2.004 Dynamics and Control II Spring 2008
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1 MT OpenCourseWare Dynamics and Control Spring 200 For information about citing these materials or our Terms of Use, visit:
2 Massachusetts nstitute of Technology Department of Mechanical Engineering 200 Dynamics and Control Spring Term 200 ecture 9 1 Reading: Nise: Sec 25 1 Thévenin 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 Z, may be modeled as either (a) A voltage source V th with a series impedance Z o (a Thevénin equivalent circuit), or (b) A current source N with a parallel impedance Z o (a Norton equivalent circuit) J D O J A 6 D A A E 1 H J The arbitrary system shown above can be represented in either form, where Z o is the system output impedance, and is the same in each case V th is the Thévenin source voltage circuit terminal voltage found by removing Z o and measuring the open 1 copyright c DRowell
3 O J A J D F A? E H? K E J J = C A N is the Norton source current found by short-circuiting the output and measuring the current O J A 1 D H J? E H? K E J? K H H A J To find the output impedance Z o 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 Z o in the circuit below: J < 1 R Z o = Cs R = CRs 1 The Thévenin source voltage is found by recognizing the left-hand circuit as a voltagedivider and finding the terminal voltage: 1/(Cs) 1 V th = V s = V s R 1/(Cs) RCs 1 The complete Thévenin equivalent for this system is: 9 2
4 (b) To set a current source to zero, remove it from the circuit as shown below: 1 1 n this case the output impedance seen at the output terminals is 1 Z o = Cs and the Norton short circuit current is simply N = (Note that R does not appear in the system formulation) The Norton equivalent is: 1 Example 1 Use Thévenin and Norton source models to find the transfer function of the following system when the load is an inductor V o (s) H(s) = V (s) J J Thévenin Model: For the source model on the left 1/(Cs) 1 V th = = R 1/(Cs) RCs 1 1 R Z o = Cs R = CRs 1 Then for the full system on the right 9 3
5 Using a voltage divider relationship The transfer function is V o (s) = s V th (s) s R/(CRs 1) = s(rcs 1) 1 V (s) RCs 2 s R RCs 1 V o (s) H(s) = = V (s) s RCs 2 s R Norton Model: For the Norton model E? 1 1 R N = i sc = V, Z o = R = R Cs RCs 1 and the equivalent system model is 1 Then the transfer function is found from ( V o (s) = Z o 1 ) N (s) Cs Rs 1 = V (s) s R/(RCs 1) R s = V RCs 2 s R 9
6 As with the Thévenin method, the transfer function is V o (s) H(s) = = V (s) s RCs 2 s R 2 Modeling Mechanical Systems We now move our attention to deriving models 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 A? E J E A B H? A = H A A = K H M E J D H A F A? J J = E A H J E = H A B A H A? A B H = A For a mass element m dp dv m F m = = 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 E = P v dt = mv which is a function of velocity v and is stored as kinetic energy Note: nertial forces and velocities must be measured with respect to a nonaccelerating inertial reference frame n this course we will assume a reference velocity v ref = 0 9 5
7 N N (b) The spring (compliance) element: et x spring = (x 2 x 1 ) x o where x o is the rest length of the spring Then dx spring dx 2 dx 1 v spring = = dt dt dt = v 2 v 1 From Hooke s law, F = Kx spring where K is the stiffness (N/m) df K dt K = Kv K or F K (s) = v K (s) s The energy stored in a moving mass element is t 1 2 E = P v dt = F 2K K which is a function of velocity F K and is stored as potential energy (c) The viscous friction (dissipative) element: (Also known as the dashpot or damper element) et V B = v 2 v 1, the elemental equation is F B = Bv B or F B (s) = Bv B (s) The power flow is P = F v = Bv 2 0 so that the power flow is uni-directional and cannot be recovered (3) deal Sources: We define a pair of ideal sources (a) Force Source: The force source maintains a prescribed force F s regardless of the velocity at which it travels 9 6
8 = B H? A K H? A B H? A E A F A J B J D A A? E J O H A G K E H J = E J = E J D A? K H H A J H? A =? J J =?? A A H = J A J D A = J J D A D A B J D A = H H M E J D A F E J E A A? E J E H A? J E (b) Velocity Source: The velocity source maintains a prescribed velocity V s regardless of the force required to maintain that velocity = A? E J O K H? A A? E J O E A F A J B J D A B H? A = F F E J J D A ) H H M E E J D E H A? J E B J D A = K A? E J H F nterconnection Rules: (a) Continuity: Consider a mass element with n external forces acting J or n Fi = m dv m dt i=1 n Fi m dv m = 0 dt i=1 9 7
9 * * *! * * f we substitute a fictitious force F m = m dv m / dt (known as a d Alembert force) we van write n Fi F m = 0 and state the following i=1 The sum of all forces acting on a mass element (including the d Alembert force) is zero f we represent a point in space containing a mass element as a node on a graph, and the forces acting on the node as branches, where an arrow pointing at the node means that positive force acts to accelerate the node in the reference direction! A A A A H A F H A A J J D A B J D A = A A A J J E We can write F 1 F 2 F 3 = F m = 0 The node represents a point in a mechanical system with a distinct velocity The continuity condition for a mechanical system is equivalent to Kirchoff s current law in an electrical system Example 2 The system * * may be drawn as a graph: J A = > H =? D H A F H A A J E C = = A A A J J D A = H H M = M = O F E J J J D A E A H J E = H A B A H A? A 9
10 Notice the arrow directions on the branches - as implied by the right hand figure above The graph implies a pair of continuity equations at the two nodes: dv 1 F K F B F m1 = 0 or F K F B = m 1 dt dv 2 F K F B F m2 = 0 or F K F B = m 2 dt For a massless node (that is at the interconnection of n ideal massless elements) n Fi = 0 For example i=1 * A = F H E = F A H = H A = K J > A = A * * implies F K F B = 0 at the junction 9 9
2.004 Dynamics and Control II Spring 2008
MIT OpenCourseWare http://ocwmitedu 00 Dynamics and Control II Spring 00 For information about citing these materials or our Terms of Use, visit: http://ocwmitedu/terms Massachusetts Institute of Technology
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