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

p1 RJM 18/02/03 EG1C2 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 i1 + 0.5 i3 24 = 0.5 i3 + 4 i2i1 + i2 = i3

i) Express these equations as one matrix equation. (2 Marks)ii) Use Gaussian Elimination to find the values of the currents.

(8 Marks)

i22

0.510V

4

24V

i1

i3

Figure Q7-1


b) Figure Q7-2 shows a two port network.

i22

0.5v1

4

v2

i1

Figure Q7-2

i) Derive the A matrix for the network such that:

2

2

1

1i-

v

i

vA

ii) Suppose v1 = 10V and i1 = 3.5A. Use Cramer’s theorem, or matrix inversion, to find v2 and i2. (4 marks)

(6 marks)


Answer to 7a)

Matrix Equation is

0

24

10

i

i

i

111

5.040

5.002

3

2

1 [2 Marks]

Hence Augmented matrix is:

0111

245.040

105.002[1 Marks]

Row3 = Row1 – 2*Row3 gives:

105.220

245.040

105.002 [2 Marks]


Last row means 5.5i3 = 44, so i3 = 8A. [1 Mark]

Second row means 4i2 = 24-0.5i3 = 20, so i2 = 5A [1 Mark]

Top row means 2i1 = 10 –0.5i3 = 6, so i1 = 3A [1 Mark]

Row3 = Row2 + 2*Row3 gives:

445.500

245.040

105.002[2 Marks]

Sensible to check by substituting in original equations2 i1 + 0.5 i3 = 6 + 4 = 10

0.5 i3 + 4 i2 = 4 + 20 = 24

i1 + i2 3 + 5 = 8 = i3


Answer to 7b)By Inspection

2

2

2

2

2

2

5.01

1

1

i-

v

92

225

i-

v

10

41

12

25

i-

v

10

41

1

01

10

21

i

v

6 Marks

Cramer’s theorem solution: clearly |A| = 5*9 –22*2 = 1

V13779095.3

2210 vThus, 2 2 Marks

2 MarksA5.2205.175.32

105i and 2


Could solve by matrix inversion

52

229

52

229

444511-A

A5.2i

V13v so

5.2

13

5.1720

7790

5.3

10

52

229

i-

v

2

2

2

2

Thus


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.

x

m

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

vdtdx and kx-Fv

dtdv

m

F is a constant associated with air resistance; k is the stiffness of the pole.


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

v

x e wher

dtd

yAyy

(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)


Answer to Question 8

Ayy

v

x

69

10

v

x

2.02.1

2.08.1

10

dtdvdtdx

dtd

[3 marks]

To find the eigenvalues we then find the characteristic equation:

0)3(9669

1 22

So the repeated eigenvalue is = -3. [4 Marks]

For the eigenvector, when = -3, then (A-I)y = 0 gives:

0

v

x

39

13

v

x

369

130


Thus, 3x + v = 0. Let x = 1, then v = -3, so

3

1

[3 marks]

Finding the vector V such that (A – I)V = , is done by:

3

1

v

x

39

13

v

x

369

130

Thus, 3x+v=1. Let x = 1, then v = -2, so V is:

2

1 [3 marks]

3t-2

3t-1 e

2

1t

3

1*ce

3

1*c=

v

x

y

[2 marks]


The particular solution is found using the initial conditions

2

10*3-2

0*3-1 c

c

23

11e

2

10

3

1*ce

3

1*c

0

1=y

By Cramer: The determinant of the square matrix is -2--3 = 1.

21

021/

20

11c1

3

130

1/03-

11c2

3t-3t-3t- e0

1t

9

3e

6

3t

9

3e

6

2=

v

x

y

So [2 marks]

[2 marks]

[1 mark]


Extra. Suppose k=1N/m, and m and F unchanged:

-6-5-

1-;

6-5-

10IAA

|A-I| = 2 + 6 + 5 = ( + 1)( +5) = 0; so = -5 or -1

5

1 so,

0

0

v

x

565

15)-( -5,For ΛyIA

1

1 so,

0

0

v

x

165

11)-( -1,For ΛyIA

t-2

5t-1 e

1

1*ce

5

1*c=

v

x Therefore,

y

Suppose at t=0, x = 0 m and v = 4 m/s:By inspection: c1 = -1, c2 = 1, so

2

121 c

c

15

11

1

1*c

5

1*c=

4

0

t-5t- e1

1e

5

1

y

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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