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.

Input signal x(t)

Sensor or system

Output signal y(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 d m −1 x (t ) dx (t ) d n −1 y (t ) dy (t ) d m x (t ) d n y (t ) + + L + + ( ) = + + L + b + b0 x (t ) a a a y t b b an n − m m − 1 1 0 1 1 n n −1 m m −1 dt dt dt dt dt dt

m≤n y(t) x(t) t a’s and b’s

F(t) = forcing function

Where

= output from the system = input to the system = time = system physical parameters, assumed constant

y(0) The solution

x(t)

Measurement system

y(t)

y (t ) = y ocf + y opi

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

Complementary-Function Solution The solution yocf is obtained by calculating the n roots of the algebraic characteristic equation Characteristic equation

an D n + an −1 D n −1 + ... + a1 D + a0 = 0

Roots of the characteristic equation:

D = s1 , s2 ,..., sn

Complementary-function solution: Ce st

1. Real roots, unrepeated: 2. Real roots, repeated: each root s which appear p times 3. Complex roots, unrepeated: the complex form: a ± ib

(C

2 p −1 st + C t + C t + ... + C t e ) 0 1 2 p −1

Ce at sin(bt + φ ) [C0 sin(bt + φ0 ) + C1t sin(bt + φ1 ) + C2t 2 sin(bt + φ2 )

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

+ ... + C p −1t p −1 sin(bt + φ p −1 )]e at

Particular Solution Method of undetermined coefficients: y opi = Af (t ) + B f ′(t ) + C f ′′(t ) + ... Where f(t) = the function that describes input quantity A, B, C = constant which can be found by substituting yopi into ODEs

Important Notes

9 •After a certain-order derivative, all higher derivatives are zero. a certain-order derivative, all higher derivatives have the same 9 •After functional form as some lower-order derivatives. 2 •Upon repeated differentiation, new functional forms continue to arise.

Zero-order Systems All the a’s and b’s other than a0 and b0 are zero.

y (t ) = Kx (t )

a0 y (t ) = b0 x(t )

where K = static sensitivity = b0/a0

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

xm Vr

+ x=0

y=V -

x here, K = Vr / xm V = Vr ⋅ xm Where 0 ≤ x ≤ xm and Vr is a reference voltage

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

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

dy (t ) + a 0 y (t ) = b0 x (t ) dt

τ

dy (t ) + y (t ) = Kx (t ) dt

y K (D) = x τD + 1

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

First-Order Systems

Surface area A

Ti

Consider a thermometer based on a mass m =ρV with 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

q

Assume no heat loss from the thermometer

Sensor m, Ttf, C

UA(Ti − Ttf )dt − 0 = ρVCdTtf

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

dTtf dt

+ UATtf = UATi

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

τ

dTif dt

+ Tif = Ti

First-Order Systems: Step Response Assume for t < 0, y = y0 , at time = 0 the input quantity, x increases instantly by an amount A. Therefore t > 0 0 t ≤ 0 x(t ) = AU (t ) =  A t > 0

τ

dy (t ) + y (t ) = KAU (t ) dt y (t ) = Ce − t /τ + KA

The complete solution: 2

U(t)

yocf

Transient response

1

yopi Steady state response

Applying the initial condition, we get C = y0-KA, thus gives 0

-1

0

1

2 Time, t

3

4

5

y (t ) = KA + ( y0 − KA)e − t /τ

First-Order Systems: Step Response Here, we define the term error fraction as y (t ) − KA y (t ) − y (∞) = = e −t / τ y0 − KA y ( 0) − y ( ∞ )

1.0

1.0

.8

.8 Error fraction, em

Output Signal, (y(t)-y0)/(KA-y0)

em (t ) =

0.632

.6

y (t ) − y0 = 1 − e −t / τ KA − y0

.4

.6 .4

0.368

.2

.2 0.0

y (t ) − KA = e −t /τ y (0) − KA

0

1

2

3 t/τ

4

5

0.0

0

1

2

3 t/τ

Non-dimensional step response of first-order instrument

4

5

Determination of Time constant y (t ) − KA = e −t /τ y (0) − KA

ln em = 2.3 log em = −

t

τ

1 y (t ) − KA = e −t /τ y (0) − KA

0.368

Error fraction, em

em =

.1

Slope = -1/τ

.01

.001

0

1

2

3 t

4

5

First-Order Systems: Ramp Response Assume that at initial condition, both y and x = 0, at time = 0, the input quantity start to change at a constant rate q&is Thus, we have

Therefore The complete solution:

t≤0 0 x(t ) =  q&is t t > 0 dy (t ) + y (t ) = Kq&is tU (t ) τ dt y (t ) = Ce − t /τ + Kq&is (t − τ ) Transient Steady state response response

Applying the initial condition, gives Measurement error

em = x(t ) −

y (t ) = Kq&is (τe − t /τ + t − τ ) y (t ) = − q&isτe −t /τ + q&isτ K Transient error

Steady state error

First-Order Systems: Ramp Response 10

Input x(t)

Output signal, y/K

8

y(t)/K

6

Steady state time lag = τ

4

Steady state error = q&isτ

2 0

0

2

4

6

8

10

t/τ

Non-dimensional ramp response of first-order instrument

First-Order Systems: Frequency Response From the response of first-order system to sinusoidal inputs, x(t ) = A sin ωt we have τ

dy + y = KA sin ωt dt

The complete solution:

(τD + 1) y(t ) = KA sin ωt

y (t ) = Ce −t /τ +

KA 1 + (ωτ )

Transient response

2

(

sin ωt − tan −1 ωτ

Steady state response

=

) Frequency response

If we do interest in only steady state response of the system, we can write the equation in general form y (t ) = Ce −t /τ + B(ω ) sin[ωt + φ (ω )]

B(ω ) =

KA

[1 + (ωτ ) ]

2 1/ 2

φ (ω ) = − tan −1 ωτ

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

First-Order Systems: Frequency Response M (ω ) =

The amplitude ratio

[

(ωτ ) 2 + 1

-2

-3 dB

.6

-4

.4

-6 -8 -10 Cutoff frequency

.1

-20 1 ωτ

10

100

-20 Phase shift, φ(ω)

.8 0.707

φ (ω ) = − tan −1 (ωτ )

-10

Decibels (dB)

0

0.0 .01

The phase angle is 0

1.0

.2

]

Dynamic error

1.2

Amplitude ratio

1

M (ω ) =

B 1 = KA 1 + (ωτ )2 1/ 2

-30 -40 -50 -60 -70 -80 -90 .01

.1

Frequency response of the first order system

1

10

100

ωτ

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

First-Order Systems: Frequency Response Ex: Inadequate frequency response

Suppose we want to measure x(t ) = sin 2t + 0.3 sin 20t

x(t)

With a first-order instrument whose τ is 0.2 s and static sensitivity K Superposition concept: For ω = 2 rad/s: B (2 rad/s) = A

y(t)/K

For ω = 20 rad/s: B (20 rad/s) = A

K ∠ − 21.8o = 0.93K∠ − 21.8o 0.16 + 1 K ∠ − 76o = 0.24 K∠ − 76o 16 + 1

Therefore, we can write y(t) as y (t ) = (1)(0.93K ) sin(2t − 21.8o ) + (0.3)(0.24 K ) sin(20t − 76o ) y (t ) = 0.93K sin(2t − 21.8o ) + 0.072K sin(20t − 76o )

Second-Order Systems In general, a second-order measurement system subjected to arbitrary input, x(t) 2

a2

d y (t ) dy (t ) + + a0 y (t ) = b0 x(t ) a 1 2 dt dt

 D 2 2ζ   2 + D + 1 y (t ) = Kx(t )  ωn ωn 

1 d 2 y (t ) 2ζ dy (t ) + + y (t ) = Kx(t ) 2 2 ωn dt ωn dt The essential parameters K=

ζ =

b0 a0

a1 2 a0 a2

ωn =

a0 a2

= the static sensitivity = the damping ratio, dimensionless = the natural angular frequency

Second-Order Systems Consider the characteristic equation 1 2 2ζ D + D +1 = 0 2

ωn ωn This quadratic equation has two roots:

S1, 2 = −ζω n ± ωn ζ 2 − 1

Depending on the value of ζ, three forms of complementary solutions are possible  −ζ + ζ 2 −1 ω t   n

Overdamped (ζ > 1):

yoc (t ) = C1e

Critically damped (ζ = 1):

yoc (t ) = C1e −ωnt + C2te −ωnt

Underdamped (ζ< 1): :

yoc (t ) = Ce −ζω nt sin ωn 1 − ζ 2 t + Φ

(

+ C2 e

 −ζ − ζ 2 −1 ω t   n

)

Second-Order Systems Case I Underdamped (ζ< 1):

Case 2 Overdamped (ζ > 1):

yHtL

)

(

S1, 2 = −ζω n ± ωn ζ 2 − 1

S1, 2 = − ζ ± ζ 2 − 1 ωn

= σ ± jωd

Case 3 Critically damped (ζ = 1):

Ae

S1, 2 = −ωn

−σt

t

sin(ωd t + φ )

yHtL

ζ =1 ζ >1 t

Second-order Systems Example: The force-measuring spring consider a spring with spring constant Ks under applied force fi and the total mass M. At start, the scale is adjusted so that xo = 0 when fi = 0; Σforces=(mass)(acceleration)

dxo d 2 xo fi − B − K s xo = M dt dt 2 ( MD 2 + BD + K s ) xo = f i the second-order model:

K=

1 Ks

m/N

Ks rad/s M B ζ = 2 Ks M

ωn =

Second-order Systems: Step Response For a step input x(t)

 D 2 2ζ   2 + D + 1 y (t ) = KAU (t )  ωn ωn 

1 d 2 y 2ζ dy + + y = KAU (t ) 2 2 ωn dt ωn dt

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

The complete solution: Overdamped (ζ > 1):

ζ + ζ 2 − 1  −ζ + y (t ) =− e 2 KA 2 ζ −1

Critically damped (ζ = 1):

y (t ) = −(1 + ωnt )e −ωnt + 1 KA

Underdamped (ζ< 1): :

y (t ) e −ζω nt =− sin 1 − ζ 2 ωnt + φ + 1 KA 1− ζ 2

(

ζ 2 −1 ω n t 

+

ζ − ζ 2 −1 2 ζ −1 2

)

e

 −ζ − ζ 2 −1 ω t   n

(

+1

φ = sin −1 1 − ζ 2

)

Second-order Systems: Step Response Ringing period

Output signal, y(t)/KA

2.0

Ringing frequency:

ζ=0 0.25

1.5

2π

ωd

ωd = ωn 1 − ζ 2

Rise time decreases ζ with but increases ringing

0.5 1.0

Optimum settling time can be obtained from ζ ~ 0.7

.5

1.0

Practical systems use 0.6< ζ 1) or critical damped (ζ = 1), there is neither overshoot nor steadystate 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. 1.4

arctan(−ω d / δ )

Rise time:

tr =

Maximum overshoot:

M p = exp −πζ / 1 − ζ 2

Peak time:

tp =

ωd

π ωd

Resonance frequency:

ω r = ω n 1 − 2ζ 2

Resonance amplitude:

Mr =

1 2

)

Output signal, y(t)/KA

1.2

(

2ζ 1 − ζ

Td

overshoot

1.0 .8 peak time

.6 .4

settling time

.2

rise time

0.0

0

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

5

10 Time, t (s)

15

20

Dynamic Characteristics Speed 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 its maximum 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 Frequency Response describe how the ratio of output and input changes with the input frequency. (sinusoidal input)

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

tr ≈

0.35 fc

Dynamic Characteristics 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: Define

M (ω ) =

1

ω 2τ 2 + 1

  1  − 1 ×100% Dynamic error = (M (ω ) − 1)× 100% =  2 2  ω τ +1  1 0 . 95 ≤ ≤ 1.05 From the condition |Dynamic error| < 5%, it implies that 2 2 ω τ +1 But for the first order system, the term 1 / ω 2τ 2 + 1 can not be greater than 1 so that the constrain becomes 1

0.95 ≤

Solve this inequality give the range

≤1

ω τ +1 0 ≤ ωτ ≤ 0.33 2 2

The largest allowable time constant for the input frequency 100 Hz is τ = The phase shift at 50 and 100 Hz can be found from

φ = − arctan ωτ

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

0.33 = 0.52 ms 2π 100 Hz

Dynamic Characteristics

Amplitude ratio M(ω)

1.05 0.95

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

ωτ

Dynamic Characteristics Example: 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 2π Solution: y (t ) = 325 + 21.2 sin( t − 0.56) o C 60

x(t), y(t)

x(t) 350oC

350oC

325oC

325oC

300oC

300oC

T

TL

t

T

t

Dynamic Characteristics Example: 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. t d = 6.7s

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. Solution:

y (t ) = 0.8[1 − e −3t sin( 29.85t + 1.47)] V

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. Solution:

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

Dynamic Characteristics Example: 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/s

Amplitude ratio M(ω)

Amplitude ratio M(ω)

Solution:

1.05 0.95

ω/ ωn

1.05 0.95

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. Let x(t)

∞

x(t ) = A0 + ∑ ( An cos nω0t + Bn sin nω0t ) n =1

The frequency response of the measurement system ∞

y (t ) = KA0 + ∑ n =1

(

)

An2 + Bn2 KM (nω0 ) sin (nω0t + φM (nω0 ) + φn (nω0 ) )

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

Response of a General Form of System of a Periodic Input yHt L

xHt L

x(t)

t

Linear system

y(t)

|KM(ω)|

|x(ω)|

|Y(ω)|

=

X

ω0 φn

2ω0 3ω0 4ω0 5ω0

ω φM

ω0

ω

ω

2ω0 3ω0 4ω0 5ω0

φY(ω)

=

+ ω

t

ω

ω

Response of a General Form of System to a Periodic Input 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. qi(t)

 1 Qo (iω ) = ∑  π n =1  n  4

+1 -0.02

-0.01

0.01

0.02

t, sec

-1

 1 qo (t ) = ∑  π n =1  n  4

qoH tL

n=7 n=5 n=3

0.5

0.01

-1

∞

 ∠φn  2 ( nω oτ ) + 1  1

Where n = odd number

 sin(nω ot + φn )  and φ n 2  ( nω oτ ) + 1  1

= − arctan(nω oτ )

qoH tL

1

-0.5

∞

0.02

0.03

0.04

1

n = 25

0.5

t

0.01 -0.5 -1

0.02

0.03

0.04

t