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19 Pages

### EG260 Lagrange

Course: ENGINEERIN EG 260, Spring 2010
School: Swansea UK
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Word Count: 1916

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Slone EG-260 A.K. Dynamics (1) EG-260 DYNAMICS I Lagrange's Equation 1. 2. 3. 4. 5. 6. 7. 8. Introduction ......................................................2 An overview of the procedure. ........................3 Useful energy expressions................................4 A mass-spring example ....................................4 A trolley mounted pendulum example ...........7 Two pendulum...

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Slone EG-260 A.K. Dynamics (1) EG-260 DYNAMICS I Lagrange's Equation 1. 2. 3. 4. 5. 6. 7. 8. Introduction ......................................................2 An overview of the procedure. ........................3 Useful energy expressions................................4 A mass-spring example ....................................4 A trolley mounted pendulum example ...........7 Two pendulum example.................................13 Damping ..........................................................16 Advantages of Lagrange's method ...............19 a.k.slone 2006 1 of 19 A.K. Slone EG-260 Dynamics (1) 1. Introduction Lagrangian methods use an energy approach to obtain the equations of motion in terms of generalised coordinates. The equation of motion in terms of the kinetic energy (T), potential energy (U) and generalised coordinated qi for an n degree of freedom system is given by d T T U - + = Qi( n ) i = 1,2,....... n q q q dt &i i i q (1) & where qi = t is the generalised velocity and Qi(n) is the non-conservative force corresponding to the generalised co-ordinates qi .The forces represented by Qi(n) may be either damping, or dissipative forces or other external forces. If Fxk , Fyk and Fzk represent the external forces acting on the kth mass if the system in the x, y and z directions, respectively, the generalised force is given by: x y z Qi(n ) = Fxk k + Fyk k + Fzk k qi qi qi (2) where xk , yk , zk are the displacements of the kth. mass in the x , y and z directions, respectively. For a torsional system the force Fxk is replaced by the moment Mxk about the x axis and the displacement xk is replaced by the angular displacement qxk about the x axis a.k.slone 2006 2 of 19 A.K. Slone EG-260 Dynamics (1) If the Lagrangian, L, is defined as L = T- U, then equation (1) may be expressed as: d L dt q i & L - q = Qi i (3) For a conservative system Qi(n) = 0, so that equation (1) becomes: d T dt qi & T U - n q + q = 0 i = 1,2 ,....... (4) i i 2. An overview of the procedure. An overview of the procedure is given below. Determine a set of generalized co-ordinates, denoted by q, which are an independent set of variables to record position. Express the kinetic and potential energy in terms of the generalised co-ordinates q. Express the work done by forces and moments. Derive the equation of motion using equations of the type expressed by equations (1) or (3). a.k.slone 2006 3 of 19 A.K. Slone EG-260 Dynamics (1) 3. Useful energy expressions KE of a mass T= 1 2 & mx 2 (5) KE of body rotating about centre of gyration 1 &2 T = I G 2 PE of a spring (6) U= PE due to gravity 1 2 k (extension ) 2 (7) (8) U = mgh 4. A mass-spring example k1 m1 k2 m2 k3 Figure 1 MDOF mass-spring system a.k.slone 2006 4 of 19 A.K. Slone EG-260 Dynamics (1) Figure 1 shows a 2 degree of freedom mass-spring system, the kinetic energy is given by: T= 1 1 & &2 m1q12 + m2q2 2 2 (9) The potential energy is given by: U= 1 2 1 1 2 2 k1q1 + k2 (q2 - q1 ) + k3 q2 2 2 2 (10) The work done by external forces is given by: f1 (q1 ) + f 2 (q2 ) (11) From equation (9) the partial derivatives of the kinetic energy are given by: T & = m1q1 & q1 T T = =0 q1 q2 T & = m2 q2 & q2 (12) and from equation(10) the derivatives of potential energy are: U = k1q1 - k 2( q2 - q1 ) q1 U = k 2( q 2 - q1 ) + k3q2 q2 (13) a.k.slone 2006 5 of 19 A.K. Slone EG-260 Dynamics (1) Applying Lagrange's equation (1) to each mass in turn gives: d T T U q - q + q = Q1 dt &1 1 1 (14) Substituting from equations (12)and (13) gives: d & (m1q1 ) - 0 + (k1 + k 2 )q1 - k2 q2 = Q1 dt (15) which gives the equation of motion for mass m1 as: && m1q1 + (k1 + k2 )q1 - k 2q2 = Q1 (16) and for mass m2 Lagrange's equation is : d T T U q - q + q = Q2 dt & 2 2 2 (17) Substituting from equations (12)and (13) gives: d & (m2 q2 ) - k2q1 + (k 2 + k3 )q2 = Q2 dt (18) which gives the equation of motion for mass m2 as: && m2q2 - k2q1 + (k2 + k3 )q2 = Q2 (19) Thus from equations (16): && m1 q1 + (k1 + k 2 )q1 - k 2 q 2 = f1 and a.k.slone 2006 6 of 19 (20) A.K. Slone EG-260 Dynamics (1) && m2 q2 - k2 q1 + (k2 + k3 )q 2 = f 2 which for q1 = x1 and q2 = x2 are the same equations of motion as given previously. (21) 5. A trolley mounted pendulum example Figure 2 shows a trolley mounted pendulum. The surface over which the trolley moves is horizontal and the surface between the rollers and the surface is assumed to be friction free. Find the natural frequency of the system. x M q l l cos(q) m l sin(q) Figure 2 Trolley mounted pendulum a.k.slone 2006 7 of 19 A.K. Slone EG-260 Dynamics (1) Solution Kinetic energy of trolley is: Tt = 1 1 & & Mx 2 = Mq12 2 2 (22) To calculate the kinetic energy of pendulum, first calculate the velocity in terms of the generalised co-ordinates x and q: vx = d & (x + l sin( )) = x + l& cos dt vy = d (- l cos( )) = l& sin dt (23) (24) Thus the kinetic energy of pendulum is: Tp = 1 2 m v2 + vy x 2 1 & & & = m l 2 2 sin 2 + x + l cos 2 ( ) ( ( )) 2 (25) Simplifying gives: Tp = 1 & & m l 2& 2 + x 2 + 2 xl& cos 2 ( ) (26) The total kinetic energy is T = Tp + Tt and from equations (22) and (25) is: a.k.slone 2006 8 of 19 A.K. Slone EG-260 Dynamics (1) T = 1 1 & & & & M x 2 + m l 2& 2 + x 2 + 2 xl cos 2 2 ( ) (27) T= 1 & & (M + m )x 2 + 1 ml 2& 2 + mlx& cos 2 2 (28) Next consider the potential energy of the system. The pendulum has moved a vertical distance of l l cos (q) and the trolley has no movement in the vertical direction thus the potential energy of the system is: U = mg (l - l cos ) = mgl (1 - cos ( )) (29) The external forces are zero i.e. Qi(n) = 0 so for Lagrange's equation (1) with q1 = x d T T U + =0 - & dt x x x From equation (28) T & & = (M + m )x + ml cos & x (30) (31) and T =0 x (32) From equation (29) a.k.slone 2006 9 of 19 A.K. Slone EG-260 Dynamics (1) U =0 x (33) Substituting equations (31), (32) and (33) in equation (30) gives: d & (M + m )x + ml& cos + 0 + 0 = 0 dt which gives: & (M + m)&& + ml& cos - ml& 2 sin = 0 x For Lagrange's equation (1) with = q2 q. d T T U + =0 &- dt ( ) (34) (35) (36) From equation (28) T & & = ml 2 + mlx cos & (37) and T && = -mlx sin (38) From equation (29) U = mgl sin ( ) (39) a.k.slone 2006 10 of 19 A.K. Slone EG-260 Dynamics (1) Substituting equations (37), (38) and (39) in equation (36) gives: d & & & ml 2 + mlx cos - - mlx& sin + mgl sin = 0 (40) dt Hence: & & && ml 2& + ml&& cos - mlx sin + mlx sin + mgl sin = 0 (41) x Hence: ( ) ( ) & ml 2& + ml&&cos + mgl sin = 0 x For small angular displacements: sin(q) = q and cos(q) = 1 Substituting the small angle approximation in equation (35) gives: (42) & (M + m)&& + ml& - ml& 2 = 0 x (43) & Since the last term is quadratic in , linearising equation (43) gives & (M + m)&& + ml& = 0 x (44) Similarly, substituting the small angle approximation in equation (42) gives: & && mlx + ml 2& + mgl = 0 a.k.slone 2006 (45) 11 of 19 A.K. Slone EG-260 Dynamics (1) Equations (44) and (45) may be written in matrix notation: && M + m ml x 0 0 x 0 ml 2 & & + 0 mgl = 0 ml 4 244 4 4 (46) 1 4 3{ 1 2 3 { Mass M && x Stiffneass K x To find the natural frequency use: det - 2 M + K = 0 So (47) - 2 (M + m ) - ml 2 - 2 ml - ml + mgl 2 2 =0 (48) Hence - 2 (M + m) - 2 ml 2 + mgl - 2 ml Simplifying gives: ( ) ( ) 2 =0 (49) 2 - (M + m ) - 2 ml 2 + mgl - 2 m 2 l 2 = 0 So either w2 = 0, i.e. rigid body motion or: { ( ) } (50) - (M + m ) - 2 ml 2 + mgl - 2 m 2 l 2 = 0 a.k.slone 2006 ( ) (51) 12 of 19 A.K. Slone EG-260 Dynamics (1) Simplifying: (M + m )(- 2l + g ) + 2 ml = 0 - Ml + Mg + mg = 0 2 - 2 Ml + Mg - 2 ml + mg + 2 ml = 0 So (52) 2 = (m + M )g Ml (53) g M 2 = , which is the result for a simple As l pendulum. 6. Two pendulum example Figure 3 shows two pendulums of length l, separated by a spring of stiffness k N/m2 q1 a q2 l m1 a.k.slone 2006 m1 Figure 3 Double pendulum 13 of 19 A.K. Slone EG-260 Dynamics (1) The generalised co-ordinates are q1 = q1 and q2 =q2 . The kinetic energy of the system is: T= 2 1 1 & 2 m1 l&1 + m 2 l 2 2 2 1 &2 1 &2 = m1 l 21 + m 2 l 2 2 2 2 ( ) ( ) (54) So that: T & = m1l 21 & 1 T & = m1l 2 2 & 2 (55) and (56) and T T = =0 1 2 U = m1 gl(1 - cos 1 ) + m2 gl( 1 - cos 2 ) + 1 2 2 k (sin 1 - sin 2 ) 2 (57) The potential energy is: (58) a.k.slone 2006 14 of 19 A.K. Slone EG-260 Dynamics (1) U = m1 gl sin 1 + k 2 cos 1 (sin 1 - sin 2 ) 1 U = m2 gl sin 2 - k 2 cos 2 (sin 1 - sin 2 ) 2 (59) (60) Substituting equations (55) and (59) in Lagrange's equation: d T & dt i gives for q1: T U - + = 0 i i (61) d m1 l 2&1 + m1 gl sin1 + k 2 cos 1 (sin1 - sin 2 ) = 0 (62) dt ( ) hence: & m1l 2& + m1 gl sin1 + k 2 cos1 (sin1 - sin 2 ) = 0 (63) 1 and for q2 d & m2 l 2 2 + m2 gl sin 2 - k 2 cos 2 (sin 2 - sin 2 ) = 0 (64) dt ( ) hence: & m2 l 2&2 + m2 gl sin 2 - k 2 cos 2 (sin1 - sin 2 ) = 0 (65) a.k.slone 2006 15 of 19 A.K. Slone EG-260 Dynamics (1) Using the small angle assumption, equations (63) and (65) may be expressed as & m1l 2& + m1 gl1 + k 2 (1 - 2 ) = 0 (66) 1 & m2 l 2&2 + m2 gl 2 - k 2 (1 - 2 ) = 0 (67) Collecting terms in q1 and q2 gives: & m1l 2& + m1 gl + k 2 1 - k 2 2 = 0 1 & m2 l 2&2 - k 21 + m2 gl + k 2 2 = 0 ( ( ) ) (68) (69) Equations (68)and (69) may be expressed in matrix form as: m1l 2 0 & 0 &1 m1 gl + k 2 + 2 && m2 l 2 - k 2 1 0 = 2 2 0 m 2 gl + k - k 2 (70) From which the frequencies and mode shapes may be calculated in the normal manner. 7. Damping Lagrange's equations are based on the principle of conservation of energy and therefore are not applicable to non-conservative systems such as those including damping. However, if damping is expressed in terms of Rayleigh & dissipation function, i.e. damping is a function of q , such that the force due to damping is given by: a.k.slone 2006 16 of 19 A.K. Slone EG-260 Dynamics (1) 1 2 Fd = c(velociy) 2 then Lagrange's equation may used. (71) For Lagrange's equation as given by equation (1) the generalized force is given by F Qi = - d (72) & qi k1 c1 k2 k3 c2 c3 Figure 4 MDOF mass-spring-damper system For the spring mass-damper system shown in Figure 4, the damping force is given by: 1 1 & & & 2 &2 Fd = 1 c1 q12 + c2 (q2 - q1 ) + c3 q 2 2 2 2 (73) The generalised force for Lagrange's equation is then given by: a.k.slone 2006 17 of 19 A.K. Slone EG-260 Dynamics (1) Q1 = - Fd & & & = -{c1 q1 - c2 (q 2 - q1 )} &1 q & & = -(c1 + c2 )q1 + c2 q 2 Fd & & & = -{c 2 (q 2 - q1 ) + c3 q2 } &2 q & & = c2 q1 - (c2 + c3 )q 2 (74) and Q2 = - (75) The generalised forces are given by equations (74) and (75). Equations (20)and (21) give the equations of motion for the undamped system as : && m1 q1 + (k1 + k 2 )q1 - k 2 q 2 = f1 (76) and && m2 q2 - k2 q1 + (k2 + k3 )q 2 = f 2 (77) Substituting for the friction force as given equations (74) by equations (76) gives: && & & m1 q1 + (k1 + k 2 )q1 - k 2 q 2 = - (c1 + c 2 )q1 + c 2 q 2 hence the equation of motion for mass m1 is: && & & m1 q1 + (c1 + c2 )q1 - c2 q 2 + (k1 + k 2 )q1 - k 2 q 2 = 0 (79) Substituting for the friction force as given equations (75) by equations (77) gives: (78) a.k.slone 2006 18 of 19 A.K. Slone EG-260 Dynamics (1) && & & m2 q2 - k 2 q1 + (k 2 + k 3 )q 2 = c2 q1 - (c2 + c3 )q2 Hence the equation of motion for mass m2 is: && & & m2 q2 - c2 q1 + (c2 + c3 )q2 - k 2 q1 + (k 2 + k 3 )q2 = 0 (80) (81) In matrix form the equations of motion with q1 = x1 and q2 = x2 are: m1 0 0 &&1 (c1 + c2 ) x + m 2 &&2 - c2 x & - c 2 x1 (k1 + k 2 ) + & (c2 + c3 ) x2 - k 2 - k2 =0 (k 2 + k 3 ) (82) 8. Advantages of Lagrange's method Equations contain only scalar quantities One equation for each degree of freedom. The resulting equations are independent of the choice of coordinate system since kinetic and potential energy does not depend on coordinates a.k.slone 2006 19 of 19
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Swansea UK - ENGINEERIN - EG 260
A.K. SloneEG-260 Dynamics (1)EG-260 DYNAMICS I MeasurementIntroduction.2 1. Measurement .21.1. The logarithmic decrement method. .3 1.1.1. Measurement over more than one cycle.5Example 1 .5 1.2. The added mass method. .6 Example 2 .6a.k.slone 20091
Swansea UK - ENGINEERIN - EG 260
A.K. SloneEG-260 Dynamics (1)EG-260 DYNAMICS I Modelling and Energy Methods 1. Kinetic Energy. 3 2. Potential Energy . 4 2.1. Potential Energy in Elastic springs. 4 2.2. Gravitational PE . 5 3. Relating Potential and Kinetic Energy. 5 4. A vertical spri
Swansea UK - ENGINEERIN - EG 260
A.K. SloneEG-260 Dynamics (1)EG-260 DYNAMICS I Multiple Degrees of FreedomEG-260 DYNAMICS I Multiple Degrees of Freedom . 1 1. 2. 2.1. 2.2. 2.3. 2.3.1. 3. 3.1. 4. 5. 5.1. 5.1.1. 5.1.2. 6. Introduction . 2 Two degrees of freedom. 2 Finding eigenvalues a
Swansea UK - ENGINEERIN - EG 260
EG-260 Dynamics 1 Pre-course Revision Questions 1. a. State what is meant by angular velocity. b. A stone is tied to one end of a cord and rotated in a horizontal circle about a point C with the cord horizontal, as shown in Figure 1. The stone has speed v
Swansea UK - ENGINEERIN - EG 260
Swansea UK - ENGINEERIN - EG 260
A.K. SloneEG-260 Dynamics (1)EG-260 DYNAMICS I Stiffness1. 2. 3.Introduction .2 Material Properties.2 Examples of spring constants.23.1. Longitudinal vibration. .3 3.2. Transverse vibration. .5 3.3. Torsional vibration.6 3.4. Summary of spring consta
Swansea UK - ENGINEERIN - EG 260
A.K. SloneEG-260 Dynamics (1)EG-260 DYNAMICS I Harmonic Excitation 1EG-260 DYNAMICS I Harmonic Excitation 1.1 Introduction to harmonic excitation .2 1. Equation of motion .2 2. Harmonic excitation of an undamped system.4 2.1. Example.8 2.2. The phenome
Swansea UK - ENGINEERIN - EG 260
Swansea UK - ENGINEERIN - EG 260
EG-260 DYNAMICS COURSE REVISIONSTANDARD DIFFERENTIALS AND INTEGRALS. 2 Some Series . 3 1. The exponential series. 3 1.1. Sine and Cosine. 3 2. Trigonometrical formulae . 3 Euler Relationship .
Swansea UK - ENGINEERIN - EG 260
1st Printing Errata If you have the second printing, the following corrections have been made. If you have a first printing you will need to print and refer the following corrections. To tell which printing you have: Look on the page, 4 in from the front
Swansea UK - ENGINEERIN - EG 260
A.K. SloneEG-260 Dynamics (1)EG-260 DYNAMICS I GlossarySymbol k m c ccr e z fn T wn wd w rc c crEquationDescription stiffness coefficient mass damping coefficient critical damping coefficient eccentricity damping ratio natural frequency period of os
Swansea UK - ENGINEERIN - EG 260
EG-260EGD260 Dynamics - Example Sheet 1 Single degree of freedom systems1.Formulate the equation of motion for a simple pendulum of length l. If a grandfather clock requires the pendulum to have a period of 2 seconds, what is the required length? Compu
Swansea UK - ENGINEERIN - EG 260
EG-260EG -260 Dynamics I - Example Sheet 5 Multi degree of freedom systems1. Write down the equations of motions for the torsional system of two circular plates with moments of inertia I1 and I2, and torsional stiffness k 1 and k 2. Use variables 1 and
Swansea UK - ENGINEERIN - EG 260
EG-260EG260 Dynamics - Example Sheet 2 Single degree of freedom systems1. Develop the equations of motion and hence obtain the natural angular frequencies of the following systems: i) A mass-spring-bar system (shown below left), assuming that the bar is
Swansea UK - ENGINEERIN - EG 260
EG-260EG-260 Dynamics I - Example Sheet 3 Damped single degree of freedom systems1. A weight of 2 N is attached to a damped spring having a stiffness of 400 N/m and a damping constant of 9.03 kg/s. Determine a) b) c) critical damping constant undamped n
Swansea UK - ENGINEERIN - EG 260
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Swansea UK - ENGINEERIN - EG 260
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Swansea UK - ENGINEERIN - EG 260
EG-260 Examples Sheet 1 1. This is a rotational problem, so we express Newton's Law in the form&amp; C = I&amp;where C is the applied moment and I is the moment of inertia. In the pendulum caseC = -mgl sin I = ml 2So that&amp; - mgl sin = ml 2&amp;For small angles
Swansea UK - ENGINEERIN - EG 260
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Swansea UK - ENGINEERIN - EG 260
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Swansea UK - ENGINEERIN - EG 260
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NAME: Project 1 KEY Super Bowl Point Spreads The following data represent the number of points by which the winning team won Super Bowls I to XXXIX: 25 10 10 1 15 19 4 29 13 7 9 18 22 35 27 16 17 36 17 3 3 419 23 27 21 12 32 10 3 7 17 4 14 3 17 5 45 7Yo
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Name _KEY_ ID# _ Project4 _ Chapter 6 I have included a very detailed KEY that is probably at the level of your solutions but not necessarily at the level of your communications. Highlighted in yellow are the details.Use complete sentences when appropria
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KEY P roject F iveP roblem Talk T ime On a Cell Phone: Suppose the talk t ime in digital mode on a MotorolaTimeport P8160 is approximately normally distributed with mean 324 minutes and standard deviation of 24 minutes. A.) What proportion of the t ime
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Project 13 Project Thirteen_ Chapter 13/20 pointsA MINITAB output is required for this exercise. One of the selling points of golf balls is their durability. An independent testing laboratory is commissioned to compare the durability of three different b
University of Toronto - MATH - 246
Project 14 _/20 Chapter 14READ EVERYTHING CAREFULLY: Show all your work. Indicate clearly the methods you use, because you will be graded on the correctness of your methods as well as on the accuracy of your results and explanation. Use complete sentence
King Fahd University of Petroleum & Minerals - EE - EE390
King Fahd University of Petroleum &amp; Minerals Electrical Engineering Department EE-390; Exam-2 (0xx);Prob.1 Prob.2 Prob.3 TotalName :Section :I.D.Answer all the questions. All questions carry equal marks.1 Write an effective assembly language program
King Fahd University of Petroleum & Minerals - EE - EE390
King Fahd University of Petroleum &amp; Minerals Electrical Engineering Department EE-390; Exam-1 (0xx)Prob.1 Prob.2 Prob.3 TotalName :Section :I.D.Answer all three questions. All questions carry equal marks.1(a). Assume BX=1FF8H. Write a 1-line assembl
King Fahd University of Petroleum & Minerals - EE - EE390
King Fahd University of Petroleum &amp; Minerals Electrical Engineering Department EE-390; Exam-2(081); 14th January, 2009Prob.1 Prob.2 Prob.3 TotalName :Section :1I.D.Answer all three questions, which carries equal marks.Exam TIME: 8:00 to 9:30 PM1(a
King Fahd University of Petroleum & Minerals - EE - EE390
King Fahd University of Petroleum &amp; Minerals Electrical Engineering Department EE-390; Exam-1Prob.1 Prob.2 Prob.3 TotalAnswer the THREE question. All questions carry equal marks. Name : Section : I.D.1(a) If the data segment is starts from PA=1234H, wh
King Fahd University of Petroleum & Minerals - EE - EE390
King Fahd University of Petroleum &amp; Minerals Electrical Engineering Department EE-390; Exam-2(071); 6th January, 2008Prob.1 Prob.2 Prob.3 TotalName :Section :I.D. Exam TIME: 6:30 to 8:00 PMAnswer all three questions, which carries equal marks.1(a).
King Fahd University of Petroleum & Minerals - EE - EE390
King Fahd University of Petroleum &amp; Minerals Electrical Engineering Department EE-390; Exam-1(071); 4th October, 2007Prob.1 Prob.2 Prob.3 TotalAnswer all questions. Time 1.20 hourName : Section : I.D.1(a). Write a program that will add the contents of
King Fahd University of Petroleum & Minerals - EE - EE390
King Fahd University of Petroleum & Minerals - EE - EE390
EE390 QZ 9Name : Section : I.D.1.Assume you have an 8255A chip that is used as an isolated I/O with port A at address 5000H, Port B at address 5001H, Port C at address 5002H, and the Control Register at address 5003H. Port C is connected to eight diffe
King Fahd University of Petroleum & Minerals - EE - EE390
Name:.
King Fahd University of Petroleum & Minerals - EE - EE390
&quot;TITLE &quot;PALINDROME CHECK MODEL SMALL. STACK 100. DATA. CODE. MOV DL,00H MOV AL,10100101B MOV CL,4 LOP: RCL AL,1 RCR DL,1 DEC CL JNZ LOP AND AL,0F0H CMP AL,DL JE EQL MOV BL,00H JMP EXT EQL: MOV BL,0AAH EXT: NOP END
King Fahd University of Petroleum & Minerals - EE - EE390
Quiz 6Name:. ID:. King Fahd University of Petroleum &amp; MineralsEE390
King Fahd University of Petroleum & Minerals - EE - EE390
Quiz 5Name:. ID:. King Fahd University of Petroleum &amp; MineralsEE390Execute the following program and find the contents of the required registers and the memory contents of the stack segment. (Assume Label=1111H)Title &quot;Quiz5&quot; .MODEL small .STACK 32 .D