Chapter 9 - Solved Problems

Transcription

1 Capter 9 - Solved Problems Solved Problem 9.. Consider an internally stable feedback loop wit S o (s) = s(s + ) s + 4s + Determine weter Lemma 9. or Lemma 9. of te book applies to tis system. Solutions to Solved Problem 9. () Solved Problem 9.. Consider a linear plant aving a nominal model aving relative degree and no poles in te open RHP. Compute a lower bound for te sensitivity peak using Lemma 9. from te book and assuming tat te following specifications are met: S o (jω) < ɛ for ω T o (jω) < ɛ ω for ω > were ɛ =., = [rad/s] and = 3 [rad/s]. Solutions to Solved Problem 9. Solved Problem 9.3. Consider a plant aving a linear nominal model of relative degree wit no poles in te open RHP. Assume tat te nominal model as a zero at s = c =.5. Furter assume tat te following design specifications are met: S. S o (jω) < ɛ ω l for ω S. T o (jω) < ɛ ω for ω > were ɛ =., = [rad/s] and = 3 [rad/s]. Find a lower bound for te complementary sensitivity peak. Solutions to Solved Problem 9.3 Solved Problem 9.4. Consider a stable plant wit NMP zeros located at s = ± jα. Furter assume tat we require S o (jω) ɛ < for all ω [, ]. Using Lemma 9.5 from te book investigate te effect of te imaginary part, α of te NMP zeros on te lower bound for te sensitivity peak. Solutions to Solved Problem 9.4 Solved Problem 9.5. Consider a linear system aving te nominal model G o (s) = 4( s + 4) ( s + )(s + 5)(s + ) Determine a lower limit for te sensitivity peak, assuming tat we require tat () Solutions to Solved Problem 9.5 S o (jω). ω [, ] (3) T o (jω). ω [8, ] (4)

2 Capter 9 - Solutions to Solved Problems Solution 9.. Since te feedback loop is internally stable, no unstable pole-zero cancellation can occur. Tus, all open loop poles in te RHP will appear in te sensitivity numerator. Since no zero of S o (s) is located in te open RHP, ten Lemma 9. applies. We can calculate te open loop transfer function H ol (s) as H ol (s) = S o (s) = (s + ) s(s + ) (5) Note tat H ol (s) as relative degree, n r, equal to and tat We can tus apply equation (9..3) from te book to obtain κ = lim s H ol(s) = (6) dω = κ π = π (7) Solution 9.. We use (7), splitting te integration interval as follows: [; ] = [; ] ( ; ] (, ). Ten = < dω = ω ln ɛ dω + ln S max dω + Te last integral on te rigt is ln + ɛ ω dω = From te above equations we conclude tat ω dω + dω + dω (8) ln + ɛ ω dω (9) ( ln + ɛ ) ω dω = ɛ arctan ɛ ω ln( + ɛ) = f(ɛ) ɛ () ln S max > ln ɛ ɛ 4.4 () Note tat te lower limit of te sensitivity peak increases as ɛ and/or Solution 9.3. To compute te lower bound for te complementary sensitivity peak, we can use Lemma 9.4 of te book (wit τ = ) splitting te integration interval, as in Solved Problem 9.. Note tat specification [S.] implies tat te open loop transfer function H ol (s) = G o (s)c(s) as one or more poles at te origin. Tus equation (9.3.) of te book is satisfied. Actually, S. would seem to imply tat wen k v is computed from (9.3.4) te result is k v =. To compute tis integral use eiter a matematical symbolic package suc as MAPLE or MATHEMATICA, or te MATLAB symbolic toolbox, wic is a MAPLE subset.

3 Tus = < ω ln T o (jω) ln T o (jω) ln T o (jω) dω = dω + dω + ω ) ( ln + ɛ ω ω ( ) ωl dω + ln T max ω l dω ɛ ω + ln dω I } {{ } I } {{ } I 3 Tese integrals can be computed using, for instance, MAPLE. Tis yields ln T o (jω) dω () (3) I = ln( + ɛ) + ɛ arctan ɛ (4) I = (5) I 3 = ln(ɛ) (6) We can now use equation (9.3.) of te book wit τ =, M = and k v =. Hence ln T max > π c( ) (ln( + ɛ) ɛ arctan ɛ) ω + + ( + ln(ɛ) ) 46.4 (7) It is interesting to note te factors wic will cause tis bound to grow. Tey are tends to one, i.e. te design specification demands a very sarp transition from te pass band to te stop band in T o (jω). Te parameter ɛ tends to zero. Tis means tat te design specification requires an excessively flat frequency response. Te ratio c tends to zero. Tis is in agreement wit time domain analysis previously given in Capter 8 of te book. It is also interesting to note tat te most significant contribution to tis lower limit is te NMP zero. It accounts for rougly 8% of te limit. Solution 9.4. We apply Lemma 9.5 from te book, wit M =, c = + jα and c = jα, and te rigt and side in equation (9.4.) of te book equal to (since tere are no unstable poles). Tus, for c ωl + (α ω) dω = + (α ω) dω (α ω) dω + dω (8) + (α ω) 3

4 We can now substitute by its upper bound on every interval S max ω [, ] max S o (jω) = ɛ ω [, ] S max ω [, ] (9) Ten ωl + (α ω) dω < ln S max + (α ω) dω + + ln(ɛ) I + (α ω) dω I + ln S max We can now compute te integrals I, I and I 3 using MAPLE. Tis yields + (α ω) dω I 3 () I = π arctan(α + ) () I = arctan( α + ) + arctan(α + ) () I 3 = π arctan( α + ) (3) Combining te above expressions we ave tat were ln S max > ln(ɛ) f(α) (4) f(α) = π arctan( α + ) + arctan(α + ) > for all finite and all finite α (5) In conclusion, te following observations are seen to apply: Te function f(α) is an even function of α. Tus te above result is also valid for te NMP zero c. Te maximum lower bound occurs wen α =. Tis can be proved by differentiating f(α) wit respect to α Wen α >>, we ave tat te lower bound reaces a minimum, since ten f(α) goes to infinity. Solution 9.5. We use te Poisson formula as in Lemma 9.5 to compute a lower bound for S max. We first identify te following parameters and expressions: ɛ =.; = ; = 8 (6) B p (s) = s s + ; B z(s) = s (7) s + 4

5 Ten, using equation (9.4.) of te book ln S max > Ω(4, 8) Ω(4, ) [ π ln B p(4) + (ln.)ω(4, ) (π Ω(4, 8)) ln(.)] (8) Te lower limit for S max can ten be computed using te MATLAB routine smax (provided on te CD-ROM in te book). Tis gives S max >