Engineering principle

8/2/2019 Engineering principle

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PART A

a) The figure 1 shows the circuit of a series resonance filter.

Figure 1

i) Use the following parameters to plot/measure the magnitude and phase frequencyresponse.

L C R min max

Measurement 1 100 mH 100 nF 200 500 rad/s 50000 rad/s




Below is a diagram showing the oscillation of the first measurement at 80 Hz.


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Measurement 1

Frequency V output

80 Hz 1.4

250 Hz 5

500 Hz 10750 Hz 18

1000 Hz 28

2000 Hz 55

3000 Hz 20

4000 Hz 14

5000 Hz 10

6000 Hz 8

8000 Hz 6

Measurement 2

Frequency V output

80 Hz 0.7

250 Hz 2.2

500 Hz 5

750 Hz 9

1000 Hz 14

2000 Hz 28

3000 Hz 10

4000 Hz 7

5000 Hz 5

6000 Hz 4

8000 Hz 3

Measurement 3

Frequency V output

80 Hz 0.36

250 Hz 1.1

500 Hz 2.2

750 Hz 4.21000 Hz 7

2000 Hz 15

3000 Hz 5

4000 Hz 3.2

5000 Hz 2.4

6000 Hz 2

8000 Hz 1.4


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Measurement 4

Frequency V output

80 Hz 0.18

250 Hz 0.55

500 Hz 1.1

750 Hz 2.2

1000 Hz 3.8

2000 Hz 7

3000 Hz 2.6

4000 Hz 1.7

5000 Hz 1.2

6000 Hz 1

8000 Hz 0.7

The first five measurement have the same Fo but the Q factor is different.

The graph below shows the result from measurement 1-5

Determine the series resonant frequency and Q factor for each measurement, where,

0

10

20

30

40

50

60

0 2000 4000 6000 8000 10000

Series1

Series2

Series3

Series4

Series5


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Series resonant frequency occurs at the maximum value of V-out and is definedas fo------------------------1

o = 2 fo -----------------2 Q= 2 foL/R--------------3

Solution

1. fo= .L= 100mH

C=100nF

= 12 x 0.001=10.0006283=1591.6

2. o= 2 x 1591.6 = 10000.323. Measurement 1Q = (2 x 1591.6 x 0.1)200= 1000.032200 = 5.00016

Measurement 2 Q when R= 100, Q= (2 x 1591.6 x 0.1) 100 =

10.0003

Measurement 3 Q when R = 50, Q= (2 x 1591.6 x 0.1) 50 = 20.0006

Measurement 4 Q when R = 25, Q= (2 x 1591.6 x 0.1) 25 = 40.001

ii) Use the following parameters t plot/measure the magnitude and phase frequencyresponse for

L C R min max





Measurement (5) will get the same result with measurement (1).

Measurement 6

Frequency V output

40Hz 1.4

80Hz 2.9

150Hz 5.5

200Hz 7.6


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300Hz 12.4

400Hz 18.9

600Hz 46.7

800Hz 141.2

1000Hz 56.2

2000Hz 13.3

3000Hz 8.03

4000HZ 5.8

The graph below shows the result for measurement 6

Measurement 7

Frequency V output

20Hz 1.4

60Hz 4.5

80Hz 5.92

100Hz 7.6

150Hz 12.4

200Hz 18.9

300Hz 46.7

400Hz 141.2

700Hz 23.4

900Hz 15.4

1000Hz 13.3

2000Hz 5.9

The graph below show the result for measurement 7


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Measurement 8

Frequency V output

10Hz 1.4

40Hz 5.9

60Hz 9.4

80Hz 13.5

150Hz 46.7

200Hz 141.2

300Hz 32.6

400Hz 18.5

500Hz 13.3

600Hz 10.5

700Hz 8.7

800Hz 7.5

900Hz 6.6

1000Hz 5.9

The graph below shows the result for measurement 8


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iii) Determine the series resonant frequency and Q factor for each measurement. Measurement 6

Fo=

o =2fo = 2 x 795.78 = 5000.03 = 5000

Measurement7

= Wo =2fo= 2 x 397.887 = 2499.998

Q=

Measurement 8

Fo =

Wo =2fo =2 x 198.94= 1249.97

Q=

Given

0

5

10

15

20

25

30

35

40

45

0 1000 2000 3000 4000 5000 6000

M5

M6

M7

M8


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To get rid of the j we take the square root

H (jw) =

We take frequency to be 1000 Hz

From measurement 1.Fo= 1591.6

Q= 5.0

* +

For measurement 2Fo = 1591.6

F= 1000

Q= 10

* +

For measurement 3Fo =1591.6

F= 1000

Q= 20

* +

For measurement 4Fo =1591.6

F= 1000

Q= 40

* +

Measurement 5 will get the same result with measurement 1 because it has the same

values.

For Measurement 6


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Fo = 795.78

F = 1000

Q= 5

* +

For Measurement 7Fo = 397.887

F = 1000

Q= 4.99

* +

=0.0943

For measurement 8Fo = 198.94

F = 1000Q= 4.9999

* +

= 0.041

Comparison frequency is 1000 HzMeasurement Measured value(Vout/Vin) Calculated value

1 0.28 0.2033

2 0.14 0.1033

3 0.07 0.051

4 0.038 0.0259

5 0.28 0.2033

6 0.562 0.398

7 0.133 0.10

8 0.059 0.041

The result shows that the measured value is close to the calculated value.


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8. Derive equation 4 from first principles. Hint: Vout =

( )

( )

PART B

a) Investigate and explain what is meant by Bows NotationBows notation is a graphical process ofrepresenting coplanar forces and stresses, using

alphabetical letters, in the solution of stresses or in determining the resultant of a

system of concurrent forces.

Solution

Sum of upward forces = sum of downward forces

R1+R2= 24+30+20 ------ (a)

4 x R2= 30 x (3) + 20 x (2) + 24 x (1)

4R2= 24+40+90

4R2= 154

R2=

To get R1, substitute value of R2 into ---- (a)

R1 +38.5 = 24+30+20

R1= 74 38.5

R1= 35.5

Check for equilibrium 1= 1

35.5 + 38.5 = 24+ 30 + 20

74= 74


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b) Figure 2 shows a free body diagram with all adjacent members of the truss spaced at 60degrees.

24KN 30Kn

C

D

B F

E G

2 A 2

R1 Y 20KN X R2

Diagram of split drawing;

1. B 2. C

E B F

3. 30KN 4. D

G D X R2

5. E G

Y X


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B

X

F C

E Y

D

G

c) Figure 3 shows two masses connected via a frictionless pulley.The system held stationary by the force located at F when M1=7kg and M2 = 9kg

Calculate the following when F is removed; then the tension is in the connecting line.

M1= T-7g = 7A

T-(7x9.81) =7A

T= 68.67 +7A

M2 = 9g-T = 9A

(9x9.81)-T=9A

88.29-T=9A

T=88.29-9A

From these equations the tension is still unknown.

A=

T=68.67+(7x1.226)=68.67 + 8.582=77.252N

M1

M2


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The tension in the connecting line =77.252

The acceleration and the velocity of M2 after 500msThe acceleration is the same. The acceleration of M2 after 500s= 1.226m

V=A x T=1.226x0.5= 0.613mThe velocity of M2 after 500ms= 0.613 m/s

The distance moved by M1 after 200msD=0.5 x A x = 0.5 x 1.226 x 0. = 0.5 x 1.226 x 0.04 = 0.5 x 0.04904 = 0.02452mThe distance moved by M1 after 200s = 0.02452m

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Engineering principle