Dynamic Characteristics

Dynamic characteristics tell us about how well a sensor responds to changes in its input. For dynamic signals, 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 understanding the input signal correctly.

General Model For A Measurement System

nth Order ordinary linear differential equation with constant 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

tdxbdt

txdbdt

txdbtyadt

tdyadt

tydadt

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:

Particular Solution

Method of undetermined coefficients:

•After a certain-order derivative, all higher derivatives are zero.•After a certain-order derivative, all higher derivatives 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 xx

= ⋅ =

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 txbtya

dttdya =+

)()()( tKxtydt

tdy=+τ 1

)(+

=DKD

xy

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

iifif TT

dtdT

=+τ

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

yocf yopi

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

τ/0 )()( teKAyKAty −−+=

>≤

==000

)()(tAt

tAUtx

)()()( tKAUtydt

tdy=+τ

Transient response

Steady state response

KACety t += − τ/)(

First-Order Systems: Step Response

t/τ

0 1 2 3 4 5

Out

put S

igna

l , (y

(t)-y

0)/(K

A-y 0)

0.0

.2

.4

.6

.8

1.0

0.632

t/τ

0 1 2 3 4 5

Erro

r fra

ctio

n, e

m

0.0

.2

.4

.6

.8

1.0

0.368

Non-dimensional step response of first-order instrument

τ/

)0()( te

KAyKAty −=

−−

τ/

0

0 1)( teyKAyty −−=

−−

Here, we define the term error fraction as

τ/

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

m eyyyty

KAyKAtyte −=

∞−∞−

=−−

=

Determination of Time constant

t

0 1 2 3 4 5

Erro

r fra

ctio

n, e

m

.001

.01

.1

1

τ/

)0()( te

KAyKAty −=

−−0.368

τtee mm −== log3.2ln

Slope = -1/τ

τ/

)0()( t

m eKAyKAtye −=

−−

=

First-Order Systems: Ramp Response

The complete solution:

Applying the initial condition, gives

>≤

=0 0 0

)(ttqt

txis&

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

Steady state response

ττ τis

tism qeq

Ktytxe && +−=−= − /)()(

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 instrument

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

tKAydtdy ωτ sin=+

tAtx ωsin)( =

( ) tKAtyD ωτ sin)(1 =+

The complete solution: ( )ωτωωτ

τ 1

2

/ tansin)(1

)( −− −+

+= tKACety t

Transient response

Steady state 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=

ωτ

.01 .1 1 10 100

Am

plitu

de ra

tio

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 100Ph

ase

shift

, φ(ω

)-90

-80

-70

-60

-50

-40

-30

-20

-10

0

First-Order Systems: Frequency Response

The phase angle isThe amplitude ratio1)(

1)(2 +

=ωτ

ωM )(tan)( 1 ωτωφ −−=

( )[ ] 2/121

1)(ωτ

ω+

==KABM

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

First-Order Systems: Frequency Response

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

)rad/s 2( −∠=−∠+

= KKAB

oo 7624.076116

)rad/s 20( −∠=−∠+

= KKAB

)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

The essential parameters

= the static sensitivity

= the damping ratio, dimensionless

= the natural angular frequency

0

0

bKa

=

0

2n

aa

ω =

1

0 22aa a

ζ =

Second-Order Systems

)()()()(0012

2

2 txbtyadt

tdyadt

tyda =++ )()(122

2

tKxtyDD

nn

=

++

ωζ

ω

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

)()()(2)(12

2

2 tKxtydt

tdydt

tyd

nn

=++ωζ

ω

Consider the characteristic equation

Second-Order Systems

0121 22 =++ DD

nn ωζ

ωThis quadratic equation has two roots:

122,1 −±−= ζωζω nnS

Overdamped (ζ > 1):

Critically damped (ζ = 1):

Underdamped (ζ< 1): :

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

( )Φ+−= − tCety nt

ocn 21sin)( ζωζω

tt

ocnn eCeCty

ωζζωζζ

−−−

−+−

+=1

2

1

1

22

)(

ttoc

nn teCeCty ωω −− += 21)(

t

yHtLtAe σ−

)sin( φω +td

Second-Order Systems

Case 2 Overdamped (ζ > 1):

Case 3 Critically damped (ζ = 1):

Case I Underdamped (ζ< 1):

d

nn

jS

ωσζωζω

±=

−±−=

12

2,1 ( ) nS ωζζ 122,1 −±−=

nS ω−=2,1

t

yHtL1=ζ

1>ζ

Second-order Systems

Example: The force-measuring spring

2

2o o

i s odx d xf B K x Mdt dt

− − =

forces=(mass)(acceleration)Σ

the second-order model:

consider a spring with spring constant Ks under applied force fiand the total mass M. At start, the scale is adjusted so that xo = 0 when fi = 0;

2( )s o iMD BD K x f+ + =

1 m/Ns

KK

=

rad/ssn

KM

ω =

2 s

BK M

ζ =

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 (ζ > 1):

Critically damped (ζ = 1):

Underdamped (ζ< 1): :

)(212

2

2 tKAUydtdy

dtyd

nn

=++ωζ

ω

112

112

1)( 1

2

21

2

2 22

+−

−−+

−

−+−=

−−−

−+− tt nn ee

KAty ωζζωζζ

ζ

ζζ

ζ

ζζ

)()(122

2

tKAUtyDD

nn

=

++

ωζ

ω

1)1()(++−= − t

nnet

KAty ωω

( ) 11sin1

)( 2

2++−

−−=

−

φωζζ

ζω

teKA

tyn

tn ( )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

Second-order Systems: Step Response

Non-dimensional step response of second-order instrument

ζ = 0

0.25

0.5

1.02.0

21 ζωω −= ndRinging frequency:

Rise time decreases ζ with but increases ringing

Optimum settling time can be obtained from ζ ~ 0.7

Practical systems use 0.6< ζ <0.8

ddT

ωπ2

=Ringing period

Time, t (s)0 5 10 15 20

Out

put s

igna

l , y(

t)/K

A

0.0

.2

.4

.6

.8

1.0

1.2

1.4

settling time

rise time

overshoot

100% ± 5%

Typical response of the 2nd order system

Second-order Systems: Step Response

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:

)(212

2

2 ttUqKydtdy

dtyd

isnn

&=++ωζ

ω)()( ttUqtx is&=

−

−−+−+

−

−−−+−=

−+−

−−−

t

t

n

isis

n

n

e

eqtqKty

ωζζ

ωζζ

ζζ

ζζζ

ζζ

ζζζωζ

1

2

22

1

2

22

2

2

141212

141212

12)( &&

+−−= − tn

n

isis

netqtqKty ωω

ω)

11(12)( &

&

( )

+−

−−−=

−

φωζζζω

ζ ζω

teqtqKty

n

t

n

isis

n2

21sin

1212)( &

&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 = 0.30.6

1.02.0

Ramp input

Steady state error = n

isqωζ&2

Steady state time lag =

nωζ2

Second-order Systems: Ramp Response

where

Second-order Systems: Frequency Response

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

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

)()( 2/1222

ωφωωζωωω

++−

+= tKAtytynn

ocf

[ ])(sin)()(steady ωφωω += tBty

The steady state response of a second-order to a sinusoidal input

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

)(nn

KABωζωωω

ω+−

=ωωωω

ζωφ//

2tan)( 1

nn −−= −

ωωωωζωφ

//2tan)( 1

nn −−= −

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

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

1)(nn

KABM

ωζωωωω

+−==

Second-order Systems: Frequency Response

Magnitude and Phase plot of second-order Instrumentω/ωn

.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

ζ = 0.1

0.3

0.5

1.0

2.0

ω/ωn

.01 .1 1 10 100

Phas

e sh

ift, φ(ω)

-180

-160

-140

-120

-100

-80

-60

-40

-20

0 ζ0 = 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)/K

A

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.

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

arctan( / )dr

d

t ω δω−

=

pd

t πω

=

( )2exp / 1pM πζ ζ= − −

Rise time:

Peak time:

Maximum overshoot:

2

12 1

rMζ 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

Dynamic CharacteristicsSpeed of response: indicates how fast the sensor (measurement system) reacts to changes in the input variable. (Step input)

Rise time: the length of time it takes the output to reach 90% of full response when a step is applied to the input

Time constant: (1st order system) the time for the output to change by 63.2% of itsmaximum possible change.

Settling time: the time it takes from the application of the input step until the output has settled within a specific band of the final value.

Transfer Function: a simple, concise and complete way of describing the sensor or system performance

H(s) = Y(s)/X(s) where Y(s) and X(s) are the Laplace Transforms of the input and output respectively. Sometimes, the transfer function is displayed graphically as magnitude and phase plots VS frequency (Bode plot).

Dynamic Characteristics

Cutoff frequency: the frequency at which the system response has fallen to 0.707 (-3 dB) of the stable low frequency.

cr f

t 35.0≈

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

Bandwidth the frequency band over which M(ω) ≥ 0.707 (-3 dB in decibel unit)

Frequency Response describe how the ratio of output and input changes with the input frequency. (sinusoidal input)

Dynamic CharacteristicsExample: 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 +

=τω

ωMDefine

( ) %10011

1%1001)(error Dynamic22

×

−

+=×−=

τωωM

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

195.022

≤+

≤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 ≤≤ωτThe 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

195.022

≤+

≤τω

ms 52.0Hz 1002

33.0==

πτ

Dynamic Characteristics

Am

plitu

de ra

tio M

(ω)

1.05

0.95

ωτ

M(ω) ≥ 0.95 region or δ(ω) ≤ 0.05 region

Dynamic CharacteristicsExample: A temperature measuring system, with a time constant 2 s, is used to measured temperature of a heating medium, which changes sinusoidal 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

T

325oC

350oC

300oC

x(t)

t

T

325oC

350oC

300oC

x(t), y(t)

t

TL

Solution: C)56.0602sin(2.21325)( o−+= tty π

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. s7.6=dt

Dynamic Characteristics

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 y(t)/x(t) = 1.152 and phase shifts –50.2o.

Solution:

Solution:

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

Dynamic CharacteristicsExample: An Accelerometer is to 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 ra

tio M

(ω)

ω/ ωn

1.05

0.95

1.05

0.95A

mpl

itude

ratio

M(ω

)

ω/ ωn

Response of a General Form of System to a Periodic Input

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

( )∑∞

=

++=1

000 sincos)(n

nn tnBtnAAtx ωωLet x(t)

The frequency response of the measurement system

Where KM(ω) = Magnitude of the frequency response of the measurement system and φM(ω) = Phase shift = tan-1(An/Bn)

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

220 ωφωφωω nntnnKMBAKAty nM

nnn ++++= ∑

∞

=

Response of a General Form of System of a Periodic Input

x(t) Linear system

y(t)t

xHtLt

yHtL

|x(ω)|

ω

ω

φnφM

ω

ω

|KM(ω)|

X

+

|Y(ω)|

φY(ω)ω0 2ω0 3ω0 4ω0 5ω0

ω0 2ω0 3ω0 4ω0 5ω0

ω

ω

=

=

0.01 0.02 0.03 0.04t

-1

-0.5

0.5

1

qoHtL

0.01 0.02 0.03 0.04t

-1

-0.5

0.5

1

qoHtL

Example: If qi(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 Qo(iω) and qo(t) for the periodic steady state.

-1

-0.02

qi(t)

t, sec-0.01 0.01 0.02

+1 ( )21

4 1 1( ) 1

o nn

o

Q in n

ω φπ ω τ

∞

=

= ∠ +

∑

arctan( )n onφ ω τ= −( )2

1

4 1 1( ) sin( ) 1

o o nn

o

q t n tn n

ω φπ ω τ

∞

=

= + +

∑

Where n = odd number

and

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

Response of a General Form of System to a Periodic Input