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# Effects of poles and zeros affect control system

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### Effects of poles and zeros affect control system

1. 1. Effects of poles and zeros on performance of control systems1 Dominantly ﬁrst order systems. Effects of additional poles and zeros on performance of a ﬁrst order control system1.1 Step response of a ﬁrst order systemConsider a ﬁrst order system of the form 1 x = −ax + r, ˙ or, equivalently, X(s) = R(s). s+aLet r(t) be a unit step input, i.e, R(s) = 1 . Then, assuming x(0) = 0, s 1 1/a 1/a X(s) = = − , s(s + a) s s+aand 1 1 x(t) = (1 − e−at ), and x(t) → as t → ∞. a aThe output of the system monotonically approaches to the ﬁnal value of 1/a; the rate of con-vergence is exponential and is determined by the pole of the system at s = −a. Note that thetime constant of the transient is τ = 1/a is also related to the location of the pole. To achieve adesired rate of convergence, the designer must place the pole at a corresponding location; e.g.,using the state or output feedback; see Figure 1.Example Suppose the plant is quite slow, its time constant is τ orig = 100 sec, 1 x = −0.01x + u, ˙ or, equivalently, X(s) = U (s). s + 0.01To speed up it to τdesired = 1 sec, use the feedback controller u = −kx + r.Then, the closed loop system is 1 x = −(0.01 + k)x + r, ˙ or, equivalently, X(s) = R(s). s + 0.01 + kTo achieve the desired time constant of 1 sec, the pole of the closed loop system must be placedat s = −1/τdesired = −1, i.e., we need a = (0.01 + k) = 1, which is achieved using the gain ofk = 0.99. 1
2. 2. R U Y + Plant − KPSfrag replacements Feedback Controller Figure 1: Static feedback control system. 1.2 The Effect of an Additional Pole A ﬁrst order system often serves as an approximation of a system of higher order. Therefore, it is important to study the error of approximation. To this end, we examine the effect of additional poles and zeros on a ﬁrst-order system. This will explain the difference between an original system and its ﬁrst-order approximation. We now examine the step response of a system whose transfer function is given by Y (s) p 1 = = 1 . (1) R(s) (s + 1)(s + p) (s + 1)( p s + 1) That is, we are concerned with a 2nd order system which is approximated by the ﬁrst order approximation s+a , where a = 1. 1 Note on the form of the system. Eq. (1) is selected because its value at s = 0 equals 1 for any p. We will see that the step response of the system will approach the value of 1 for large values of t, independent of the value of p. This is convenient, since we wish to examine the step response for different values of p. However, the conclusions drawn from the example will apply to any second order system with two real poles. Let R(s) = 1/s and perform a partial-fraction expansion on the resulting step response transform Y (s): p 1 p 1 Y (s) = = − p−1 + p−1 . s(s + 1)(s + p) s s+1 s+p The step response y(t) is given by p −t 1 −pt y(t) = 1 − e + e . (2) p−1 p−1 p We will consider this response as the sum of two terms. The ﬁrst term is given by 1 − p−1 e−t , and the second term is p−1 e−pt . 1 Let us examine the step response of Eq. (2) for different values of p: 2
3. 3. 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 p=infinity p=10 0.2 p=1 p=0.5 0.1 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Figure 2: Step response of the system with the additional pole at −p.pv=[.5 1 10 inf];t=[0:.1:5];yv=[];for i=4:-1:1 p=pv(i); G=tf(1,conv([1 1],[1/p 1])); [y,x]=step(G,t); yv=[yv y];endplot(t,yv(:,1),’-’,t,yv(:,2),’--’,t,yv(:,3),’-.’,t,yv(:,4),’.’)legend(’p=infinity’,’p=10’,’p=1’,’p=0.5’);gridThis code produces the step response of Eq. (2) for p = 0.5, 1, 10 and ∞; see Figure 2. Forp = ∞, the step response is the same as the step response of the system with the ﬁrst-ordertransfer function s+1 . As p reduces, the plot moves away from the step response of this ﬁrst- 1order system. Indeed p/(p − 1)e−t → e−t as p → ∞, and 1/(p − 1)e−pt → 0 as p → ∞. In the latter limit,two factors matter: 1/(p − 1) p/(p − 1), and also e−pt has a time constant 1/p decreases asp → ∞, hence this is a fast term. The conclusion can be drawn from Figure 2 that if p 1, theterm p−1 e corresponds to the fast part of the step response. The effect of this term (and hance 1 −ptthe effect of the additional pole at s = −p is negligibly small as t → ∞. Furthermore, the term p1 − p−1 e−t describes the dominant part of the steop response. Since the response is similar tothe response of the ﬁrst-order system, the system with p 1 is said to be dominantly ﬁrst-order.Also, we say that the pole at s = −1 dominates on the pole at s = −p; see Figure 3 3
4. 4. Im Re -p -1 Dominant pole Figure 3: Pole locus of a dominantly ﬁrst-order system. Additional zero Im Re -p -z -1 Dominant pole Figure 4: Zero-pole locus of the system with the additional zero at s = −z.1.3 The Effect of a Zero on a Dominantly First-order SystemWe have seen that an additional pole retards a dominantly ﬁrst-order system as the pole movesalong the negative real axis. We now examine the effect of moving a zero in along the negativereal axis. The system to be studied is given by 1 Y (s) s+1 = z 1 . (3) R(s) (s + 1)( 10 s + 1)If we set z = ∞, then we obtain the system of the previous section in which p = 10. We haveseen that for the example in the previous section, this value of the parameter p was sufﬁcentlylarge in order to say that the system was a dominantly ﬁrst-order system. The zero-pole locusof the system (3) is shown in Figure 4. We are again interested in the step response, that isR(s) = 1/s, 1 z s+1 Y (s) = 1 . s(s + 1)( 10 s + 1)Take the partial fraction expansion of Y (s): 10 z−1 1 z−10 1 Y (s) = − 9 z + 9 z . s s+1 s + 10The corresponding step response is then found to be: 10 z − 1 −t 1 z − 10 −10t y(t) = 1 − e + e . 9 z 9 z 4
5. 5. 4.5 z=∞ 4 z=10 z=2. z=1. 3.5 z=.5 z=.2 3 2.5 2 1.5 1 0.5 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5Figure 5: Step response of the dominantly ﬁrst-order system with the additional zero at s = −z.Effects of the additional zero:z 10. We can see that if z 10, then 10 z−1 9 z ≈ 10/9, 1 z−10 9 z ≈ 1/9, and 10 −t y(t) ≈ yd (t) = 1 − e for sufﬁciently large t. 9 Thus, the pole at s = −1 remains dominant.z = 10. This zero cancels the pole at s = −10. The system becomes a ﬁrst-order system.1 < z < 10. In this case, the additional zero speeds up the system; see Figure 5.z = 1. This cancels the pole at s = −1.z < 1. The additional zero becomes dominant. It speeds up the system and at the same time leads to occuring an overshoot; see Figure 5.From this analysis, one can see the general effect that the speed of the response increases as zeromoves from +∞ to 0 along the negative real axis. When zero becomes dominant, an overshootoccurs. To obtain Figure 5, the following Matlab code was used:zv=[.2 .5 1. 2. 10 inf];t=[0:.1:5];yv=[];for i=6:-1:1 z=zv(i); G=tf([1/z 1],conv([1 1],[1/10 1])); 5
6. 6. [y,x]=step(G,t); yv=[yv y];endplot(t,yv(:,1),’--’,t,yv(:,2),’-’,t,yv(:,3),’-.’,t,yv(:,4),’-o’,t,yv(:,5),’-x’,t,yv(:,6),’-*’)legend(’z=infty’,’z=10’,’z=2.’,’z=1.’,’z=.5’,’z=.2’);grid;1.3.1 The Effect of a Right Half-plane ZeroWe now examine the step response for the system given by Eq. (3) for the case when z is negative.A negative value for z corresponds to a zero in the right half-plane. The found expression fot the step response does not depend on whether z is negative orpositive. Let us ﬁnd the derivative y(t) at t = 0: ˙ dy 10 z − 1 −t 10 z − 10 −10t 10 = e − e = . dt t=0 9 z 9 z t=0 z When z is positive, corresponding to a left half-plane zero, the derivative is also positive;when z is negative, corresponding to a right half-plane zero, the derivative is negative, indicatingthat the response starts off in the negative direction. As the magnitude of a positive z decreases,corresponding to a zero moving along the negative real axis toward the origin, the increasingmagnitude of the derivative indicates, ﬁrst a more rapid response caused by the zero and, then,an increase in overshoot. For right half-plane zeros moving along the positive real axis towardthe origin, the increase in the magnitude of the derivative indicates an increase in the magnitudeof the overshoot going in the opposite direction (“undershoot”) and an increase in the delaybefore the response approaches its ﬁnal value. This conclusion is illustrated in Figure 6. Thecorresponding Matlab code is the following:zv=[-.2 -.5 -2. -inf];t=[0:.1:5];yv=[];for i=4:-1:1 z=zv(i); G=tf([1/z 1],conv([1 1],[1/10 1])); [y,x]=step(G,t); yv=[yv y];endplot(t,yv(:,1),’-’,t,yv(:,2),’--’,t,yv(:,3),’-.’,t,yv(:,4),’-x’)legend(’z=-infty’,’z=-2.’,’z=-0.5’,’z=-0.2’);grid;1.4 Summary 1. We have seen that if there is a single pole that is signiﬁcantly closer to the origin than other poles and zeros of a transfer function with all its poles in the left halfplane, the time constant of that pole closest to the origin dominates the response of the system. 6
7. 7. 1 0.5 0 −0.5 −1 −1.5 −2 z=−∞ −2.5 z=−2. z=−0.5 z=−0.2 −3 −3.5 −4 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Figure 6: Effect of an additional zero in the right half-plane.2. The response from the dominant pole is modiﬁed from a pure ﬁrst-order system response by the presence of other poles and zeros. Additional poles delay the response of the system while left half-plane zeros speed up the response. Right halfplane zeros cause the response to start off in the wrong direction before recovering. The effect increases as either a pole or zero moves toward the origin.3. If there is a zero and a pole modifying the effect of another dominant pole and the modi- fying zero is closer to the origin than the modifying pole, the response from the dominant pole is modiﬁed more by the zero than the pole and the response is slightly advanced or sped up in time. If the modifying pole is closer to the origin than the modifying zero, the response is modiﬁed more by the pole than the zero and the response is slightly delayed. 7
8. 8. 2 Effects of additional poles and zeros on performance of a second order underdamped systemIt often happens that the poles closest to the origin are a complex pole pair. Hence, the dominantresponse will be from those poles which correspond to a second-order system. We will see inthis section that additional poles and zeros modify the response of a second-order system in asimilar way to how the response of a ﬁrst-order system is modiﬁed. The discussion is based upon the second-order differential equation d2 y dy 2 2 2 + 2ζωn + ωn y(t) = ωn r(t), (4) dt dtwhere r(t) is a step-function input and the initial conditions are assumed to be zero. The constantζ is called the damping ratio, and ωn is referred to as the undamped natural frequency. There are numerous examples in which the above differential equation occurs such as a serieselectrical circuit containing resistors, inductors and capacitors, a system consisting of a spring,mass, and viscous friction etc.2.1 The derivation of the unit step response of the prototype second order system with two complex conjugate polesThe transfer function of the system of Eq. (4) is 2 Y (s) ωn = 2 2 (5) R(s) s + 2ζωn s + ωnThe transfer function (5) has poles located at s = −ζωn ± ωn (ζ 2 − 1) (6)These poles are real or complex conjugate depending on ζ. If ζ < 0, the poles are in the right halfplane, hence the system is unstable. If ζ > 1 then the system has two real poles. This case has already been studied. We knowthat the step response is determined to a large extent by a dominant pole in this case. The systemis “overdamped”, its step response approaches the ﬁnal value. If ζ = 1 then the system has a double pole at −ζωn . This is a critical case (“critically dampedsystem”). The step response has the form R1 + R2 e−ζωn + R3 te−ζωn ;R1 , R2 , R3 are constants. If 0 < ζ < 1, poles are complex conjugate and located in the left half-plane: s = −ζωn ± jωn (1 − ζ 2 ) (7) We focus on the last case. The location of the poles with respect to the damping ratio and theundamped natural frequency is indicated in Fig. 7. To investigate the response of the system, we 8
9. 9. Pole jω s−plane ωn √ ωd = ω n 1 − ζ 2 θ 0 Re ωn ζPSfrag replacements Pole Figure 7: Location of the poles with respect to ζ and the ωn chose the input as a step input. For R(s) = 1/s, Y(s) is given by 2 ωn Y (s) = . s [(s + ζωn )2 + ωn (1 − ζ 2 )] 2 The partial fraction expansion is found for the case of the pair of complex conjugate poles and a single real pole: R1 R R¯ Y (s) = + √ + √ s s + ζωn + jωn 1 − ζ 2 s + ζωn − jωn 1 − ζ 2 We use the residues method: R1 = 1 2 ωn R = √ √ s(s + ζωn − jωn 1 − ζ 2 ) s=−ζωn −jωn 1−ζ 2 −j = √ √ 2(ζ + j 1 − ζ 2 ) 1 − ζ 2 1 ζ = − 1 + j√ 2 1 − ζ2 1 √ 2 ejtan (ζ/ 1−ζ ) −1 = − √ 2 1 − ζ2 1 = − √ ej(π/2−θ) 2 1 − ζ2 9
10. 10. where θ = cos−1 ζ.We now use the table of Laplace transforms and write y(t) as 1 π y(t) = 1 − √ 2 e−ζωn t cos(ωn t 1 − ζ 2 + θ − ) 1−ζ 2 1 = 1− √ 2 e−ζωn t sin(ωn t 1 − ζ 2 + θ) 1−ζ √The value ωd = ωn 1 − ζ 2 is the actual frequency of oscillation in radians per second, it isreferred to as the damped frequency. The period of oscillation, T d , associated with the dampedfrequency ωd is 2π 2π Td = = √ . ωd ωn 1 − ζ 2 Typical responce of the system is shown in Fig. 8. In addition to the period and frequency of Step Response 1.5 Maximum overshoot Td 105% 1 95% 90% Amplitude 0.5 Settling time 10% 0 0 5 10 15 20 25 Time (sec) Rise time Figure 8: Typical unit step response of an underdamped (0 < ζ < 1) second-order system.damped oscillations, other important transient performance characteristics of a stable 2nd orderunderdamped system are 10
11. 11. − √ πζ • Percent maximum overshoot, P O = e 1−ζ 2 × 100%; • Rise time shows how long it takes for the response to rise from from 10% of the ﬁnal value to 90% of the ﬁnal value. There is no exact equation to express the rise time. • Settling time shows how long it takes for transients to settle. The settling time is measured as a time required for the response to settle to within ±5%, or in some cases ±2% of the ﬁnal value. In the ﬁrst case, the settling time is approximately equal to t setlle = ζωn . 3Effect of varying the damping ratio ζ Let ωn = 1 be ﬁxed. Step responces for variousdamping ratios are given in the Figure 9. The overshoot decreases as ζ increases. Howeverthe decrease in the overshoot is at the expense of rise time. While the response becomes moresluggish, its settles quicker as ζ increases.Effect of varying the undamped natural frequency ωn Varying the undamped natural fre-quency ωn simply scales the time axis, see the Figure 10. It affects the period of damped oscilla-tions, rise and settling time, but has no effect on the percent maximum overshoot.2.2 The effect of additional poles and zeros on a dominantly second order systemThe effects that added poles and zeros have on dominantly second-order systems are similar tothe effects that added poles and zeros have on dominantly ﬁrst-order systems.2.2.1 An additional zero in the closed loop transfer functionFigure 11 displays the step responses of the system given by ωn 2 Y (s) (s + z) = 2 z 2 R(s) s + 2ζωn s + ωnThis system has the same poles as the system in equation (5). Also, it has a single zero at s = −z. The effect of additional left-halfplane zero can be seen from the following analysis. The unitstep response of the above transfer function can be written as follows ωn2 z (s + z) Y (s) = s(s2 2 + 2ζωn s + ωn ) ωn 2 2 s ωn z = 2 + 2ζω s + ω 2 ) + 2 + 2ζω s + ω 2 ) s(s n n s(s n n 2 ωn dy2nd order (t) y(t) = y2nd order (t) + . z dtHence, the additional zero in the left half-plane speeds up transients, making rises and fallssharper. Smaller values of z make this effect more prominent. As a result, an additional zero 11
12. 12. Pole−Zero Map 1.2 0.74 0.6 0.42 0.2 0.83 ζ=0.1 1 ζ=0.3 ζ=0.5 0.8 0.91 ζ=0.707 0.6 Imag Axis 0.96 0.4 0.99 0.2 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 −0.2 −1.6 −1.4 −1.2 −1 −0.8 −0.6 −0.4 −0.2 0 Real Axis 1.8 ζ=0.1 1.6 ζ=.3 ζ=.5 ζ=.707 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 5 10 15 20 25 Step responsesFigure 9: Second order system with various damping ratios. Locations of poles on the complexplane are obtained using Matlab function pzmap. The (ζ, ω n )-grid can be added using grid orsgrid. 12
13. 13. Pole−Zero Map 1.5 0.74 0.6 0.42 0.22 ω =2 0.84 n 1 System: sys Pole: −0.707 + 0.707i 0.91 Damping: 0.707 Overshoot (%): 4.32 Imag Axis Frequency (rad/sec): 1 ω =1 n 0.96 0.5 ωn=0.5 0.99 2 1.75 1.5 1.25 1 0.75 0.5 0.25 0 −2 −1.8 −1.6 −1.4 −1.2 −1 −0.8 −0.6 −0.4 −0.2 0 Real Axis 1 0.8 0.6 ωn=0.5 ωn=1. ωn=2. 0.4 0.2 0 0 5 10 15 20 25Figure 10: Second order system with various undamped natural frequencies. 13
14. 14. zv=[0.5 1 2 10 inf];t=[0:.1:5];omega_n=1;zeta=sqrt(2)/2;yv=[];for i=5:-1:1 z=zv(i); G=tf([omega_nˆ2/z omega_nˆ2],[1 2*zeta*omega_n omega_nˆ2]); [y,x]=step(G,t); yv=[yv,y];endplot(t,yv(:,1),’-’,t,yv(:,2),’--’,t,yv(:,3),’-.’,t,yv(:,4),’-o’,t,yv(:,5),’-grid;legend(’z=infty’,’z=10’,’z=2’,’z=1’,’z=0.5’); Effect of an addtitional zero, ζ=0.707, ωn=1 1.5 1 0.5 z=∞ z=10 z=2 z=1 z=0.5 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Figure 11: Effect of an additional left halfplane zero on a dominantly second-order system 14
15. 15. in the left half-plane makes the system faster and more oscillatory. This can be seen from thesimulations. As the zero moves along the negative real axis toward the origin, the time to the ﬁrst peak ofthe step response decreases monotonically while the percent overshoot increases monotonically.Also, it takes longer for the system to settle to the ﬁnal value of the response. The zero in the right half-plane retards the system and produces an undershoot. The persentundershoot decreases as the zero moves along the positive real axis toward the inﬁnity, see Fig-ure 12. Again the system oscillates for a longer time.zv=[-0.5 -1 -2 -10 -inf];t=[0:.1:5];omega_n=1;zeta=sqrt(2)/2;yv=[];for i=5:-1:1 z=zv(i); G=tf([omega_nˆ2/z omega_nˆ2],[1 2*zeta*omega_n omega_nˆ2]); [y,x]=step(G,t); yv=[yv,y];endplot(t,yv(:,1),’-’,t,yv(:,2),’--’,t,yv(:,3),’-.’,t,yv(:,4),’-o’,t,yv(:,5),’-grid;legend(’z=-infty’,’z=-10’,’z=-2’,’z=-1’,’z=-0.5’); Effect of an addtitional zero in the right half−plane, ζ=0.707, ωn=1 1.2 1 0.8 0.6 0.4 0.2 0 z=−∞ z=−10 −0.2 z=−2 z=−1 z=−0.5 −0.4 −0.6 −0.8 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5Figure 12: Effect of an additional zero in the right half-plane on a dominantly second-ordersystem 15