Routh criterion based on regular Routh table: 1)d(s) is A.S. iff 1 st col have same sign 2) # of sign changes in 1 st col = # of roots in right half plane Routh Criteria

Special case 1: one whole row = 0 – use prev. row to form aux. poly. A(s) – use coeff of A’(s) to replace 0-row – continue as usual Fact: Roots of A(s) are also roots of d(s) A(s) is a factor of d(s) Special case 2: 1 st col = 0 but whole row != 0 –Replace 0 by positive  –Continue as usual Fact : cannot be A.S.

Useful case: variable parameters in d(s) 1) form table as usual 2) set 1 st col. >0 3) solve for parameter range for A.S. 2’) set s 1 row =0 3’) solve for parameters 4’) use s 2 row for sustained oscillation

Steady-state tracking & sys. types G(s) C(s) + - r(s) e y(s) plant

Assuming closed-loop system stability

r(t)=u s (t) r(s)=1/s r(t)=tu s (t) r(s)=1/s 2 r(t)= ½t 2 u s (t) r(s)=1/s 3 type 0 (N=0 a 0 ≠0) K p =b 0 /a 0 e ss =1/(1+K p ) K v =0 e ss =∞ K a =0 e ss =∞ type 1 (N=1 a 0 =0 a 1 ≠0 b 0 ≠0 ) K p = ∞ e ss =0 K v =b 0 /a 1 e ss =1/K v K a =0 e ss =∞ type 2, N=2 a 0 =a 1 =0 a 2 ≠0,b 0 ≠0 K p = ∞ e ss =0 K v = ∞ e ss =0 K p =b 0 /a 2 e ss =1/K a type≥3, N ≥ 3 a 0 =a 1 =a 2 =0 b 0 ≠0 K p = ∞ e ss =0 K v = ∞ e ss =0 K a = ∞ e ss =0 sys. type ref. input

G1(s) + - r(s) e G2(s) d(s) A A

G1(s) + - r(s) e G2(s) d(s) A A

G1(s) + - r(s) e G2(s) d(s) A

K p s+K I s + - r(s) e ω n 2 s(s+2ζ ω n ) 1 Ts+1 d 1 (s)d 2 (s) AB

The root locus technique 1.Obtain closed-loop TF and char eq d(s) = 0 2.Re-arrange to get 3.Mark zeros with “o” and poles with “x” 4.High light segments of x-axis and put arrows 5.Decide #asymptotes, their angles, and x-axis meeting place: 6.Determine jw-axis crossing using Routh table 7.Compute breakaway: 8.Departure/arrival angle:

rlocus([1 3], conv([ ],[ ]))

Example: motor control The closed-loop T.F. from θ r to θ is:

What is the open-loop T.F.? The o.l. T.F. of the system is: But for root locus, it depends on which parameter we are varying. 1.If K P varies, K D fixed, from char. poly.

The o.l. T.F. for K P -root-locus is the system o.l. T.F. In general, this is the case whenever the parameter is multiplicative in the forward loop. 2.If K D is parameter, K P is fixed From

What if neither is fixed? Multi-parameter root locus? –Some books do this Additional specs to satisfy? –Yes, typically –Then use this to reduce freedom The o.l. T.F. of the system is: It is type 1, tracks step with 0 error Suppose ess to ramp must be <=1

Since ess to ramp = 1/Kv Kv = lim_s->0 {sGol(s)} =2KP/(4+2KD) Thus, ess=(4+2KD)/2KP Design to just barely meet specs: ess=1 (4+2KD)/2KP = 1 4+2KD=2KP

More examples 1. No finite zeros, o.l. poles: 0,-1,-2 Real axis: are on R.L. Asymp: #: 3

-axis crossing: char. poly:

>> rlocus(1,[ ]) >> grid >> axis equal

Example: Real axis: (-2,0) seg. is on R.L.

For

Break away point:

-axis crossing: char. poly:

>> rlocus(1, conv([1 2 0], [1 2 2])) >> axis equal >> sgrid

Example: in prev. ex., change s+2 to s+3

-axis crossing: char. poly:

>> rlocus(1, conv([1 3 0], [1 2 2])) >> axis equal >> sgrid