MIT OpenCourseWare http://ocw.mit.edu 2.004 Dynamics and Control II Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.H ( s ) H ( s )12H ( s)H ( s)12U ( s )Y ( s )U ( s )Y ( s )X ( s )H ( s)12++H ( s ) + H ( s )12Y ( s )Y ( s )U ( s )U ( s )H ( s)� Massachusetts Institute of Technology Department of Mechanical Engineering 2.004 Dynamics and Control II Spring Term 2008 Lecture 41 Reading: Nise: Secs. 5.1–5.3 • 1 Block Diagram Algebra (Interconnection Rules) a)Series (Cascade) Connection: Since the output of the first block is X(s) = H1(s)U(s), Y (s) = H2(s)X(s) = H1(s)H2(s)U(s) Note: This is only true if the connection of Hs(s) to H1(s) does not alter the output of H1(s) – known as the “non-loading” condition. b)Parallel Connection In this case the input U(s) is applied to both inputs and the outputs are summed: Y (s) = H1(s)U(s) + H2(s)U(s) = (H1(s) + H2(s))U(s) Example 1 Express 6 H(s) = s2 + 5s + 6 1copyright cD.Rowell 2008 4–1U ( s )Y ( s ) 3s + 3 2s + 2o r d e r o f b l o c k s i s a r b i t r a r y-+Y ( s )U ( s ) 6s + 2 6s + 3+Y ( s )U ( s )1U ( s )2H ( s)-+Y ( s )U ( s )1U ( s )2H ( s)H ( s)+U ( s )Y ( s )H ( s)1H ( s)2U ( s )Y ( s )H ( s)2H ( s)1as (a) a series connection, and (b) a parallel connection of first-order blocks a) Series: 6 3 2 H(s) = = s2 + 5s + 6 s + 3 × s + 2 b) Parallel: Using partial fractions we find 6 6 H(s) = s + 2 − s + 3 Notes: a) These two systems are equivalent. b) A partial fraction expansion is effectively a parallel implementation. c) A factored representation of H(s) is effectively a series implementation. c)Associative Rule: Y (s) = (U1(s) + U2(s))H(s) ≡ U1(s)H(s) + U2(s)H(s) d)Commutative Rule: The order does not matter in a series connection. 4–2G ( s)cG ( s)pE ( s)R ( s)C ( s)c o n t r o l l e rp l a n te r r o r+-f e e d b a c k p a t hr e f e r e n c ei n p u tc o n t r o l l e do u t p u tG ( s)cG ( s)pE ( s)V ( s)c o n t r o l l e rc a r d y n a m i c se r r o r+-f e e d b a c k p a t hd e s i r e ds p e e da c t u a l s p e e dV ( s)dq( s )2 The “Closed-Loop” Transfer Function a) Unity feedback Notes: (a) The term unity feedback means that the actual output value is used to generate the error signal (the feedback gain is 1). (b) In control theory transfer functions in the “forward“ path are often designated by G(s) (see below). (c) It is common to use R(s) to designate the reference (desired) input, and C(s) to desig-nate the controlled (output) variable. From the block diagram: C(s) = (Gp(s)Gc(s))E(s) and E(s ) = R(s) − C(s) or C(s) = Gp(s)Gc(s)(R(s) − C(s) Rearranging: C(s) Gc(s)Gps Gcl(s) = = R(s) 1 + Gc(s)Gps is the unity feedback closed-loop transfer function. Example 2 Find the closed-loop transfer function for the automobile cruise control example: 4–3For the car mv˙ + Bv = Fp = Keθ so that V (s) KsGp(s) = = θ(s) ms + B For the controller θ(s) = KcE(s) θ(s)Gc(s) = = KcE(s) Then from above V (s) Gc(s)Gp(s)Gcl(s) = = Vd(s) 1 + Gc(s)Gp(s) KcKe ms+B KcKeGcl(s) = = , 1 + KcKe ms + (B + KcKe)ms+B and by inspection the closed-loop differential equation is mv˙ + (B + KcKe)v = KcKevd. Aside: Use the Laplace transform final value theorem to find the steady state velocity to a step input vd(t) = vd For the step input vd vd(s) = s and in the Laplace domain KcKe vd v(s) = Gc1(s)Vd(s) = ms + (B + KcKe) s The F.V. theorem states limt→∞f(t) = lims 0sF (s) so that →KcKe vd vss = limt→∞v(t) = lims→0sms + (B + KcKe) s KcKe vss = B + KcKe which is same as we obtained before. 3 Closed-Loop Transfer Function With Sensor Dynamics: Until now we have assumed that the output variable y(t) is measured instantaneously, and without error. Frequently the sensor has its own dynamics - for example the sensor might be temperature measuring device modeled as a first-order system: 4–4RCf r o m s e n s o rt o c o n t r o l l e rVs e n s o rVo u tf i l t e r 1t s + 1ss y s t e mo u t p u ts e n s o rf e e d b a c ks i g n a lttwhere τs is the sensor time constant. The closed-loop block diagram is G ( s)cG ( s)pE ( s)R ( s)C ( s)c o n t r o l l e rp l a n te r r o r+-r e f e r e n c ei n p u ta c t u a lo u t p u tH ( s)s e n s o ri n d i c a t e d o u t p u twhere H(s) is the transfer function of the sensor. In this case: C(s) = (Gc(s)Gp(s))E(s) but now E(s) is the indicated error (as opposed to the actual error): E(s) = R(s) − H(s)C(s) so C(s) = Gc(s)Gp(s)(R(s) − H(s)C(s)) or C(s)(1 + Ge(s)Gp(s)H(s)) = Ge(s)Gp(s)H(s). C(s) Gc(s)Gp(s)Gcl(s) = = R(s) 1 + Gc(s)Gp(s)H(s) is the modified closed-loop transfer function. Example 3 Suppose that velocity sensor in the cruise control is “noisy”, and a simple elec-trical filter is used to smooth the output. Find the effect of the filter on the closed-loop dynamics. 4–5cE ( s)V ( s)c o n t r o l l e rp l a n te r r o r+-r e f e r e n c ei n p u ta c t u a ls p e e df i l t e ri n d i c a t e d s p e e dK Km s + Be 1R C s + 1dV ( s)Using Kirchoff’s Voltage Law (KVL) we find RC v˙out + vout = vsensor so that Vout(s) 1 H(s) = = Vsensor(s) RCs + 1 Then the closed-loop transfer function is KcKeV (s) Gc(s)Gp(s) ms+B = = =Gcl(s) Va(s) 1 + Gc(s)Gp(s)H(s) 1 + KcKe (ms+B)(RCs+1) KcKe(RCs + 1) = (ms + B)(RSs + 1) + KcKe KcKe(RCs + 1) Gcl(s) = …
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