Transient Response First order system transient response

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Presentation transcript:

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

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

First order system step resp Normalized time t/t

Simple 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

In every τ seconds, the error is reduced by 63.2%

We know how this responds to input 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

Step response by MATLAB: >> n = [ b1 b0 ] >> d = [ 1 p ] >> step ( n , d ) Other MATLAB commands to explore: plot, hold, axis, xlabel, ylabel, title, text, gtext, semilogx, semilogy, loglog, subplot

Unit ramp response:

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

Unit impulse response:

Prototype 2nd order system:

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

cosq = z q = cos-1z d s

To find y(t) max:

For 5% tolerance Ts ~= 3/zwn

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+0.9z)/wn

Useful Range Tr=4.5(z-0.2)/wn Or about 2/wn

Putting all things together: Settling time: = (3 or 4 or 5)/s

2) When ζ = 1, ωd = 0

The tracking error:

3) Over damped: ζ > 1

Transient Response Recall 1st order system step response: 2nd order:

Pole location determines transient

All closed-loop poles must be strictly in the left half planes Transient dies away Dominant poles: those which contribute the most to the transient Typically have dominant pole pair (complex conjugate) Closest to jω-axis (i.e. the least negative) Slowest to die away

Typical design specifications Steady-state: ess to step ≤ # % ts ≤ · · · Speed (responsiveness) tr ≤ · · · td ≤ · · · Relative stability Mp ≤ · · · %

These specs translate into requirements on ζ, ωn or on closed-loop pole location : Find ranges for ζ and ωn so that all 3 are satisfied.

Find conditions on σ and ωd.

In the complex plane :

Constant σ : vertical lines σ > # is half plane

Constant ωd : horizontal line ωd < · · · is a band ωd > · · · is the plane excluding band

Constant ωn : circles ωn < · · · inside of a circle ωn > · · · outside of a circle

Constant ζ : φ = cos-1ζ constant Constant ζ = ray from the origin ζ > · · · is the cone ζ < · · · is the other part

If more than one requirement, get the common (overlapped) area e.g. ζ > 0.5, σ > 2, ωn > 3 gives Sometimes meeting two will also meet the third, but not always.

Try to remember these:

When given unit step input, the output looks like: Example: + - When given unit step input, the output looks like: Q: estimate k and τ.

Effects of additional zeros Suppose we originally have: i.e. step response Now introduce a zero at s = -z The new step response:

Effects: Increased speed, Larger overshoot, Might increase ts

When z < 0, the zero s = -z is > 0, is in the right half plane. Such a zero is called a nonminimum phase zero. A system with nonminimum phase zeros is called a nonminimum phase system. Nonminimum phase zero should be avoided in design. i.e. Do not introduce such a zero in your controller.

Effects of additional pole Suppose, instead of a zero, we introduce a pole at s = -p, i.e.

L.P.F. has smoothing effect, or averaging effect Effects: Slower, Reduced overshoot, May increase or decrease ts