ME411Engineering Measurement
& Instrumentation
Winter 2017 – Lecture 5
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Model of a Measurement System
• Consider the lumped parameter model
• Coefficients represent physical parameters – obtained by considering governing equations
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Example
• With this knowledge, we could predict the output with knowledge of the input!!
• This will then allow us to better design our measuring system…
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Model of a Measurement System
• Fortunately, most measurement systems can be modelled using zero, 1st or 2nd order linear ODEs
• Even complicated systems can usually be thought of as a combination of these simpler cases
• Zero-order system
• 1st order system
• 2nd order system
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Zero-order system
• System responds instantly!
• K is the static sensitivity (gain)
• Used to model the non-time
dependent system response to
static inputs!
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K = 1/ao = Static sens
Zero-order system
6From Doebelin - Measurement Systems Application and Design 5th ed
1st-order system
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• Measurement systems that have a storage system
• = a1/ao is called the time constant – will always have a dimension of time
• Let’s now consider some special types of input signals and observe how the system responds…
K = 1/ao = Static sens
1st-order system – Step input
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From Doebelin - Measurement Systems Application and Design 5th ed
1st-order system – Step input
• Time constant, • Time required for system to achieve 63.2% of
• Property of the system!
• Error fraction
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inf 0y y
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Assumptions made in thermometer example
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1st-order system – Ramp input
14From Doebelin - Measurement Systems Application and Design 5th ed
1st-order system – Example
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1st-order system – Impulse
16From Doebelin - Measurement Systems Application and Design 5th ed
1st-order system – General Periodic
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where C is based on initial conditions!
NOTE: Both B and are frequency dependent!
Dynamic error:
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2nd-order system
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• Measurement systems that have inertia
K = 1/ao = Static sens
2nd-order system
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• As usual, we can’t solve a non-homogenous 2nd order ODE without first solving the homogenous case determines the transient response of the system
• damping ratio internal energy dissipation
• Depending on the value of , three solutions are possible:
2nd-order system
• < 1: oscillation!
• 1: no oscillation
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Complex roots
Real, repeated roots
Real, unrepeated roots
2nd-order system – Step input
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With initial conditions:
(0) (0) 0y y
2nd-order system – Step input
• 2nd order time-constant
• Rise time: Time to first reach 90% of (KA-y0)
• Settling time: Time for output to reach ±10% of KA
• Best damping?? 0.6 – 0.8, but depends…
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“ringing frequency” independent of input signal
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n
2nd-order system – Ramp input
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From Doebelin - Measurement Systems Application and Design 5th ed
2nd-order system – Terminated ramp input
25From Doebelin - Measurement Systems Application and Design 5th ed
2nd-order system – Impulse
26From Doebelin - Measurement Systems Application and Design 5th ed
2nd-order system – General Periodic
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( ) sin( )F t A t
2nd-order system – General Periodic
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2nd-order system – General Periodic
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Image credits
• All images from Figliola and Beasley, Mechanical Measurements 5th edition unless otherwise stated
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