Dynamic Response Unit step signal Step response ysHss

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Dynamic Response • Unit step signal: • Step response: y(s)=H(s)/s, y(t)=L-1{H(s)/s} Time domain response

Dynamic Response • Unit step signal: • Step response: y(s)=H(s)/s, y(t)=L-1{H(s)/s} Time domain response specifications • Defined based on unit step response • Defined for closed-loop system

Transient Response • First order system transient response – Step response specs and relationship

Transient Response • First order system transient response – Step response specs and relationship to pole location • Second order system transient response – Step response specs and relationship to pole location • Effects of additional poles and zeros

Prototype first order system E U(s) + - 1 τs Y(s)

Prototype first order system E U(s) + - 1 τs Y(s)

First order system step resp Normalized time t/t

First order system step resp Normalized time t/t

Prototype first order system • • • No overshoot, tp=inf, Mp = 0 Yss=1,

Prototype first order system • • • No overshoot, tp=inf, Mp = 0 Yss=1, ess=0 Settling time ts = [-ln(tol)]/p Delay time td = [-ln(0. 5)]/p Rise time tr = [ln(0. 9) – ln(0. 1)]/p • All times proportional to 1/p= t • Larger p means faster response

The error signal: e(t) = 1 -y(t)=e-ptus(t) Normalized time t/t

The error signal: e(t) = 1 -y(t)=e-ptus(t) Normalized time t/t

In every τ seconds, the error is reduced by 63. 2%

In every τ seconds, the error is reduced by 63. 2%

General First-order system We know how this responds to input Step response starts at

General First-order system We know how this responds to input Step response starts at y(0+)=k, final value kz/p 1/p = t is still time constant; in every t, y(t) moves 63. 2% closer to final value

Unit ramp response:

Unit ramp response:

Note: In step response, the steady-state tracking error = zero.

Note: In step response, the steady-state tracking error = zero.

Unit impulse response:

Unit impulse response:

Prototype nd 2 order system:

Prototype nd 2 order system:

xi=[0. 7 1 2 5 10 0. 1 0. 2 0. 3 0. 4

xi=[0. 7 1 2 5 10 0. 1 0. 2 0. 3 0. 4 0. 5 0. 6]; x=['zeta=0. 7'; 'zeta=1 '; 'zeta=2 '; 'zeta=5 '; 'zeta=10 '; 'zeta=0. 1'; 'zeta=0. 2'; 'zeta=0. 3'; 'zeta=0. 4'; 'zeta=0. 5'; 'zeta=0. 6']; T=0: 0. 01: 16; figure; hold; for k=1: length(xi) n=[1]; d=[1 2*xi(k) 1]; y=step(n, d, T); plot(T, y); if xi(k)>=0. 7 text(T(290), y(310), x(k, : )); else text(T(290), max(y)+0. 02, x(k, : )); end grid; end text(9, 1. 65, 'G(s)=w_n^2/(s^2+2zetaw_ns+w_n^2)') title('Unit step responses for various zeta') xlabel('w_nt (radians)') Can use omega in stead of w

annotation Create annotations including lines, arrows, text arrows, double arrows, text boxes, rectangles, and

annotation Create annotations including lines, arrows, text arrows, double arrows, text boxes, rectangles, and ellipses xlabel, ylabel, zlabel Add a text label to the respective axis title Add a title to a graph colorbar Add a colorbar to a graph legend Add a legend to a graph

For example: “help annotation” explains how to use the annotation command to add text,

For example: “help annotation” explains how to use the annotation command to add text, lines, arrows, and so on at desired positions in the graph ANNOTATION('textbox', POSITION) creates a textbox annotation at the position specified in normalized figure units by the vector POSITION ANNOTATION('line', X, Y) creates a line annotation with endpoints specified in normalized figure coordinates by the vectors X and Y ANNOTATION('arrow', X, Y) creates an arrow annotation with endpoints specified Example: ah=annotation('arrow', [. 9. 5], [. 9, . 5], 'Color', 'r'); th=annotation('textarrow', [. 3, . 6], [. 7, . 4], 'String', 'ABC');

Unit step response: 1) Under damped, 0 < ζ < 1

Unit step response: 1) Under damped, 0 < ζ < 1

d =Im cosq = z =-Re/|root| q= cos-1(Re/|root|) q= tan-1(-Re/Im) s =-Re

d =Im cosq = z =-Re/|root| q= cos-1(Re/|root|) q= tan-1(-Re/Im) s =-Re

To find y(t) max:

To find y(t) max:

z=0. 3: 0. 1: 0. 8; Mp=exp(-pi*z. /sqrt(1 -z. *z))*100 plot(z, Mp) grid; Then

z=0. 3: 0. 1: 0. 8; Mp=exp(-pi*z. /sqrt(1 -z. *z))*100 plot(z, Mp) grid; Then preference -> figure… ->powerpoint -> apply to figure Then copy figure

For 5% tolerance Ts ~= 3/zwn

For 5% tolerance Ts ~= 3/zwn

 • Delay time is not used very much • For delay time, solve

• Delay time is not used very much • For delay time, solve y(t)=0. 5 and solve for t • For rise time, set y(t) = 0. 1 & 0. 9, solve for t • This is very difficult • Based on numerical simulation:

Useful Range td=(0. 8+0. 9 z)/wn

Useful Range td=(0. 8+0. 9 z)/wn

Useful Range tr=4. 5(z-0. 2)/wn Or about 2/wn

Useful Range tr=4. 5(z-0. 2)/wn Or about 2/wn

Putting all things together: Settling time:

Putting all things together: Settling time: