Root Locus

Root Locus - ME375 Handouts Root Locus • Motivation •...

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Unformatted text preview: ME375 Handouts Root Locus • Motivation • Sketching Root Locus • Examples School of Mechanical Engineering Purdue University ME375 Root Locus - 1 1 ME375 Handouts Servo Table Example DC Motor Position Control The block diagram for position control of the servo table is given by: by: Position Controller θD 0.09 θDV KP θV Plant Gp(s) Ei 168 s(0.3s + 1) θ 0.09 See how the closed-loop poles move as proportional gain KP varies from 0 to ∞. closedFind closed-loop characteristic equation: closed- School of Mechanical Engineering Purdue University ME375 Root Locus - 2 2 ME375 Handouts Servo Table Example (cont.) School of Mechanical Engineering Purdue University ME375 Root Locus - 3 3 ME375 Handouts Motivation – Example 1 Revisit the DC motor positioning system with proportional control. control. corresponding block diagram is: is: Its Plant Gp(s) R(s) + Controller KP − U(s) Y(s) 27.6 s( s + 57.5) Sketch the closed-loop poles as the controller gain KP varies from 0 to ∞. closedFind closed-loop characteristic equation: closed- s 2 + 57.5 s + 27.6 K P = 0 School of Mechanical Engineering Purdue University ME375 Root Locus - 4 4 ME375 Handouts Example 1 Formulate an expression for the roots of the characteristic equation: equation: Find the roots for KP = 0 and KP → ∞: Find KP when the roots are repeated. repeated. School of Mechanical Engineering Purdue University ME375 Root Locus - 5 5 ME375 Handouts Example 1 Sketch the root locus: Img. Axis 30 20 10 Real Axis 0 -60 -50 -40 -30 -20 -10 0 -10 -20 -30 School of Mechanical Engineering Purdue University ME375 Root Locus - 6 6 ME375 Handouts Example 2 Using the same plant as in Example 1, try a different controller choice: choice: Plant Gp(s) R(s) + Controller Kd(s + 80) − U(s) Y(s) 27.6 s( s + 57.5) Sketch the root locus of the closed-loop poles as the controller gain Kd varies from closed0 to ∞. Find closed-loop characteristic equation: closed- School of Mechanical Engineering Purdue University ME375 Root Locus - 7 7 ME375 Handouts Example 2 Formulate an expression for the roots of the characteristic equation: equation: Find the roots for Kd = 0 and Kd → ∞: Find repeated roots. roots. School of Mechanical Engineering Purdue University ME375 Root Locus - 8 8 ME375 Handouts Example 2 Repeated roots (cont.): School of Mechanical Engineering Purdue University ME375 Root Locus - 9 9 ME375 Handouts Example 2 Sketch the root locus: Imag Axis 40 30 20 10 Real Axis 0 -10 -20 -30 -40 -140 -120 -100 -80 -60 -40 -20 School of Mechanical Engineering Purdue University 0 ME375 Root Locus - 10 10 ME375 Handouts ClosedClosed-Loop Characteristic Roots (CL Poles) Reference Input R(s) + Error − E(s) Control Input KP U(s) Output Gp(s) Y(s) Plant H(s) The closed-loop transfer function GCL(s) is: closed- Img. jω GCL ( s ) = The closed-loop characteristic equation is: closed- Real − jω School of Mechanical Engineering Purdue University ME375 Root Locus - 11 11 ME375 Handouts Definitions • Root Locus Root Locus plotting is the method of determining the roots of the following equation on the complex plane when the parameter K varies from 0 → ∞: 1 + K ⋅ G OL ( s ) = 0 or 1+ K ⋅ N (s) =0 D (s) where N(s) and D(s) are known polynomials in factorized form: N ( s ) = ( s − z1 )( s − z 2 ) L ( s − z N z ) D ( s ) = ( s − p1 )( s − p2 ) L ( s − p N P ) The NZ roots of the polynomial N(s) , z1 , z2 , …, zNz , are called the finite openopenloop zeros. The NP roots of the polynomial D(s) , p1 , p2 , …, pNp , are called the finite openopen-loop poles. School of Mechanical Engineering Purdue University ME375 Root Locus - 12 12 ME375 Handouts Root Locus Methods of obtaining root locus: • Given a value of K, numerically solve the 1 + K GOL(s) = 0 equation for a set of roots. Repeat this for a set of K values and plot the corresponding roots on roots. the complex plane. (This is what we did in the last in-class exercise.) plane. in• Use MATLAB. In MATLAB use the commands rlocus and rlocfind. rlocus rlocfind You You can use on-line help to find the usage for these commands. on1 + K P ⋅ 0 .03 ⋅ 16 =0 s ( 0 .0174 s + 1) ⇒ 1 + KP ⋅ 0 . 48 =0 0 .0174 s 2 + s >> op_num=[0.48]; >> op_den=[0.0174 1 0]; >> Gol=tf(op_num,op_den); >> rlocus(Gol) rlocus(Gol) >> [K, poles]=rlocfind(Gol); • Apply root locus sketching rules to obtain an approximate root locus plot. approximate School of Mechanical Engineering Purdue University ME375 Root Locus - 13 13 ME375 Handouts Root Locus Sketching Rules Reading material: Palm book 11.1-11.2 material: 11. 11. Motivation: In the design of feedback controllers, a qualitative understanding is needed on the influence of control parameters on the output performance of the controlled system. We have seen earlier the role played by the poles of the closed-loop transfer functions closed(CLFTs) on the transient and steady-state response of the output response. The influence steadyof control parameters on the placement of the CLTF poles in the complex plane can been control parameters on the placement of the CLTF poles in the complex plane can been seen by directly solving the characteristic equation of the CLTF. However, this process does not allow for a qualitative understanding. Here, we will use a set of rules for sketching the so-called root locus plots based on soinformation known about the zeros and poles of the open-loop system. From this, we can enbetter understand the trends of parameter influence on response and use this in the design of control systems. School of Mechanical Engineering Purdue University ME375 Root Locus - 14 14 ME375 Handouts Rules for sketching root locus 1. The number of branches in the root locus plot equals the number of open-loop poles; openpoles; 2. The locus starts (K=0) at an open-loop pole, and as K → ∞ , m branches ends at open (K=0 openloop zeros, and n-m branches will end at infinity and asymptotically approach straight lines. (n: number of open-loop poles; m: number of open loop zeros) lines. (n: openpoles; 3. The locus is symmetric about the real axis. axis. 4. The locus, on the real axis, includes all segments left of an odd number of open loop poles and zeros. zeros. K → ∞ , n-m branches will approach straight lines (asymptotes). These straight lines 5. As intersect the real axis at σ 0 with an angle θ k , where: where: σ0 = ∑ pi − ∑ zi n−m , ( pi : OL poles; zi : OL zeros.) θ k = (2k − 1) 180o , k = 0,1,..., n − m − 1 n−m 6. A locus leaves (“break away”) or enters (“breaks in”) the real axis at values of K for which repeated roots exist in the closed loop characteristic equation. These points and values of K at these points satisfy: satisfy: dK =0 ds School of Mechanical Engineering Purdue University ME375 Root Locus - 15 15 ME375 Handouts Rules for sketching root locus 7. The departure angle for an OL pole (or the arrival angle for an OL zero) can be calculated from: rom: φl ,dep = ∑ψ i − ∑ φi − 180o i ≠l ψ l ,arr = ∑ φi − ∑ψ i + 180o i ≠l where φi is the angle formed by the positive direction of the real axis and the vector connecting another pole to the pole(or zero) of interest; ψ i is the angle formed by interest; the positive direction of the real axis and the vector connecting another zero to the pole(or zero) of interest. interest. 8. If the locus passes through the imaginary axis (boundary between stable and unstable responses), the crossing point and corresponding values of K can be found by: - Replacing s by jω in the closed-loop characteristic equation, and closed- Setting the real and imaginary parts to zero and solving for K and jω School of Mechanical Engineering Purdue University ME375 Root Locus - 16 16 ME375 Handouts Steps for sketching root locus Step 1. Determine the closed-loop transfer function; closedfunction; Step 2. Express the characteristic equation for the closed-loop system in the standard form: closedorm: 1+ K N (s) =0 D( s) Step 3. Determine the open-loop poles and zeros using, respectively: D(s)=0 and N(s)=0. openrespectively: D(s)=0 N(s)=0 Mark the open-loop poles with an “x” and zeros with an “o” in the complex plane. openplane. Step 4. Apply Rule 4. Step 5. Find the number of asymptotes and the corresponding values of σ 0 and θ k . (Rule 5) Step 6. If necessary, find the break-away and break-in points. (Rule 6) breakbreak- points. Step 7. If necessary, find the departure and arrival angles. (Rule 7) angles. Step 8. Sketch the root locus. locus. School of Mechanical Engineering Purdue University ME375 Root Locus - 17 17 ME375 Handouts Example 3 A feedback control system is proposed. The corresponding block diagram is: proposed. is: Plant Gp(s) Controller R(s) + − K (s + 4) U(s) 1 s(s + 2) Y(s) Sketch the root locus of the closed-loop poles as the controller gain K varies from closed0 to ∞. Find closed-loop characteristic equation: closed- School of Mechanical Engineering Purdue University ME375 Root Locus - 18 18 ME375 Handouts Example 3 Step 1: Formulate the (closed-loop) characteristic equation into the standard form for (closedsketching root locus: ocus: Step 2: Find the open-loop zeros, zi , and the open-loop poles, pi : openopen- School of Mechanical Engineering Purdue University ME375 Root Locus - 19 19 ME375 Handouts Example 3 Step 3: Determine locations of repeated roots, if any. any. School of Mechanical Engineering Purdue University ME375 Root Locus - 20 20 ME375 Handouts Example 3 Step 4: Determine the imaginary axis crossings, if any. School of Mechanical Engineering Purdue University ME375 Root Locus - 21 21 ME375 Handouts Example 3 Step 5: Use the information from Steps 1-4 to sketch the root locus. 1Imag Axis 4 3 2 1 Real Axis 0 -1 -2 -3 -4 -6 -5 -4 -3 -2 -1 School of Mechanical Engineering Purdue University 0 ME375 Root Locus - 22 22 ME375 Handouts Example 4 The transfer function for a simple spring-mass-dashpot oscillator is known to be: spring-massG( s) = 1 Ms + Cs + K 2 Using the root locus rules, sketch the root locus for this system: a. in terms of the stiffness parameter K with M = 2 and C = 8 . with b. in terms of the damping parameter C with M = 2 and K = 8 . with School of Mechanical Engineering Purdue University ME375 Root Locus - 23 23 ME375 Handouts Example 4 a. Root locus for varying K School of Mechanical Engineering Purdue University ME375 Root Locus - 24 24 ME375 Handouts Example 4 b. Root locus for varying C School of Mechanical Engineering Purdue University ME375 Root Locus - 25 25 ME375 Handouts Example 4 b. Root locus for varying C School of Mechanical Engineering Purdue University ME375 Root Locus - 26 26 ME375 Handouts Revisit PID Control Examples Disturbance D(s) Reference Input R(s) Controller + Error − E(s) Control Input GC (s) Plant GP (s) + + 4 ( 2 s + 1)(0.5s + 1) U (s ) Output Y(s) Sensor H (s) 1 (A) Proportional (P) control: GC (s) = KP Proportional GYR ( s ) = 4K P (2 s + 1)(0.5s + 1) + 4K P (B) Add Integral action, ⇒ Proportional-Plus-Integral (PI) control: Add Proportional-PlusGC ( s ) = K P + K I 1 KPs + KI = s s GYR ( s ) = 4( KPs + KI ) s(2 s + 1)(0.5s + 1) + 4 ( K P s + K I ) (C) Add Derivative action, ⇒ Proportional-Plus-Integral-Plus Derivative (PID) control: Add Proportional-Plus-IntegralK s2 + K P s + K I 1 GC ( s ) = K P + K I + K D s = D s s GYR ( s ) = School of Mechanical Engineering Purdue University 4( KDs2 + KPs + KI ) s (2s + 1)(0.5s + 1) + 4 ( K D s 2 + K P s + K I ) ME375 Root Locus - 27 27 ME375 Handouts Effect of Proportional Control Closed Loop Characteristic Equation: s 2 + 2.5s + (1 + 4K P ) = 0 Rewrite the characteristic equation: OpenOpen-loop poles and zeros: 5 4 Repeated roots: Imaginary part of s 3 2 1 0 -1 -2 -3 -4 -5 -8 -7 -6 -5 -4 -3 -2 -1 Real part of s School of Mechanical Engineering Purdue University 0 1 2 ME375 Root Locus - 28 28 ME375 Handouts Effect of Proportional Control Root Locus for K P Locus for 1.5 Imaginary Axis 1 0.5 0 -0.5 -1 -1.5 -2.5 -2 -1.5 -1 -0.5 Real Axis School of Mechanical Engineering Purdue University 0 0.5 ME375 Root Locus - 29 29 ME375 Handouts Effect of PI Control Closed Loop Characteristic Equation: s 3 + 2.5s 2 + (1 + 4K P ) s + 4 K I = 0 Rewrite the characteristic equation: OpenOpen-loop poles and zeros: 5 4 Repeated roots: roots Imaginary axis crossings: Imaginary part of s 3 2 1 0 -1 -2 -3 -4 -5 -8 -7 -6 -5 -4 -3 -2 -1 Real part of s School of Mechanical Engineering Purdue University 0 1 2 ME375 Root Locus - 30 30 ME375 Handouts Effect of PI Control Locus or RootRoot LocusforKK I Locus ffor I Locus for Imaginary A xis 30 20 Imaginary Axis 30 20 10 10 0 0 -10 -10 -20 -30 -30 -20 -30 -30 -20 -20 -10 -10 0 0 Real Axis 10 10 20 20 30 Real Axis School of Mechanical Engineering Purdue University ME375 Root Locus - 31 31 ME375 Handouts Effect of PID Control Closed Loop Characteristic Equation: s 3 + ( 2.5 + 4 K D ) s 2 + (1 + 4K P ) s + 4 K I = 0 Rewrite the characteristic equation: OpenOpen-loop poles and zeros: 5 4 Repeated roots: Repeated roots Imaginary axis crossings: Imaginary part of s 3 2 1 0 -1 -2 -3 -4 -5 -8 -7 -6 -5 -4 -3 -2 -1 Real part of s School of Mechanical Engineering Purdue University 0 1 2 ME375 Root Locus - 32 32 ME375 Handouts Effect of PID Control Root Locus for K D Locus for Imaginary Axis 5 0 -5 -8 -6 -4 Real Axis -2 School of Mechanical Engineering Purdue University 0 ME375 Root Locus - 33 33 ME375 Handouts Controller Design using Root Locus Objective: Design a system that has zero stead state error for step inputs with %OS < 10% dy %OS 10% and TS (2%) < 6 [sec]. (2%) [sec]. Disturbance D(s) Reference Input Controller + R(s) Error Control Input GC (s) − E(s) Plant GP (s) + U (s ) + Output 4 ( 2 s + 1)(0.5s + 1) Y(s) Sensor H (s) 1 GC ( s ) = K P + K I 1 KPs + KI = s s Closed loop transfer functions: 4 ⎛ KPs + KI ⎞ ⎜ ⎟⋅ 4( KPs + KI ) GC ( s ) ⋅ GP ( s ) s ⎝ ⎠ (2s + 1)(0.5s + 1) = = GYR ( s ) = 4 1 + GC ( s ) ⋅ GP ( s ) ⋅ H ( s ) s (2 s + 1)(0.5s + 1) + 4 ( K P s + K I ) ⎛ K s + KI ⎞ 1+ ⎜ P ⎟⋅ s ⎝ ⎠ (2 s + 1)(0.5s + 1) School of Mechanical Engineering Purdue University ME375 Root Locus - 34 34 ME375 Handouts Design PI Control Closed Loop Characteristic Equation: Loop Characteristic Equation s (2 s + 1)(0.5s + 1) + 4 ( K P s + K I ) = 0 let let ζ = 0.707 and ζωn = 1 and ζω s 3 + 2.5s 2 + (1 + 4K P ) s + 4 K I = 0 School of Mechanical Engineering Purdue University ME375 Root Locus - 35 35 ME375 Handouts Step Response with PI Control Step Response 1.4 1.2 Amplitude 1 0.8 0.6 0.4 0.2 0 0 2 4 6 Time (sec) School of Mechanical Engineering Purdue University 8 10 ME375 Root Locus - 36 36 ME375 Handouts PID Control Design Objective: Design a system that has zero stead state error for step inputs with %OS < 10% dy %OS 10% and TS (2%) < 6 [sec]. (2%) [sec]. Disturbance D(s) Reference Input Controller + R(s) Error − E(s) Control Input GC (s) Plant GP (s) + U (s ) + 4 ( 2 s + 1)(0.5s + 1) Output Y(s) Sensor H (s) 1 GC ( s ) = K P + K I K s2 + K P s + K I 1 + KDs = D s s Closed loop transfer functions: K Ds2 + KP s + K I 4 ⋅ 4( KDs2 + KPs + KI ) GC ( s ) ⋅ GP ( s ) s (2 s + 1)(0.5s + 1) GYR ( s ) = = = K s2 + K Ps + K I 4 1 + GC ( s ) ⋅ GP ( s ) ⋅ H ( s ) s (2s + 1)(0.5s + 1) + 4 ( K D s 2 + K P s + K I ) 1+ D ⋅ s (2s + 1)(0.5s + 1) School of Mechanical Engineering Purdue University ME375 Root Locus - 37 37 ME375 Handouts Design PID Control Closed Loop Characteristic Equation: Loop Characteristic Equation s (2 s + 1)(0.5s + 1) + 4 ( K D s + K P s + K I ) = 0 let let ζ = 0.707, ζωn = 1, and a = 5 ζω and 2 s 3 + ( 2.5 + 4 K D ) s 2 + (1 + 4K P ) s + 4 K I = 0 School of Mechanical Engineering Purdue University ME375 Root Locus - 38 38 ME375 Handouts Step Response with PID Control Step Response 1.4 1.2 Amplitude 1 0.8 0.6 0.4 0.2 0 0 2 4 6 Time (sec) School of Mechanical Engineering Purdue University 8 10 ME375 Root Locus - 39 39 ME375 Handouts Additional Examples School of Mechanical Engineering Purdue University ME375 Root Locus - 40 40 ...
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