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A core Course on Modeling Introduction to Modeling 0LAB0 0LBB0 0LCB0 0LDB0 c.w.a.m.v.overveld@tue.nl v.a.j.borghuis@tue.nl P.5

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Dynamic Models: dealing with time Examples of recursive functions for equal-interval sampling and infinitesimal time. Contents: Equal intervals Infinitesimal intervals Equal intervals revisited

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t i K(t) K i u(t) u i u’(t) (u(t+ )-u(t))/ = (u i+1 -u i )/ v’(t) (v(t+ )-v(t))/ = (v i+1 -v i )/ So u i+1 = u i + v i v i+1 = v i + a i = v i+1 = v i + a i = = v i + K i /m = = v i + K i /m = = v i + C(u rest -u i ) /m = v i + C(u rest -u i ) /m Recursive function Q curr =F(Q prev,P prev ): u curr =F 1 (u prev,v prev )=u prev + v prev v curr =F 2 (v prev,u prev )= =v prev + C(u rest -u prev ) /m, =v prev + C(u rest -u prev ) /m, with suitable u 0 and v 0 Equal intervals : a mass-spring system with damping. From physics, we know K=ma where K=C(u rest -u) and a=u’’. Write v=u’, then a=v’. Here, K=K(t), u=u(t). Let us approximate by sampling. For 0: u(t+ )= =u(i + ) =u((i+1) ) =u i+1, and similar for v Remember: order of evaluation F 1, F 2 is irrelevant out

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t i K(t) K i u(t) u i u’(t) (u(t+ )-u(t))/ = (u i+1 -u i )/ v’(t) (v(t+ )-v(t))/ = (v i+1 -v i )/ So u i+1 = u i + v i v i+1 = v i + a i = v i+1 = v i + a i = = v i + K i /m = = v i + K i /m = = v i + C(u rest -u i ) /m = v i + C(u rest -u i ) /m Recursive function Q curr =F(Q prev,P prev ): u curr =F 1 (u prev,v prev )=u prev + v prev v curr =F 2 (v prev,u prev )= =v prev + C(u rest -u prev ) /m, =v prev + C(u rest -u prev ) /m, with suitable u 0 and v 0 Equal intervals : a mass-spring system with damping. From physics, we know K=ma where K=C(u rest -u) and a=u’’. Write v=u’, then a=v’. Here, K=K(t), u=u(t). Let us approximate by sampling. For 0: -v-v-v-v ( - v prev ) ( - v i ) out

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Equal intervals WARNING: 0 is not the same as ' =small'. 0 is not the same as ' =small'. The substitutions for u’(t) and v’(t) are approximations only. As we will see in a later example, aliasing errors are worse when the sampled signal changes rapidly, compared to the sampling rate. Concrete: unless is much smaller than the period of the oscillation (= m/C), the approximations are bad – with the risk of instability.

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out Infinitesimal intervals the mass-spring system with damping revisited. From physics, we know K=ma where K=C(u rest -u) and a=u’’. For some purposes the numerical solution (sampling) is not acceptable. Interested in outcome? use better numerical methods Interested in insight? try to use symbolic methods, i.e.: analysing differential equations Only possible in very few special cases In particular: linear (sets of) DE’s.

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Infinitesimal intervals the mass-spring system with damping revisited. From physics, we know K=ma where K=C(u rest -u) and a=u’’. Here, K=K(t), u=u(t). Let us try a solution of the form u=u rest +A sin(t/T) u’’= -AT -2 sin(t/T) Substitute back: -mAT -2 sin(t/T)=C(u rest -u rest -Asin(t/T) ), mT -2 =C, or T= m/C ( =:T 0 ) out

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Infinitesimal intervals the mass-spring system with damping revisited. From physics, we know K=ma where K=C(u rest -u) and a=u’’. Here, K=K(t), u=u(t). try u=u rest +Ae t, then C+ +m 2 =0, or 1,2 =(- ( 2 -4mC))/2m 1,2 =(- ( 2 -4mC))/2m Let 0 = (4mC). For = 0 critical damping; for < 0 oscillations; T=T 0 / (1- ( / 0 ) 2 ) for > 0 super critical damping (further details: lectures ‘dynamical systems’) out -v-v-v-v By substituting u = Ae t into C(u rest -u)- u’=mu’’ and using u’=A e t ; u’’=A 2 e t. C(u rest -u rest -Ae t )- A e t =mA 2 e t Must hold for any t, so indeed C + + m 2 =0

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Equal intervals revisited playing a CD. Sound on a CD is represented as a series of 44100 samples/sec of the amplitude. Playing the sound: take a sample P i Do a calculation to process the sound (e.g., Q i+1 = Q i +(1- )P i ) Next, output Q i+1 to the speaker. So again Q i+1 =F(Q i,P i ) If is (close to) 0, we get the original samples; if is closer to 1, Q i+1 is more similar to Q i, so no fast changes in Q: higher frequencies are reduced: low-pass filtering

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Equal intervals revisited The reconstructed sound will contain a pitch that was not present initially. The reconstructed sound will contain a pitch that was not present initially. (sampling aliasing) (sampling aliasing) samples taken every 1/44100 sec original sound the reconstructed signal A remedy to aliasing, is to apply a low- pass filter on the input signal. This is also a dynamic system of the form Q curr =F(Q prev,P prev ) (details fall beyond the scope of this lecture)

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Summary From the purpose, propose clever exposed and hidden quantities From the purpose, determine what time-concept you need: partial order or total order If total order: unknown or known intervals? What determines (sampling) intervals? If emphasis on verification or specification: consider statecharts and process models If emphasis on simulation or prediction: use Q curr =F(Q prev,P prev ) interested in outcomes? use simplest possible numerical techniques, be aware for accuracy (perhaps decrease performance !!) interested in insight? try symbolic methods

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Topics 1 Specific topics to be covered are: Discrete-time signals Z-transforms Sampling and reconstruction Aliasing and anti-aliasing filters Sampled-data.

Topics 1 Specific topics to be covered are: Discrete-time signals Z-transforms Sampling and reconstruction Aliasing and anti-aliasing filters Sampled-data.

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