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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.s y s t e mZLZZoLZZoLINT h e v e n i nN o r t o nt hV Massachusetts Institute of Technology Department of Mechanical Engineering 2.004 Dynamics and Control II Spring Term 2008 Lecture 91 Reading: Nise: Sec. 2.5 • 1 Th´evenin and Norton Source Models We state the following without proof: Any linear system (regardless of its internal complexity) containing a single source (voltage or current), and with an external load element ZL, may be modeled as either (a) A voltage source Vth with a series impedance Zo (a Thev´enin equivalent circuit), or (b) A current source IN with a parallel impedance Zo (a Norton equivalent circuit). The arbitrary system shown above can be represented in either form, where Zo – is the system output impedance, and is the same in each case. Vth 1copyright – is the Th´evenin source voltage circuit” terminal voltage. c� D.Rowell 2008 – found by removing Zo and measuring the “open-9–1s y s t e ms h o r t - c i r c u i tc u r r e n tNI CV ( t )+-sRZLZ = oC+-R RR C s + 1 `ZL RR C s + 1 1R C s + 1Vst hV s y s t e mo p e n - c i r c u i tv o l t a g eIN – is the Norton source current – found by “short-circuiting” the output and measuring the current. To find the output impedance Zo – set the internal source to zero, and measure the input impedance at the system’s output terminals. Some care must be taken in setting any source to zero: (a) To set a voltage source to zero, short-circuit it, for example to find Zo in the circuit below: 1 R Zo = Cs�R = CRs + 1 The Th´evenin source voltage is found by recognizing the left-hand circuit as a voltage-divider and finding the terminal voltage: 1/(Cs) 1 Vth = Vs = VsR + 1/(Cs) RCs + 1 The complete Th´evenin equivalent for this system is: 9–2CsRZLICsRIZ = o 1C sZLIN 1C sCV ( t )+-sRCV ( t )+-sRLVo(b) To set a current source to zero, remove it from the circuit as shown below: In this case the output impedance seen at the output terminals is 1 Zo = Cs and the Norton short circuit current is simply IN = I (Note that R does not appear in the system formulation.) The Norton equivalent is: Example 1 Use Th´evenin and Norton source models to find the transfer function of the following system when the load is an inductor L. Vo(s)H(s) = V (s) Th´evenin Model: For the source model on the left 1/(Cs) 1 Vth = = R + 1/(Cs) RCs + 1 1 R Zo = Cs�R = CRs + 1 Then for the full system on the right 9–3CR i =s cVR+-VCRZ = o RR C s + 1+-VL sI = NVR RR C s + 1L s RR C s + 1 1R C s + 1VVo0sUsing a voltage divider relationship Ls Vo(s) = Vth(s)Ls + R/(CRs + 1) Ls(RCs + 1) 1 = . V (s). RLCs2 + Ls + R RCs + 1 The transfer function is Vo(s) Ls H(s) = = V (s) RLCs2 + Ls + R Norton Mo del: For the Norton model 1 1 R IN = isc = V, Zo = R� = R Cs RCs + 1 and the equivalent system model is Then the transfer function is found from Vo(s) = �Zo� 1 Cs � IN (s) RLs 1 = V (s)Ls + R/(RCs + 1) R Ls = V RLCs2 + Ls + R 9–4FmvV e l o c i t i e s a n d f o r c e s a r em e a s u r e d w i t h r e s p e c t t oa n i n e r t i a l r e f e r e n c e f r a m eAs with the Th´evenin method, the transfer function is Vo(s) Ls H(s) = = V (s) RLCs2 + Ls + R 2 Modeling Mechanical Systems We now move our attention to deriving mo dels of mechanical systems with motion in one dimension. The approach will be to draw analogies with the electrical modeling methods so as to have a unified technique that can be applied without regard to the energy domain. (1) Choice of Modeling Variables: As in the electrical domain, we select a pair of variables whose product is power, that is we choose F – force (N), and v – velocity (m/s) since P = F v (2) Modeling Elements: As in the electrical domain we find there are two energy storage elements, and one dissipative element. (a) The mass element. For a mass element m dp dvmFm = = m dt dt where p is the momentum. We will use the elemental relationship dv 1 1 = F or V (s) = F (s)dt m ms The energy stored in a moving mass element is t� 1 2E = P v dt = mv 2−∞ which is a function of velocity v and is stored as kinetic energy. Note: Inertial forces and velocities must be measured with respect to a non-accelerating inertial reference frame. In this course we will assume a reference velocity vref = 0. 9–5FFv1v2(b) The spring (compliance) element: KFFxxvv1221Let xspring = (x2 − x1) − xo where xo is the “rest length” of the spring. Then dxspring dx2 dx1 vspring =dt =dt − dt = v2 − v1 From Hooke’s law, F = Kxspring where K is the stiffness (N/m). dFK K = KvK or FK (s) = vK (s)dt s The energy stored in a moving mass element is t� 1 E = P v dt = FK 2 2K−∞ which is a function of velocity FK and is stored as potential energy. (c) The viscous friction (dissipative) element: (Also known as the dashpot or damper element) Let VB = v2 − v1, the elemental equation is FB = BvB or FB (s) = BvB (s) The power flow is P = F v = Bv2 ≥ 0 so that the power flow is uni-directional and cannot be recovered. (3) Ideal Sources: We define a pair of ideal sources (a) Force Source: The force source maintains a prescribed force Fs regardless of the velo city at which it travels. 9–6vV0a v e l o c i t y s o u r c ev e l o c i t y i s i n d e p e n d e n t o ft h e f o r c e a p p l i e d t o t h el o a d .FsVsA r r o w i s i n t h e d i r e c t i o no f t h e a s s u m e d v …


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MIT 2 004 - Thevenin and Norton Source Models

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