Week 05 Systems and Response 123

7/26/2019 Week 05 Systems and Response 123

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Mechatronics (06-92-412)

System and Measurement System

Dynamics


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General Background

Each electromechanical system responds differently todifferent types of input signals

Different systems respond to a given input signal differently

A particular system may not be suitable for measuringcertain signals


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General Background


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Electromechanical Actuator DC Motor

Example

Electrical systemMechanical system

)(in teKIRIL baaaa

Lam TIKbJ

Complete DC motor model3rd-order, coupled linear system


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Dynamic Characteristics of System

A measurement system can be expressed as OrdinaryDifferentiate Equations (ODE):

where

n = order of the system,x = input,

y = output,

t = time,

as= constant coefficientswhich depend on the characteristics of the measurement system.

A more general form (when subjected to a general forcing function F(t)) is

a d y

dta

d y

dta

dy

dta y bxn

n

n n

n

n o

1

1

1 1

a d y

dt

a d y

dt

a dy

dt

a y F t n

n

n nn

n

n o

1

1

1 1...


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Zero-order Systems

Zero order system is independent of time, i.e., n = 0

Output responds instantaneously with any input change

aoy = F(t)

y = (1/ao) F(t)

Output = Constant x Input

a d y

dta

d y

dta

dy

dta y F t n

n

n nn

n

n o

1

1

1 1...


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Zero-order Systems

Example #1 Linear potentiometer

Eo= (l/L) Ei


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Zero-order Systems

Responds instantaneously with respect to any changes in theinput.

Independent of time

Mostly used for static measurements

Example #2

A tire pressure gage with negligible inertia or piston mass.


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First order systems

Measurement systems that do not respond instantaneously tochanges in input and contains a storage elements

Example: Bulb-thermometer: The temperature of the bulbdoes not increase instantaneously. First the bulb exchange

energy with its environment until it goes to equilibrium

System with storage or dissipative behavior but no inertialforces can be modeled

a d y

dta

d y

dta

dy

dta y F t n

n

n nn

n

n o

1

1

1 1...


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First order systems

a1y + aoy = F(t), where y signifies dy/dt, y is Dividing by ao

y + y = K F(t)

where is called the time constant= a1/ao

a d y

dta

d y

dta

dy

dta y F t n

n

n nn

n

n o

1

1

1 1...


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First order systems: Examples

Simple R-C (resistance-capacitance) circuits. Temperature sensors which works by using thermal capacitance

and resistance.

Mechanical systems that have friction and springs, but with

negligible inertial effects of mass

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First order systems

a1y + aoy = F(t) Dividing by ao [a1/ ao] y + y = [1 / ao] F(t)

y + y = Kx = a1/ao, and K = 1/ao Where, time constant

Time constant shows how quickly a first order system will

response with change in input


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First order systems


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First-order System: Step Input

Consider step input is given.

Step function, AU(t) is expressed as

AU(t) = 0 when t 0-,= A when t 0+.


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General governing equation y + y = K F(t)

Setting F(t) = AU(t)

y + y = K AU(t) Solving the above first order differential equation Boundary condition y(0)=y0

Solving for t>0

+

Y(t) = KA + (y0 - kA) e

-t/

Y(t) = C e-t/+ KA


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The solution of y(t) gives the time response of the system toa step change in input

The value y(t) is the output value indicated by the displaystage of the measurement system

Y(t) = KA + (y0 - kA) e-t/

Time response Steady response Transient response


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Time response of first order system to a step input function


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At t = , Y()= KA

Error fraction of the output signal

(t) = [y(t)y

] / [y0y

] = e (-t/)

Y(t) = KA + (y0 - kA) e-t/


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Time constant shows how quickly a first ordersystem will response with change in input

Time is the time required for a first order system

to achieve 63.2% of the step change magnitude


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First-order System: Periodic Function

Input

Recall periodic function: example: sine wave Consider in a first-order measuring system a input of

periodic signal in the form of F(t) = A sin t is applied

y + y = K F(t)

y + y = K A sin t


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Input

This could be alternatively written as

y(t) = C e-t/+ B() sin ( t + )

where B() = K A / [1 + ()2]1/2

() = - tan-1() (phase shift)

Magnitude ratio,

M() = B/(KA) = 1/[1 + ()2]1/2


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First-order System:

Periodic Function Input

Input

Output

Phase lag

Amplitude lag


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Input

Magnitude ratio Phase shift


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Filtering

Filteringprocess of attenuating unwanted components of ameasurand while permitting the desired components topass.

The two basic classes of filters are:

1) active- uses powered components, commonlyconfigured of op amps.

2) passive- made up of some form of RLCarrangement.


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Filters Types

Passive filters are circuits made up of resistors, capacitors,and inductors.

Active filters incorporate operational amplifiers.

The sharp cut off of an ideal filter can not be realized.

Roll off designated in decibels per decade

Phase shift between input and output

Filter design is based on its cut off frequency, which is the

frequency where the signal power is reduced to , which isequivalent to m(w) = 0.707

Decibels; db = 20 log m(w) = 20 log (0.707) = -3 db


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Filters Types


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Butterworth Low Pass Filter A simple passive low-pass filter can be constructed by using the

resistor and capacitor (RC) circuit.

Capacitor

block low-frequency currentsand pass high-frequency

currentsshort-circuit the high-frequency components of the inputsignal.


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Filters Types


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High-Pass RC Filter

We can interchange the resistor and capacitor to convertfrom low-pass to high-pass RC filter.

Signal Conditioning: Filtering


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Signal Conditioning: Filtering

L = Inductance


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Filter example

Design a one-stage Butterworth RC low-pass filter with acutoff frequency of 100 Hz at 3 dB if the source and load

impedances are 50 . Calculate the expected dynamic errorand attenuation at 192 Hz in the realized filter.


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Solution

A single-stage low-pass Butterworth RC filter circuit is just afirst-order system with time constant = RC. With therelation to = 2f, and the magnitude ratio given by


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Second order system

Second order systems are modeled by second orderdifferential equations.

a2d2y/dt2+ a1dy/dt + aoy = F(t)

Divide by ao

(a2/ao) d2y/dt2+ (a1/ao) dy/dt + y = (1/ao) F(t)

a d ydt

a d y

dta dy

dta y F t n

n

n nn

n

n o

1

1

1 1...


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Second order system

(a2/ao) d2y/dt2+ (a1/ao) dy/dt + y = (1/ao) F(t)

the undamped natural frequency,n= (ao/a2),the damping ratio, = a1/ [2(aoa2)]

1 d y

dt

2 dy

dty Kx

2

2

n

n2

F(t)


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Second order system

Consider the characteristics equation

The roots of the quadratic equation

And depending on the value for three forms ofhomogenous solutions are possible


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2ndorder electrical system

0

LC IeC

)(in teeRIIL CLL


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Second order system- spring mass

damper

Newtons 2ndlaw, F = m (d2y/dt2),

m d2y/dt2+ c dy/dt + ky = F(t).


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Second order system

Comparing with the governing equation

1 d y

dt

2 dy

dty Kx

2

2

n

n2

m d2y/dt2+ c dy/dt + ky = F(t).

K = 1/k,n= (k/m), = c / 2(km).


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Second order system- step function

2ndorder system

Where, K = 1/k,

n= (k/m), = c / 2(km).

1 d y

dt

2 dy

dty Kx

2

2

n

n2


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Second order system

>1 (over-damped)

=1 (critically damped)


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Second-order Systems: Step Response

For a step input

>1 (over-damped)

=1 (critically damped)


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For under damped system the transient system is oscillatoryabout the steady state value

Ringing period

Ringing frequency

Optimum settling time

can be obtained at

Practical use


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Second-order Systems: Periodic Input

Simple periodic function input

The response is given by

Frequency dependent phase shift

Steady state response for sinusoidal input

F(t)


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Week 05 Systems and Response 123

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