Dynamic Response Steady State Response: the part of response when t → ∞ Transient response: the part of response right after the input is being applied. Both are part of the total response total resp = z.i. resp + z.s.resp. z.i. resp = “Output due to i.c. when input ≡ 0” z.s. resp = “Output due to input excitation when all i.c. are set=0 at t=0” Both z.i. resp and z.s. resp have their own transient resp and their own steady state resp. Here ignore i.c. and consider z.s. resp.

Typical test signal Unit step signal: Unit impulse:δ(t) 0 u s (t) 1 δ(t) t

Unit ramp: Unit acc. signal: r(t) a(t) t t

Exponential signal: sinusoidal: 0 1 t

Unit step response: In Matlab: step Unit impulse resp: Matlab: impulse H(s) u(s)= 1s1s y(s)=H(s) 1s1s H(s) u(s)=1 y(s)=H(s)

Defined based on unit step response Defined for closed-loop system Steady-state value y ss Steady-state error e ss Settling time t s = time when y(t) last enters a tolerance band Time domain response specifications

If yss = 1 tptp

By final value theorem In MATLAB: num = [ ]; den=[……]; b0 = num(length(num)), or num(end) a0 = den(length(den)), or den(end) yss=b0/a0

If numerical values of y and t available, from [y, t]=step(num, den); abs(y – y ss ) < tolmeans inside band abs(y – y ss ) ≥ tolnot inside e.g.t_out = t(abs(y – y ss ) >= tol)contains all those time points when y is not inside the band. Therefore, the last value in t_out will be the settling time. ts=t_out(end)

Peak time t p = time when y(t) reaches its maximum value. Peak value y max = y(t p ) Hence: y max = max(y); t p = t(y = y max ); Overshoot:OS = y max - y ss Percentage overshoot:

If t50 = t(y >= 0.5*y ss ), this contains all time points when y(t) is ≥ 50% of y ss so the first such point is t d. td=t50(1); Similarly,t10 = t(y >= 0.1*y ss ) &t90 = t(y >= 0.9*y ss ) can be used to find t r. tr=t90(1)-t10(1)

t p ≈0.9sec 10%y ss 90%y ss t r ≈0.45 t d ≈0.35 tsts

t r ≈0.35 t s =0.45 y ss =1 e ss =0 O.S.=0 M p =0 t p =∞ t d ≈0.2

t r ≈0.1 t d ≈0.2 t s ≈0.92 t p =0.35 O.S.=0.4 M p =40% y ss =1 e ss =0

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 1 τ s U(s) E + - Y(s)

First order system step resp Normalized time t/ 

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=  Larger p means faster response

The error signal: e(t) = 1-y(t)=e -pt u s (t) Normalized time t/ 

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 y(0+)=k, final value kz/p 1/p =  is still time constant; in every , y(t) moves 63.2% closer to final value

Step response by MATLAB: >> p =.. >> 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: