a. Closedloop system; b. equivalent transfer function Then the CLTF () T is s the poles of () T are s from a contribution of a


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1 Root Locus Simple definition Locus of points on the s plane that represents the poles of a system as one or more parameter vary. RL and its relation to poles of a closed loop system RL and its relation to transient response and stability RL to select a parameter (such as K) to meet closed loop transient response specifications a. Closedloop system; b. equivalent transfer function NG () s NH () s If G() s and () H s D ()() s D s Then the CLTF () T is s G KNG ()() s DH s T () s D ()()()() s D s KN s N s G H G H The zeros of () T are s from ()() Nands D s the poles of () T are s from a contribution of a lot factors. Also the transient response is affected by the poles and zeros of () G and s (). H s The RL give a good representaion of the poles of T() as s varies. K G H H
2 Vector representation of complex numbers: a. s = + j; b. (s + a); c. alternate representation of (s + a); d. (s + 7) s5 + j Consider a function of the form F() s m j s zi i n s p j The parameter m is the number of zeros and n is the number of poles which are complex factors. The magnitude M of F() to s any point iss m s zi i M n s p j j The term is the magnitude of the vector from the zeros of F() at s to z the point. s Similarly, s z i s p j i is the magnitude of the vector from the poles of () F at s p to the point. s j The angle of () F s to any point s is = zero angles  poles angles s zi s p j
3 The zero angle is measured in the positive sense from the vector starting at the zero at  z on the s plane to the s point in question. i The pole angle is measured in the positive sense from the vector starting at the pole  p on the s plane to the s point in question. j Find the vector representation M of F(s) to the point +j M? F() s s s s The vector from zero at  to the s point is The vector from pole at to the s point is (8 tan) 6.6 (8 tan) The vector from pole at  to the s point is Using the Eq. m s zi i M n s zi s pi s pi i (8 tan) 7. M which is evaluating F() at s the point + j
4 Courtesy of ParkerVision. a. CameraMan Presenter Camera System automatically follows a subject who wears infrared sensors on their front and back (the front sensor is also a microphone); tracking commands and audio are relayed to Camera Man via a radio frequency link from a unit worn by the subject. b. block diagram. c. closedloop transfer function. Pole location as a function of gain for the system a. Pole plot from Table b. root locus
5 Root Locus Definition of the RL The root locus of a closed loop TF is a representation of a continuous path of the closed loop poles on the splane as the gain K or other parameter is varied from  to +. For this course, the parameter K is Properties of the RL Consider the CLTF () T s KG() s ()() KG s H s A pole s exists when the characteristic polynomial in the denominator becomes zero. Therefore, ()() KG s H s 5
6 That pole can be represented by a vector that has magnitude and an angle. Therefore, a value of s is a closed loop pole if KG()() s H s This is called the magnitude criterion. KG()()( s H s )8 k ;, k,,... i.e. an odd multiple of 8. This is called the angle criterion. A pole s exist when the char. eqn becomes zero or KG()() s H s ( )8 k The value of K can be evaluated as K G()() s H s Since the magnitude of KG(s)H(s) is unity, K can be solved as above once the pole value is substituted. So satisfying the angle and magnitude criteria of KG(s)H(s) indicates that the s value is a pole on the root locus. Example Prove whether that the s point +j is on the RL of a open loop system as KG()() s H s K( s )( s ) ( s )( s ) If the s point +j is on the RL of the system, then the magnitude and angle criteria are satisfied. 6
7 a. Example system; b. polezero plot of G(s) Vector representation of G(s) from + j Test point +j Using the angle criterion i j s z s p (k )8 ; k,,,... Using the previous figure, we need to evaluate all the angles from the zeros and poles to the point in question ( j) and observe whether the result is an odd multiple of 8. The result of has to be evaluated. 7
8 This image cannot currently be displayed. tan() 56. tan() tan() ( )8 k ; k,,,... +j Therefore, the point j is not on the root locus of K( s )( s ). ( s )( s ) The point j is a point on the RL so the angle add up to a 8 (check). Now the gain K has to be evaluated at this point of j using the magnitude criterion pole vector lengths i K = n G()() s H s M zero vector lengths m j L. L L..77 L L L K. L L. ( )( ) The point j is a point on the RL of ( )( ) with a gain of.. K K s s s s 8
9 Sketching the root locus Number of branches the number of branches of the root locus equals the number of closed loop poles Symmetry root locus is symmetrical about the real axis Real axis segment on the real axis, for K> the root locus exists to the left of an odd number of real axis, finite openloop poles and/or finite open loop zeroes example Complete root locus for the system The number of paths to infinity = n (poles)  m (zeros) none 9
10 From Matlab.8 Root Locus System: sys Gain:. Pole: i Damping:.9 Overshoot (%):. Frequency (rad/sec):..6.. Imag Axis Stable for all K Real Axis Start and end points of the RL T () s KN G ()() s DH s D ()()()() s D s KN s N s G H G H The RL begins at K and ends at K As K ; () T s KN G ()() s DH s D ()() s D s G H As approaches zero due to K, the closed loop poles of () T s becomes the poles of D ( s) D (). s G H This implies that the RL commences at the poles of ()(). D s D s G H The RL therefore begins at the poles of the loop transfer function at K. Analytically, this can also be seen from D ()()()() s D s KN s N s G H G H As K ;()() D sd s G H RL end T () s T () s KNG ()() s DH s D ()()()() s D s KN s N s G H G H KNG ()() s DH s KN ()() s N s G H As K ; The poles of T() is s therefore the ze ros of N()() s N s G H
11 KNG ()() s DH s T () s D ()()()() s D s KN s N s D ()()()() s D s KN s N s Dividing by K K DG ()() s DH s NG ()() s NHs K N ()() s N s G H G H G H G H G H The RL ends at K, at the zeros of the open loop transfer function N()(). s N s G H Therefore, the RL starts at the poles of ()() G sand H sends at the zeros of ()() G s(the H sopen loop TF). Behavior at infinity If there are n poles of P(s) and m finite zeros of P(s), the number of loci that approaches infinity as K approaches infinity is nm. They will approach infinity along asymptotes with angles of 8 (nm=); +9 (nm=); 8 and +6 (nm=), or +5 and +5 (nm=). Angles 8 k.6 k, k,,,, n m n m Real axis intercept p p pn z z zm c, n m n m
12 Example The real axis intercept a The angle of the lines that intersect at / are θ a = π/ for k = θ a = π for k = θ a = 5π/ for k = Root locus and asymptotes for the system G() s K( s ) s 7s s 8s # of Paths to infinity n m = # of zeros # of poles Root Locus Imag Axis System: sys Gain: 9.5 Pole: i Damping: .87 Overshoot (%): Frequency (rad/sec): Stable up to a limiting K value Real Axis
13 Root Locus Imag Axis System: sys Gain:.5 Pole: .5 Damping: Overshoot (%): Frequency (rad/sec): K at the breakaway point Real Axis Root locus example showing real axis breakaway ( ) and breakin points ( ) Variation of gain along the real axis for the previous root locus
14 The previous plot show that the gain reaches a maximum between the poles (where K starts off at ). This occurs at the breakaway point. The gain is a minimum as the RL plot comes back on the real axis and goes towards the zeros (K becomes infinite). This occurs at the breakin point. Therefore, we can use basic calculus to find the breakaway and breakin points first method. Repeating part (d), Recall Recall K G()() s H s Subst s in the above K G()() H dk() Differentiating with respect to, with d we can solve for the values of (or valu s es) where the RL leaves and arrives on the real axis (the breakaway point and breakin points). Example K s s K s s KG()() s H s ( )( 5)( 8 5) ( s )( s ) s s Subst. s K ( 8 5) Making K the subject of the fromula K 8 5 Diff. with respect to and equating to dk d 6 6 ( 8 5) 6 6 Solving for gives.5 and =.8 the breakaway and breakin ponts
15 Second method breakaway and breakin point without differentiation (transition method). These points satisfy the relationship: m n p where zi and pi are the negative of the zero and pole values From the example zi and.8 i Data for breakaway and breakin points for the root locus j crossing CLTF () T s K( s ) s 7s (8) s K s K Completing the Routh array 5
16 Only s can from a row of zeros Solving K 65K 7 K 9.65 Using the auxillary equation of the s term and K 9.65 (9) K K8.5 s.7 s j.59 The RL crosses the imaginary axis at s j.59 at a gain K At this gain, marginal stability occurs. Also, the system is stable for K 9.65 Angles of departure from poles of P(s) and angles of arrival at finite zeros of P(s) can determined by application of the angle criterion to a point selected arbitrarily close to the departure or arrival point Example Given a unity feedback system that has a forward TF K( s ) G() s, do the following ( s s ) (a) Obtain a RL using Matlab Roots: +i (b) Find the complex poles that crosses on the imaginary axis (c) Determine the gain at the j crossing (d) Determine the breakin point Discuss plant stability 6
17 Repeating part (d), Recall Recall K G()() s H s Subst s in the above K G()() H dk() Differentiating with respect to, with d we can solve for the values of (or valu s es) where the RL leaves and arrives on the real axis (the breakaway point and breakin points). 7
18 K( s ) KG()() s H s Subst s ; K dk d ( s s ) ( )( )( ), 7 ( ) The breakin point occurs at s 7. Example Given a unity feedback system that has a forward TF K( s )( s ) G() s, do the following ( s 6s 5) (a) Obtain a RL using Matlab (b) Find the complex poles that crosses on the imaginary axis (c) Determine the gain K at the j crossing (d) Determine the breakin point (e) Find the point where the RL crosses the.5 damping ratio line (f) Find the gain at the point where the RL crosses the.5 damping ratio line (g) Find the range of gain K, for which the system is stable 8
19 Root Locus System: sys Gain: Pole:.6 +.5i Damping: .65 Overshoot (%): Frequency (rad/sec):.5 Imag Axis Real Axis Root Locus Imag Axis System: sys Gain: 5. Pole: e8i Damping:  Overshoot (%): Inf Frequency (rad/sec): Real Axis 9
20 System: sys Gain:.8 Pole: i Damping:.5 Overshoot (%): 6. Frequency (rad/sec):.8 Root Locus Real Axis Root Locus System: sys Gain:.8 Pole: i Damping:.5 Overshoot (%): 6. Frequency (rad/sec):.8 Imag Axis Real Axis Repeat part (f) using the magnitude criterion. This is to be done at the point. j.8 First from the poles ( j) L L (.)(.8 ).67 (.)(.8) 8. Now from the zeros (, ) L (.) L (.) Imag Axis L L K L L.7 Transient Response via Gain Adjustment nd order approximation must be upheld since the RL provides various damping ratios, settling time, peak time etc. Higher order poles are much farther from the dominant secondorder pair. Closed loop zeros near the closed loop second poles are canceled or nearly cancelled by the close proximity of other higher order closed loop poles.
21 Closed loop zeros not cancelled by the close proximity of other higher order poles are far removed from the closed loop second order dominant pair. Design procedure for higher order systems Sketch the root locus for the given system Assume the system is second order with no zeroes. Find the gain to meet the desired spec Justify second order approximation If it is not justified, then perform constrol simulations to ensure that the specs are met Secondorder approximations Example: For the system as shown, determine the value of K to give a.5% overshoot. Evaluate the settling time, peak time and SSE. The system is third order with a zero at s = .5
22 First get the Root locus Assuming the system can be nd order approx draw the damping ratio line Searching for the closed loop poles at =.8 Root Locus.5 Imag Axis.5.5 System: sys Gain:.7 Pole: i System: sys Damping:.8 Gain: 7.6 Overshoot (%):.5 Pole: i Frequency (rad/sec): Damping:.9.8 Overshoot (%):.5 Frequency (rad/sec): Real Axis Root Locus 8 6 System: sys Gain: 9. Pole: i Damping:.8 Overshoot (%):.5 Frequency (rad/sec): 5.7 Imag Axis Real Axis
23 The point where the RL crosses the =.8 is at points yielding sets of closed loop poles at.87 j.66,.9 j.9 and.6 j.5. The respective gains from the RL plot are 7.6,.79 and 9.6. For each point the settling time, time to first peak can be evaluated from Ts n ; T p d The third closed loop pole must be obtained for each dominant set having the same corresponding gain. Searching for the third pole on the RL at each of the corresponding gains 7.6,.79 and 9.6. Note that the third pole cannot be complex as the CLTF is third order, ie. the third pole must be on the real axis. The poles are at s = 9.5, 8.6 and.8 respectively. Root Locus 8 6 Imag Axis System: System: sys sys Gain: 7. Gain:.8 Pole: 9.5 Pole: 8.6 Damping: Damping: Overshoot Overshoot (%): (%): Frequency Frequency (rad/sec): (rad/sec): System: sys Gain: 9. Pole: .8 Damping: Overshoot (%): Frequency (rad/sec): Real Axis
24 Using Ts ; T n p and the real and imaginary parts of the dominant pole. The velocity error constant is sk( s.5)(.5) K Kv lim() sg slim s s s( s )( s )()() d Subst. various values of K gives the K values in the Table below. v Second and thirdorder responses a. Case ; b. Case Cases and have the third pole far away from the complex pair. However, there is no approx. pole zero cancellation. Case, the third pole is closer to the zero, so a nd order approx can be considered valid. The plots are relatively close. A step input is used to show the second order dynamics and validity of the second order approximation. We will now reevaluate for the third pole analytically knowing the gain at the corresponding dominant pair. As an example, Case will be used.
25 In case, K 9.6. Using the magnitude criterion assuming the third pole p exists somewhere between the pole at and the zero at.5. The point in question is p. L p L p L p L p.5 ()( p p ) p K 9.6 ( p.5) Solving for p gives p.795 which is the third pole. The same can be done for the finding the other third poles. Note the RL does NOT exist on portions of the real axis where the sum of poles and zeros to the left is even. The closed loop TF is using K 9.6 C() s 9.6 s 59.6 R S s s s () Since ()(discuss R s Matlab) s The figure shows that there is a steady state error e() of (.5.).7. NOTE THAT THIS IS = K For Case, K 9.6 K 5.9 e().7 which matches the Matlab plot. 5.9 v v 6 Step Response 5 System: sys Time (sec):.5 Amplitude:. Amplitude 5 6 Time (sec) 5
26 Example Given a unity feedback system that has a forward TF K G() s, do the following ( s )( s )( s 6) (a) Obtain a RL using Matlab (b) Using a nd order approx. determine the value of K to give a % overshoot for a unit step input (c) Determine settling and peak times (d) The natural frequency (e) The SSE for the value of K. (f) Determine the validity of the nd order approx. Repeat part (b) using the magnitude criterion. This is to be done at the point.8 j.768 From the poles L (.8 )(.768).768 L (.8 )(.768).99 L (.8 6)(.768).8 There are no zeros L L L K
27 (d) n (e) The system is Type, the position error constant is K 5.55 lim() G s K p.99 s * *6 8 Therefore, e().5 K p (f) In this case, K Using the magnitude criterion assuming the third pole p exists somewhere to the left of the pole at 6. The point in question is p. L p 6 L p L p ( p 6)( p )( p ) K 5.55 Solving for p gives p 7.9 which is the third pole. Since this third close loop pole is NOT 5 times or more the magnitude of the real part of the closed loop dominant pole, the second approx. NOT valid. Root Locus 8 6 Imag Axis System: sys Gain:.6 Pole: 7.9 Damping: Overshoot (%): Frequency (rad/sec): Real Axis 7
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