Measurement System Behavior

Measurement System Behavior

2141-375 Measurement and Instrumentation

Dynamic Characteristics

Dynamic characteristics tell us about how well a sens or responds to changes in its input. For dynamic signal s, the sensor or the measurement system must be able to respond fast enough to keep up with the input signals.

Sensor or

system

Input signalx(t)

Output signaly(t)

In many situations, we must use y(t) to infer x(t), therefore a qualitative understanding of the operation that the sensor or measurement system performs is imperative to understand ing the input signal correctly.

General Model For A Measurement System

n th Order ordinary linear differential equation with constan t coefficient

Where m ≤ ny(t) = output from the systemx(t) = input to the system

t = timea’s and b’s = system physical parameters, assumed constant

)()()()(

)()()()(

011

1

1011

1

1 txbdt

tdxb

dt

txdb

dt

txdbtya

dt

tdya

dt

tyda

dt

tyda

m

m

mm

m

mn

n

nn

n

n ++++=++++ −

−

−−

−

− LL

Measurement system

x(t) y(t)

y(0)

F(t) = forcing function

The solution

Where yocf = complementary- function part of solutionyopi = particular-integral part of solution

opiocf yyty +=)(

Complementary-Function Solution

The solution yocf is obtained by calculating the n roots of the algebraic characteristic equation

11 1 0... 0n n

n na D a D a D a−−+ + + + =Characteristic equation

Roots of the characteristic equation: 1 2, ,..., nD s s s=

1. Real roots, unrepeated:

2. Real roots, repeated: each root s which appear p times

3. Complex roots, unrepeated:the complex form: a ± ib

4. Complex roots, repeated:each pair of complex root which appear p times

stCe

( )2 10 1 2 1... p st

pC C t C t C t e−−+ + + +

sin( )atCe bt φ+

20 0 1 1 2 2

11 1

[ sin( ) sin( ) sin( )

... sin( )]p atp p

C bt C t bt C t bt

C t bt e

φ φ φ

φ−− −

+ + + + +

+ + +

Complementary-function solution:

Complementary-Function Solution

Case 1: Real roots, unrepeated:ex: -1.7, 3.2, 0

Case 2: Real roots, repeated:ex: -1, -1, 2, 2, 2, 0, 0

Case 3: Complex number, unrepeated:ex: -3 ± j4, 2 ± j5, 0 ± j7

Case 4: Complex number, repeated:ex: -3 ± j2, -3 ± j2, -3 ± j2

Particular Solution

Method of undetermined coefficients:

•After a certain-order derivative, all higher deriva tives are zero.•After a certain-order derivative, all higher deriva tives have the same functional form as some lower-order derivatives.

•Upon repeated differentiation, new functional forms continue to arise.

Where f(t) = the function that describes input quantityA, B, C = constant which can be found by substituting yopi into ODEs

Important Notes

...)()()( +′′+′+= tfCtfBtAfyopi

��

�

Zero-order Systems

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

A linear potentiometer used as position sensor is a zero-order sensor.

Vr

xm

x = 0y = V

-

+ here, /r r m

m

xV V K V x

x= ⋅ =

Where 0 ≤ x ≤ xm and Vr is a reference voltage

All the a’s and b’s other than a0 and b0 are zero.

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

Where K = b0/a0 is the static sensitivityτ = a1/a0 is the system’s time constant (dimension of time)

All the a’s and b’s other than a1, a0 and b0 are zero.

First-Order Systems

)()()(

001 txbtyadt

tdya =+

)()()(

tKxtydt

tdy=+τ

q

Sensorm, Ttf, C

Surfacearea A Ti

Thermometer based on a mass,m with specified heat, C

Consider a thermometer based on a mass m =ρVwith specified heat C (J/kg.K), heat transmission area A, and (convection heat transfer coefficient U(W/m2.K).

(Heat in) – (Heat out) = Energy stored

Assume no heat loss from the thermometer

( ) 0i tf tfUA T T dt VCdTρ− − =

tftf i

dTVC UAT UAT

dtρ + =

Therefore, we can immediately define K =1 and τ = ρ VC/UA

First-Order Systems

itftf TT

dt

dT=+τ

itftf TT

dt

dT

UA

VC=+

ρ

Time, t

-1 0 1 2 3 4 5

U( t

)

0

1

2

First-Order Systems: Step Response

The complete solution

Assume for t < 0, y = y0 , at time = 0 the input quantity, x increases instantly by an amount A. Therefore t > 0

yocfyopi

Applying the initial condition, we get C = y0-KA, thus gives

τ/0 )()( teKAyKAty −−+= 0>t

)()()(

tKAUtydt

tdy=+τ

Transient response

Steady state response

KACety t += − τ/)(


t/τ

0 1 2 3 4 5

Out

put S

igna

l, (

y(t)

-y0)

/(K

A-y

0)

0.0

.2

.4

.6

.8

1.0

0.632

t/τ

0 1 2 3 4 5

Err

or fr

actio

n, ΓΓ ΓΓ

0.0

.2

.4

.6

.8

1.0

0.368

Non-dimensional step response of first-order instru ment

τ/

)0(

)( teKAy

KAty −=−−

τ/

0

0 1)( te

yKA

yty −−=−−

Here, we define the term error fraction as

τ/

0 )()0(

)()()( teyy

yty

KAy

KAty −=∞−∞−

=−−

=Γ


Time constant, ττττ : the time required for a system to response 63.2% of a step change

Rise time, tr : the time required for a system to response 90% of a step change. (sometimes defined time for 10 to 90% o f a step change) ~ 2.3 ττττ

99.35

95.03

90.02.3

86.52

63.21

% responset/ττττ

First-order system response

Determination of Time constant

t/τ

0 1 2 3 4 5

Err

or fr

actio

n,

e m

.001

.01

.1

1

τ/

)0(

)( teKAy

KAty −=−−

0.368

τt

−=Γ=Γ log3.2ln

Slope = -1/2.3τ

τ/

)0(

)( teKAy

KAty −=−−

=Γ

Advantage- Check whether the system is

1st order- Avoid the influence of an

error in any one data point- eliminate the need to

accurately determine the 0.368 point

First-Order Systems: Ramp Response

The complete solution:

Applying the initial condition, gives

>

≤=

0

0 0)(

ttq

ttx

is&

Therefore

)()( / ττ τ −+= − teqKty tis&

Measurement error

Transient error

Steady state error

Assume that at initial condition, both y and x = 0, at time = 0, the input quantity start to change at a constant rate Thus, we haveisq&

)()()(

ttUqKtydt

tdyis&=+τ

)()( / ττ −+= − tqKCety ist

&

Transient response


ττ τis

tism qeqtx

K

tye && −=−= − /)(

)(

Initial condition: y(0) = 0

First-Order Systems: Ramp Response

t/τ

0 2 4 6 8 10

Out

put s

igna

l, y

/K

0

2

4

6

8

10

Non-dimensional ramp response of first-order instru ment

Steady state error = τisq&

τSteady state time lag =

Input x(t)

y(t)/K

First-Order Systems: Frequency Response

From the response of first-order system to sinusoidal inputs, we have

tKAydt

dyωτ sin=+

tAtx ωsin)( =

The complete solution: ( )ωτωωτ

τ 1

2

/ tansin)(1

)( −− −+

+= tKA

Cety t

Transient response


If we do interest in only steady state response of the system, we can write the equation in general form

[ ])(sin)()( / ωφωωτ ++= − tBCety t

[ ] 2/12)(1)(

ωτω

+=

KAB

ωτωφ 1tan)( −−=

Where B(ω) = amplitude of the steady state response and φ(ω) = phase shift

Frequency response=

Time (sec)

0.00 .01 .02 .03 .04 .05 .06

Inpu

t x(

t) a

nd O

utpu

t y(

t)

-2.0

-1.5

-1.0

-.5

0.0

.5

1.0

1.5

2.0 InputOutput


[ ])(sin)()( / ωφωωτ ++= − tBCety t

tAtx ωsin)( =

A

B

td

ωτ

.01 .1 1 10 100

Am

plitu

de r

atio

0.0

.2

.4

.6

.8

1.0

1.2

Dec

ibel

s (d

B)

0

-2

-4

-6-8-10

-20

ωτ

.01 .1 1 10 100

Pha

se s

hift,

φ (ω

)

-90

-80

-70

-60

-50

-40

-30

-20

-10

0


The phase angle isThe amplitude ratio1)(

1)(

2 +=

ωτωM )(tan)( 1 ωτωφ −−=

( )[ ] 2/121

1)(

ωτω

+==

KA

BM

Frequency response of the first order system

0.707-3 dB

Cutoff frequency

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

Dynamic error


Ex: Inadequate frequency responseSuppose we want to measure

With a first-order instrument whose τ is 0.2 s and static sensitivity K

Superposition concept:

For ω = 2 rad/s:

For ω = 20 rad/s:

Therefore, we can write y(t) as

tttx 20sin3.02sin)( +=

oo 8.2193.08.21116.0

1)rad/s 2( −∠=−∠

+=M

oo 7624.076116

1)rad/s 20( −∠=−∠

+=M

)7620sin()24.0)(3.0()8.212sin()93.0)(1()( oo −+−= tKtKty

)7620sin(072.0)8.212sin(93.0)( oo −+−= tKtKty

x(t)

y(t)/K


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

Solution:1

1)(

22 +=

τωωMDefine

( ) %10011

1%1001)(error Dynamic

22×

−

+=×−=

τωωM

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

195.0

22≤

+≤

τω

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

1/1 22 +τω

Solve this inequality give the range 33.00 ≤≤ωτ

The largest allowable time constant for the input frequency 100 Hz is

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

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

11

195.0

22≤

+≤

τω

ms 52.0Hz 1002

33.0==

πτ


Am

plitu

de r

atio

M(ωω ωω

)

1.05

0.95

ωωωωττττ

M(ωωωω) ≥≥≥≥ 0.95 region or δδδδ(ωωωω) ≤≤≤≤ 0.05 region


Example: A temperature measuring system, with a time constant 2 s, is used to measured temperature of a heating medium, which changes sinusoidally between 350 and 300oC with a periodic of 20 s. find the maximum and minimum values of temperature, as indicated by the measuring system and the time lag between the output and input signals

−+= o36.3120

2sin3.21325)( tty

πSolution: s 75.1=dt

T

325oC

350oC

300oC

x(t)

t

T

325oC

350oC

300oC

x(t), y(t)

t

td

Dynamic CharacteristicsExample: The approximate time constant of a thermometer is determined by immersing it in a bath and noting the time it takes to reach 63% of the final reading. If the result is 28 s, determine the delay when measuring the temperature of a bath that is periodically changing 2 times per minute.

Solution: s 69.6=dt

The essential parameters

= the static sensitivity

= the damping ratio, dimensionless

= the natural angular frequency

0

0

bK

a=

0

2n

a

aω =

1

0 22

a

a aζ =

Second-Order Systems

)()()()(

0012

2

2 txbtyadt

tdya

dt

tyda =++

In general, a second-order measurement system subjected to arbitrary input, x(t)

)()()(2)(1

2

2

2tKxty

dt

tdy

dt

tyd

nn

=++ωζ

ω

Consider the characteristic equation


0121 2

2=++ DD

nn ωζ

ωThis quadratic equation has two roots:

122,1 −±−= ζωζω nnD

Overdamped ( ζζζζ > 1):

Critically damped ( ζζζζ = 1):

Underdamped ( ζζζζ< 1): :

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

( )Φ+−= − tCety nt

ocn 21sin)( ζωζω

tt

oc

nn

eCeCtyωζζωζζ

−−−

−+−

+=1

2

1

1

22

)(

ttoc

nn teCeCty ωω −− += 21)(

Unrepeated real roots

Repeated real roots

Complex roots

t

H L

tAe σ−

)sin( φω +td


Case 2 Overdamped ( ζζζζ > 1):

Case 3 Critically damped ( ζζζζ = 1):

Case I Underdamped ( ζζζζ< 1):

d

nn

j

D

ωσ

ζωζω

±−=

−±−=

122,1 ( ) nD ωζζ 12

2,1 −±−=

nD ω−=2,1

t

H L

1=ζ

1>ζ

y(t)

y(t)

The complementary function solution determines the transient responses of the system

Second-order Systems: Step Response

For a step input x(t)

With the initial conditions: y = 0 at t = 0+, dy/dt = 0 at t = 0

The complete solution:

Overdamped ( ζζζζ > 1):

Critically damped ( ζζζζ = 1):

Underdamped ( ζζζζ< 1): :

)(21

2

2

2tKAUy

dt

dy

dt

yd

nn

=++ωζ

ω

KAeeKAtytt nn

+

−

−−+

−

−+−=

−−−

−+− ωζζωζζ

ζ

ζζ

ζ

ζζ 1

2

21

2

2 22

12

1

12

1)(

KAetKAty tn

n ++−= −ωω )1()(

( ) KAte

KAty n

tn

+

+−

−−=

−

φωζζ

ζω2

21sin

1)( ( )21 1sin ζφ −= −

ωnt

0 2 4 6 8 10

Out

put s

igna

l, y(

t)/K

A

0.0

.5

1.0

1.5

2.0


Non-dimensional step response of second-order instr ument

ζ = 0

0.25

0.5

1.0

2.0

21 ζωω −= ndRinging frequency:

Rise time decreases ζζζζ with but increases ringing

Optimum settling time can be obtained from ζζζζ ~ 0.7

Practical systems use 0.6< ζζζζ <0.8

ddT

ωπ2

=Ringing period


Time, t (s)

0 5 10 15 20

Out

put s

igna

l, y

(t)/

KA

0.0

.2

.4

.6

.8

1.0

1.2

1.4

settling time

rise time

overshoot

100% ±±±± 5%

Typical response of the 2 nd order system

Second-order Systems: Ramp Response

For a ramp input

With the initial conditions: y = dy/dt = 0 at t = 0+

The possible solutions:

Overdamped:

Critically damped:

Underdamped:

)(21

2

2

2ttUqKy

dt

dy

dt

ydis

nn

&=++ωζ

ω)()( ttUqtx is&=

−

−−+−+

−

−−−+−=

−+−

−−−

t

t

n

isis

n

n

e

eqK

tqKty

ωζζ

ωζζ

ζζ

ζζζ

ζζ

ζζζωζ

1

2

22

1

2

22

2

2

14

1212

14

12121

2)(

&&

+−−= − tn

n

isis

netqK

tqKty ωωω

)1

1(12

)(&

&

( )

+−

−−−=

−

φωζζζω

ζ ζω

teqK

tqKty n

t

n

isis

n2

21sin

121

2)(

&&

12

12tan

2

21

−

−= −

ζζζ

φ

Time, t (s)

0 2 4 6 8 10

Out

put s

igna

l, y

( t)/

K

0

2

4

6

8

10


Typical ramp response of second-order instrument

ζ = 0.3

0.61.0

2.0

Ramp input

Steady state error = n

isq

ωζ&2

Steady state time lag =

nωζ2

where

Second-order Instrument: Frequency Response

The response of a second-order to a sinusoidal inpu t of the form x(t) = Asinωωωωt

( )[ ] ( ){ } [ ])(sin/2/1

)()( 2/1222

ωφωωζωωω

++−

+= tKA

tyty

nn

oc

[ ])(sin)()(steady ωφωω += tBty

The steady state response of a second-order to a si nusoidal input

( )[ ] ( ){ } 2/1222 /2/1

)(

nn

KAB

ωζωωωω

+−=

−−= −

ωωωωζ

ωφ//

2tan)( 1

nn

−−= −

ωωωωζ

ωφ//

2tan)( 1

nn

Where B(ω) = amplitude of the steady state response and φ(ω) = phase shift

( )[ ] ( ){ } 2/1222 /2/1

1)(

nnKA

BM

ωζωωωω

+−==

Second-order Systems: Frequency Response

Magnitude and Phase plot of second-order Instrumentω/ωn

.01 .1 1 10 100

Am

plitu

de r

atio

0.0

.5

1.0

1.5

2.0

Dec

ibel

(dB

)

6

3

0

-3

-6-10-15

ζζζζ = 0.1

0.3

0.5

1.0

2.0

ω/ωn

.01 .1 1 10 100

Pha

se s

hift

, φ(ω)

-180

-160

-140

-120

-100

-80

-60

-40

-20

0 ζζζζ0 = 0.1

0.3

0.5

1.0

2.0

The phase angleThe amplitude ratio

( )[ ] ( ){ } 2/1222 /2/1

1)(

nn

Mωζωωω

ω+−

=ωωωω

ζωφ

//

2tan)( 1

nn −−

= −

Time, t (s)

0 5 10 15 20

Out

put s

igna

l, y

(t)/

KA

0.0

.2

.4

.6

.8

1.0

1.2

1.4

For overdamped (ζ >1) or critical damped (ζ = 1), there is neither overshoot nor steady-state dynamic error in the response. (Time response)

In an underdameped system (ζ < 1) the steady-state dynamic error is zero, but the speed and overshoot in the transient are related. (Time response)

pd

tπω

=

( )2exp / 1pM πζ ζ= − −

Rise time:

Peak time:

Maximum overshoot:

2

1

2 1rM

ζ ζ=

−Resonanceamplitude:

2 2where = , 1 , and arcsin( 1 )n d nδ ζω ω ω ζ φ ζ= − = −

Resonancefrequency:

21 2r nω ω ζ= −

Second-order Systems

Td

settling time

rise time

overshoot

peak time

10 ;5.28.0

~ <<+

ζω

ζ

nrt


Example: A pressure transducer has a natural frequency of 30 rad/s, damping ratio of 0.1 and static sensitivity of 1.0 µV/Pa. A step pressure input of 8x105 N/m2 is applied. Determine the output of a transducer.

Example: A second order instrument is subjected to a sinusoidal input. Undamped natural frequency is 3 Hz and damping ratio is 0.5. Calculate the amplitude ratio and phase angle for an input frequency of 2 Hz.

Amplitude ratio M(ω)= 1.152 and phase shifts = -0.875 rad =–50.2o.

Solution:

Solution:

V )]47.185.29sin(1[8.0)( 3 +−= − tety t


Example: An Accelerometer is selected to measure a time-dependent motion. In particular, input signal frequencies below 100 Hz are of prime interest. Select a set of acceptable parameter specifications for the instrument, assuming a dynamic error of ±5% and damping ratio ζ =0.7

ωn ≥ 1047 rad/sSolution:

Am

plitu

de r

atio

M(ωω ωω

)

ωωωω/ ωωωωn

1.05

0.95

1.05

0.95

Am

plitu

de r

atio

M(ωω ωω

)

ωωωω/ ωωωωn

Response of a General Form of System to a Periodic Input

The steady state response of any linear system to t he complex periodic signal can be determined using the frequency respon se technique and principle of superposition.

( )∑∞

=

++=1

000 sincos)(n

nn tnBtnAAtx ωωLet x(t)

The frequency response of the measurement system

Where KM(ωωωω) = Magnitude of the frequency response of the measur ement system and φφφφM(nωωωω0) = Phase shift of the measurement system at nωωωω0φφφφn(nωωωω0) = tan -1(An/Bn)

( ) ( ))()(sin)()( 00001

220 ωφωφωω nntnnKMBAKAty nM

nnn ++++= ∑

∞

=

( )( )∑∞

=

+++=1

0022

0 sin)(n

nnn ntnBAAtx ωφω

Response of a General Form of System of a Periodic Input

x(t) Linear system

y(t)

|x(ωωωω)|

ωωωω

ωωωω

φφφφnφφφφM

ωωωω

ωωωω

|KM(ωωωω)|

X

+

|Y(ωωωω)|

φφφφY(ωωωω)ωωωω0 2ωωωω0 3ωωωω0 4ωωωω0 5ωωωω0

ωωωω0 2ωωωω0 3ωωωω0 4ωωωω0 5ωωωω0

ωωωω

ωωωω

=

=

t

H L

t

H L

0.01 0.02 0.03 0.04t

-1

-0.5

0.5

1

H L

0.01 0.02 0.03 0.04t

-1

-0.5

0.5

1

qoHtL

Example: If x(t) as shown in Figure below is the input to a first-order system with a sensitivity of 1 and a time constant of 0.001 s, find y(t) for the periodic steady state.

-1

-0.02

x(t)

t, sec-0.01 0.01 0.02

+1

n = 3n = 5n = 7 n = 25

Response of a General Form of System to a Periodic Input

( )( )τωωφωφω

τωπ oon

oo

o

nnntnnn

ty arctan)( and )(sin(1

114)(

12

−=

+

+= ∑

∞

=

( )number odd n where)(

1

114)(

12

=

∠

+= ∑

∞

=no

o

nnn

iy ωφτωπ

ω

y(t) y(t)

Measurement System Behavior

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