDAL FORCING OF (MacFarlane, 1979). A FIRST-ORDER PROCESS We introduce a simplified procedure to calculate the frequency response characteristics from the transfer We start with the response properties of a first-order function model of any linear process. Two concepts, process when forced by a sinusoidal input and show the Bode and Nyquist stability criteria, are generally how the output response characteristics depend on applicable for feedback control systems and stability the frequency of the input signal. This is the origin of 244 14.1 Sinusoidal Forcing of a First-Order Process 245 the term frequency response.
The responses for first- process example discussed earlier. In Chapter 4, the and second-order processes forced by a sinusoidal input transfer function model for the stirred-tank blending were presented in Chapter 5. Recall that these responses system was derived as consisted of sine, cosine, and exponential terms. Specifi- K1 K2 K3 cally, for a first-order transfer function with gain K and X ′ (s) = X ′ (s) + W ′ (s) + W ′ (s) (4-84) τs + 1 1 τs + 1 2 τs + 1 1 time constant τ, the response to a general sinusoidal input, x(t) = A sin ωt, is Suppose flow rate w2 is varied sinusoidally about KA a constant value, while the other inlet conditions y(t) = (ωτe−t∕τ − ωτ cos ωt + sin ωt) (5-23) are kept constant at their nominal values; that is, ω2 τ2 + 1 w′1 (t) = x′1 (t) = 0.
Because w2 (t) is sinusoidal, the output where y is in deviation form. composition deviation x′ (t) eventually becomes sinu- If the sinusoidal input is continued for a long time, soidal according to Eq. However, there is a phase the exponential term (ωτe−t/τ ) becomes negligible. The shift in the output relative to the input, as shown in remaining sine and cosine terms can be combined via a Fig.1, owing to the material holdup of the tank.
If trigonometric identity to yield the flow rate w2 oscillates very slowly relative to the KA residence time τ(ω ≪ 1/τ), the phase shift is very small, y𝓁 (t) = √ sin (ωt + ϕ) (14-1) ω2 τ2 + 1 approaching 0∘ , whereas the normalized amplitude ̂ ratio (A/KA) is very nearly unity. For the case of a where ϕ = −tan−1 (ωτ). The long-time response y𝓁 (t) is called the frequency response of the first-order system low-frequency input, the output is in phase with the and has two distinctive features (see Fig. input, tracking the sinusoidal input as if the process model were G(s) = K.
The output signal is a sine wave that has the same On the other hand, suppose that the flow rate is frequency, but its phase is shifted relative to the varied rapidly by increasing the input signal frequency. input sine wave by the angle ϕ (referred to as the For ω ≫ 1/τ, Eq. 14-1 indicates that the phase shift phase shift or the phase angle); the amount of phase approaches a value of −π/2 radians (−90∘ ). The pres- shift depends on the forcing frequency ω.
ence of the negative sign indicates that the output lags 2. The sine wave has an amplitude  that is a function behind the input by 90∘ ; in other words, the phase lag of the forcing frequency: is 90∘. The amplitude ratio approaches zero as the fre- KA quency becomes large, indicating that the input signal  = √ (14-2) is almost completely attenuated; namely, the sinusoidal ω2 τ2 + 1 deviation in the output signal is very small. Dividing both sides of Eq.
14-2 by the input signal These results indicate that positive and negative devi- amplitude A yields the amplitude ratio (AR) ations in w2 are essentially canceled by the capacitance  K of the liquid in the blending tank if the frequency is high AR = =√ (14-3a) enough. High frequency implies ω ≫ 1/τ. Most pro- A ω2 τ2 + 1 cesses behave qualitatively similar to the stirred-tank which can, in turn, be divided by the process gain blending system, when subjected to a sinusoidal input. to yield the normalized amplitude ratio (ARN ): For high-frequency input changes, the process output AR 1 deviations are so completely attenuated that the cor- ARN = =√ (14-3b) K ω τ2 + 1 2 responding periodic variation in the output is difficult (perhaps impossible) to detect or measure.
Next we examine the physical significance of the pre- Input–output phase shift and attenuation (or amplifi- ceding equations, with specific reference to the blending cation) occur for any stable transfer function, regardless of its complexity. In all cases, the phase shift and P amplitude ratio are related to the frequency ω of the sinusoidal input signal. In developments up to this A A 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 Output, y Time be obtained experimentally.
By performing a series of shift, Δt Input, x tests in which a sinusoidal input is applied to the pro- Time, t cess, the resulting amplitude ratio and phase shift can Figure 14.1 Attenuation and time shift between input and be measured for different frequencies. In this case, the output sine waves. The phase angle ϕ of the output signal is frequency response is expressed as a table of measured given by ϕ = Δt/P × 360∘, where Δt is the time shift and P is amplitude ratios and phase shifts for selected values the period of oscillation. However, the method is very time-consuming 246 Chapter 14 Frequency Response Analysis and Control System Design because of the repeated experiments for different val- The shortcut method can be summarized as follows: ues of ω.
Thus other methods, such as pulse testing (Ogunnaike and Ray, 1994), are utilized, because only a Step 1. single test is required., express G(jω) as the In this chapter, the focus is on developing a powerful sum of real (R) and imaginary (I) parts R + jI, analytical method to calculate the frequency response where R and I are functions of ω, using com- for any stable process transfer function. Later in this plex conjugate multiplication. chapter, we show how this information can be used to Step 3.
The √ output sine wave has amplitude design controllers and analyze the properties of the ̂ −1 A = A R2 + I 2 and phase angle√ϕ = tan (I/R). closed loop system responses. 2 2 The amplitude ratio is AR = R + I and is independent of the value of A.2 SINUSOIDAL FORCING OF AN nTH-ORDER PROCESS This section presents a general approach for deriving the EXAMPLE 14.1 frequency response of any stable transfer function. The Find the frequency response of a first-order system, with physical interpretation of frequency response is not valid for unstable systems, because a sinusoidal input 1 G(s) = (14-9) produces an unbounded output response instead of a τs + 1 sinusoidal response.
A rather simple procedure can be employed to find the sinusoidal response. SOLUTION After setting s = jω in G(s), by algebraic manipulation we can separate the expression into real (R) and imagi- First substitute s = jω in the transfer function nary (I) terms (j indicates an imaginary component): 1 1 G(jω) = = (14-10) G(jω) = R(ω) + jI(ω) (14-4) τjω + 1 jωτ + 1 Then multiply both numerator and denominator by the Similar to Eq. 14-1, we can express the long time complex conjugate of the denominator, that is, −jωτ + 1 response for a linear system (cf. 14-1) as −jωτ + 1 −jωτ + 1 y𝓁 (t) =  sin(ωt + ϕ) (14-5) G(jω) = (jωτ + 1)(−jωτ + 1) = 2 2 ω τ +1  and ϕ are related to I(ω) and R(ω) by the following 1 (−ωτ) = +j 2 2 = R + jI (14-11) relations (Seborg et al., 2004): ω2 τ2 + 1 ω τ +1 √ where  = A R2 + I 2 (14-6a) 1 R= (14-12a) −1 ω2 τ2 + 1 ϕ = tan (I∕R) (14-6b) and −ωτ Both  and ϕ are functions of frequency ω.
A simple I= (14-12b) ω2 τ2 + 1 but elegant relation for the frequency response can be From Eq. 14-7, derived, where the amplitude ratio is given by √ √ ( )2 ( )2 Â AR = |G(jω|) = 1 −ωτ AR = = |G| = R2 + I 2 (14-7) ω2 τ2 + 1 + ω2 τ2 + 1 A The absolute value denotes the magnitude of G, and Simplifying, the phase shift (also called the phase angle or argument √ (1 + ω2 τ2 ) 1 of G, ∠G) between the sinusoidal output and input is AR = = √ (14-13a) (ω2 τ2 + 1)2 ω2 τ2 + 1 given by ϕ = ∠G = tan−1 (I∕R) (14-8) ϕ = ∠G(jω) = tan−1 (−ωτ) = −tan−1 (ωτ) (14-13b) Because R(ω) and I(ω) (and hence AR and ϕ) can be If the process gain had been a positive value K instead of 1, obtained without calculating the complete transient K response y(t), these characteristics provide a convenient AR = √ (14-14) ω2 τ2 + 1 shortcut method to determine the frequency response of transfer functions. and the phase angle would be unchanged (Eq. Both the amplitude ratio and phase angle are identi- Equations 14-7 and 14-8 can calculate the frequency cal to those values calculated in Section 14.1 using the response characteristics of any stable G(s), including time-domain derivation.
those with time-delay terms.3 Bode Diagrams 247 From this example, we conclude that direct analysis of the complex transfer function G(jω) is computation- Combining these expressions via Eqs. 14-17a and 14-17b ally easier than solving for the actual long-time output yields response j𝓁 (t). The computational advantages are even |Ga (jω)| AR = |G(jω)| = greater when dealing with more complicated processes, |G1 (jω)‖G2 (jω)| as shown in the following. Start with a general transfer K = √ √ (14-18a) function in factored form ω2 τ21 + 1 ω2 τ22 + 1 G (s)Gb (s)Gc (s) · · · G(s) = a (14-15) G1 (s)G2 (s)G3 (s) · · · ϕ = ∠G(jω) = ∠Ga (jω) − (∠G1 (jω) + ∠G2 (jω)) G(s) is converted to the complex form G(jω) by the sub- = −tan−1 (ωτ1 ) − tan−1 (ωτ2 ) (14-18b) stitution s = jω: G (jω)Gb (jω)Gc (jω) · · · G(jω) = a (14-16) G1 (jω)G2 (jω)G3 (jω) · · · 14.