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P1 RJM 18/02/03EG1C2 Engineering Maths: Matrix Algebra Revision 7 (a) Figure Q7-1 shows an electronic circuit with two batteries and three resistors. The.

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Presentation on theme: "P1 RJM 18/02/03EG1C2 Engineering Maths: Matrix Algebra Revision 7 (a) Figure Q7-1 shows an electronic circuit with two batteries and three resistors. The."— Presentation transcript:

1 p1 RJM 18/02/03EG1C2 Engineering Maths: Matrix Algebra Revision 7 (a) Figure Q7-1 shows an electronic circuit with two batteries and three resistors. The current through each resistor is to be found Revision : April 2001 Questions The following three equations relate these currents: 10 = 2 i 1 + 0.5 i 3 24 = 0.5 i 3 + 4 i 2 i 1 + i 2 = i 3 i) Express these equations as one matrix equation. (2 Marks) ii) Use Gaussian Elimination to find the values of the currents. (8 Marks) Figure Q7-1

2 p2 RJM 18/02/03EG1C2 Engineering Maths: Matrix Algebra Revision b) Figure Q7-2 shows a two port network. Figure Q7-2 i) Derive the A matrix for the network such that: ii) Suppose v 1 = 10V and i 1 = 3.5A. Use Cramer’s theorem, or matrix inversion, to find v 2 and i 2. (4 marks) (6 marks)

3 p3 RJM 18/02/03EG1C2 Engineering Maths: Matrix Algebra Revision Answer to 7a)

4 p4 RJM 18/02/03EG1C2 Engineering Maths: Matrix Algebra Revision Last row means 5.5i 3 = 44, so i 3 = 8A. [1 Mark] Second row means 4i 2 = 24-0.5i 3 = 20, so i 2 = 5A [1 Mark] Top row means 2i 1 = 10 –0.5i 3 = 6, so i 1 = 3A [1 Mark] Sensible to check by substituting in original equations 2 i 1 + 0.5 i 3 = 6 + 4 = 10 0.5 i 3 + 4 i 2 = 4 + 20 = 24 i 1 + i 2 3 + 5 = 8 = i 3

5 p5 RJM 18/02/03EG1C2 Engineering Maths: Matrix Algebra Revision Answer to 7b) By Inspection 6 Marks Cramer’s theorem solution: clearly |A| = 5*9 –22*2 = 1 2 Marks

6 p6 RJM 18/02/03EG1C2 Engineering Maths: Matrix Algebra Revision Could solve by matrix inversion Thus

7 p7 RJM 18/02/03EG1C2 Engineering Maths: Matrix Algebra Revision 8) Figure Q8-1 shows a sign comprising a heavy load of mass m at the top of a pole. This sign is subject to winds, which make the mass move. Figure Q8-1 If x is the distance the mass has moved from the vertical and v is its velocity, then the system can be defined by the following equations and F is a constant associated with air resistance; k is the stiffness of the pole.

8 p8 RJM 18/02/03EG1C2 Engineering Maths: Matrix Algebra Revision a) Suppose that m = 0.2kg, k = 1.8 N/m and F = 1.2 Ns/m. Write down the equations as one matrix equation of the form (3 marks) b) Find the repeated eigenvalue,, and eigenvector, , for the system. Find also the vector V such that (A – I)V =  (10 Marks) c) Hence write down the general expression showing the variation of y with time. (2 Marks) d) Suppose also that x = 1 m and v = 0 m/s at time t = 0. Find the variation of y given these initial conditions. (5 Marks)

9 p9 RJM 18/02/03EG1C2 Engineering Maths: Matrix Algebra Revision Answer to Question 8 [3 marks] To find the eigenvalues we then find the characteristic equation: So the repeated eigenvalue is = -3. [4 Marks] For the eigenvector, when = -3, then (A- I)y = 0 gives:

10 p10 RJM 18/02/03EG1C2 Engineering Maths: Matrix Algebra Revision Thus, 3x + v = 0. Let x = 1, then v = -3, so [3 marks] Finding the vector V such that (A – I)V = , is done by: Thus, 3x+v=1. Let x = 1, then v = -2, so V is: [3 marks] [2 marks]

11 p11 RJM 18/02/03EG1C2 Engineering Maths: Matrix Algebra Revision The particular solution is found using the initial conditions By Cramer: The determinant of the square matrix is -2--3 = 1. So [2 marks] [1 mark]

12 p12 RJM 18/02/03EG1C2 Engineering Maths: Matrix Algebra Revision Extra. Suppose k=1N/m, and m and F unchanged: |A- I| = 2 + 6 + 5 = ( + 1)( +5) = 0; so = -5 or -1 Suppose at t=0, x = 0 m and v = 4 m/s: By inspection: c 1 = -1, c 2 = 1, so


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