MECE 3320

MECE 3320 – Measurements & Instrumentation

Measurement System Behavior

Dr. Isaac ChoutapalliDepartment of Mechanical Engineering

University of Texas – Pan American

MECE 3320Measurement System Behavior+

+pioneer.netserv.chula.ac.th/~tarporn

MECE 3320Response of a Seismic Accelerometer

The response can be modeled by the following differential equation:

kxdtdxcky

dtdyc

dtydm 2

2

Most measurement systems can be modeled as: Zero-Order First-Order Second Order

MECE 3320

The behavior is characterized by its static sensitivity, K and remains constant regardless of input frequency (ideal dynamic characteristic).

In zero-order systems, the measurement system responds to an input instantly.

It is useful for static inputs or static calibration.

Dynamic signals can also be measured but only at equilibrium conditions.

Zero-order system model is represented by:

where K = static sensitivity = b0/a0)()( 00 txbtya )()( tKxty

Zero-Order Systems

MECE 3320Pencil-Type Pressure Gauge

AppKy atm )(

))(/( atmppKAy Zero-order response equation is given by:

MECE 3320

forcing function)()()( tKxtydt

tdy

First-Order Systems: Step Input

A first-order system is a measurement system that cannot respond to a change in input instantly.

ycf ypi

000

)()(tAt

tAUtx

)()()( tKAUtydt

tdy

Transient response

Steady state response

KACety t /)(

MECE 3320

/

)0()( te

KAyKAty

Suppose we rewrite the equation as

/

0 )()0()()()()( t

m eyyyty

KAyKAtyte

First-Order Systems: Step ResponseSteady-state

responseError Fraction

63.2% response to input response

90% 99%

MECE 3320

/

)0()( te

KAyKAty

tee mm log3.2ln

Slope = -1/

/

)0()( t

m eKAyKAtye

First-Order Systems: Step Response

Is called the time-constant – time it takes for the measurement system to respond to 63.2% of the input signal.

MECE 3320

/)0()(gives,equation aldifferenti above theSolving

)()(obtain weRewriting,

)()(

t

s

v

sv

eTTTtT

TtTdt

tdThAmc

tTThAdt

tdTmc

QdtdE

First-Order Systems: Step Response

Thermometer & Energy Balance: Energy Balance:

Assumptions: Uniform temperature within the bulb (lumped analysis) Constant mass

MECE 3320

Periodic signals are encountered in many applications. Some examples are:

First-Order Systems: Frequency Response

Vehicle Suspension System

Reciprocating Pumps

Pulsed Detonation Engines

MECE 3320

Consider a first-order measuring system to which an input represented by the following equation is applied.

tKAydtdy sin

tAtx sin)(

The complete solution:

1

2

/ tansin)(1

)(

tKACety t

Transient response

Steady state response

)(sin)()( / tBCety t

2/12)(1)(

KAB 1tan)(

Frequency response=

First-Order Systems: Frequency Response

Amplitude of steady state response

Phase-Shift

MECE 3320

The steady-state response of any system to which a periodic input of frequency, , is applied is known as the frequency response of that system.

First-Order Systems: Frequency Response

Time-lag

Amplitude-lag input

output

MECE 3320

The phase angle is )(tan)( 1 2/121

1)(

KABM

Dynamic error, () = M() - 1: a measure of an inability of a system to adequately reconstruct the amplitude of the input for a particular frequency.

First-Order Systems: Frequency Response

The ratio of the output signal amplitude to the signal amplitude is called magnitude ratio.

0.707 -3dB

dynamic error

Cut-off frequency

MECE 3320

A first order instrument is to measure signals with frequency content up to 100 Hz with an accuracy of 5%. What is the maximum allowable time constant? What will be the phase shift at 50 and 100 Hz?

%10011

1error Dynamic22

From the condition |Dynamic error| < 5%, it implies that 05.11

195.022

But for the first order system, the term can not be greater than 1 so that the constrain becomes

1/1 22

Solve this inequality give the range 33.00 The largest allowable time constant for the input frequency 100 Hz is

The phase shift at 50 and 100 Hz can be found from arctan

This gives = -9.33o and = -18.19o at 50 and 100 Hz respectively

11

195.022

ms 52.0Hz1002

33.0

First-Order Systems

MECE 3320

where

= the static sensitivity

= the damping ratio, dimensionless

= the natural angular frequency

0

0

bKa

0

2n

aa

1

0 22aa a

)()()()(0012

2

2 txbtyadt

tdyadt

tyda

Second order systems are modeled by second order differential equations.

)()()(2)(12

2

2 tKxtydt

tdydt

tyd

nn

Second-Order Systems

MECE 3320

The solution to the second order differential equation depends on the roots of the characteristic equation.

0121 22 DD

nn

This quadratic equation has two roots:

122,1 nnS

Overdamped ( > 1):

Critically damped ( = 1):

Underdamped (< 1):

Depending on the value of , three forms of complementary solutions are possible

tCety nt

ocn 21sin)(

tt

ocnn eCeCty

1

2

1

1

22

)(

ttoc

nn teCeCty 21)(

Second-Order Systems

MECE 3320Second-Order Systems

Animation courtesy of Dr. Dan Russell, Kettering University

Undamped & Underdamped Second Order-Systems

MECE 3320

For a step input x(t)

Initial conditions: y = 0 at t = 0, dy/dt = 0 at t = 0

Solution:

Overdamped ( > 1):

Critically damped ( = 1):

Underdamped (< 1):

)(212

2

2 tKAUydtdy

dtyd

nn

112

112

1)( 1

2

21

2

2 22

tt nn ee

KAty

1)1()( t

nnet

KAty

11sin1

)( 2

2

teKA

tyn

tn 21 1sin

Second-Order Systems: Step Input

MECE 3320

21 ndRinging frequency:

ddT

2

Ringing period:

Second-Order Systems: Step Response

Underdamped systems ( < 1): Small rise time (90 % of input value), Large settling time

(10% of steady-state value, KA) Overdamped systems ( > 1): Large rise time, Small settling time Most measurement systems have damping ratios between 0.6 and 0.8.

MECE 3320

Typical response of the 2nd order system

Second-Order Systems: Step Response

MECE 3320

where

Let the input signal to the second-order system be of the form x(t) = Asint

)(sin/2/1

)()( 2/1222

tKAtytynn

c

)(sin)()(steady tBtyThe steady state response is given by:

2/1222 /2/1

)(nn

KAB

//

2tan)( 1

nn

//2tan)( 1

nn

The magnitude ratio is given by:

2/1222 /2/1

1)(nn

KABM

Second-Order Systems: Frequency Response

MECE 3320

n

.01 .1 1 10 100

Phas

e sh

ift,

-180

-160

-140

-120

-100

-80

-60

-40

-20

0

n

.01 .1 1 10 100

Ampl

itude

ratio

0.0

.5

1.0

1.5

2.0

Dec

ibel

(dB

)

6

3

0

-3-6-10-15

= 0.1

0.3

0.5

1.0

2.0

0 = 0.1

0.3

0.5

1.0

2.0

The phase shiftThe magnitude ratio

2/1222 /2/1

1)(nn

M

//

2tan)( 1

nn

Second-Order Systems: Frequency Response