Electronic InstrumentationExperiment 3

* Part A: Instrumented Beam as a Harmonic Oscillator

* Part B: RC and RLC Circuits – Examples of AC Circuits

2

Part A

Harmonic Oscillators Analysis of Cantilever Beam Frequency

Measurements

3

Macroscopic Examples of Harmonic Oscillators Spring-mass combination Violin string Wind instrument Clock pendulum Playground swing LC or RLC circuits Others?

4

Large Scale Examples

Petronas (452m) and CN (553m) Towers

5

Harmonic Oscillator

Equation

Solution x = Asin(ωt)

x is the displacement of the oscillator while A is the amplitude of the displacement

022

2

xdt

xd

6

Spring

Spring Force

F = ma = -kx Oscillation Frequency

This expression for frequency hold for a massless spring with a mass at the end, as shown in the diagram.

m

k

7

Spring Model for the Cantilever Beam

Where l is the length, t is the thickness, w is the width, and mbeam is the mass of the

beam. Where mweight is the applied mass and

a is the length to the location of the applied mass.

8

Spring Model -- Continued

For a beam loaded with a mass at the end, a is equal to l. For this case:

where E is Young’s modulus of the beam. Since real beams have finite mass, it is

necessary to use the equivalent mass at the end that would produce the same frequency response. This is 0.23mbeam.

3

3

4l

Ewtk

9

Spring Model -- Continued

Now we can apply the expression for the ideal spring mass frequency to the beam.

Where the mass used now is given by

k

mf

T o ta l

22

m m mT o ta l b ea m ex tra 0 2 3.

10

Atomic Force Microscopy -AFM

This is one of the key instruments driving the nanotechnology revolution

Dynamic mode uses frequency to extract force information Note Strain Gage

11

AFM on Mars

Redundancy is built into the AFM so that the tips can be replaced remotely.

12

AFM on Mars

Soil is scooped up by robot arm and placed on sample. Sample wheel rotates to scan head. Scan is made and image is stored.

13

AFM Image of Human Chromosomes

There are other ways to measure deflection.

14

AFM Optical Pickup

On the left is the generic picture of the beam. On the right is the optical sensor.

15

Our Experiment

For our beam, we must deal with the beam mass, the extra mass of the magnet and its holder (for the magnetic pick up coil), and any extra load we add to the beam to observe how its performance depends on load conditions.

16

Our Experiment

For simplicity, call m the combination of 0.23 times the beam mass plus the magnet and its holder.

To obtain a good measure of k and m, we will make 4 measurements of oscillation frequency using 3 extra masses.

m m mb ea m m a g n e t h o ld er 0 2 3.

17

Our Experiment

We will then have 4 equations and 2 unknowns.

One we obtain values for k and m we can plot to see how we did.

k m f o 22 k m m f 1 1

22

k m m f 2 2

22 k m m f 3 3

22

fk

m m lo a d

1

2

18

Our Experiment One method is to guess values for k and m

and then keep adjusting them until we obtain a good fit, obtaining something like:

5 10 15 200

50

100

150

200

250

300

350

400

450

500

19

Our Experiment

How to guess values for k and m?• Estimate the mass by measuring the beam dimensions

and assuming the beam, magnet and holder are made from something reasonable.

• Use your guess for m with your largest load mass to obtain a guess for k.

• Plot f as a function of load mass

• Change values of k and m until your function and data match.

20

Our Experiment

Can you think of other ways to more systematically determine k and m?

Experimental hint: make sure you keep the center of any mass you add as near to the end of the beam as possible. It can be to the side, but not in front or behind the end.

Magnet

C-Clamp

21

MEMS Accelerometer

An array of cantilever beams can be constructed at very small scale to act as accelerometers.

Note Scale

22

23

24

25

Hard Drive Cantilever

The read-write head is at the end of a cantilever. This control problem is a remarkable feat of engineering.

26

More on Hard Drives

A great example of Mechatronics.

27

Modeling Damped Oscillations

v(t) = A sin(ωt)

Time

0s 5ms 10ms 15msV(L1:2)

-400KV

0V

400KV

28

Modeling Damped Oscillations

v(t) = Be-αt

29

Modeling Damped Oscillations

v(t) = A sin(ωt) Be-αt = Ce-αtsin(ωt)

Time

0s 5ms 10ms 15msV(L1:2)

-200V

0V

200V

30

Finding the Damping Constant

Choose two maxima at extreme ends of the decay.

31

Finding the Damping Constant

Assume (t0,v0) is the starting point for the decay.

The amplitude at this point,v0, is C. v(t) = Ce-αtsin(ωt) at (t1,v1):

v1 = v0e-α(t1-t0)sin(π/2) = v0e-α(t1-t0)

Substitute and solve for α: v1 = v0e-α(t1-t0)

32

Part B

LC and RLC Circuits Intro to AC Circuits

33

Inductors

An inductor is a coil of wire through which a current is passed. The current can be either AC or DC.

34

Inductors

This generates a magnetic field, which induces a voltage proportional to the rate of change of the current.

dt

dILV L

L

35

Inductors

Inductors are also electromagnets. The magnetic field produced by the

inductor current can do work. There is, therefore, energy stored in the

magnetic field of the inductor.

W L IL 1

22

36

Combining Inductors

Inductances add like resistances Series

Parallel

L L L L N 1 2 . . .

1 1 1 1

1 2L L L L N

. . .

37

Inductor Impedance

Note that this behavior is exactly the opposite of capacitors.

frequencyfrequency

approachesimpedance approaches

looks like called

low zero zero   short circuit

high infinity infinity   open circuit

38

Filters

Even simple circuits can pass some frequencies and reject others. Thus, by selecting the right kind of circuit, certain frequencies can be filtered out.

39

Oscillator Analysis

Spring-Mass W = KE + PE KE = kinetic

energy=½mv² PE = potential

energy(spring)=½kx² W = ½mv² + ½kx²

Electronics W = LE + CE LE = inductor

energy=½LI² CE = capacitor

energy=½CV² W = ½LI² + ½CV²

40

Oscillator Analysis

Take the time derivative

Take the time derivative

dt

dxxk

dt

dvvm

dt

dW22 2

12

1

dt

dxkx

dt

dvmv

dt

dW

dt

dVVC

dt

dIIL

dt

dW cc22 2

12

1

dt

dVCV

dt

dILI

dt

dW cc

41

Oscillator Analysis

W is a constant. Therefore,

Also

W is a constant. Therefore,

Also

0dt

dW0

dt

dW

dt

dxv

2

2

dt

xd

dt

dva

dt

dVCI c

C

I

dt

dVc

2

2

dt

VdC

dt

dI c

42

Oscillator Analysis

Simplify Simplify

kxvdt

xdmv

2

2

0

02

2

xm

k

dt

xd

C

ICV

dt

VdLIC c

c 2

2

0

01

2

2

cc V

LCdt

Vd

43

Oscillator Analysis

Solution Solution

m

k

LC

1

x = Asin(ωt) Vc= Asin(ωt)

44

Using Conservation Laws

Please also see the write up for Experiment 3 for how to use energy conservation to derive the equations of motion for the beam and voltage and current relationships for inductors and capacitors.

Almost everything useful we know can be derived from some kind of conservation law.

45

Thevenin Voltage Equivalents+

Recall that the function generator (aka waveform generator) has an internal impedance of 50 Ohms. Thus, the circuit representation of this device looks like:

+Required for HW #3

Vs

FREQ = VAMPL = VOFF =

0

Rs

50

46

Thevenin

Such a simple combination of a voltage source and resistance is called a Thevenin source.

It is generally possible to simplify any complex source into such a combination.

In fact, the function generator is much more complex, but it can be most simply represented this way.

47

Thevenin

ExampleVs

FREQ = VAMPL = VOFF = RL

Rs

0

Load ResistorRth

Vth

FREQ = VAMPL = VOFF =

0

RL

48

Thevenin

We might also see a circuit with no load resistor, like the voltage divider.

R2

0

R1

Vs

FREQ = VAMPL = VOFF =

49

Thevenin Method

Find Vth (open circuit voltage)• Remove load if there is one so that load is open• Find voltage across the open load

Find Rth (Thevenin resistance)• Set voltage sources to zero (current sources to open) –

in effect, shut off the sources• Find equivalent resistance from A to B

Vth

FREQ = VAMPL = VOFF =

0

RL

Rth

A

B

50

Thevenin Method Example

Remove Load

RL

0

Vo

0Vdc

R4R2

R1 R3

0

Vo

0Vdc

R4R2

R1 R3

A AB B

51

Thevenin Method Example

Let Vo=12, R1=2k, R2=4k, R3=3k, R4=1k

V Bk

k kV

1

1 31 2 3

V Ak

k kV

4

4 21 2 8

V th V A V B V 8 3 5

52

Thevenin Method Example

Short out the voltage source (turn it off) & redraw the circuit for clarity.

0

R4R2

R1 R3

A BR3 R4

R2R1A

B

53

Thevenin Method Example

First find the parallel combinations of R1 & R2 and R3 & R4.

Then find the series combination of the results.

RR R

R R

k k

k k

kk1 2

1 2

1 2

4 2

4 2

8

61 3 3

.

RR R

R R

k k

k k

kk3 4

3 4

3 4

1 3

1 3

3

40 7 5

.

54

Thevenin Method Example

The Thevenin Voltage Source is Then:

R th R R k k

1 2 3 44

3

3

42 1.

0

Vth

5V

Rth

2.1k

55

Thevenin Method

Note• When a short goes across a resistor, that resistor

is replaced by a short.• When a resistor connects to nothing, there will

be no current through it and, thus, no voltage across it.

R

56

Thevenin Applet (see webpage)

Test your Thevenin skills using this applet from the links for Exp 3

57

Thevenin

To confirm that the Thevenin method works, add a load and check the voltage across and current through the load to see that the answers agree whether the original circuit is used or its Thevenin equivalent.

If you know the Thevenin equivalent, the circuit analysis becomes much simpler.

58

Thevenin Method Example

Checking the answer with PSpice

Note the identical voltages across the load.

R3

3kVth

5Vdc

0

7.448V

0V

4.132V

3.310VVo

12VdcRload

10k

R1

2k

12.00V

RL

10kR2

4k

5.000V Rth

2.1k

R4

1k

0