Chapter 3: Dynamic Response

1. Chapter 3: Dynamic Response Part C: Transient-response analysis with MATLAB.

2. Introduction • The practical procedure for plotting time response curves of systems higher than second-order is through computer simulation. • In this part, computational approach to the transient-response analysis with MATLAB is presented through various examples.

3. Representation of a Linear System A linear system can be representedeither: • In state-variable form: with the values of the matrices F, G, H and the constant J. Or • By its transfer function: Either in numerator-denominator polynomial form, Or in pole-zero form Or in partial expansion form

4. Example 1: Standard State-Variable Form • Consider a linear system described by:

5. F = [0 1;0 -0.05]; G = [0;0.001]; H = [0 1]; J = 0; step(F,G,H,J) % defines state variable matrices % generates plot of unit-step response (with Time (sec)and Amplitudelabels on x- and y-axis respectively, andStep responsetitle ) Note: time vector is automatically determined when t is not explicitly included in the step command. Unit-Step Response for a Control System defined in State-Variable form

6. F = [0 1;0 -0.05]; G = [0;0.001]; H = [0 1]; J = 0; sys = ss(F, 50*G, H, J); step(sys) % defines state variable matrices % defines system by its state-space matrices % generates plot of 50-step response vs t Note:State-variable form is also called state-space form 50-Step Response for a system defined in State-Variable form

7. F = [0 1;0 -0.05]; G = [0;0.001]; H = [0 1]; J = 0; sys = ss(F, G, H, J); t = 0:0.2:100; y=step(sys,t); plot(t,y) % defines state variable matrices % defines system by its state-space matrices % setup time vector ( dt = 0.2 sec) % plots unit step response versus time ranging from 0 to 100 sec (with x- and y-labels) Unit-Step Response on specific time interval for a system defined in State-Variable form

8. F = [0 1;0 -0.05]; G = [0;0.001]; H = [0 1]; J = 0; sys = ss(F, G, H, J); impulse(sys) % defines state variable matrices % defines system by its state-space matrices % generates plot of impulse response (with labels & title) Note: an alternative use of impulse command is: impulse(F,G,H,J) Impulse Response for a system defined in State-Variable form

9. Example 2: Initial Conditions • Consider a linear system such as: • In state-variable form, it is described by:

10. F = [0 1;-10 -5]; G = [0;0]; H = [1 0]; J = 0; t = 0:0.5:3; y=initial(F,G,H,J,[2;1],t); plot(t,y) % defines state variable matrices % set up time vector % computes initial condition response % generates plot of response Note: Initial conditions are defined between [ ]. Initial Condition Response for a system defined in State-Variable form

11. Example 3: Transfer function in numerator-denominator form • Consider a linear system whose the transfer function is:

12. num = [0 0 25]; den = [1 4 25]; step(num,den) % defines numerator % defines denominator % generates plot of unit-step response (with labels and title) Unit-Step Response for a system Transfer Function defined in num/den polynomial form

13. num = [0 0 50]; den = [1 0.2 1]; step(num,den) % defines numerator % defines denominator % generates plot of 50-step response (with labels and title) 50-Step Response for a system Transfer Function defined in num/den polynomial form

14. t = 0:0.2:10; zeta = [0 0.2 0.4 0.6 0.8 1]; for n = 1:6; num = [0 0 1]; den = [1 2*zeta(n) 1]; [y(1:51,n),x, t] = step(num,den,t); end plot(t,y) % setup time vector % defines zeta, numerator and denominator % generates 2-D plot of the n unit-step responses (on same graph) Unit-Step Responses for system Transfer Functions defined by

15. t = 0:0.2:10; zeta = [0 0.2 0.4 0.6 0.8 1]; for n = 1:6; num = [0 0 1]; den = [1 2*zeta(n) 1]; [y(1:51,n),x, t] = step(num,den,t); end mesh(t,zeta,y’) % setup time vector % defines zeta, numerator and denominator % generates 3-D plot of the n unit-step responses (on same graph) Unit-Step Responses for system Transfer Functions defined by

16. num = [0 0 0 1]; den = [1 1 1 0]; step(num,den) % defines numerator % defines denominator % generates plot of unit-step response (with x- and y-labels) Unit Step Response for a 3rd order system defined by its Transfer Function

17. num = [0 0 1]; den = [1 0.2 1]; sys=tf(num,den); impulse(sys) % defines numerator % defines denominator % defines system by its transfer function % generates plot of impulse response Note: an alternative use of impulse command is: impulse(num,den) Impulse Response for a system Transfer Function defined in num/den polynomial form

18. num = [0 1 0]; den = [1 0.2 1]; step(num,den) % defines numerator of sG(s) % defines denominator % generates plot of impulse response (with x- and y-labels) Alternative approach to obtain Impulse Response

19. Example 4: Transfer function in standard 2nd order system • Consider a standard second order system: natural undamped frequency damping ratio

20. w0 = 5; damping_ratio = 0.4; [num0,den] = ord2(w0,damping_ratio); num = 5^2*num0; printsys(num,den,’s’) % defines natural undamped frequency % defines damping ratio % defines numerator % prints num/den as a ratio of s-polynomials num/den = MATLAB Description of Standard Second Order System

21. Example 5: Transfer function in pole-zero form • Consider a linear system whose the transfer function is:

22. num = conv([1 2],[1 4]); den = conv([1 1 0],[1 3]); step(num,den) % defines zero ratios % defines pole ratios % plots unit-step response Unit-Step Response for a system Transfer Function defined in pole-zero form

23. Example 6: Transfer function in Partial Expansion Form • Consider a linear system whose the transfer function is:

24. r = [8/3 -3/2 -1/6]; p = [0 -1 -3]; K = [] ; [num,den] = residue(r,p,K) step(num,den) % defines residues % defines poles % define additive constant % convert partial expansion form to polynomial form % plots unit-step response Note: to see ratio use printsys(num,den,’s’) Unit-Step Response for a system Transfer Function defined in partial expansion form

25. State-variable form Transfer function: In num-den polynomial form In zero-pole form In partial expansion form Convertion [num,den] = ss2tf(F,G,H,J) [z,p,k]=tf2zp(num,den) [r,p,K]=residue(num,den) [z,p,k] = ss2zp(F,G,H,J)

26. State-variable form Transfer function: In num-den polynomial form In zero-pole form In partial expansion form Convertion [F,G,H,J] = tf2ss(num,den) [num,den]=zp2tf(z,p,k) [num,den]=residue(r,p,K) [F,G,H,J] = zp2ss(z,p,k)

27. Title, grid, labels, text on graphical screen, symbols, … Cosmetic

28. title (‘Step-response’); grid; sys = …; t = 0:0.2:100; y = step(sys,t); plot (t,y); xlabel(‘t (sec)’); ylabel(‘response’) % writes the title Step-response % draws a grid between ticks % defines system by … % setup time vector ( dt = 0.2 sec) % computes step response % plots step response % writes label t (sec) on x-axis. % writes label response on y-axis. Title, Grid & Labels on the graphical screen

29. text(3.4, -0.06, ‘Y111’); text(4.1,1.86,’\zeta’); gtext(‘blabla’) % writes Y111 beginning at the coordinates x=3.4, y=-0.06. % writes  at x=4.1, y=1.86 % waits until the cursor is positioned (using the mouse) at the desired position in the screen and then writes on the plot at the cursor’s location the text enclosed in simple quotes. Note: any number of gtext command can be used in a plot. Writing Text on the Graphical Screen

30. num = [0 0 25]; den = [1 6 25]; t = 0:0.5:5; y = step(num,den,t); plot(t,y,’o’,t,1,’-’); % defines numerator % defines denominator % defines time vector % computes unit-step response % plot of unit step response y and unit step input 1 using oooo and ---- symbols respectively. Use of Symbols in graph

31. num = [0 0 25]; den = [1 6 25]; t = 0:0.5:5; y = step(num,den,t); plot(t,y,’x’,t,y,’-’); % defines numerator % defines denominator % defines time vector % computes step response % plot of unit step response y using -x-x-x-x- symbols Use of Symbols in graph (cont’d)

32. Computing roots using MATLAB Plotting pole(s) and zero(s) in the s-plane using MATLAB Plotting Step-response versus a parameter range Obtaining rise time, peak time, maximum overshoot and settling time using MATLAB Additional Convenient MATLAB Commands

33. pol= [1 4 3 2 1 4 4]; roots(pol) ans = -3.2644 -0.6046 + 0.9935i -0.6046 - 0.9935i 0.6797 + 0.7488i 0.6797 - 0.7488i -0.8858 Computing Roots

34. den= [1 5 11 23 28 12]; roots(den) ans = -3.0000 0.0000 + 2.0000i 0.0000 - 2.0000i -1.0000 + 0.0000i -1.0000 - 0.0000i Stability Analysis by Computing Roots 2 poles are in the RHP

35. num=[0 2 1]; den= [2 3 2]; zmap(num,den) Plotting Poles and Zero in the s-domain poles as crosses zero as circle

36. xlabel('Time (sec)'); ylabel('Amplitude'); Title('Step-Response versus K parameter'); grid; text(7.1,3.8,'K=6.5'); text(7.5,3.15,'7'); text(7.15,2.65,'7.5'); text(7.1,2.3,'8'); text(6.65,1.37,'10'); text(6.4,0.75,'12.5'); t=0:0.1:10; K=[6.5 7 7.5 8 10 12.5]; for n=1:6 num=[K(n) K(n)]; den=[1 5 K(n)-6 K(n)]; [y(1:101,n),x,t]=step(num,den,t); end plot(t,y); Step-Response versus a Parameter Range

39. Reminder: Rise Time • The rise time is the time requiredfor the responseto rise from 0% to 100% of its final value. 1 t d 0.5 0 t r Note: for overdamped systems, the 10% to 90% rise time is commonly used.

40. num= [0 0 25]; den=[1 6 25]; t=0:0.001:5; [y,x,t]=step(num,den,t); r=1; while y(r) <1.0001; r=r+1; end; rise_time=(r-1)*0.001 rise_time = 0.5540 Computing Rise Time using MATLAB No ;

41. Reminder: Peak Time • The peak time is the time requiredfor the responseto reach the first peak of the overshoot. t p 1 t d 0.5 0 t r

42. num= [0 0 25]; den=[1 6 25]; t=0:0.001:5; [y,x,t]=step(num,den,t); [ymax,tp]=max(y); peak_time=(tp-1)*0.001 peak_time = 0.7850 Computing Peak Time using MATLAB No ;

43. Reminder: Maximum Overshoot • The maximum overshoot is the relative maximum peak value of the response curve measured from the final value. t p M p 1 t d 0.5 0 t r Note: the maximum overshoot directly indicates the relative stability of the system.

44. num= [0 0 25]; den=[1 6 25]; t=0:0.001:5; [y,x,t]=step(num,den,t); [ymax,tp]=max(y); peak_time=(tp-1)*0.001 max_overshoot=ymax-1 peak_time = 0.7850 max_overshoot = 0.0948 Computing Maximum Overshoot using MATLAB

45. Reminder: Settling Time • The settling time is the time required for the response curve to reach and stay within a range about1% or 2% of the final steady-state value. t p M p ±1% 1 t d 0.5 0 t t r s Note:t is the time it takes the system transients to decay. s

46. num= [0 0 25]; den=[1 6 25]; t=0:0.001:5; [y,x,t]=step(num,den,t); s=5001; while y(s)>0.98 & y(s)<1.02; s=s-1; end; settling_time=(s-1)*0.001 settling_time = 1.1880 Computing Settling Time using MATLAB(based on +/-2%)

47. Results given by MATLAB are: rise_time = 0.5540; peak_time = 0.7850, max_overshoot = 0.0948, settling_time = 1.1880