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
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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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First-order System: Step Input
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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First-order System: Step Input
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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First-order System: Step Input
Time response of first order system to a step input function
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First-order System: Step Input
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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First-order System: Step Input
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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First-order System: Periodic Function
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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First-order System: Periodic Function
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
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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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Second-order Systems: Step Response
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Second-order Systems: Step Response
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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