Linear Control Theory (EE 326) PDF

Linear Control Theory (EE 326): 3. Time Response Analysis Dharavath Kishan Dept. of Electrical and Electronics Engineering NITK Surathkal First Order System Step Response 1 𝑎 1 𝑎 𝑋 𝑠 = 𝐺 𝑠 =...

Linear Control Theory (EE 326): 3. Time Response Analysis Dharavath Kishan Dept. of Electrical and Electronics Engineering NITK Surathkal First Order System Step Response 1 𝑎 1 𝑎 𝑋 𝑠 = 𝐺 𝑠 = 𝑌 𝑠 = 𝑠 𝑠+𝑎 𝑠 𝑠+𝑎 Taking Partial Fraction and solving for 𝑦 𝑡 we get 𝑦 𝑡 = 1 − 𝑒 −𝑎𝑡 𝑢 𝑡 1 𝑦 𝑡→∞ =1 𝑦 𝑡= = 1 − 𝑒 −1 = 0.632 𝑎 1 𝑡 = 𝑎 is called the time constant of the system: time it takes for the step response to rise to 63% of its final value (or the exponent 𝑒 −𝑎𝑡 to decay to 37% of the initial value of unity) Time constant = 1/ (magnitude of the pole) Pole is also called as exponential frequency (as it has the unit of frequency) d𝑦 𝑡=0 Initial slope: = 𝑎 𝑑𝑡 EE326 Dept. of E & E, NITK Surathkal 2 First Order System Step Response 1 𝑎 1 𝑎 𝑋 𝑠 = 𝐺 𝑠 = 𝑌 𝑠 = 𝑠 𝑠+𝑎 𝑠 𝑠+𝑎 Taking Partial Fraction and solving for 𝑦 𝑡 we get 𝑦 𝑡 = 1 − 𝑒 −𝑎𝑡 𝑢 𝑡 4 𝑦 𝑡= = 1 − 𝑒 −4 = 0.981 𝑎 Settling time (four time constants) to reach within ±2 % tolerance Rise time: time required for 10 % to 90% transition 2.2 𝑇𝑟 = 𝑎 EE326 Dept. of E & E, NITK Surathkal 3 First Order System Step Response 𝑎 Transfer function 𝐺 𝑠 = 𝑠+𝑎 1 Time Constant 𝑇 𝑎 seconds 2.2 Rise time 𝑇𝑟 seconds 𝑎 4 ±2 % tolerance seconds 𝑎 Settling time 𝑇𝑠 3 𝑎 seconds ±5 % tolerance 5 seconds we assume that steady-state has reached 𝑎 EE326 Dept. of E & E, NITK Surathkal 4  First Order System Step Response Slope = 𝑎 𝑇𝑟 EE326 Dept. of E & E, NITK Surathkal 5 Second Order Systems 1 𝐾 1 𝐾 𝑋 𝑠 = 𝐺 𝑠 = 2 𝑌 𝑠 = 𝑠 𝑠 + 𝑎𝑠 + 𝑏 𝑠 𝑠2 + 𝑎𝑠 + 𝑏 Find 𝐾 such that the final value is 1 (for critically stable systems) 𝐾 𝑦 ∞ = lim 𝑠𝑌(𝑠) lim =1 ⇒𝐾=𝑏 𝑠→0 𝑠→0 𝑠2 + 𝑎𝑠 + 𝑏 9 𝐺 𝑠 = 2 𝑠 + 𝑎𝑠 + 9 𝑎 ⇒ 0: 0.5: 10 EE326 Dept. of E & E, NITK Surathkal 6  9 Second Order Systems 𝐺 𝑠 = 2 𝑠 + 𝑎𝑠 + 9 Undamped 𝑎 = 0 Underdamped 𝑎 =2 overdamped 𝑎 = 8 EE326 Dept. of E & E, NITK Surathkal 7  9 Second Order Systems 𝐺 𝑠 = 2 𝑠 + 𝑎𝑠 + 9 Underdamped: fast but with oscillations Critically damped: fastest possible without oscillations 𝑎 =? Overdamped: no oscillations but slow EE326 Dept. of E & E, NITK Surathkal 8  9 Second Order Systems 𝐺 𝑠 = 2 𝑠 + 𝑎𝑠 + 9 Analyse step by step Value of a Roots of the ch. eqn Natural response Case I: 𝑎 = 0 ±𝑗3 𝑐 sin 3𝑡 + 𝜙 Case II: 𝑎 = 9 −1.146, −7.854 𝑐1 𝑒 −1.146𝑡 + 𝑐2 𝑒 −7.854𝑡 Case III: 𝑎 = 2 𝛼 𝑒 −𝑡 sin 8𝑡 + 𝜃 −1 ± 𝑗 8 What is critically damped (i.e., fastest possible without oscillations) Hint: exponential is slow, sinusoidal is oscillatory, polynomial may be? ☺ Case IV: 𝑎 = 6 −3, −3 𝑎1 𝑡𝑒 −3𝑡 + 𝑎2 𝑒 −3𝑡 EE326 Dept. of E & E, NITK Surathkal 9  Step Response: Second Order Systems 9 𝐺 𝑠 = Case I: 𝒂 = 𝟎 Case II: 𝒂 = 𝟐 𝑠2 + 𝑎𝑠 + 9 Undamped Underdamped Case IV: 𝒂 =6 Case III: 𝒂 = 𝟗 Critically Overdamped damped Not to be confused with critically stable!! EE326 Dept. of E & E, NITK Surathkal 10  The General Second Order System 𝑏 Some definitions 𝐺 𝑠 = 2 𝑠 + 𝑎𝑠 + 𝑏 Natural frequency 𝝎𝒏 Frequency of oscillation of the system without damping Ex. Tank circuit 𝜔𝑛 =1/ 𝐿𝐶 rad/sec Take 𝑎 = 0 ⇒ poles = ±𝑗 𝑏 Exponential decay frequency 1 Natural period Damping ratio 𝜻 (zeta) = = 𝜔𝑛 𝑎 Τ2 2𝜋 Time constant ⇒ 𝜔𝑛 Thus 𝑎 = 2𝜁𝜔𝑛 A general second order system T. F. is given by 𝜔𝑛2 𝐺 𝑠 = 2 𝑠 + 2𝜁𝜔𝑛𝑠 + 𝜔𝑛2 EE326 Dept. of E & E, NITK Surathkal 11  The General Second Order System 𝜔𝑛2 𝐺 𝑠 = 2 𝑠 + 2𝜁𝜔𝑛𝑠 + 𝜔𝑛2 Poles: 𝜁𝜔𝑛 ± 𝜔𝑛 𝜁 2 − 1 Damping ratio and the type of step response of the system 𝜁=0→ ←0 5 2 (neglect the term 𝑠 + 15 ) 360  𝐺2 𝑠 = (𝑠+4)(𝑠 2 +2𝑠+90) 4 0) Vary 0.1 ≤ 𝑎 ≤ 10) zoom EE326 Dept. of E & E, NITK Surathkal 35 Effect of additional zeros Lets take an example: 9 𝑠+𝑎 𝐺 𝑠 = 𝑎 𝑠 2 + 2𝑠 + 9 First consider only LHS zeros (i.e., 𝑎 > 0) Vary 0.1 ≤ 𝑎 ≤ 10) EE326 Dept. of E & E, NITK Surathkal 36 Effect of additional zeros Lets take an example: 9 𝑠+𝑎 𝐺 𝑠 = 𝑎 𝑠 2 + 2𝑠 + 9 First consider only LHS zeros (i.e., 𝑎 > 0) Vary 0.1 ≤ 𝑎 ≤ 10) Error as a function of value of a (when the term 𝑠 + 𝑎 is neglected) EE326 Dept. of E & E, NITK Surathkal 37 Effect of additional zeros More general explanation (for LHS zeros only) ℒ −1 𝑠 + 𝑎 𝑌(𝑠) 𝑦1 𝑡 = 𝑦ሶ 𝑡 + 𝑎𝑦(𝑡) ←𝑧 𝑠 − plane 𝑦ሶ 𝑡 ≪ 𝑎𝑦(𝑡)  Step response is affected when the derivative term dominates  Since initial slope is usually +ve ◦ Higher overshoot is expected for lower values of “a” 𝑎 = 0.1 EE326 Dept. of E & E, NITK Surathkal 38  Effect of additional zeros More general explanation (for RHS zeros only) 𝑠 − plane 𝑧→ 𝑦1 𝑡 = 𝑦(𝑡) ሶ + 𝑎𝑦 𝑡 ; 𝑎 < 0 EE326 Dept. of E & E, NITK Surathkal 39  Effect of additional zeros More general explanation (for RHS zeros only) 𝑦1 𝑡 = 𝑦(𝑡) ሶ + 𝑎𝑦 𝑡 ; 𝑎 < 0 Normalized: 9/𝑎  When a is small +ve ◦ Derivative is dominating ◦ The step response will follow derivative  When a increases ◦ Step response go negative ◦ In contrast to LHS zeros Nonminimum phase systems EE326 Dept. of E & E, NITK Surathkal 40 P1 For each of the systems, find 𝜁, 𝜔𝑛, 𝑇𝑠, 𝑇𝑝, 𝑀𝑝 (𝑖𝑛 %) 16 (i) 𝐺 𝑠 = 2 𝜔𝑛 = 4 rad/s 𝜁 = 0.375 𝑠 + 3𝑠 + 16 𝑇𝑠 = 2.667 s 𝑇𝑃 = 0.847 s 𝑀𝑃 = 0.28 (ii) 𝐺 𝑠 = 0.04 𝑠2 + 0.02𝑠 + 0.04 𝜔𝑛 = 0.2 rad/s 𝜁 = 0.05 𝑇𝑠 = 400 s 𝑇𝑃 = 15.727 s 𝑀𝑃 = 0. 85 EE326 Dept. of E & E, NITK Surathkal 41 P1 (iii) For each of the systems, find 𝜁, 𝜔𝑛, 𝑇𝑠, 𝑇𝑝, 𝑀𝑝 (𝑖𝑛 %) 14.145 𝐺 𝑠 = 2 (𝑠 + 0.842𝑠 + 2.829)(𝑠 + 5) 𝜔𝑛 = 1.682 rad/s 𝜁 = 0.25 𝑇𝑠 = 9.501 s 𝑇𝑃 = 1.929 s 𝑀𝑃 = 0. 432 Note: 5 = 11.87 > 5 0.842 14.145 2 14.145 (𝑠2 + 0.842𝑠 + 2.829)(𝑠 + 5) 5 2 (𝑠 + 0.842𝑠 + 2.829) Step response plot with 2nd order approximation EE326 Dept. of E & E, NITK Surathkal 42 P2 For each of the pair of the second order specifications, find the location of the complex conjugate poles (i) Percentage overshoot = 12, 𝑇𝑠 = 0.6 seconds 𝜁 = 0.559 𝜔𝑛 = 11.917 rad/s poles = −6.661 ± 𝑗9.88 (ii) Percentage overshoot = 10, 𝑇𝑃 = 5 seconds 𝜁 = 0.591 𝜔𝑛 = 0.779 rad/s poles = −0.460 ± 𝑗0.628 EE326 Dept. of E & E, NITK Surathkal 43 P3 For the system shown find the percentage overshoot, settling time (𝑇𝑠) and peak time (𝑇𝑃) of 𝜃2(𝑡) when a step torque 𝑇 𝑡 = 𝑢 𝑡 is applied as shown. 𝑇 𝑠 = 1.07𝑠 2 + 1.53𝑠 𝜃1 𝑠 − 1.53𝑠𝜃2 (𝑠) 0 = −1.53𝑠𝜃1 𝑠 + 1.53𝑠 + 1.92 𝜃2 (𝑠) 1.07𝑠 2 + 1.53𝑠 𝑇 𝜃2 𝑠 −1.53𝑠 0 1.53 = = 𝑇 𝑠 1.07𝑠 2 + 1.53𝑠 −1.53𝑠 1.6371𝑠 2 + 2.0544𝑠 + 2.9376 −1.53𝑠 1.53𝑠 + 1.92 ⇒ 𝐷 𝑠 = 𝑠 2 + 1.255𝑠 + 1.794 𝜔𝑛 = 1.3394 rad/s 𝜁 = 0.4685 𝑇𝑠 = 6.374 s 𝑇𝑃 = 2.6549 s 𝑀𝑃 = 0.189 EE326 Dept. of E & E, NITK Surathkal 44 P4 (Matlab): Home Work Problems The matlab code given below simulates a system with Transfer function 1. Try using the two different inputs (one sinusoidal corrupted by 𝑠 2 +0.2𝑠+4 noise) and the other pure noise. What do you observe? %Code Q7 clc; clear all; t=linspace(0,50,1001); Sys=tf(1,[1 0.2 4]); x=sin(2*t)+2*sin(.1*t)+0.2*cos(10*t)+rand(1,length(t)); % x=rand(1,length(t))-0.5; %commented, uncomment to check for this input y=lsim(Sys,x,t); plot(t,x,'b--',t,y,'r'); EE326 Dept. of E & E, NITK Surathkal 45 P4 (Matlab): Home Work Problems The matlab code given below simulates a system with Transfer function 1. Try using the two different inputs (one sinusoidal corrupted by 𝑠 2 +0.2𝑠+4 noise) and the other pure noise. What do you observe? EE326 Dept. of E & E, NITK Surathkal 46 P4 (Matlab): Home Work Problems Plotting the frequency response of the system and the spectrum of the input as well as the output F=linspace(0,4,100); G=bode(Sys,2*pi*F); figure(2) plot(F,db(abs(squeeze(G)))); xlabel('Frequency (Hz)') ylabel('20log_{10}|G(\omega)|') X=fft(x); Y=fft(y); F1=(0:1000)*20/1000; figure(3) plot(F1(1:200),db(X(1:200))); xlabel('Frequency (Hz)') ylabel('20log_{10}|X(\omega)|') figure(4) plot(F1(1:200),db(Y(1:200))); xlabel('Frequency (Hz)') ylabel('20log_{10}|Y(\omega)|') EE326 Dept. of E & E, NITK Surathkal 47 P4 (Matlab): Home Work Problems 1 Frequency response of 𝑠 2 +0.2𝑠+4 × 2𝜋 = 2.03 rad System poles: −0.1 ± 𝑗2 System modes: 𝑒 −0.1±𝑗2 𝑡 Natural response: 𝐶 𝑒 −0.1𝑡 sin 2𝑡 + 𝑝ℎ EE326 Dept. of E & E, NITK Surathkal 48 P4 (Matlab): Home Work Problems Input signal (rand signal) spectrum EE326 Dept. of E & E, NITK Surathkal 49 P4 (Matlab): Home Work Problems Output signal spectrum EE326 Dept. of E & E, NITK Surathkal 50  P4 (Matlab): Home Work Problems 1 𝑠 2 +0.2𝑠+4 Can be viewed in two ways: Filtering only a component with 2 rad/sec (signal processing perspective) Responding only to 2 rad/sec (system perspective) EE326 Dept. of E & E, NITK Surathkal 51 Linear Control Theory (EE 326): 2.1 Review of Laplace Transform Dharavath Kishan Dept. of Electrical and Electronics Engineering NITK Surathkal Definition Laplace Transform ∞ ℒ 𝑓(𝑡) = න 𝑓 𝑡 𝑒 −𝑠𝑡 𝑑𝑡 𝑠 ∈ ROC = 𝐹(𝑠) 0− Lower limit is 0− → integrate even when 𝑓 𝑡 is discontinuous at 𝑡 = 0 E.g. 𝛿(𝑡) Solving differential equation: does require only initial conditions (i.e., at 0− ) No need of auxiliary conditions (i.e., at 0+ ) Remember, initial conditions are discontinuities at 𝑡 = 0 Inverse Laplace Transform 1 𝜎+𝑗∞ ℒ −1 𝐹(𝑠) = න 𝑓 𝑡 𝑒 −𝑠𝑡 𝑑𝑡 = 𝑓 𝑡 𝑢(𝑡) 2𝜋𝑗 𝜎−𝑗∞ Here 𝑢 𝑡 is the unit step function Since it is unilateral Laplace Transfrom EE326 Dept. of E & E, NITK Surathkal 2 Definition Laplace Transform Table Sl. 𝑓(𝑡) 𝐹(𝑠) No. 1 𝛿(𝑡) 1 2 𝑢(𝑡) 1 𝑠 3 𝑡𝑢(𝑡) 1 𝑠2 4 𝑡 𝑛 𝑢(𝑡) 𝑛! 𝑠 𝑛+1 5 𝑒 −𝑎𝑡 𝑢(𝑡) 1 𝑠+𝑎 6 sin 𝜔𝑡 𝑢(𝑡) 𝜔 𝑠2 + 𝜔2 7 cos 𝜔𝑡 𝑢(𝑡) 𝑠 𝑠2 + 𝜔2 EE326 Dept. of E & E, NITK Surathkal 3 Definition Laplace Transform Properties Sl. Properties Name No. ∞ 1 Definition ℒ 𝑓(𝑡) = 𝐹 𝑠 = න 𝑓 𝑡 𝑒 −𝑠𝑡 𝑑𝑡 0− 2 ℒ 𝛼𝑓(𝑡) = 𝛼𝐹 𝑠 Linearity 3 ℒ 𝑓1 𝑡 + 𝑓2 (𝑡) = 𝐹1 𝑠 + 𝐹2 (𝑠) Linearity 4 ℒ 𝑒 −𝑎𝑡 𝑓(𝑡) = 𝐹 𝑠 − 𝑎 Frequency Shift 5 ℒ 𝑓(𝑡 − 𝑇) = 𝐹 𝑠 𝑒 −𝑠𝑇 Delay 6 1 𝑠 Scaling ℒ 𝑓(𝑎𝑡) = 𝐹 𝑎 𝑎 7 𝑑𝑓 𝑡 Differentiati ℒ = 𝑠𝐹 𝑠 − 𝑓 0− on 𝑑𝑡 𝑛 8 𝑑𝑛 𝑓 𝑡 𝑑 𝑘−1 𝑓 0− Differentiati ℒ = 𝑠𝑛𝐹 𝑠 − ෍ 𝑠 𝑛−𝑘 on 𝑑𝑡 𝑛 𝑑𝑡𝑘−1 𝑘=1 EE326 Dept. of E & E, NITK Surathkal 4 Definition Laplace Transform Properties, continued… Sl. Properties Name No. 𝑡 9 𝐹 𝑠 Integration ℒ න 𝑓 𝜏 𝑑𝜏 = 0− 𝑠 10 𝑓 ∞ = lim 𝑠𝐹 𝑠 Final Value 𝑡→0 11 𝑓 0+ = lim 𝑠𝐹 𝑠 Initial Value 𝑡→∞ EE326 Dept. of E & E, NITK Surathkal 5 Problems: Partial Fraction Expansion P1: Find the inverse Laplace Transform of 2 𝐹 𝑠 = ⇒ 𝑓 𝑡 = 2𝑒 −𝑡 − 2𝑒 −2𝑡 𝑢 𝑡 𝑠+1 𝑠+2 P2: Solve the differential equation using Laplace transform (with zero initial condition) 𝑑2𝑦 𝑑𝑦 + 12 + 32𝑦 = 32𝑢 𝑡 ⇒ 𝑦 𝑡 = 1 − 2𝑒 −4𝑡 + 𝑒 −8𝑡 𝑢 𝑡 𝑑𝑡 2 𝑑𝑡 P3: Find inverse Laplace Transform of (repeated roots) 2 𝐹 𝑠 = 2 ⇒ 𝑓 𝑡 = 2𝑒 −𝑡 − 2𝑡𝑒 −2𝑡 − 2𝑒 −2𝑡 𝑢 𝑡 𝑠+1 𝑠+2 P4: Find inverse Laplace Transform of (Complex conjugate roots) 3 3 3 −𝑡 𝐹 𝑠 = ⇒𝑓 𝑡 = − 𝑒 4 cos 2𝑡 + 2 sin 2𝑡 𝑢 𝑡 2 𝑠 2 + 2𝑠 + 5 5 20 EE326 Dept. of E & E, NITK Surathkal 6  Components Passive Linear Components Impedance Admittance Component Voltage-current Current-Voltage 𝑉 𝑠 𝐼 𝑠 𝑍 𝑠 = 𝑌 𝑠 = 𝐼(𝑠) 𝑉(𝑠) Capacitor 1 𝑡 𝑑𝑣 𝑡 1 𝑣 𝑡 = න 𝑖 𝑡 𝑑𝑡 𝑖 𝑡 =𝐶 𝐶𝑠 𝐶 0 𝑑𝑡 𝐶𝑠 Resistor 𝑣 𝑡 1 𝑣 𝑡 = 𝑅𝑖(𝑡) 𝑖 𝑡 = 𝑅 𝑅 𝑅 𝑑𝑖 𝑡 1 𝑡 1 Inductor 𝑣 𝑡 =𝐿 𝑖 𝑡 = න 𝑣 𝑡 𝑑𝑡 𝐿𝑠 𝑑𝑡 𝐿 0 𝐿𝑠 EE326 Dept. of E & E, NITK Surathkal 7 Problems: Electrical Networks P1: Write the differential Equation using mesh 𝑉𝑐 𝑠 analysis and derive the transfer function 𝑉 𝑠 𝐿𝑑𝑖 𝑡 1 𝑡 𝐼 𝑠 𝑣 𝑡 = + 𝑅𝑖 𝑡 + න 𝑖 𝑡 𝑑𝑡 ⇒ 𝑉 𝑠 = 𝐿𝑠𝐼 𝑠 + 𝑅𝐼 𝑠 + 𝑑𝑡 𝐶 0 𝐶𝑠 1 = 𝐿𝑠 + 𝑅 + 𝐼 𝑠 1 𝐶𝑠 𝑉𝐶 𝑠 𝐿𝐶 Also, = 𝑉 𝑠 𝑅 1 𝑉𝐶 𝑠 𝐶𝑠 = 𝐼(𝑠) 𝑠 2 + 𝐿 𝑠 + 𝐿𝐶 1 𝑉 𝑠 𝑉𝐶 𝑠 𝐿𝐶 𝑉𝐶 𝑠 = 𝑉 𝑠 𝑅 1 𝑠 2 + 𝐿 𝑠 + 𝐿𝐶 EE326 Dept. of E & E, NITK Surathkal 8  Problems: Electrical Networks P2: Write the Algebraic Equation in Laplace transformed domain using mesh analysis and derive the 𝑉𝑐 𝑠 transfer function 𝑉 𝑠 𝑉 𝑠 = 𝑅1 𝐼1 𝑠 + 𝐿𝑠 𝐼1 𝑠 − 𝐼2 𝑠 ⇒ 1 𝑅1 + 𝐿𝑠 −𝐿𝑠 𝐼1 (𝑠) 𝑉(𝑠) 𝐼2 𝑠 1 = −𝐿𝑠 𝑅2 + 𝐿𝑠 + 𝐼2 𝑠 0 0 = 𝑅2 𝐼2 𝑠 + + 𝐿𝑠 𝐼2 𝑠 − 𝐼1 𝑠 ⇒ 2 𝐶𝑠 𝐶𝑠 Using Cramer’s Rule 𝑅1 + 𝐿𝑠 𝑉(𝑠) 𝐼2 (𝑠) = −𝐿𝑠 0 𝐼2 𝑠 𝐿𝑠 𝑅1 + 𝐿𝑠 −𝐿𝑠 ⇒ = 𝑉 𝑠 1 2 1 𝑅1 + 𝐿𝑠 𝑅2 + 𝐿𝑠 + − 𝐿𝑠 −𝐿𝑠 𝑅2 + 𝐿𝑠 + 𝐶𝑠 𝐶𝑠 𝑉𝑐 𝑠 1 𝐼2 𝑠 ⇒ = 𝑉 𝑠 𝐶𝑠 𝑉 𝑠 EE326 Dept. of E & E, NITK Surathkal 9  Problems: Electrical Networks 𝑉𝐿 𝑠 P3: Find the TF 𝑉 𝑠 1 1 𝑉1 𝑠 1 + + 1 − 𝑉𝐿 𝑠 = 𝑉(𝑠) ⇒ 1 2+ −1 𝑉(𝑠) 𝑠 𝑠 𝑉1 (𝑠) = 1 2 𝑉𝐿 𝑠 𝑉 𝑠 1 1 1 −1 +1 𝑠 −𝑉1 𝑠 + + + 1 𝑉𝐿 𝑠 = 𝑉(𝑠) ⇒ 2 𝑠 𝑠 𝑠 𝑠 Using Cramer’s Rule 1 𝑉𝐿 𝑠 𝑠 2 + 2𝑠 + 1 2+ 𝑉(𝑠) 𝑠 ⇒ = 2 1 𝑉 𝑠 𝑠 + 5𝑠 + 2 −1 𝑉𝐿 (𝑠) = 𝑠𝑉 𝑠 1 2+𝑠 −1 Can we calculate the step response? 2 −1 𝑠+1 EE326 Dept. of E & E, NITK Surathkal 10  Components Translation System Impedance Component Force-Velocity Force-Displacement 𝐹 𝑠 𝑍𝑀 𝑠 = Spring 𝑡 𝑋(𝑠) 𝑓 𝑡 = 𝐾 න 𝑣 𝑡 𝑑𝑡 𝑓 𝑡 = 𝐾𝑥(𝑡) 𝐾 0 Viscous Damper 𝑑𝑥 𝑡 𝑓 𝑡 = 𝐵𝑣(𝑡) 𝑓 𝑡 =𝐵 𝐵𝑠 𝑑𝑡 𝐵 Mass 𝑑𝑣 𝑡 𝑑2𝑥 𝑡 𝑓 𝑡 =𝑀 𝑓 𝑡 =𝑀 𝑀𝑠 2 𝑑𝑡 𝑑𝑡 2 EE326 Dept. of E & E, NITK Surathkal 11  Simple Translational mechanical system P1: Write the differential Equation of motion using Newton’s Law and derive 𝐹 𝑠 the transfer function 𝑋 𝑠 Transfer Function 1 𝑋 𝑠 𝑀 = 𝐹 𝑠 𝐵 𝑘 𝑠2 + 𝑀 𝑠 + 𝑀 Free Body Diagram Remember: 1 𝐼 𝑠 𝐿𝐶 = 𝑉 𝑠 𝑅 1 𝑠 2 + 𝐿 𝑠 + 𝐿𝐶 EE326 Dept. of E & E, NITK Surathkal 12 Simple Translational mechanical system  Different components and their units: ◦ Spring constant ,K (N/m) ◦ Coefficient of friction, B (N-s/m) ◦ Mass, M (N s2/m or kg)  Steps involved: ◦ Drawing the free body diagram ◦ Writing the differential equation ◦ Converting to frequency domain ◦ Find the transfer function (by solving the linear eqn)  Number of independent motion  The Degrees of Freedom (DoF/DOF) EE326 Dept. of E & E, NITK Surathkal 13  Simple Translational mechanical system P2: Write the differential Equation of motion using Newton’s Law Two DOF Free body diagram of 𝑀1 Two components: (1) due to its own movement 𝑥1 & (2) due to the movement of 𝑥2 𝐹(𝑠) 𝐾1 𝑋1 (𝑠) 𝐾2 𝑋2 (𝑠) 𝐵3 𝑠𝑋1 (𝑠) 𝐵1 𝑠𝑋1 (𝑠) 𝑀1 𝑀1 𝐾2 𝑋1 (𝑠) 𝑀1 𝑠 2 𝑋1 (𝑠) 𝐵3 𝑠𝑋2 (𝑠) EE326 Dept. of E & E, NITK Surathkal 14  Simple Translational mechanical system Two DOF Free body diagram of 𝑀1 Net effect by using superposition 𝐾1 𝑋1 (𝑠) 𝐾2 𝑋1 𝑠 − 𝑋2 𝑠 𝐹(𝑠) 𝐵3 𝑠 𝑋1 𝑠 − 𝑋2 𝑠 𝑀1 𝐵1 𝑠𝑋1 (𝑠) 𝑀1 𝑠 2 𝑋1 (𝑠) EE326 Dept. of E & E, NITK Surathkal 15  Simple Translational mechanical system Two DOF Free body diagram of 𝑀2 Two components: (1) due to its own movement 𝑥2 & (2) due to the movement of 𝑥1 𝐾3 𝑋2 (𝑠) 𝐾2 𝑋2 (𝑠) 𝐾2 𝑋1 (𝑠) 𝐵3 𝑠𝑋2 (𝑠) 𝑀2 𝑀2 𝐵2 𝑠𝑋2 (𝑠) 𝑀2 𝑠 2 𝑋2 (𝑠) 𝐵3 𝑠𝑋1 (𝑠) EE326 Dept. of E & E, NITK Surathkal 16  Simple Translational mechanical system Two DOF Free body diagram of 𝑀2 Net effect by using superposition 𝐾2 𝑋2 𝑠 − 𝑋1 𝑠 𝐾3 𝑋2 (𝑠) 𝐵3 𝑠 𝑋2 𝑠 − 𝑋1 𝑠 𝑀2 𝐵2 𝑠𝑋2 (𝑠) 𝑀2 𝑠 2 𝑋2 (𝑠) EE326 Dept. of E & E, NITK Surathkal 17  Simple Translational mechanical system 𝑋2 𝑠 Find G 𝑠 = 𝐹 𝑠 𝑠 2 + 3𝑠 + 1 − 3𝑠 − 1 𝑋1 (𝑠) = 𝐹 𝑠 −3𝑠 − 1 𝑠 2 + 4𝑠 + 1 𝑋2 𝑠 0 Applying Cramer’s rule 𝑠 2 + 3𝑠 + 1 𝐹(𝑠) 3𝑠 + 1 𝑋2 𝑠 = 2 −3𝑠 − 1 0 G 𝑠 = 𝑠 + 3𝑠 + 1 − 3𝑠 − 1 𝑠 𝑠 3 + 7𝑠 2 + 5𝑠 + 1 −3𝑠 − 1 𝑠 2 + 4𝑠 + 1 EE326 Dept. of E & E, NITK Surathkal 18 Rotational mechanical systems Component Torque-Angular Displacement Unit N-m/rad N-m-s/rad N-m-s2/rad or kg-m2 EE326 Dept. of E & E, NITK Surathkal 19  Rotational mechanical system Write the equation(s) of motion for the system Lumped system modeling Free body diagram of 𝐽1 Two components: (1) due to its own movement 𝜃1 & (2) due to the movement of 𝜃2 𝑇(𝑠) 𝑇(𝑠) 𝐷1𝑠𝜃1 (𝑠) 𝐷1𝑠𝜃1 (𝑠) 𝐽1 + 𝐽1 = 𝐽1 𝐾𝜃2 (𝑠) 𝐾𝜃1 (𝑠) 𝐽1 𝑠2𝜃1 (𝑠) 𝐾[𝜃1 𝑠 − 𝐽1 𝑠2𝜃 1 (𝑠) 𝜃2 (𝑠)] EE326 Dept. of E & E, NITK Surathkal 20  Rotational mechanical system Write the equation(s) of motion for the system Lumped system modeling Free body diagram of 𝐽2 Two components: (1) due to its own movement 𝜃2 & (2) due to the movement of 𝜃1 𝐷2𝑠𝜃2 (𝑠) 𝐷2𝑠𝜃2 (𝑠) 𝐽2 + 𝐽2 = 𝐽2 𝐾𝜃1 (𝑠) 𝐾𝜃2 (𝑠) 𝐽2 𝑠2𝜃2 (𝑠) 𝐾[𝜃2 𝑠 − 𝐽2 𝑠2𝜃 2 (𝑠) 𝜃1 (𝑠)] EE326 Dept. of E & E, NITK Surathkal 21  Rotational mechanical system Write the equation(s) of motion for the system Lumped system modeling Equations 𝐽1𝑠 2 + 𝐷1𝑠 + 𝐾 − 𝐾 𝜃1 (𝑠) 𝑇 𝑠 = −𝐾 𝐽2𝑠 2 + 𝐷2𝑠 + 𝐾 𝜃2 𝑠 0 EE326 Dept. of E & E, NITK Surathkal 22  Rotational mechanical system Write the equation(s) of motion for the system 𝜃2 (𝑡) 𝜃1 (𝑡) 𝑠 2 + 1𝑠 + 1 − 1𝑠 − 1 𝜃1 (𝑠) = 𝑇 𝑠 −1𝑠 − 1 2(𝑠 + 1) 𝜃2 𝑠 0 𝜃2 𝑠 𝑠+1 1 = = 𝑇 𝑠 2 𝑠2 + 𝑠 + 1 𝑠 + 1 − 𝑠 + 1 2 2𝑠2 + 𝑠 + 1 EE326 Dept. of E & E, NITK Surathkal 23 Rotational mechanical system with gears Simple Gear arrangement 2𝜋𝑟 𝜃𝑟 𝜃 Assuming No Backlash 𝜃1𝑟1 = 𝜃2𝑟2 𝑁1 𝑟1 𝜃2 = = 𝑁1 ∝ 𝑟1& 𝑁2 ∝ 𝑟2 𝑁2 𝑟1 𝜃1 Also Assuming gear to be lossless Energy remains the same 𝑁1 𝑇1 = 𝑇1𝜃1 = 𝑇2𝜃2 𝑁2 𝑇2 EE326 Dept. of E & E, NITK Surathkal 24  Rotational mechanical system with gears Equation connecting the number of teeth with torque and angular displacement 𝑁1 𝑇1 𝜃2 = = 𝑁2 𝑇2 𝜃1 Example-1 Writing the total equation EE326 Dept. of E & E, NITK Surathkal 25  Rotational mechanical system with gears Equation connecting the number of teeth with torque and angular displacement 𝑁1 𝑇1 𝜃2 = = Example-1 𝑁2 𝑇2 𝜃1 Removing the gear Writing eqn w.r.t. (2) 𝑁2 𝜃2 𝑇1 𝑁2 𝑁1 𝑇1 𝑠 = Js2 + Ds + K 𝜃2(𝑠) 𝑁1 𝑁1 Writing eqn w.r.t. (1) × 𝑁2 𝑁1 E.g.: Js2𝜃2(𝑠) Js2𝜃2(𝑠) × 𝑁1 2 From 2 to 1 𝑁2 𝐷 𝑁2 2 𝑁1 𝐽 𝑁 𝑁2 𝑇1 𝑠 = Js2 + Ds + K 𝜃2(𝑠) × 𝑁1 2 2 𝑁1 𝐾 𝑁2 𝑁1 2 𝑇1 𝑠 = Js2 + Ds + K 𝜃1(𝑠) × EE326 Dept. of E & E, NITK Surathkal 𝑁2 26  Rotational mechanical system with gears P1: Find the quantities 𝐽𝑒 , 𝐷𝑒 & 𝐾𝑒 so that the systems are similar. In Fig 𝜃 𝑠 (a) and (b). Also find the resulting transfer function 𝑇2 1 𝑠 2 𝑁2 𝑁2 2 𝐽𝑒 = 𝐽2 + 𝐽1 𝐷𝑒 = 𝐷2 + 𝐷1 𝐾𝑒 = 𝐾2 𝑁1 𝑁1 𝜃2 𝑠 𝑁2 1 = 𝑇1 𝑠 𝑁1 𝐽𝑒 𝑠 2 + 𝐷𝑒 𝑠 + 𝐾2 EE326 Dept. of E & E, NITK Surathkal 27  Rotational mechanical system with gears P2: write 𝜃4 in terms of 𝜃1 𝑁2 𝜃1 𝑡 = 𝜃 𝑡 𝑁1 2 𝑁2 𝑁4 𝜃1 𝑡 = 𝜃 𝑡 𝑁1 𝑁3 3 𝑁2 𝑁4 𝑁6 𝜃1 𝑡 = 𝜃 𝑡 𝑁1 𝑁3 𝑁5 4 EE326 Dept. of E & E, NITK Surathkal 28  Rotational mechanical system with gears P3: Find the quantities 𝐽𝑒 , 𝐷𝑒 such that the two systems are similar. Also 𝜃1 𝑠 find the resulting transfer function 𝑇 𝑠. Note that the moment of inertia and the friction are not neglected for the gearing arrangement. 2 2 𝑁3 𝑁1 𝐽1 + (𝐽4 + 𝐽5 ) +(𝐽2 + 𝐽3 ) = 𝐽𝑒 𝑁4 𝑁2 2 𝑁1 𝜃1 𝑠 1 𝐷1 + 𝐷2 = 𝐷𝑒 = 𝑁2 𝑇1 𝑠 𝐽𝑒 𝑠 2 + 𝐷𝑒 𝑠 EE326 Dept. of E & E, NITK Surathkal 29  Rotational mechanical system with gears P4:write the equation for the system shown below. 50 𝜃1 (𝑡) 𝜃21 𝑡 = 𝜃 (𝑡) 25 2 2 𝑇 𝑠 = 𝑠 2 + 𝑠 𝜃1 𝑠 − 𝑠𝜃21 (𝑠) 𝑁2 𝑁1 0 = 4𝜃2 𝑠 + 1𝑠 𝜃2 𝑠 − 𝜃 (𝑠) 𝑁1 𝑁2 1 𝑁2 𝑇 𝑠 = 𝑠2 + 𝑠 𝜃1 𝑠 − 𝑠𝜃2 (𝑠) 0 = 4𝜃2 𝑠 + 4𝑠 𝜃2 𝑠 − 0.5 𝜃1 (𝑠) 𝑁1 𝑇 𝑠 = 𝑠 2 + 𝑠 𝜃1 𝑠 − 2𝑠𝜃2 (𝑠) 0 = −2𝑠𝜃1 𝑠 + 4𝑠 + 4 𝜃2 𝑠 EE326 Dept. of E & E, NITK Surathkal 30 Armature controlled DC motor 𝑣𝑏 : back emf 𝑑𝜃𝑚 𝑣𝑏 𝑡 = 𝐾𝑏 𝑑𝑡 𝐾𝑏 : back emf constant (V-s/rad) 𝑇𝑚 (𝑡) = 𝐾𝑡 𝑖𝑎 (𝑡) 𝐾𝑡 : Torque constant Electrical Ckt (N-m/A) 𝑑𝑖𝑎 𝑡 𝑒𝑎 𝑡 = 𝑅𝑎 𝑖𝑎 𝑡 + 𝐿𝑎 + 𝑣𝑏 (𝑡) 𝑑𝑡 𝑇𝑚 𝑠 𝐸𝑎 𝑠 = 𝑅𝑎 + 𝐿𝑎 𝑠 + 𝐾𝑏 𝑠Θm (𝑠) 𝐾𝑡 ? 𝐸𝑎 𝑠 Θm (𝑠) EE326 Dept. of E & E, NITK Surathkal 31 Armature controlled DC motor 𝑇𝑚 𝑠 𝐸𝑎 𝑠 = 𝑅𝑎 + 𝐿𝑎 𝑠 + 𝐾𝑏 𝑠Θ𝑚 (𝑠) 𝐾𝑡 ? 𝐸𝑎 𝑠 Θ(𝑠) Θ𝑀 𝑠 𝐾𝑡 Τ𝑅𝑎 𝐽𝑚 = 𝐸𝑎 𝑠 1 𝐾𝑡 𝐾𝑏 𝑠 𝑠 + 𝐽 𝐷𝑚 + 𝑅 𝑚 𝑎 Motor Load Mechanical Model 𝑇𝑚 𝑠 = 𝐽𝑚 𝑠 2 + 𝐷𝑚 𝑠 Θ𝑚 (𝑠) What is the composition of 𝐽𝑚 & 𝐷𝑀 EE326 Dept. of E & E, NITK Surathkal 32 Armature controlled DC motor Θ𝑀 𝑠 𝐾𝑡 Τ𝑅𝑎 𝐽𝑚 = 𝐸𝑎 𝑠 1 𝐾𝑡 𝐾𝑏 𝑠 𝑠 + 𝐽 𝐷𝑚 + 𝑅 𝑚 𝑎 Composition of 𝐽𝑚 & 𝐷𝑀 2 𝑁1 𝐽𝑚 = 𝐽𝑎 + 𝐽𝐿 𝑁2 2 𝑁1 𝐷𝑚 = 𝐷𝑎 + 𝐷𝐿 𝑁2 Mechanical Constants EE326 Dept. of E & E, NITK Surathkal 33  Armature controlled DC motor Θ𝑀 𝑠 𝐾𝑡 Τ𝑅𝑎 𝐽𝑚 = 𝐸𝑎 𝑠 1 𝐾𝑡 𝐾𝑏 𝑠 𝑠 + 𝐽 𝐷𝑚 + 𝑅 𝑚 𝑎 2 𝑁1 𝐽𝑚 = 𝐽𝑎 + 𝐽𝐿 𝑁2 2 𝑁1 𝐷𝑚 = 𝐷𝑎 + 𝐷𝐿 𝑁2 Finding the electrical Constants 𝑇𝑚 Torque-Speed Curve We have 𝑇𝑚 𝑠 𝑇stall 𝐸𝑎 𝑠 = 𝑅𝑎 + 𝐿𝑎 𝑠 + 𝐾𝑏 𝑠Θ𝑚 (𝑠) 𝑒𝑎1 > 𝑒𝑎2 𝐾𝑡 𝑒𝑎1 At steady state while taking inv Laplace transform and keeping 𝐿𝑎 = 0 𝑒𝑎2 𝑇𝑚 𝐾𝑡 𝐾𝑏 𝐾𝑡 𝑒𝑎 = 𝑅𝑎 + 𝐾𝑏 𝜔𝑚 𝑇𝑚 = − 𝜔 + 𝑒 𝐾𝑡 𝑅𝑎 𝑚 𝑅𝑎 𝑎 𝜔no−load𝜔𝑚 𝐾𝑡 𝑒𝑎 𝑇stall = 𝑒 𝜔no−load = 𝑅𝑎 𝑎 𝐾𝑏 EE326 Dept. of E & E, NITK Surathkal 34  Armature controlled DC motor Θ P1: Given the system and torque-speed curve find the transfer function 𝐸 𝐿(𝑠) 𝑠 𝑎 𝑁1 2 𝐾𝑡 Θ𝑀 𝑠 𝐾𝑡 Τ𝑅𝑎 𝐽𝑚 𝐽𝑚 = 𝐽𝑎 + 𝐽𝐿 𝑇stall = 𝑒 = 𝑁2 𝑅𝑎 𝑎 𝐸𝑎 𝑠 1 𝐾𝑡 𝐾𝑏 𝑠 𝑠 + 𝐽 𝐷𝑚 + 𝑅 2 𝑚 𝑎 𝑁1 𝑒𝑎 𝐷𝑚 = 𝐷𝑎 + 𝐷𝐿 𝜔no−load = 𝑁2 𝐾𝑏 Θ𝐿 𝑠 0.0417 = 𝐸𝑎 𝑠 𝑠 𝑠 + 1.667 EE326 Dept. of E & E, NITK Surathkal 35  Electrical Series analogy (translational) Eqn (in terms of velocity): 1 𝐾 𝐿𝑠 + 𝑅 + 𝐼 𝑠 = 𝐸(𝑠) 𝑀𝑠 + 𝐵 + 𝑉 𝑠 = 𝐹(𝑠) 𝐶𝑠 𝑠 𝑀 𝐵 Mass 𝑀 Inductor M 𝐻 Viscous Damper 𝐵 Resistor B Ω 1 𝐹(𝑠) Spring Const 𝐾 1 𝑉(𝑠) 𝐾 Capacitor K F Force 𝑓(𝑡) Voltage 𝑓 𝑡 𝑉 Velocity 𝑣(𝑡) Current 𝑣 𝑡 𝐴 EE326 Dept. of E & E, NITK Surathkal Force-Voltage analogy 36  Electrical Series analogy (translational) P1: Draw the Electrical Series Analogy of the given translational system 1 1 𝑀1 𝐾1 𝐵1 𝑀2 𝐾2 𝐹(𝑠) 𝐵3 𝑉1 (𝑠) 𝑉2 (𝑠) 𝐵2 1 𝐾3 EE326 Dept. of E & E, NITK Surathkal 37  Electrical parallel analogy (translational) Eqn (in terms of velocity): 1 1 𝐾 𝐶𝑠 + + 𝐸 𝑠 = 𝐼(𝑠) 𝑀𝑠 + 𝐵 + 𝑉 𝑠 = 𝐹(𝑠) 𝑅 𝐿𝑠 𝑠 Mass 𝑀 Capacitor M 𝐹 𝑉(𝑠) 1 Viscous Damper 𝐵 Resistor B Ω 𝐹(𝑠) 1 Spring Const 𝐾 1 𝑀 1 𝐾 Inductor K H 𝐵 Force 𝑓(𝑡) Current 𝑓 𝑡 𝐴 Velocity 𝑣(𝑡) Voltage 𝑣 𝑡 𝑉 EE326 Dept. of E & E, NITK Surathkal Force-Current analogy 38  Electrical Parallel analogy (translational) P2: Draw the Electrical Parallel Analogy of the given translational system 1 𝐵3 𝑉1 (𝑠) 𝑉2 (𝑠) 𝐹(𝑠) 1 1 1 1 1 𝑀1 𝐾2 𝑀2 𝐵1 𝐾1 𝐵2 𝐾3 EE326 Dept. of E & E, NITK Surathkal 39  Electrical analogy (rotational) HW: Draw the Electrical Series and Parallel Analogy of the following rotational system EE326 Dept. of E & E, NITK Surathkal 40  Azimuth Angle Control System 𝜃𝑖 𝑡 Desired azimuth potentiometer angle input 𝜃0 𝑡 Azimuth angle output Differential amplifier potentiometer and Power amplifier motor Detailed Layout Functional Block Diagram 𝜃𝑖 𝑡 + error Signal and 𝜃0 𝑡 Motor, load Potentiometer Power and gears − amplifiers Potentiometer EE326 Dept. of E & E, NITK Surathkal 41  Azimuth +𝑉 Angle Control System 𝑛- turn potentiometer 𝜃𝑖 𝑡 −𝑉 𝐽𝑎 𝐾1 𝐷𝑎 𝜃𝑚 𝑡 𝑁1 𝑠+𝑎 𝐾𝑏 𝐾𝑡 𝑁2 +𝑉 𝜃0 𝑡 𝐷𝐿 𝐽𝐿 𝑛 -turn 𝑁3 potentiometer −𝑉 𝑉 = 10 𝑉 20 1 𝑛 = 10 Potentiometer 10 turns = 20V ⇒ 𝐾𝑝𝑜𝑡 = = 20𝜋 𝜋 𝐾1 = 100 Motor Load 2 𝑁1 𝑎 = 100 𝐽𝑚 = 𝐽𝑎 + 𝐽𝐿 = 0.02 + 0.1 2 × 1 = 0.03 𝐷𝑚 = 0.02 𝑅𝑎 = 8 Ω 𝑁2 𝐽𝑎 = 0.02 𝑘𝑔 − 𝑚2 𝜃𝑚 𝑠 𝐾𝑡 Τ𝑅𝑎 𝐽𝑚 2.083 𝐷𝑎 = 0.01 𝑁 − 𝑚 𝑠Τ𝑟𝑎𝑑 = = 𝐸𝑎 𝑠 1 𝐾𝐾 𝑠 𝑠 + 1.708 𝐾𝑏 = 0.5 𝑉 𝑠Τ𝑟𝑎𝑑 𝑠 𝑠+ 𝐷𝑚 + 𝑡 𝑏 𝐽𝑚 𝑅𝑎 𝐾𝑡 = 0.5 𝑁 − 𝑚Τ𝐴 TF 𝑁1 = 25, 𝑁2 = 𝑁3 = 250 𝜃0 𝑠 0.2083 = 𝐽𝐿 = 1 𝑘𝑔 − 𝑚2 𝐸𝑎 𝑠 𝑠 𝑠 + 1.708 𝐷𝐿 = 1 𝑁 − 𝑚 𝑠Τ𝑟𝑎𝑑 EE326 Dept. of E & E, NITK Surathkal 42  Azimuth Angle Control System Power Motor and Potentiometer Preamplifier amplifier Load Gear 𝜃𝑚 (𝑠) 𝜃0 (𝑠) + 100 𝐸𝑎 (𝑠) 2.083 1Τ𝜋 1 0.1 Desired 𝑉𝑝 (𝑠) 𝑠 + 100 𝑠(𝑠 + 1.708) Azimuth − 𝑉𝑒 (𝑠) angle 𝜃𝑖 (𝑠) 1Τ𝜋 Assumptions: Potentiometer All components are linear. The amplifier is assumed to have no saturation. Dynamics of the pre-amplifier is neglected. 𝐿 𝐽 Electrical time constant 𝑅𝑎 is neglected (∵≪ 𝐷𝑚 ). 𝑎 𝑚 𝜃0 𝑠 20.83 = 𝜃𝑖 𝑠 𝑠 𝑠 + 100 𝑠 + 1.7088 𝜋 + 20.83 EE326 Dept. of E & E, NITK Surathkal 43 Components of a block diagram (linear System) 𝑅(𝑠) 𝐶(𝑠) 𝑅(𝑠) 𝐶(𝑠) 𝐺(𝑠) input Output Signals System 𝑅(𝑠) 𝑅1 (𝑠) 𝐶 𝑠 = 𝑅1 𝑠 + 𝑅2 𝑠 − 𝑅3 (𝑠) 𝑅(𝑠) + 𝑅(𝑠) + 𝑅2 (𝑠) − 𝑅(𝑠) 𝑅3 (𝑠) Summing Junction Pickoff Point EE326 Dept. of E & E, NITK Surathkal 44 Familiar Configurations 1) Cascade Form 𝑋1 𝑠 = 𝑋2 𝑠 = 𝑅 𝑠 𝐺1 𝑠 𝐺2 𝑠 𝑅(𝑠) 𝑅 𝑠 𝐺1 (𝑠) 𝐺1 (𝑠) 𝐺2 (𝑠) 𝐶 𝑠 = 𝐺3 (𝑠) 𝑅 𝑠 𝐺1 𝑠 𝐺2 𝑠 𝐺3 (𝑠) 𝐺𝑒 𝑠 = 𝐺1 𝑠 𝐺2 𝑠 𝐺3 (𝑠) 2) Parallel Form 𝐺1 (𝑠) 𝑅(𝑠) ± 𝐶 𝑠 = 𝐺2 (𝑠) ± 𝑅 𝑠 𝐺1 𝑠 ± 𝐺2 𝑠 ± 𝐺3 (𝑠) ± 𝐺3 (𝑠) 𝐺𝑒 𝑠 = 𝐺1 𝑠 ± 𝐺2 𝑠 ± 𝐺3 (𝑠) EE326 Dept. of E & E, NITK Surathkal 45 Familiar Configurations 3) Feedback Form 𝑅(𝑠) + 𝐶(𝑠) 𝐺(𝑠) ∓ 𝐻(𝑠) 𝐺 𝑠 𝐺𝑒 𝑠 = 1 ± 𝐺 𝑠 𝐻(𝑠) EE326 Dept. of E & E, NITK Surathkal 46  Moving Blocks to Create Familiar Configurations 1) Basic Block moves left or right past summing junctions 𝑅(𝑠) Left 𝑅(𝑠) 𝐶(𝑠) + 𝐶(𝑠) + 𝐺(𝑠) ≡ 𝐺(𝑠) ∓ ∓ 𝑋(𝑠) 𝐺(𝑠) 𝑋(𝑠) 𝑅(𝑠) 𝐶(𝑠) 𝑅(𝑠) + Right + 𝐶(𝑠) 𝐺(𝑠) 𝐺(𝑠) ∓ ≡ 𝑋(𝑠) ∓ 1 𝐺(𝑠) EE326 Dept. of E & E, NITK Surathkal 𝑋(𝑠) 47  Moving Blocks to Create Familiar Configurations 2) Basic Block moves left or right past pickoff points 𝑅 𝑠 𝐺(𝑠) 𝑅 𝑠 𝐺(𝑠) Left 𝐺(𝑠) 𝑅(𝑠) ≡ 𝑅(𝑠) 1 𝑅(𝑠) 𝐺(𝑠) 𝑅(𝑠) 𝐺(𝑠) 𝑅(𝑠) 1 𝑅(𝑠) 𝐺(𝑠) 𝑅 𝑠 𝐺(𝑠) 𝐺(𝑠) 𝑅 𝑠 𝐺(𝑠) Right 𝑅(𝑠) 𝑅 𝑠 𝐺(𝑠) 𝑅(𝑠) 𝐺(𝑠) ≡ 𝐺(𝑠) 𝑅 𝑠 𝐺(𝑠) 𝑅 𝑠 𝐺(𝑠) 𝐺(𝑠) 𝑅 𝑠 𝐺(𝑠) EE326 Dept. of E & E, NITK Surathkal 48 Block Diagram Reduction Problems P1 𝑅(𝑠) 𝐶(𝑠) 𝐺1 (𝑠) 𝐺2 (𝑠) 𝐺3 (𝑠) + + + − − + 𝐻1 (𝑠) + − 𝐻2 (𝑠) − + 𝐻3 (𝑠) Step-1: Feedback TF → − 𝐻1 − 𝐻2 + 𝐻3 and 𝐺2 𝐺3 (cascade) 𝑅(𝑠) + 𝐶(𝑠) 𝐺1 (𝑠) 𝐺2 𝑠 𝐺3 (𝑠) − 𝐻1 𝑠 − 𝐻2 𝑠 + 𝐻3 (𝑠) EE326 Dept. of E & E, NITK Surathkal 49 Block Diagram Reduction Problems P1 𝑅(𝑠) + 𝐶(𝑠) 𝐺1 (𝑠) 𝐺2 𝑠 𝐺3 (𝑠) − 𝐻1 𝑠 − 𝐻2 𝑠 + 𝐻3 (𝑠) Step-2: Feedback TF cascaded with 𝐺1 (𝑠) 𝐺1 𝑠 𝐺2 𝑠 𝐺3 𝑠 1 + 𝐺2 𝑠 𝐺3 𝑠 𝐻1 𝑠 − 𝐻2 𝑠 + 𝐻3 (𝑠) EE326 Dept. of E & E, NITK Surathkal 50 Block Diagram Reduction Problems P2 Step-1 EE326 Dept. of E & E, NITK Surathkal 51 Block Diagram Reduction Problems P2 Step-2 EE326 Dept. of E & E, NITK Surathkal 52 Block Diagram Reduction Problems P2 Step-3 EE326 Dept. of E & E, NITK Surathkal 53 Block Diagram Reduction Problems P2 Step-4 Step-5: Answer EE326 Dept. of E & E, NITK Surathkal 54 Block Diagram Reduction Problems P2 MATLAB CODE 1 2 3 5 6 4 1 1 Let 𝐺1 = 𝐺2 = 𝐺3 = 𝑠+1 and 𝐻1 = 𝐻2 = 𝐻3 = 𝑠 Code: G1= tf(1,[1 1]); G2=G1; G3 = G1; H1 = tf(1,[1 0]); H2=H1;H3=H1; Sys = append(G1,G2,G3,H1,H2,H3); inp=1;out=3; Q = [1 -4 0 0 0; T=connect(Sys,Q, inp, out); 2 1 -5 0 0; T=tf(T); 3 1 -5 2 -6 ; 4 2 0 0 0; 5 2 0 0 0; 6 3 0 0 0]; EE326 Dept. of E & E, NITK Surathkal 55 Block Diagram Reduction Problems P3 − 𝐶(𝑠) 𝑅(𝑠) + 1 + 𝑠 𝑠 𝑠 − + 1 𝑠 𝑠 − 𝐶(𝑠) 𝑅(𝑠) 𝑠3 + 1 1 + 𝑠 𝑠 − 𝑠 EE326 Dept. of E & E, NITK Surathkal 56 Block Diagram Reduction Problems P3 𝑅(𝑠) 𝐶(𝑠) 𝑠3 + 1 1 + 𝑠3 + 𝑠 + 1 𝑠 − 𝑠 𝑅(𝑠) 𝐶(𝑠) 𝑠3 + 1 2𝑠 4 + 𝑠 2 + 2𝑠 EE326 Dept. of E & E, NITK Surathkal 57 Components of a Signal Flow Graphs 𝑅(𝑠) 𝐶(𝑠) 𝑅(𝑠) 𝐶(𝑠) 𝐺(𝑠) input Output Signals System 𝑅(𝑠) 𝐶(𝑠) 𝑅(𝑠) 𝐺(𝑠) 𝐶(𝑠) 𝑅(𝑠) 𝑅1 (𝑠) 𝐶 𝑠 = 𝑅1 𝑠 + 𝑅2 𝑠 − 𝑅3 (𝑠) 𝑅(𝑠) + 𝑅(𝑠) + 𝑅2 (𝑠) − 𝑅(𝑠) 𝑅3 (𝑠) 𝑅1 (𝑠) 𝐺1 (𝑠) 𝐶(𝑠) 1 𝑅(𝑠) 𝐺3 (𝑠) 𝑅(𝑠) 1 𝑅(𝑠) Summing Junction 𝑅3 (𝑠) Pickoff Point EE326 Dept. of E & E, NITK Surathkal 58 Familiar Configurations 1) Cascade Form – In block Diagram 𝑋1 𝑠 = 𝑋2 𝑠 = 𝑅 𝑠 𝐺1 𝑠 𝐺2 𝑠 𝑅(𝑠) 𝑅 𝑠 𝐺1 (𝑠) 𝐺1 (𝑠) 𝐺2 (𝑠) 𝐶 𝑠 = 𝐺3 (𝑠) 𝑅 𝑠 𝐺1 𝑠 𝐺2 𝑠 𝐺3 (𝑠) 𝐺𝑒 𝑠 = 𝐺1 𝑠 𝐺2 𝑠 𝐺3 (𝑠) In Signal Flow Graph 𝑅(𝑠) 𝐺1 (𝑠) 𝑋2 (𝑠) 𝐺2 (𝑠) 𝑋1 (𝑠) 𝐺3 (𝑠) 𝐶(𝑠) EE326 Dept. of E & E, NITK Surathkal 59 Familiar Configurations 2) Parallel Form 𝐺1 (𝑠) 𝑅(𝑠) ± 𝐶 𝑠 = 𝐺2 (𝑠) ± 𝑅 𝑠 𝐺1 𝑠 ± 𝐺2 𝑠 ± 𝐺3 (𝑠) ± 𝐺3 (𝑠) 𝐺𝑒 𝑠 = 𝐺1 𝑠 ± 𝐺2 𝑠 ± 𝐺3 (𝑠) In Signal Flow Graph ±𝐺1 (𝑠) 𝑅(𝑠) C(𝑠) ±𝐺2 (𝑠) ±𝐺3 (𝑠) EE326 Dept. of E & E, NITK Surathkal 60 Familiar Configurations 3) Feedback Form 𝑅(𝑠) + 𝐶(𝑠) 𝐺(𝑠) ∓ 𝐻(𝑠) 𝐺 𝑠 𝐺𝑒 𝑠 = 1 ± 𝐺 𝑠 𝐻(𝑠) In Signal Flow Graph 𝑅(𝑠) 1 𝐸(𝑠) 𝐺(𝑠) 𝐶(𝑠) ∓𝐻(𝑠) EE326 Dept. of E & E, NITK Surathkal 61 Block Diagram Reduction Problems P1: Convert this block diagram to signal flow graph EE326 Dept. of E & E, NITK Surathkal 62 Mason’s Rule Demo Example Terminology Loop Gains: The product of branch gains found by traversing a path that starts at a node and ends at the same node, following the direction of the signal flow, without passing through any other node more than once 𝐺2 𝐻1 𝐺4 𝐻2 𝐺4 𝐺5 𝐻3 𝐺4 𝐺6 𝐻3 EE326 Dept. of E & E, NITK Surathkal 63 Mason’s Rule Demo Example Terminology Forward-Path Gain: The product of gains found by traversing a path from the input node to the output node of the signal-flow graph in the direction of signal flow 𝐺1 𝐺2 𝐺3 𝐺4 𝐺5 𝐺7 𝐺1 𝐺2 𝐺3 𝐺4 𝐺6 𝐺7 EE326 Dept. of E & E, NITK Surathkal 64 Mason’s Rule Demo Example Terminology Non-touching Loops: Loops that do not have any nodes in common. Non-touching Loop Gain: The product of loop gains from non-touching loops taken two, three, four, or more at a time. Non-touching Loop 𝐺2 𝐻1 𝐺4 𝐺5 𝐻3 Gains 𝐺4 𝐻2 𝐺2 𝐻1 𝐺4 𝐺5 𝐻3 𝐺4 𝐺6 𝐻3 𝐺2 𝐻1 𝐺4 𝐻2 𝐺2 𝐻1 𝐺4 𝐺6 𝐻3 Set of touching loops EE326 Dept. of E & E, NITK Surathkal 65 Mason’s Rule Loop Gains: 𝐺2 𝐻1 𝐺4 𝐻2 𝐺4 𝐺5 𝐻3 𝐺4 𝐺6 𝐻3 Demo Example Forward-Path Gain: 𝐺1 𝐺2 𝐺3 𝐺4 𝐺5 𝐺7 𝐺1 𝐺2 𝐺3 𝐺4 𝐺6 𝐺7 Non-touching Loop Gains 𝐺2 𝐻1 𝐺4 𝐺5 𝐻3 𝐺2 𝐻1 𝐺4 𝐻2 𝐺2 𝐻1 𝐺4 𝐺6 𝐻3 𝐺 𝑠 = 𝐶 𝑠 = σ𝑘 𝑇𝑘 Δ𝑘 𝑘 :no of forward-paths Transfer Function: 𝑅 𝑠 Δ 𝑇𝑘 :forward-path gains Δ :1 − Σ loop gains + Σ nontouching-loop gains taken two at a time − Σ nontouching-loop gains taken three at a time +… Δk : formed by eliminating those loops from Δ which touches the 𝑘th forward path Δ = 1 −(𝐺2 𝐻1 + 𝐺4 𝐻2 + 𝐺4 𝐺5 𝐻3 + 𝐺4 𝐺6 𝐻3 ) +(𝐺2 𝐻1 𝐺4 𝐻2 + 𝐺2 𝐻1 𝐺4 𝐺5 𝐻3 + 𝐺2 𝐻1 𝐺4 𝐺6 𝐻3 ) 𝑇1 = 𝐺1 𝐺2 𝐺3 𝐺4 𝐺5 𝐺7 𝑇2 = 𝐺1 𝐺2 𝐺3 𝐺4 𝐺6 𝐺7 Δ1 = 1 Δ2 = 1 EE326 Dept. of E & E, NITK Surathkal 66 Mason’s Rule Demo Example 𝐶 𝑠 σ𝑘 𝑇𝑘 Δ𝑘 𝐺 𝑠 = = 𝑅 𝑠 Δ Δ = 1 −(𝐺2 𝐻1 + 𝐺4 𝐻2 + 𝐺4 𝐺5 𝐻3 + 𝐺4 𝐺6 𝐻3 ) +(𝐺2 𝐻1 𝐺4 𝐻2 + 𝐺2 𝐻1 𝐺4 𝐺5 𝐻3 + 𝐺2 𝐻1 𝐺4 𝐺6 𝐻3 ) 𝑇1 = 𝐺1 𝐺2 𝐺3 𝐺4 𝐺5 𝐺7 𝑇2 = 𝐺1 𝐺2 𝐺3 𝐺4 𝐺6 𝐺7 Δ1 = 1 Δ2 = 1 G s 1 × 𝐺1 𝐺2 𝐺3 𝐺4 𝐺5 𝐺7 + 1 × 𝐺1 𝐺2 𝐺3 𝐺4 𝐺6 𝐺7 = 1 − (𝐺2 𝐻1 + 𝐺4 𝐻2 + 𝐺4 𝐺5 𝐻3 + 𝐺4 𝐺6 𝐻3 ) + (𝐺2 𝐻1 𝐺4 𝐻2 + 𝐺2 𝐻1 𝐺4 𝐺5 𝐻3 + 𝐺2 𝐻1 𝐺4 𝐺6 𝐻3 ) EE326 Dept. of E & E, NITK Surathkal 67 Mason’s Rule P1: Loop Gains: 1. 𝐺2 𝐻1 2. 𝐺4 𝐻2 3. 𝐺7 𝐻4 4. 𝐺2 𝐺3 𝐺4 𝐺5 𝐺6 𝐺7 𝐺8 Forward Paths: 1. 𝐺1 𝐺2 𝐺3 𝐺4 𝐺5 Δ = 1 − 𝐺2 𝐻1 + 𝐺4 𝐻2 + 𝐺7 𝐻4 + 𝐺2 𝐺3 𝐺4 𝐺5 𝐺6 𝐺7 𝐺8 + 𝐺2 𝐻1 × 𝐺4 𝐻2 + 𝐺2 𝐻1 × 𝐺7 𝐻4 + 𝐺4 𝐻2 × 𝐺7 𝐻4 − 𝐺2 𝐻1 × 𝐺4 𝐻2 × 𝐺7 𝐻4 𝑇1 = 𝐺1 𝐺2 𝐺3 𝐺4 𝐺5 Δ1 = 1 − 𝐺7 𝐻4 𝐺1 𝐺2 𝐺3 𝐺4 𝐺5 1 − 𝐺7 𝐻4 𝐺(𝑠) = Δ EE326 Dept. of E & E, NITK Surathkal 68 Linear Control Theory (EE 326): 1.1 Introduction Dharavath Kishan Dept. of Electrical and Electronics Engineering NITK Surathkal  Definition of a Control System Input/Stimulus Output/Response Control System Desired Response Actual Response Elevator Example 4th Floor Input Command 3rd Floor 2nd Floor Steady-State Error 1st Floor G Floor Time Transient Response Steady-State Response Loose patience Passenger Safety Uncomfortable EE326 Dept. of E & E, NITK Surathkal 2  Why we need a Control System? 1. Power Amplification 𝜃𝑟𝑒𝑓 0° 270° Amplifier Antenna knob Motor 2. Remote Control 3. Convenience of input form 25°𝐶 Deployment in Dangerous environments Thermostat Position→ Heating 4. Compensation of disturbance Examples: Wind disturbance in an antenna position. Door opening in a room changing the temperature. EE326 Dept. of E & E, NITK Surathkal 3  System Configurations Open loop Disturbance-1 Disturbance-2 Input / Input + + Output/ Controller Plant reference Transducer + + Controlled Variable Input transducer converts input/reference to a form that controller can process controller drives the plant Examples of Open loop systems 1. A bread toaster. 2. Hair dryer 3. Water tank Issues with a open loop system Never correct itself. Does not account for the disturbances (like wind on the antenna position) EE326 Dept. of E & E, NITK Surathkal 4  System Configurations Closed loop Error/Actuating Signal Disturbance-1 Disturbance-2 Input / Input + + + Output/ Controller Plant reference Transducer + + Controlled − Variable Output Transducer (sensor) Output transducer converts output/controlled variable to a form that controller can process Actuating signal (error) is the difference between the reference and the feedback (Feedback control systems ) Greater accuracy Less sensitive to noise and disturbance Transient response and steady-state error – controlled with a compensator. Expensive compared to open loop. EE326 Dept. of E & E, NITK Surathkal 5  Analysis and Design Objectives Analysis is the process by which the performance of a system is determined Design is the process by which the performance of a system is created or changed Major Objectives 1. Transient Response 2. Steady-State Response 3. Stability EE326 Dept. of E & E, NITK Surathkal 6  Analysis and Design Objectives Analysis is the process by which the performance of a system is determined Design is the process by which the performance of a system is created or changed Major Objectives 1. Transient Response 2. Steady-State Response 3. Stability Too fast - uncomfortable Elevator Example 4th Floor Input Command 3rd Floor 2nd Floor Too slow - impatient 1st Floor Optimal Speed G Floor Objectives in this course: Time Create a quantitative definition for the transient response Analyze the existing transient response and With design, yield a desired transient response EE326 Dept. of E & E, NITK Surathkal 7  Analysis and Design Objectives Analysis is the process by which the performance of a system is determined Design is the process by which the performance of a system is created or changed Major Objectives 1. Transient Response 2. Steady-State Response 3. Stability No steady state error 𝜃𝑟 Input Command 𝜃𝑟 Error in locating the satellite Time Objectives in this course: Quantitatively define this error Analyze the existing error and With design, correct the steady state behavior EE326 Dept. of E & E, NITK Surathkal 8  Analysis and Design Objectives Analysis is the process by which the performance of a system is determined Design is the process by which the performance of a system is created or changed Major Objectives 1. Transient Response 2. Steady-State Response 3. Stability E.g. RLC circuit step response Total Response Forced Response 𝑉0 Stable Natural Response Time What if the natural response is growing/unbounded? Think of the antenna control or elevator EE326 Dept. of E & E, NITK Surathkal 9 Azimuth Antenna position control 𝜃𝑖 𝑡 potentiometer Desired azimuth angle 𝜃0 𝑡 input Azimuth angle System Concept output 𝜃𝑖 𝑡 Desired potentiometer azimuth angle 𝜃0 𝑡 input Azimuth angle output Differential amplifier potentiometer and Power amplifier motor Detailed Layout EE326 Dept. of E & E, NITK Surathkal 10  Azimuth𝜃 Antenna 𝑡 position control 𝑖 Desired potentiometer azimuth angle 𝜃0 𝑡 input Azimuth angle output Differential amplifier potentiometer and Power amplifier + motor potentiometer Detailed Layout 𝜃𝑖 𝑡 − Differential Motor and power Fixed Field Gear amplifier 𝐾 DC motor Gear + 𝜃0 𝑡 Viscous Schematic Inertia damping potentiometer − Gear EE326 Dept. of E & E, NITK Surathkal 11  Azimuth Antenna position control + potentiometer 𝜃𝑖 𝑡 − Differential Motor and power Fixed Field Gear amplifier 𝐾 DC motor Gear + 𝜃0 𝑡 Viscous Inertia damping Schematic potentiometer − Gear Functional Block Diagram 𝜃𝑖 𝑡 + error Signal and 𝜃0 𝑡 Motor, load Potentiometer Power and gears − amplifiers Potentiometer EE326 Dept. of E & E, NITK Surathkal