ME411 Engineering Measurement & Instrumentation

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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