26 Feedback Example: The Inverted Pendulum

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1 6 Feedback Example: The Inverted Pendulum Solutions to Recommended Problems S6. Ld 0(t) (a) Ldz6(t) = g0(t) a(t) + Lx(t), Ld (t) dt - ga(t) = Lx(t) Taking the Laplace transform of both sides yields szlo(s) - go(s) = L, O(s) = s - g L 0(s) s- g/l (s + \g/l)(s - \ ) The pole at \g/l is in the right half-plane and therefore the system is unstable. (b) We are given that a(t) = K0(t). See Figure S6.-. x(t) o (t) Figure S6.- so, with 0(s) H + GH' H = s g/l (s)/ is given by and G = 0(s) s' - (g/l) + (K/L) The poles of the system are at s=+k-g L ' which implies that the system is unstable. Any K < g will cause the system poles to be pure imaginary, thereby causing an oscillatory impulse response. S6-

2 Signals and Systems S6- (c) Now the system is as indicated in Figure S6.-. The poles are at H(s) = g K K L L L K s+- K -- g s ± L L -K + K_ (K g) L ~ L L ' which can be adjusted to yield a stable system. A general second-order system can be expressed as Aw so, for our case, Hg(s) = s + Aws + w' K L -g Wn and,w = K L' g = 9.8 m/s L = 0.5 m f = O= 3 rad/s K, = 4.3 m/s K = 3 m/s S6. (a) Here H(s) = + + os + W G(s) = K

3 Feedback Example: The Inverted Pendulum / Solutions S6-3 The closed-loop transfer function He(s) is H _ He(s) = = " + GH s +w.s+ w + nkw! (A)n s +ws+ W( +K Wn S + (n,, Z~ where w, = Wn( + K)"/ Therefore, (s± /(i) +( s + F,+ ) w W, here - = { (A.n = wn(l + K)", (b) ( + K), A = = An +K' for K =, = VW, ands"= //. Now we want to determine the poles of the closed-loop system (c) He(s) = The poles are at -o K = +oo Im s + (,,s + (o( + K) ± /F~W - Wn( + K) s plane K=_xo= - ' Re K = +oo Figure S6. The poles start out at ± 0o, approach each other and touch at K then proceed to -fo, ±joo. = -, and

4 Signals and Systems S6-4 S6.3 (a) = H (s) = K K (Os + r)k K + K K a #s + r Os + r + K K a (b) Y(s) = H (s) K _ (Os + r)k W(s) + K K a Os + r + K K a Os + r (c) For stability we require the pole to be in the left half-plane. s= - r + KK a) r + K K a 0 >0 If /3> 0, then r/a > -KK ; if # < 0, then r/a < -K K. S6.4 H(s) = K(s + 00) K + K(s + ) s Ks + K S + 00 K(s + 00) (a) K = 0.0, H(s) = 0.0(s + 00).0(s ) (K + ) s K) K + ] The zero is at s = -00, and the pole is at s = (b) K=, H(s) = S + 00 s + The zero is at s = -00; the pole is at s = (c) K =0, H(s) = 0(s + 00) s+ 0) The zero is at s = -00; the pole is at s = -0. (d) K = 00, 00(s + 00) H(s) = 000) 0 S The zero is at s = -00; the pole is at s =

5 Feedback Example: The Inverted Pendulum / Solutions S6-5 S6.5 s-t+ (a) H(s) = K += + K s+ + K ++ The pole is at s = - - K, as shown in Figure S6.5-. Im Im K>O s plane K<O s plane -. Re Re - - Figure S6.5- The pole moves from infinity to negative infinity as K changes from negative infinity to infinity. (b) H(s) = s-i s+3 + (K + s s + 3 s + s + K - 3 (s + 3)(s - ) + K The poles are at s, = - ± \ - (K - 3), as shown in Figure S6.5-. Im K>O s plane KK 0 K =4 Im s plane Re Figure S6.5- The poles start at ± oo when K = -oo, move toward -, touch when K = 4, and proceed to - ± joo as K approaches positive infinity.

6 Signals and Systems S6-6 Solutions to Optional Problems S6.6 (a) The poles for the closed-loop system are determined by the denominator of the closed-loop transfer function so Kz + = 0 (z - )(Z - i')' (z - -)(z - ) + Kz = 0 Since we are told a pole occurs when z = -, we want to solve the equation for K: K (z -)(z-) _ 5 z z=- 8 (b) In a similar manner to that in part (a), K (z -)(z -) -3 z z= 8 (c) From the root locus diagram in Figure P6.6, we see that for K > 0 when K exceeds a critical value of K = -', as determined in part (a), one root remains outside the unit circle. Similarly, when K < -8, one root is outside the unit circle. Therefore, to ensure stability, we need - < K < ' S6.7 (a) The closed-loop transfer function is Y(s) _ H(s) _ He(s)H,(s) + G(s)H(s) + He(s)H,(s) and, therefore, from the given He(s) andh,(s), we have Ka Y(s) S + a Ka Ka + Ka s + a + Ka s + (K + )a s + a The system is stable for denominator roots in the left half of the s plane; therefore -(K + )a < 0 implies that the system is stable. Now since E(s)He(s)H,(s) = Y(s), we have E(s) s + a s + a + He(s)H,(s) s + a + Ka s + (K+ )a The final value theorem, lim, e(t) = lim,_. se(s), shows that lim s(s= 0 for -(K + )a < 0 S-0 S+ (K + )a

7 Feedback Example: The Inverted Pendulum / Solutions S6-7 Note that if x(t) = u(t), then E(s) =() (K+ )a and lims s + # 0, for -(K + )a < 0 S-0 \sj s + (K +)a K+ so limt-, e(t) # 0. ()Y(S_ He(s)H,(s) + He(s)H,(s) K+ - a s ) s + a + K, + s a (sk + K )a SS + K Kia \ K s(s + a) + (Kis + K )a S + sa(k + ) + K a The poles for this system occur at - a(k + ) (a(k+ ) - K a Note that if a(k + ) > 0 and if K a > 0, we are assured that both poles are in the left half-plane. Therefore, a(k + ) > 0 and K a > 0 are the conditions for stability. Now since then E(s) = + He(s)H,(s) s(s + a) S s + a(k + )s + K a' lim se(s) = 0 implies that lim e(t) = 0, s-0 t-co for a(ki + ) > 0 and K a > 0, so we can track a step with this stable system. S6.8 (a) = H(s)C(s) (s + )(s - )(s + We can see from this expression that the overall transfer function for the system is Y(s) (s + )(s + 3)' a stable system. In effect, the system was made stable by canceling a pole of H(s) with a zero of C(s). In practice, if this is not done exactly, i.e., if any com

8 Signals and Systems S6-8 ponent tolerances cause the zero to be slightly off from s =, the resultant system will still be unstable. (b) Y(s) C(s)H(s) K + C(s)H(s) (s + )(s - ) + K K s - s + K - The poles are at -v - (K - ) We see from this that at least one pole is in the right half-plane, i.e., there is instability for all values of K. (c) ()Y(S) X() = +K(s +_a)_(s+_)(s_-_ + a) (s + )(s - ) l+~sa(±+ )(- ) K(s + a) (s + )(s - ) + K(s + a) K(s + a) K(s + a) s - s - + Ks + Ka s + (K - )s + (Ka - ) The poles are at (K - ) K_ - -V -- - (Ka - ) Now, if Ka - > 0, the system is stable. K > /a because a > 0 is assumed. This is true for > a > 0 and > a >. For a >, the system is stable for K >. Y(S) K(s + a) (d) s + (K - )s + (Ka - )' We want K - = w,, K - = W. So (K - ) = K -, K = 3 or K= IfK=, thenw, = 0, so K= 3 implies thatw, =. S6.9 (a) E(s) S - -where + H(s) s' + G(s)' For s = 0, G(s) constant = g. K f (s - #K k= G(s) n- 7 (s - ak) k= (/s)s' s E(s) =, and lim se(s) = lim s + g s-0 s--o S + g Thus, lim e(t) = 0. = 0

9 Feedback Example: The Inverted Pendulum / Solutions S6-9 (b) E(s) = So S-I s + G(s) for =, x(t) = u- (t) lim se(s) = Constant S-0 s + G(s),=o g 5 -k -k 5 (c) E(s) = s + G(s), se(s) = s + For k >, lim se(s) = co, s-o G(s) lim e(t) = oo (d) (i) S -k E(s) = S' + G(s),G se(s) If k :5, then s' + G(s) (ii) so lim,_., e(t) = 0. If k = + and since -k+ lim se(s) = lim 0 0, = 0, s-o s-o s' + G(s) 0+ g -k E(s) = s' + G(s) (iii) then lim se(s) = lim S-0 S-O S' + G(s) g Thus, lim,me(t) = Constant. If k > +, then since S-k E(s) = S' + G(s) se(s) = Constant gl-k+ s' + G(s) lims- 0 se(s) = oc implies lim_. e(t) = 00. S6.0 E(z) (a) X(z) + H(z) z X(z) z- z(z + i) E(z) = + H(z) (z - )(z + i) + +(z - )(z + ) z +.z - Z + ±+ z Z Z' The poles are at i + -. These poles are inside the unit circle and therefore yield stable inverse z-transforms, so e[n] = b[n] + ( stable sequences). So lim-,.e[n] = 0.

10 Signals and Systems S6-0 (b) H(z) = A(z) (z - )B(z) since H(z) has a pole at z =. Now E(z) (z - )B(z) X(z) + H(z) (z - )B(z) + A(z)' Furthermore, we know that -z I )(z - )B(z) (z - )B(z) + A(z) zb(z) (z - )B(z) + A(z) for x[n] = u[n] Y(z) H(z) (z - )B(z) X(z) + H(z) (z - )B(z) + A(z) There are no poles for Iz I > because h[n] is stable. Therefore, E(z) = zb(z) (z - )B(z) + A(z) has no poles for IzI >, and lim,-e[n] = 0. (c) H(z) = - z ' E(z) z - X(z) + H(z) z E(z) = z- fz-\ z X(z) zz (z for x[n] = u[n] = =* e[n] = b[n], so e[n] = 0, n - z + z- (d) H(z) (+ -Iz ')( - z-)' E(z) ( + iz-)( - z-) X(z) + H(z) ( + iz-')( - Z-) + 4z- +,z-, ( + tz-) E(z) = ( + Iz- )( - z-) + z + - = + tz-' Therefore, e[n] = b[n] + '6[n - ] =0, n - E(z) = H H(z) X(z ()X(z) + H(z) ' E(z) For x[n] = u[n], we have X(z) = -

11 Feedback Example: The Inverted Pendulum / Solutions S6- We would like N- e[n] = a[n - k], k=o E(z) = N- akz-' k=o Therefore, H(z) = N- z) Z- T akz- k) (k=0 N- (-- ) T akz -k) (k=0 z-i + z- - z- E(z) (f) H(z) = ( + z-)( - z) X(z) + H(z) Now x[n] = (n + )u[n] and X(z) = ( - z-)> ( + z-)( - z-) ( - z-) E(z) = ( + Z- )( - z-) + z- + z- -z + z-' and e[n] = b[n] + b[n - ] =0, n >-

12 MIT OpenCourseWare Resource: Signals and Systems Professor Alan V. Oppenheim The following may not correspond to a particular course on MIT OpenCourseWare, but has been provided by the author as an individual learning resource. For information about citing these materials or our Terms of Use, visit:

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