8.0 Overview 8.1 The frequency response for a stable system

Transcription

1 8. Frequency Response Analysis 8.0 Overview 8.1 The frequency response for a stable system Simple system elements First-order system Second-order systems Systems of arbitrary order Systems in series Summary 8.2 Graphical frequency-response representation Overview Bode diagram 8.3 Stability analysis of feedback systems Bode s stability criterion Stability margins Numerical solution of frequency relationships KEH Dynamics and 8 1

2 8. Frequency Response Analysis 8.4 ler design in the frequency domain Design of PI controllers Design of PD controllers Design of PID controllers KEH Dynamics and 8 2

3 8. Frequency Response Analysis 8.0 Overview So far in this course system properties have been studied in the time domain (e.g., step response) Laplace domain (e.g., stability) In this chapter, system properties are studied in the frequency domain by studying the stationary behaviour when the system is excited by a sinusoidal input of given frequency. Stationary behaviour refers to the situation when initial effects have vanished, i.e., when the time tt. The output from a linear system will then also change sinusoidally with certain characteristic properties that depend on the system as well as the amplitude and frequency of the input sinusoidal. These properties expressed as function of the input frequency are referred to as the frequency response of the system. System analysis based on the frequency response is called frequency analysis. The study of sinusoidal inputs is useful because measurement noise and timevarying disturbances can often be approximated by sinusoidal signals. KEH Dynamics and 8 3

4 8. Frequency Response Analysis 8.1 The frequency response for a stable system In this section, the frequency response for an arbitrary linear, stable, system is derived. In order to introduce new concepts step-wise, some simple system elements that essentially lack dynamics are treated first. These elements are often part of more complex systems. After this, systems of first and second order are considered. In these systems, dynamics play a prominent role. For the derivation of the frequency response of high-order systems, the frequency response of simpler systems is utilized. The decomposition of a high-order system into low-order components connected in series is fundamental in frequency analysis. KEH Dynamics and 8 4

5 8. Frequency Response Analysis 8.1 Frequency response for a stable system Simple system elements Static system A linear static system is described by yy tt = KKKK(tt) (8.1) where uu(tt) is an input, yy(tt) is an output, and KK is the system gain. Let the input change sinusoidally as uu tt = AA sin ωωωω (8.2) where ωω 0 is the angular frequency (expressed in radians per time unit) and AA > 0 is the amplitude of the sinusoidal. The output is then yy tt = KKKK sin ωωωω (8.3) There are now two situations regarding the phase of the output: If KK > 0, it is the same as the phase of the input. If KK < 0, it is opposite to the phase of the input. In this case, yy(tt) has a maximum when uu(tt) has a minimum, and vice versa. Thus, the negative system gain causes a phase shift of ±ππ radians, or ±180, in the output. KEH Dynamics and 8 5

6 8.1.1 Simple system elements Static system It is of interest to rewrite (8.3) so that the phase shift is explicitly seen in the equation the output amplitude is always a positive quantity To accomplish this, (8.3) is written as where is the phase shift of the system. yy tt = KK AA sin(ωωωω + φφ) (8.4) 0 if KK > 0 φφ = ππ if KK < 0 The ratio between the amplitudes of the output and input signals is also of interest. In this case, the amplitude ratio is Generally, the output can be written as (8.5) AA R = KK (8.6) yy tt = AA R AA sin(ωωωω + φφ) (8.7) The stationary output of any stable linear system with an input (8.2) can be written as (8.7), where AA R is the amplitude ratio and φφ is the phase shift of the system in question. KEH Dynamics and 8 6

7 8.1 Frequency response for a stable system Simple system elements Derivative system A system with the transfer function GG ss = KKKK has an output that is the time derivative of the input amplified by the factor KK, i.e., yy tt = KK duu(tt) dtt When the input is sinusoidal as in (8.2), the output becomes yy tt = KK d(aa sin ωωωω) dtt (8.8) = KKKKωω cos ωωωω (8.9) where the relationship d(sin ωωωω)/dtt = ωω cos ωωωω has been used. Here the output is a cosine function, but by using the trigonometric identity cosωωtt = sin(ωωωω + ππ/2), (8.9) can be written as yy tt = KKωωωω sin(ωωωω + ππ/2) (8.10) Here the phase shift +ππ/2 is used, not 3ππ/2, although they both give the same output. The reason is not that +ππ/2 is closer to 0, but that the derivative yields a prediction of the input. Hence, the output can be considered to correspond to a future value of the input meaning that the phase of the output is before the phase of the input. Thus, the derivative is a phase-advancing element. KEH Dynamics and 8 7

8 8.1.1 Simple system elements Derivative system If the gain KK < 0, this has to be taken into account as for a static system. Thus, the output is given by (8.7) with AA R = KK (8.11) ππ/2 if KK > 0 φφ = ππ/2 if KK < 0 (8.12) Parallel connection of static and derivative system A system composed of a parallel connection of a static and a derivative system has the transfer function GG ss = KK(1 + TTTT) (8.13) This can be a PD controller, but other systems with zeros also have such factors in the numerator of the transfer function. In the time domain, the output is given by yy tt = KKKK tt + KKKK duu(tt) (8.14) dtt When the input is a sinusoidal as in (8.2), this gives yy tt = KKKK(sin ωωωω + TTωω cos ωωωω) (8.15) KEH Dynamics and 8 8

9 Derivative system Parallel connection of static and derivative system The right-hand side of (8.15) can be written as a single sine function by means of the trigonometric identity This gives sin ωωtt + xx cos ωωωω = 1 + xx 2 sin(ωωωω + arctan xx) (8.16) yy tt = KKKK 1 + (TTωω) 2 sin(ωωωω + arctan TTωω) (8.17) The output is thus given by (8.7) with AA R = KK 1 + (TTωω) 2 (8.18) arctan TTωω if KK > 0 φφ = arctan TTωω ππ if KK < 0 (8.19) KEH Dynamics and 8 9

10 8.1 Frequency response for a stable system Simple system elements Integrating system A system with the transfer function GG ss = KKss 1 has an output that is the time integral of the input amplified by the factor KK. This is equivalent to the time derivative of the output being equal to the input amplified by KK, i.e., The sinusoidal input (8.2) gives Because d cos ωωωω dtt dyy(tt) dtt dyy(tt) dtt = KKKK(tt) (8.20) = KKKK sin ωωωω (8.21) = ωω sin ωωωω, it is easy to verify by differentiation that yy tt = KKKKωω 1 cos ωωωω (8.22) satisfies (8.20). Because cos ωωtt = sin(ωωωω + ππ/2) = sin(ωωωω ππ/2), substitution into (8.22) yields (8.7) with AA R = KK ωω 1 (8.23) ππ/2 if KK > 0 φφ = 3ππ/2 if KK < 0 (8.24) KEH Dynamics and 8 10

11 8.1.1 Simple system elements Integrating system Parallel connection of static and integrating system A system composed of a parallel connection of a static and an integrating system has the transfer function GG ss = KK TTTT (8.25) This can, e.g., be the transfer function of a PI controller. From (8.3) and (8.22) it is clear that the sinusoidal input (8.2) gives yy tt = KKKK sin ωωωω (TTTT) 1 cos ωωωω (8.26) Taking the trigonometric identity (8.16) and the sign of KK into account yields (8.7) with AA R = KK 1 + (TTωω) 2 (8.27) φφ = arctan (TTTT) 1 if KK > 0 arctan (TTTT) 1 ππ if KK < 0 (8.28) KEH Dynamics and 8 11

12 8.1 Frequency response for a stable system Simple system elements Time delay A time delay of length LL has the transfer function GG ss = e LLLL. In the time domain, the relationship between the output and the input is The sinusoidal input (8.2) then gives yy tt = uu(tt LL) (8.29) where yy tt = AA sin ωω tt LL = AA sin(ωωωω + φφ) (8.30) φφ = LLLL (8.31) As seen from (8.30), a pure time delay has the amplitude ratio AA R = 1 (8.32) Note that the phase shift of a time delay is unlimited the higher the frequency, the more negative the phase shift. KEH Dynamics and 8 12

13 8. Frequency Response Analysis 8.1 Frequency response for a stable system First-order system A first-order system is described by the differential equation TT dyy(tt) dtt + yy tt = KKKK(tt) (8.33) where uu(tt) is an input, yy(tt) is an output, KK is the gain, and TT is the time constant of the system. Frequency response via the Laplace domain As for the previous cases, the frequency response could be derived directly in the time domain. However, it is instructive to derive the frequency response via the Laplace domain; this will help derive more general relationships for the frequency response, which are useful when systems of arbitrary order are considered. KEH Dynamics and 8 13

14 8.1.2 First-order system Frequency response via the Laplace domain Laplace transformation of (8.33) yields YY ss = GG ss UU(ss), GG ss = KK TTTT+1 (8.34) where UU(ss) is the Laplace transform of uu(tt), YY(ss) is the Laplace transform of yy(tt), and GG(ss) is the transfer function of the system. The Laplace transform of the sinusoidal input (8.2) is which substituted into (8.34) gives UU ss = AAωω ss 2 +ωω 2 (8.35) YY ss = GG ss AAωω ss 2 +ωω 2 = KK TTTT+1 AAωω ss 2 +ωω 2 (8.36) Since the second-order factor ss 2 + ωω 2 has complex-conjugated zeros, (8.36) has the partial fraction expansion YY ss = BB + CCCC+DD (8.37) TTTT+1 ss 2 +ωω 2 where the coefficients BB, CC and DD have to be determined so that (8.36) and (8.37) are equivalent. KEH Dynamics and 8 14

15 8.1.2 First-order system Frequency response via the Laplace domain It is useful to consider the inverse Laplace transform of (8.37) before BB, CC and DD are determined. The transforms 25, 38 and 39 in the Laplace transform table in Section 4.5 yield yy tt = BB TT e tt/tt + CC cos ωωωω + DD ωω 1 sin ωωωω (8.38) We are interested in the stationary solution when tt. Because TT > 0 for a stable system, and BB is finite, the first term on the right-hand side will vanish as tt. Thus, the stationary solution is yy tt lim tt yy tt = CC cos ωωωω + DDωω 1 sin ωωωω (8.39) This means that the coefficient BB does not affect the stationary solution and it is sufficient to determine only CC and DD if it can be done independently of BB. Combination of the first part of (8.36) with (8.37) gives from which BB + CCCC+DD TTTT+1 ss 2 +ωω 2 = GG ss AAωω ss 2 +ωω 2 CCCC + DD GG ss AAωω BB(ss2 +ωω 2 ) TTTT+1 (8.40) KEH Dynamics and 8 15

16 8.1.2 First-order system Frequency response via the Laplace domain The identity (8.40) has to apply for arbitrary values of ss. Choosing ss = jωω means that ss 2 + ωω 2 = 0. Then (8.40) yields CCωωj + DD GG jωω AAωω (8.41) The identity (8.41) requires that the real part and the imaginary part are satisfied independently. Because GG(jωω) is a complex number with the real part Re GG jωω and the imaginary part Im GG jωω, i.e., (8.41) yields GG jωω = Re GG jωω + Im GG jωω j CC = AA Im GG(jωω), DD = AAωω Re GG(jωω) (8.42) The first-order system (8.34) yields from which GG jωω = Re GG jωω = KK TTTTj+1 = KK 1+ TTTT KK 1 TTTTj = KK KKKKKKj (8.43) TTTTj+1 1 TTTTj 1+ TTTT 2 2, Im GG jωω = KKKKKK 1+ TTTT 2 (8.44) KEH Dynamics and 8 16

17 8.1.2 First-order system Frequency response via the Laplace domain Substitution of (8.44) into (8.42) and further into (8.39) yields the stationary solution yy tt = KKKK 2 sin ωωωω TTωω cos ωωωω (8.45) 1+(TTωω) The trigonometrical identity (8.16) applied to (8.45) yields yy tt = KKKK 1+(TTωω) 2 sin(ωωωω arctan TTTT) (8.46) Thus, the stationary solution has the same form as (8.7), i.e., For a first-order system (8.47) applies with yy tt = AA R AA sin(ωωωω + φφ) (8.47) AA R = KK 1+(TTωω) 2 (8.48) arctan(tttt) if KK > 0 φφ = arctan(tttt) ππ if KK < 0 (8.49) KEH Dynamics and 8 17

18 8. Frequency Response Analysis 8.1 Frequency response for a stable system Second-order systems A second-order system without zeros and with no time delay has the transfer function GG ss = KKωω n 2 ss 2 +2ζζωω n ss+ωω n 2 (8.50a) where KK is the static gain of the system, ζζ is its relative damping and ωω n > 0 is its undamped natural frequency. Since only stable systems are considered, ζζ > 0. If the poles of the system are real, the transfer function is often written in the form KK GG ss = (8.50b) The parameters of (8.50a) are then given by (TT 1 ss+1)(tt 2 ss+1) ζζ = TT 1+TT 2 2 TT 1 TT 2, ωω n = 1 TT 1 TT 2 (8.51) It is here sufficient to consider second-order systems without zeros because systems with zeros can be decomposed into a series connection of two (or more) subsystems and handled by the methods in Section KEH Dynamics and 8 18

19 8.1 Frequency response for a stable system Second-order systems Derivation of frequency response Analogously to (8.36), the Laplace transform of the sinusoidal input yields YY ss = GG ss AAωω ss 2 +ωω 2 = KKωω2 n ss 2 +2ζζωω n ss+ωω2 n AAωω ss 2 +ωω 2 (8.52) When ζζ > 0, there always exists a partial fraction expansion YY ss = BB 1 ss+bb 2 ss 2 +2ζζωω n ss+ωω2 + CCCC+DD (8.53) n ss 2 +ωω 2 According to our Laplace transform table, the first term on the right-hand side has an inverse transform containing the factor e ζζωω ntt, where tt denotes time. Because ζζωω n > 0, e ζζωω ntt 0 as tt and the term containing this factor will vanish. This means that also in this case the stationary solution to (8.53) is given by (8.39) the coefficients CC and DD are given by (8.42) KEH Dynamics and 8 19

20 8.1.3 Second-order systems Derivation of frequency response In this case GG jωω = KKωω n 2 (jωω) 2 +2ζζωω n jωω+ωω n 2 = KKωω n 2 ωω n 2 ωω 2 +2ζζωω n ωω j from which = KKωω2 n ωω n 2 ωω 2 2ζζωω n ωω j = KKωω n 2 (ωω 2 n ωω 2 2ζζωω n ωω j) ωω 2 n ωω 2 +2ζζωω n ωω j ωω 2 n ωω 2 2ζζωω n ωω j (ωω n 2 ωω 2 ) 2 +(2ζζωω n ωω) 2 (8.54) Re GG jωω = KKωω n 2 (ωω n 2 ωω 2 ) (ωω n 2 ωω 2 ) 2 +(2ζζωω n ωω) 2, Im GG jωω = 2KKωω 3 n ωω (8.55) (ωω 2 n ωω 2 ) 2 +(2ζζωω n ωω) 2 Substitution into (8.42) and further into (8.39) gives yy(tt) = KKωω 2 n (ωω 2 n ωω 2 ) AA sin ωωωω 2ζζωω nωω (ωω 2 n ωω 2 ) 2 +(2ζζωω n ωω) 2 ωω 2 n ωω 2 cos ωωωω (8.56) where we (for now) assume that ωω ωω n. Application of the trigonometric identity (8.16) yields the stationary response as yy(tt) = KKωω n 2 sgn(ωω 2 n ωω 2 ) AA sin(ωωωω + φφ), φφ = arctan 2ζζωω nωω (8.57) (ωω 2 n ωω 2 ) 2 +(2ζζωω n ωω) 2 ωω 2 n ωω 2 KEH Dynamics and 8 20

21 8.1.3 Second-order systems Derivation of frequency response Eq. (8.57) can be expressed in the form of (8.47) with AA R = KK ωω n 2 (ωω n 2 ωω 2 ) 2 +(2ζζωω n ωω) 2 = KK 1 ( ωω ωω n ) 2 2 +(2ζζ ωω ωω n ) 2 (8.58) arctan 2ζζωω nωω if KKωω ωω φφ = 2 n ωω 2 n > KKKK arctan 2ζζωω nωω ππ if KKωω ωω 2 n ωω 2 n < KKKK (8.59a) arctan = arctan 2ζζ ωω ωω n 1 2ζζ ωω ωω n 1 ωω ωω n 2 if KKωω n > KKKK ωω ωω n 2 ππ if KKωω n < KKKK (8.59b) For ωω = ωω n, these reduce to AA R = KK 2ζζ ππ 2 if KK > 0, φφ = 3ππ/2 if KK < 0, ωω = ωω n ( ) If ζζ < 1/ 2 0.7, the amplitude ratio has a peak (i.e., maximum) at the peak frequency ωω p, given by ωω p = ωω n 1 2ζζ 2 AA R,peak = KK 2ζζ 1 2ζζ 2 ( ) KEH Dynamics and 8 21

22 8. Frequency Response Analysis 8.1 Frequency response for a stable system Systems of arbitrary order For a stable system of arbitrary order, but no time delay, the response to a sinusoidal input can be expressed by a partial fraction expansion as in (8.37) and (8.53). As before, the stationary solution in the time-domain is given by (8.39) with CC and DD given by (8.42). To simplify the notation, we define Substitution of (8.42) into (8.39) then gives RR(ωω) Re GG(jωω), II ωω = Im GG(jωω) (8.64) yy tt = AA RR ωω sin ωωωω + II(ωω) cos ωωωω = AAAA(ωω) sin ωωωω + II(ωω) cos ωωωω (8.65) RR(ωω) Application of the trigonometrical identity (8.16) yields yy(tt) = RR(ωω) 1 + II(ωω) RR(ωω) 2 AA sin(ωωωω + φφ) where = sgn RR(ωω) RR(ωω) 2 + II(ωω) 2 AA sin(ωωωω + φφ) (8.66) φφ = arctan II(ωω) RR(ωω) (8.67) KEH Dynamics and 8 22

23 8.1 Frequency response for a stable system Systems of arbitrary order Because GG(jω) is a complex number, it can be characterized by its magnitude GG(jωω) and argument GG(jωω), also denoted arg GG(jωω). According to the theory of complex number, GG(jωω) = RR(ωω) 2 + II(ωω) 2, tan arg GG(jωω) = II(ωω) RR(ωω) (8.68a,b) The sign of sgn RR(ωω) in (8.66) has the same effect on the phase shift as the sign of the gain KK in previous sections. Substitution of (8.68) into (8.66) and (8.67) then yields yy tt = GG(jωω) AA sin(ωωωω + arg GG jωω ) (8.69) arg GG jωω = arctan II(ωω) RR(ωω) if RR(ωω) 0 arctan II(ωω) RR(ωω) ππ if RR ωω < 0 (8.70) From (8.69) it is obvious that the magnitude GG(jωω) is identical to the amplitude ratio the argument arg GG(jωω) is identical to the phase shift Thus AA R = GG(jωω), φφ = arg GG(jωω) (8.71) KEH Dynamics and 8 23

24 8.1 Frequency response for a stable system Systems of arbitrary order Because the function arctan only takes values between ππ/2 and +ππ/2, direct calculation from (8.70) gives an argument in the range 3 ππ < arg GG(jωω) < 1 ππ (8.72) 2 2 However, because of the periodicity of the trigonometric functions, (8.68b) would be satisfied for any integer multiple of 2ππ added to arg GG jωω. This means that the solution calculated from (8.70) is ambiguous by an integer multiple of 2ππ. The figure shows a sinusoidal input uu(tt) and the stationary output yy(tt), which is also sinusoidal. When the axes are normalized as indicated, the values of GG and arg GG can be directly read from the plot for the applied input frequency. Because the sinusoidal signals can be shifted by an integer multiple of 2ππ without any detectable difference, arg GG is similarly ambiguous also here. This ambiguity can be solved by the method in Section Normalized signal Normalized time KEH Dynamics and 8 24

25 8. Frequency Response Analysis 8.1 Frequency response for a stable system Systems in series Consider a system composed of NN subsystems connected in series. If the subsystems have the transfer functions GG ii (ss), ii = 1,, NN, the transfer function GG(ss) of the full system is given by GG ss = GG 1 ss GG 2 ss GG NN ss = NN ii=1 GG ii (ss) (8.73) Conversely, a system with the transfer function GG ss = KK (TT nn+1ss+1) (TT nn+mm ss+1) (TT 1 ss+1)(tt 2 ss+1) (TT nn ss+1) e LLLL (8.74) can be decomposed into the series-connected subsystems KK, 1, 1,, 1, (TT TT 1 ss+1 TT 2 ss+1 TT nn ss+1 nn+1ss + 1),, (TT nn+mm ss + 1), e LLLL. If the system has complex-conjugated poles or zeros, they can be included as second-order factors in the decomposition. Thus, a high-order transfer function can be decomposed into factors of at most second order and a possible time delay. KEH Dynamics and 8 25

26 8.1 Frequency response for a stable system Systems in series From (8.73) it follows that the frequency response of the full system is GG jωω = NN ii=1 GG ii (jωω) (8.75) where GG ii (jωω), ii = 1,, NN, are the frequency responses of the individual subsystems. According to the theory of complex numbers, GG ii (jωω) can be expressed in terms of its magnitude GG ii (jωω) and its argument arg GG ii (jωω) as GG ii jωω = GG ii (jωω) e j arg GG ii(jωω) (8.76) Substitution into (8.75) yields GG jωω = NN ii=1 GG ii (jωω) e j arg GGii(jωω) = NN ii=1 GG ii (jωω) NN ii=1 e j arg GG ii(jωω) = NN ii=1 GG ii (jωω) e j NN ii=1 arg GGii (jωω) (8.77) Naturally, GG(jωω) can also be expressed as (8.76). From that and (8.77), GG jωω = NN ii=1 GG ii (jωω) (8.78) arg GG jωω = NN ii=1 arg GG ii (jωω) (8.79) KEH Dynamics and 8 26

27 8.1 Frequency response for a stable system Systems in series Equations (8.78) and (8.79) mean that for the full system the amplitude ratio (or magnitude) is obtained as the product of the amplitude ratios (or magnitudes) of the subsystems phase shift (or argument) is obtained as the sum of the phase shifts (or arguments) of the subsystems The user is allowed to decompose the full system into subsystems as desired. However, it is advantageous to decompose into subsystems of first or second order and a possible time delay because we know the frequency responses of such systems. In particular, we know that a first-order system has a phase shift in the range ππ 2 < φφ < 0 a second-order system has a phase shift in the range ππ < φφ < 0 when the gain is positive. By adding up the phase shifts (arguments) of the individual subsystems according to (8.79), the correct phase shift (argument) for the full system is obtained. This means that a phase shift outside the range of (8.72) can be obtained, i.e., the correct integer multiple of 2ππ is obtained. KEH Dynamics and 8 27

28 8. Frequency Response Analysis 8.1 Frequency response for a stable system Summary Table 8.1. Frequency response of low-order systems. Gs () AR = G(j ω) ϕ = arg G(j ω) 1 1 π K > 0 K 0 s ω π /2 1 Ts ( Tω) arctantω 1/ s 1/ω π /2 1 1/Ts /( Tω) arctan(1/ Tω) e Ls 1 Lω 1 Ts ( Tω) 2 arctantω s 2 ωn ζωns+ ωn n ζω ωn (1 ( ω / ω ) ) + (2 / ) 2 ζω / ω arctan n 1 ( ω / ω ) 2 ζω / ω π arctan n 1 ( ω / ω ) n ω ω, 2 n n ω ω, 2 n KEH Dynamics and 8 28

29 8. Frequency Response Analysis 8.2 Graphical frequency-response representation Overview The complex-valued function GG(jωω) of a system with the transfer function GG(ss) contains all information about the frequency response of the system except for a multiple of 2ππ in the phase shift. GG(jωω) can be considered a system property GG(jωω) is called the frequency function Since GG(jωω) is a complex number, it can be represented in two ways: GG(jωω) = ReGG(jωω) + jimgg(jωω) = GG(jωω ) jarggg(jωω) e RR ωω ReGG(jωω) is the real part of GG(jωω) II ωω ImGG(jωω) is the imaginary part of GG(jωω) GG(jωω ) RR(ωω) 2 + II(ωω) 2 is the magnitude of GG(jωω) arggg(jωω) arctan[ II(ωω) RR(ωω)] ± nnnn is the argument of GG(jωω) This gives several possibilities of representing GG jωω graphically, e.g., Nyquist diagram: II ωω vs RR ωω as ωω varies Bode diagram: GG(jωω ) vs ωω and arggg(jωω) vs ωω in two diagrams KEH Dynamics and 8 29

30 8. Frequency Response Analysis 8.2 Graphical representation Bode diagram In a Bode diagram, GG(jωω ) and arggg(jωω) in two diagrams. are plotted against the frequency ωω The absolute value ) GG(jωω is plotted on a logarithmic scale, either expressed as a pure amplitude ratio or with the logarithmic unit decibel (db), defined GG(jωω ) db = 20 log 10 GG(jωω ) (8.80) In this course, the pure amplitude ratio is used. The phase shift arggg(jωω) is plotted on a linear scale, either expressed in degrees or radians, defined 1 rad = 180 ππ (8.81) In this course, degrees are used in the Bode diagram, but in calculations radians are used. The frequency ωω is expressed on a logarithmic scale in both diagrams. KEH Dynamics and 8 30

31 8.2 Graphical representation Bode diagram First-order system The transfer function of a first-order system is GG ss =. The following TTTT+1 expressions have been derived for the amplitude ratio and the phase shift amplitude ratio: AA R ωω = ) GG(jωω = KK 1+(ωωωω) 2 phase shift: φφ(ωω) = ) arggg(jωω = arctan ωωωω The Bode diagram applies to all first-order systems because the normalized amplitude ratio (obtained through division by KK) and a normalized frequency (multiplication by TT) are plotted. At high frequency slope of AA R KK is 1 φφ 90 A R K ϕ ωt KK KEH Dynamics and 8 31 ωt

32 8.2 Graphical representation Bode diagram Second-order systems The transfer function of a secondorder system is GG ss = KKωω n 2 ss 2 +2ζζωω n ss+ωω n 2. The Bode diagram displays the amplitude ratio: AA R = phase shift: arctan φφ = arctan KK 1 ( ωω ωω n ) ζζζζ ωω 2 n 2ζζζζ ωω n 1 ωω ωω 2 n 2ζζζζ ωω n 1 ωω ωω n 2 ππ if KKωω n KKωω if KKωω n KKKK resonance peak and frequency: KK AA R ωω p = at ωω 2ζζ 1 2ζζ 2 p = ωω n 1 2ζζ 2 At high frequency slope of AA R KK is 2 φφ 180 The diagram applies to all second-order systems due to normalized axes. KEH Dynamics and 8 32 A R K ϕ

33 8.2 Graphical representation Bode diagram Time delay The transfer function of a time delay is GG ss = e LLLL. The Bode diagram displays the amplitude ratio: AA R ωω = 1 phase shift: φφ ωω = ωωωω At high frequency φφ as ωω If the frequency axis ωωll were linear, the slope of the phase shift plot would be 1. A R ϕ KEH Dynamics and 8 33

34 8.2 Graphical representation Bode diagram Numerator time constant A time constant in the numerator of a transfer function corresponds to a subsystem with the transfer function GG ss = 1 + TTTT. It is also the transfer function of a PD controller (with the gain 1 and derivative time TT). The transfer function has the following characteristics amplitude ratio: AA R ωω = 1 + (ωωωω) 2 phase shift: φφ ωω = arctan ωωωω At high frequency slope of AA R is +1 φφ +90 if TT > 0, φφ 90 if TT < 0 Note the similarity (and the difference) to a first-order system. For the numerator time constant the amplitude plot is symmetrical to the normalized amplitude plot of a firstorder system with respect to the line AA R = 1 (i.e., its mirror ) if TT > 0, the phase plot is symmetrical to the phase plot of a first-order system with respect to the frequency line φφ = 0 (i.e., its mirror); if TT < 0, the plots are identical KEH Dynamics and 8 34

35 8.2 Graphical representation Bode diagram Integrating system An integrator has the transfer function GG ss = ss 1 with the following characteristics amplitude ratio: AA R ωω = 1/ωω phase shift: φφ ωω = ππ/2 At all frequencies slope of AA R is 1 φφ = 90 Derivative system A derivative system has the transfer function GG ss = ss with the following characteristics amplitude ratio: AA R ωω = ωω phase shift: φφ ωω = +ππ/2 At all frequencies slope of AA R is +1 φφ = +90 KEH Dynamics and 8 35

36 8.2 Graphical representation Bode diagram Systems in series A system composed of NN subsystems in series with the transfer functions GG 1 (ss), GG 2 (ss),, GG NN (ss), has the transfer function with the following characteristics amplitude ratio: phase shift: GG ss = NN ii=1 AA R ωω = NN ii=1 log AA R ωω = NN ii=1 φφ ωω = NN ii=1 GG ii (ss) AA R,ii (ωω) log AA R,ii (ωω) φφ ii (ωω) Because the amplitude plot in a Bode diagram is logarithmic, the logarithmic amplitude plot of the overall system is obtained by adding the logarithmic amplitude plots of the subsystems. This is a useful property if the Bode plot is drawn by hand. Because the phase plot is a Bode diagram is linear, the phase plot of the overall system is obtained by adding the phase plots of the subsystem. KEH Dynamics and 8 36

37 8. Frequency Response Analysis 8.3 Stability analysis of feedback systems Bode s stability criterion The system in the block diagram has the following transfer functions: GG p is a process to be controlled GG m is a measurement device GG v is an actuator (e.g., a valve) GG c is a controller The loop transfer function of the system is r e u y G c G v G + p An open loop. GG l = GG m GG p GG v GG c (8.82) Consider a system with GG c = KK c, GG v = 1, GG p = e 0.1ss 0.5ss+1, GG m = 1, where the time unit in GG p is minutes. The loop transfer of the system is y m G m GG l = KK ce 0.1ss 0.5ss+1 (8.83) KEH Dynamics and 8 37

38 8.3 Stability analysis of feedback systems Bode s stability criterion A thought experiment Assume that the setpoint rr changes sinusoidally with the angular frequency ωω = 17 rad/min. At stationary conditions (i.e., when the initial effects from starting the sinusoidal have vanished) the phase shift through the loop transfer is φφ = arctan r ππ = 180 This means that yy m will oscillate exactly in opposite phase to rr (and ee), as y m illustrated in the figure to the right. Assume now that the feedback loop is closed. Because of the minus sign in ee = rr yy m, yy m will be exactly in phase with rr, as shown in the figure to the right. Even if rr is set to zero at the same moment the loop is closed, ee will continue to oscillate with the same frequency because ee = yy m. r y m Before loop is closed. After loop is closed. KEH Dynamics and 8 38

39 8.3.1 Bode s stability criterion A though experiment The amplitude of yy m can be determined from the amplitude ratio of the system at the frequency 17 rad/min. In this case AA R ωω = KK c 1+(ωωωω) 2 AA R 17 = KK c 1+(17 0.5) 2 KK c 8.56 (8.84) There are now three possibilities. If the amplitude of yy m is smaller than the amplitude of rr and ee before the loop is closed (i.e., if KK c < 8.56 AA R < 1), the amplitude of ee will decrease when the loop is closed. This, in turn, will reduce the amplitude of yy m, making the amplitude of ee even smaller, and so on. Eventually, the oscillation will die out. If the amplitude of yy m is equal to the amplitude of rr and ee before the loop is closed (i.e., if KK c = 8.56 AA R = 1), the amplitude of ee will not change when the loop is closed. This means that the oscillation will continue with the same amplitude for ever. If the amplitude of yy m is larger than the amplitude of rr and ee (i.e., if KK c > 8.56 AA R > 1), the amplitude of ee will increase when the loop is closed. This, in turn, will increase the amplitude of yy m, making the amplitude of ee even larger, and so on. The amplitude of the oscillation will become larger and larger and the system is unstable. KEH Dynamics and 8 39

40 8.3 Stability analysis of feedback systems Bode s stability criterion The stability criterion The smallest frequency such that the loop transfer function GG l has a phase shift equal to 180 is determined. This frequency, ωω c, is the critical frequency of the system. The amplitude ratio of the loop transfer function at this frequency, i.e., AA R,l ωω c, is determined. The closed-loop system is stable, if AA R,l ωω c < 1 unstable, if AA R,l ωω c > 1 for the loop transfer function. Note that the stability of the closed-loop system is determined by considering the loop transfer function in open loop. KEH Dynamics and 8 40

41 8.3 Stability analysis of feedback systems Bode s stability criterion How to determine ωω c and AA R (ωω c ) The critical frequency ωω c and the amplitude ratio AA R (ωω c ) can be determined in different ways. Graphically, by drawing the Bode diagram for the loop transfer function GG l. From such a diagram, it is easy to read off ωω c at 180 for the phase curve and AA R ωω c from the amplitude curve. Numerically, by solving the equation for ωω c and calculating φφ l (ωω c ) arg GG l ωω c = ππ (8.85) AA R,l ωω c = GG l (ωω c ) (8.86) It will be shown how φφ l (ωω c ) = ππ can be solved iteratively to find ωω c. By simulation of feedback control using a P controller. The critical frequency ωω c and the maximum controller gain KK c,maxx can be determined as described in Section The amplitude ratio of the loop transfer without a controller is then AA R ωω c = 1 KK c,max. KEH Dynamics and 8 41

42 8.3 Stability analysis of feedback systems Bode s stability criterion Example Determine the critical frequency and the amplitude ratio at this frequency for a system with the loop transfer function GG l ss = GG 1 (ss)gg 2 (ss)gg 3 (ss)gg 4 (ss) where GG 1 ss = e 4ss, GG 2 ss = 1.5, 2ss+1 GG 3 ss = 2, GG 10ss+1 4 ss = 0.8 5ss+1 The solution can be found graphically by means of a Bode diagram. The diagram is constructed by plotting 1.5 AA R,l ωω = (2ωω) 2 1+(10ωω) 2 1+(5ωω) 2 and φφ l ωω = 4ωω arctan 2ωω arctan 10ωω arctan 5ωω against the frequency ωω. From this diagram it is easy to read off ωω c 0.21 rad/min at φφ l = ππ = 180 and AA R,l ωω c 0.7. AR, ϕ rad/min rad/min KEH Dynamics and 8 42

43 8.3 Stability analysis of feedback systems Bode s stability criterion Exercise A process that can be modelled as a pure time delay is controlled by a P controller. The control valve and the measurement device have negligible dynamics, but their gains are KK v = 0.5 and KK m = 0.8, respectively. When a small change in the setpoint is made, the controlled process starts to oscillate with constant amplitude and the period 10 min. a) Which is the controller gain? b) How large is the time delay? KEH Dynamics and 8 43

44 8. Frequency Response Analysis 8.3 Stability analysis of feedback systems Stability margins Gain margin The gain margin AA m is the factor by which the gain of the loop transfer function has to be changed in order to reach the stability limit. Mathematically, AA m = 1 AA R,l (ωω c ) Stability of the closed-loop system requires AA m > 1. The stability margin yields robustness not only against variations in the loop gain, but also against variations in other process parameters. (8.87) Example Consider a system with the loop transfer function GG l = KK ce 0.1ss, which was used 0.5ss+1 in the introduction of the Bode stability criterion. a) Determine a P controller to give the closed-loop system the (designed) gain margin AA m = 1.7. b) Is the closed-loop system stable with the designed P controller if the time delay changes to 0.15 min? KEH Dynamics and 8 44

45 8.3.2 Stability margins Example a) It was previously stated that an input sinusoidal with the frequency ωω = 17 rad/min resulted in a phase shift of 180. Thus, this is the critical frequency ωω c of the loop transfer function. We want a P controller such that AA m = 1.7, i.e., AA R,l ωω c = This is obtained if AA R,l ωω c = KK c = 1 KK 1+(0.5 17) c = 5.0 b) To check if the system is stable with KK c = 5 and LL = 0.15 min, a Bode diagram could be drawn using these parameters. The diagram gives the new critical frequency and the amplitude ratio at that frequency. Here, the solution is obtained numerically. The phase shift equation is ππ = 0.15ωω c arctan 0.5ωω c from which ωω c can be calculated iteratively by direct substitution in ωω c = (ππ arctan 0.5ωω c )/0.15 This converges to ωω c = 11.6 rad/min using, e.g., ωω c = 17 rad/min as starting value. The new gain margin is found by AA m = 1 AA R,l (ωω c ) = 1+( ) > 1 Because AA m > 1, the closed-loop system is stable with LL = 0.15 min. KEH Dynamics and 8 45

46 8.3 Stability analysis of feedback systems Stability margins Phase margin The phase margin φφ m denotes how much the phase shift at the cross-over frequency of the loop transfer function has to change in order to reach the stability limit. The cross-over frequency ωω g is the frequency, where the amplitude ratio of the loop transfer function is 1. Mathematically, the phase margin is defined φφ m = φφ ωω g + ππ (8.88) where ωω g is the frequency that satisfies AA R,l ωω g = 1 (8.89) Stability of the closed-loop system requires φφ m > 0. The phase margin yields robustness not only against variations in the phase shift of the loop transfer, but also against variations in other process parameters. Example Consider the same system as in Example a) Determine a P controller to yield a designed phase margin φφ m = 30. b) Is the closed-loop system stable if the time delay changes to 0.15 min? KEH Dynamics and 8 46

47 8.3.2 Stability margins Example a) We need to find ωω g such that φφ(ωω g ) = φφ m ππ = = 150 = 5ππ/6 ωω g can be read off the phase curve of the loop transfer function at 150. We can also find it numerically by solving 5ππ/6 = 0.1ωω g arctan 0.5ωω g This can be done iteratively by direct substitution in the expression ωω g = (5ππ/6 arctan 0.5ωω g )/0.1 The result is ωω g = rad/min using 11.6 rad/min as starting value. Next, KK c has to be chosen such that AA R,l ωω g = 1, i.e., KK c 1+( ) 2 = 1 KK c = 6.14 b) The stability of the system with KK c = 6.14 and LL = 0.15 min can be checked in many ways. Here, it is simplest to use the result from Ex , where LL = 0.15 min gave the critical frequency ωω c = 11.6 rad/min. (Note that the phase curve is independent of gains.) Thus, 6.14 AA R,l ω c = = 1.04 > 1 1+( ) 2 Since AA R,l > 1, the closed-loop system is unstable with LL = 0.15 min. KEH Dynamics and 8 47

48 8.3 Stability analysis of feedback systems Stability margins Example The gain and phase margins can easily be found from a Bode diagram when the controller gain is given. The figure shows the Bode diagram for the loop transfer function GG l = KK ce 0.1ss with KK c = ss+1 The gain margin is found as follows. ωω c is obtained from the phase curve at φφ l = 180. The gain curve yields the amplitude ratio AA R,l (ωω c ) at ωω c. AA m = 1 AA R,l (ωω c ). The phase margin is found as follows. ωω g is obtained from the gain curve at AA R,l = 1. φφ m is the difference between the phase shift at ωω g and 180. AR, 1 Am ϕ phase margin ϕ m rad/min rad/min KEH Dynamics and 8 48

49 8. Frequency Response Analysis 8.3 Stability analysis of feedback systems Numerical solution of frequency relationships In Example and the phase equation was solved by a simple iterative method, where the frequency appearing together with a time delay was solved out. However, there is no guarantee that the solution will converge, and if there is no time delay, the method cannot even be used. In this section, better methods for solving both the phase equation and the gain equation are developed. The system can be a general system with the transfer function GG ss = KK TT nn+1ss+1 (TT NN ss+1) TT 1 ss+1 (TT nn ss+1) ss mm e LLLL (8.90) Note that mm > 0 if a factor ss appears in the numerator, and mm < 0 if it is in the denominator. Complex-conjugated poles and zeros will also be considered. Eq. (8.90) may be any transfer function of interest, but in practice it will be the loop transfer function of the system. KEH Dynamics and 8 49

50 8.3 Stability analysis of feedback systems Numerical solution The phase equation The system (8.90) has the phase shift φφ r = mm ππ LLLL 2 ii=1 nn NN arctan TT ii ωω + ii=nn+1 arctan TT ii ωω (8.91) The phase shift φφ r, expressed in radians, is assumed to be known, and it is desired to find the frequency ωω satisfying (8.91). The phase shift can have any value relevant for the calculation, but two typical choices are φφ r = ππ, if the critical frequency ωω = ωω c is to be found φφ r = φφ m ππ, if the cross-over frequency ωω = ωω g is to be found The following function is defined: ff ωω φφ r mm ππ + LLLL + 2 ii=1 nn NN arctan TT ii ωω ii=nn+1 arctan TT ii ωω (8.92) Since ff ωω = 0 at the solution, a possibility is to calculate ωω iteratively by ωω kk+1 = ωω kk ρρ kk ff(ωω kk ) (8.93) where ωω kk is the solution at iteration step kk and ρρ kk is a parameter that depends on the iteration method. KEH Dynamics and 8 50

51 8.3.3 Numerical solution of frequency relationships The phase equation According to the Newton-Raphson method, ρρ kk = 1 fff(ωω kk ) (8.94) where ff ωω kk dff(ωω kk ) dωω kk. Differentiation of (8.92) yields ff ωω kk nn = LL + ii=1 TT ii NN 1+(TT ii ωω kk ) 2 ii=nn+1 TT ii 1+(TT ii ωω kk ) 2 (8.95) An initial guess of the frequency is required to start the iteration. This can be any frequency which might be close to the solution. If no such frequency is known, ωω 0 = 0 can be used as starting value. However, this will always result in ρρ 0 = LL + nn ii=1 NN TT ii ii=nn+1 ωω 1 = ρρ 0 (φφ r mmππ 2) TT ii 1 (8.96a) (8.96b) which thus can be used as a better starting value. If desired, the iteration can be continued with ρρ kk = ρρ 0, but better convergence is usually achieved if ρρ kk is recalculated at every step, or at least occasionally the improvement can, in fact, be quite dramatic. If divergence occurs, ρρ kk can be reduced manually, e.g., by halving ρρ kk. KEH Dynamics and 8 51

52 8.3.3 Numerical solution of frequency relationships The phase equation Complex poles and zeros If there are complex poles or zeros, they occur as complex-conjugated pairs. Such a pair has the corresponding complex-conjugated time constants TT jj and TT jj+1, which satisfy (TT jj ss + 1)(TT jj+1 ss + 1) = ss ζζωω n ss + ωω n 2 ωω n (8.97) From this is obtained TT jj + TT jj+1 = 2ζζ ωω n (8.98) arctan TT jj ωω + arctan TT jj+1 ωω = arctan 2ζζωω nωω ωω 2 n ωω 2 (8.99) which can substituted into the appropriate positions in (8.92) and (8.96a). In (8.95), the substitution TT jj 1+(TT jj ωω kk ) 2 + TT jj+1 1+(TT jj+1 ωω kk ) 2 = 2ζζωω n (ωω 2 2 n +ωω kk) ωω 4 n +2 2ζζ 2 1 ωω 2 n ωω 2 kk +ωω 4 (8.100) kk is used. Since the exact value of ρρ kk is not very important, an approximation such as might also be used. TT jj 1+(TT jj ωω kk ) 2 + TT jj+1 1+(TT jj+1 ωω kk ) 2ζζ 1+( ωω kk ωω n ) 2 2 ωω n 1+( ωω kk ωω n ) 4 (8.101) KEH Dynamics and 8 52

53 8.3 Stability analysis of feedback systems Numerical solution The gain equation The system (8.90) has the amplitude ratio AA r = KK ωω mm 1+(TT nn+1ωω) 2 1+(TT NN ωω) 2 1+(TT 1 ωω) 2 1+(TT nn ωω) 2 (8.102) The amplitude ratio AA r, expressed as a pure ratio, is assumed to be known, and it is desired to find the frequency ωω satisfying (8.102). The amplitude ratio can have any value relevant for the calculation, but a typical choices is AA r = 1, if the cross-over frequency ωω = ωω g is to be found The following function is defined: gg ωω AA r KK ωω mm 1+(TT nn+1ωω) 2 1+(TT NN ωω) 2 1+(TT 1 ωω) 2 1+(TT nn ωω) 2 (8.103) Since gg ωω = 0 at the solution, ωω can be calculated iteratively by ωω kk+1 = ωω kk σσ kk gg(ωω kk ) (8.104) where ωω kk is the solution at iteration step kk and σσ kk is a parameter that depends on the iteration method. KEH Dynamics and 8 53

54 8.3.3 Numerical solution of frequency relationships The gain equation According to the Newton-Raphson method, σσ kk = 1 gg (ωω kk ) (8.105) where gg ωω kk dgg(ωω kk ) dωω kk. Differentiation of (8.103) yields gg ωω kk nn (TT = mm + ii ωω kk ) 2 NN (TT ii ωω kk ) 2 AA r gg(ωω kk ) ii=1 1+(TT ii ωω kk ) 2 ii=nn+1 (8.106) 1+(TT ii ωω kk ) 2 ωω kk An initial guess of the frequency is required to start the iteration. This can be any frequency which might be close to the solution. However, ωω 0 = 0 can not be used as a starting value. The critical frequency ωω c, which is often known, would usually be a good starting value. If desired, the iteration can be continued with σσ kk = σσ 0, but better convergence is probably achieved if σσ kk is recalculated at every step, or at least occasionally. If divergence occurs, σσ kk can be reduced manually, e.g., by halving σσ kk. KEH Dynamics and 8 54

55 8.3.3 Numerical solution of frequency relationships The gain equation Complex poles and zeros If there are complex poles or zeros, they occur as complex-conjugated pairs. If the corresponding complex-conjugated time constants are TT jj and TT jj+1, 1 + (TT jj ωω) (TT jj+1 ωω) 2 = ζζ 2 1 is substituted into (8.103). In (8.106), the substitution (TT jj ωω kk ) 2 + (TT jj+1ωω kk ) 2 = 1+(TT jj ωω kk ) 2 1+(TT jj+1 ωω kk ) 2 ωω 2 ωω 4 + ωω n ωω n (8.107) 2ωω 2 kk [ 2ζζ 2 1 ωω 2 2 n +ωω kk] ωω 4 n +2 2ζζ 2 1 ωω 2 n ωω 2 kk +ωω 4 (8.108) kk is used. Since the exact value of σσ kk is not very important, an approximation such as (TT jj ωω kk ) 2 + (TT jj+1ωω kk ) 2 2ωω 4 kk 1+(TT jj ωω kk ) 2 1+(TT jj+1 ωω kk ) 2 ωω 4 n +ωω 4 (8.109) kk might also be used. KEH Dynamics and 8 55

56 8.3 Stability analysis of feedback systems Numerical solution Exercise Calculate KK c,max for the system below using frequency analysis. GG m = 1 ss+1, GG p = 1 5ss+1, GG v = 1 2ss+1, GG c = KK c KEH Dynamics and 8 56

57 8. Frequency Response Analysis 8.4 ler design in the frequency domain In this section it is shown how PI, PD and PID controllers can be designed to satisfy stability and performance criteria in the frequency domain. The used stability criteria are the gain margin AA m and the phase margin φφ m. Usually, AA m 2 and φφ m 45 ( ππ 4) are good values. The cross-over frequency ωω g is related to performance the higher the cross-over frequency, the better the performance. Usually, a good value is ωω g 0.3ωω c, where ωω c is the critical frequency of the uncontrolled (or P- controlled) system Design of PI controllers A PI controller has the transfer function GG PI ss = KK c TT i ss = KK c TT i ss+1 TT i ss (8.110) If the system to be controlled has the transfer function GG(ss), the loop transfer function is TT GG l ss = GG ss GG PI ss = GG(ss)KK i ss+1 c (8.111) TT i ss KEH Dynamics and 8 57

58 8.4 ler design in the frequency domain PI controller The amplitude ratio and the phase shift of the loop transfer function are AA R,l ωω = GG(jωω) KK c TT i ωω 1 + (TT iωω) 2 (8.112) φφ l ωω = arg GG(jωω) + arctan TT i ωω ππ 2 (8.113) Design for desired phase margin The integral time TT i 5 ωω g, where ωω g is the cross-over frequency, is usually a good choice for a PI controller. Based on this choice, a PI controller for a desired phase margin φφ m can be designed as follows. 1. Solve (8.113) for ωω = ωω g with φφ l = ππ + φφ m and TT i ωω g = 5, i.e., φφ m ππ 2 arctan 5 arg GG jωω g = 0 (8.114) 2. Solve (8.112) for KK c with AA R,l = 1, ωω = ωω g, and TT i ωω g = 5, i.e., 3. The integral time is TT i = 5 ωω g. KK c = 5 26 GG(jωω g) 1 (8.115) KEH Dynamics and 8 58

59 8.4 ler design in the frequency domain PI controller Example Design a PI controller for a system with the transfer function GG ss = 10ss+1 to achieve the phase margin a) φφ m = 30, b) φφ m = 60. Also calculate controller tunings by the classical methods in Section 7.4 and 7.5. a) Eq. (8.114) with the pertinent expression for arg GG jωω g (see Section 8.1.6) has to be solved with φφ m = 30 = ππ 6. This can be done iteratively according to (see Section 8.3.3) e ss ωω kk+1 = ωω kk ρρ kk ff(ωω kk ) where ff ωω kk = ππ 3 arctan 5 + ωω kk + arctan 10ωω kk ρρ kk = ωω kk 2 Using ρρ 0 = 1 11 and ωω 1 = 7ππ (11 9) 0.22 as starting values yields ρρ 1 = 0.37, ωω 2 = 0.61, ρρ 2 = 0.79, ωω 3 = 0.93, ρρ 3 = 0.90, ωω 4 = = ωω g. 1 The controller gain is KK c = / and the integral time is TT i = 5 ωω g = = KEH Dynamics and 8 59

60 8.4.1 Design of PI controllers Example b) φφ m = 60 = ππ 3 means that ff ωω kk = ππ 6 arctan 5 + ωω kk + arctan 10ωω kk A larger phase margin means a lower cross-over frequency. Based on a), we might therefore choose ωω 0 = 0.9 as starting value for the iteration. This gives ρρ 0 = 0.9, which will be used throughout the calculation to get ωω 1 = 0.48, ωω 2 = 0.53, ωω 3 = 0.514, ωω 4 = , ωω 4 = , ωω 6 = = ωω g. The controller gain is KK c = / and the integral time is TT i = 5 ωω g = The result of all considered methods is summarized below. PI Param. φφ m 30 Z-N ITAE Regul. CHR 20% Regul. CHR 0% Regul. φφ m ITAE 60 Track. CHR 20% Track. CHR 0% Track. KK c TT i KEH Dynamics and 8 60

61 8.4.1 Design of PI controllers Example The figure shows simulated setpoint responses for a) φφ m = 30 (full line), b) φφ m = 60 (dashed line). y t KEH Dynamics and 8 61

62 8. Frequency Response Analysis 8.4 Design in the frequency domain Design of PD controllers A PI controller can be designed to yield a desired phase margin, but it might require a low cross-over frequency, which means that the performance might not be good enough. A PD controller can be designed for a desired phase margin and a desired cross-over frequency. A PD controller with a derivative filter has the transfer function GG PDf ss = KK c 1 + TT dss TT f ss+1 = KK c (TT d +TT f )ss+1 TT f ss+1 (8.116) where TT f is the filter time constant. The PD controller causes a phase shift φφ PDf = arctan(tt d +TT f )ωω arctan TT f ωω = arctan TT d ωω 1+(TT d +TT f )TT f ωω 2 (8.117) For positive TT d and TT f, this phase shift is positive. The maximum phase lift is obtained at the frequency ωω = ωω max, where This gives the phase lift ωω max = (TT d + TT f )TT f 1/2 (8.118) φφ max = arctan(0.5tt d ωω max ) (8.119) KEH Dynamics and 8 62

63 8.4 ler design in the frequency domain PD controller Design for desired phase margin and cross-over frequency It is the derivative of the PD controller that produces the phase lift. The phase lift should be enough to satisfy the phase margin requirement, but excessive derivative action is not desired. This means that the maximum phase lift should be at the cross-over frequency, i.e., ωω max = ωω g. If the system to be controlled has the transfer function GG(ss), the loop transfer function is (TT GG l ss = GG ss GG PDf ss = GG(ss)KK d +TT f )ss+1 c (8.120) TT f ss+1 With the maximum phase lift at the cross-over frequency, (8.118) and (8.119) with ωω max = ωω g apply. The amplitude ratio and phase shift equations at the cross-over frequency then become GG(jωω g ) KK c TT f ωω g = 1 (8.121) φφ m ππ arg GG(jωω g ) arctan(0.5tt d ωω g ) = 0 (8.122) KEH Dynamics and 8 63

64 8.4.2 PD controller Desired phase margin and cross-over frequency A PD controller for desired phase margin and cross-over frequency can now be designed as follows. 1. Calculate the derivative time from (8.122), i.e., TT d = 2 tan φφ ωω m ππ arg GG(jωω g ) (8.123) g 2. Calculate the derivative filter time constant from (8.118), i.e., TT f = 0.5TT d TT 2 2 d + ωω g (8.124) 3. Calculate the controller gain from (8.121), i.e., KK c = TT f ωω g GG(jωω g ) 1 (8.125) KEH Dynamics and 8 64

65 8.4 ler design in the frequency domain PD controller Example Design a PD controller with a derivative filter for a system with the transfer function GG ss = e ss 10ss+1 to achieve the phase margin φφ m = 60 and the cross-over frequency 1 rad/time unit. Eq. (8.123) gives TT d = 2 ωω g tan 1 3 ππ ππ + ωω g + arctan 10ωω g Eq. (8.124) gives = 2tan 2 ππ arctan 10 = TT f = 0.5TT d TT d 2 + ωω g 2 Eq. (8.125) gives = = 0.68 KK c = TT f ωω g 1 + (10ωω g ) 2 = = 6.83 KEH Dynamics and 8 65

66 8. Frequency Response Analysis 8.4 Design in the frequency domain Design of PID controllers A PD controller can be designed for stability (phase margin) and performance (cross-over frequency), but a drawback is that the steady-state control error will not be zero due to the lack of integral action. This can be remedied by connecting a PI and a PD controller in series. However, the previous PI and PD controller designs are not necessarily optimal for that purpose. A PID controller on series form with a derivative filter has the transfer function GG PIPDf ss = KK c TT i ss 1 + TT d ss TT f ss+1 = KK c (TT i ss+1)((ttd +TTf )ss+1) TT i ss(tt f ss+1) where primed parameter symbols are used to distinguish them from the corresponding parameters in the parallel form of the PID controller. The amplitude ratio and the phase shift equations of (8.126) are AA R,PIPDf ωω = KK c TT i ωω [1+(TT i ωω) 2 ][1+ TT 2 d +TTf ωω 2 ] [1+(TT f ωω) 2 ] 1/2 (8.126) (8.127) φφ PIPDf ωω = ππ 2 + arctan TT i ωω + arctan TT d + TT f ωω arctan TT f ωω (8.128) KEH Dynamics and 8 66