Step Response for the Transfer Function of a Sensor G(s)=Y(s)/X(s) of a sensor with X(s) input and Y(s) output A) First Order Instruments a) First order transfer function G(s)=k/(1+Ts), k=gain, T = time constant Example, J thermocouple with gain at 20 0C of =51 µV/0C and time constant of approx 0.01 sec. or k=5 [10 µV/0C] and T=0.01 MATLAB program k=5; T=0.01; num=[0 k]; den=[T 1]; step(num,den);grid Fig. y(t) plot

0 0.01 0.02 0.03 0.04 0.05 0.060

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5Step Response

Time (sec)

Ampl

itude

Bode diagram MATLAB program >> bode(num,den); grid;

-30

-20

-10

0

10

20

Mag

nitu

de (d

B)

100

101

102

103

104

-90

-45

0

Phas

e (d

eg)

Bode Diagram

Frequency (rad/sec) Bandwidth Cutoff frequency, ωb ,defining the bandwidth, is defined as the frequency ω for which the Magnitude 20 log|G(jω)| drops 3 dB below its zero-frequency value 20 log|G(j0)| 20 log|G(jω)| <20 log|G(j0)| - 3 dB for ω> ωb In the above case of a first order transfer function G(s)=k/(1+Ts), k=5, T = 0.01 s G(s)=5/(1+0.01s) or G(jω)=5/(1+0.01jω) its zero-frequency value is G(j0)=5 or 20 log|G(j0)|=14 dB Cutoff frequency, ωb ,defining the bandwidth, is obtained from the equation 20 log|G(jωb)|= 20 log|G(j0)| - 3 =14 – 3 = 11 dB ωb ≈100 rad/sec fb ≈100/(2π) ≈14 Hz

Static calibration of the sensor For: Xn in unknown sensor input to the sensor and Ym is the measured output from the sensor, static calibration uses steady state sensor gain Kc for calibration, i.e for estimating the unknown input Using limit value theorem for unit step input X(s)=1/s for s tending towards zero Kc = lim s G(s) /s= lim G(s) = G(j0)=5 Consequently, unknown sensor input to the sensor is estimated as Y/ Kc = Y/5 In practice, this is considered acceptable within bandwidth, i.e for input signal frequencies ω < ωb Sinusoidal Response of the sensor x(t)=sin ωt k/(1+Ts) y(t) 1/ Kc xest(t) or for X(s)= ω/(s2+ω2) and unit impulse input δ(s)=1 δ(s)=1 ω/(s2+ω2) k/(1+Ts) y(t) 1/ Kc xest(t) k=5; T=0.01; Kc=5 MATLAB Simulations 1) ω=10 rad/s f= 1.58 Hz period = 0.63 sec MATLAB program num=[0 0 0 50]; den=[0.01 1 1 100]; impulse(num,den); Fig. y(t) plot

0 10 20 30 40 50 60-5

-4

-3

-2

-1

0

1

2

3

4

5Impulse Response

Time (sec)

Ampl

itude

2) ω=100 rad/s f= 15.8 Hz period = 0.063 sec MATLAB program num=[0 0 0 500]; den=[0.01 1 100 10000]; impulse(num,den); Fig. y(t) plot

0 1 2 3 4 5 6-4

-3

-2

-1

0

1

2

3

4Impulse Response

Time (sec)

Ampl

itude

3) ω=1000 rad/s f= 158 Hz period = 0.0063 sec MATLAB program num=[0 0 0 5000]; den=[0.01 1 10000 1000000]; impulse(num,den); Fig. y(t) plot

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-0.5

0

0.5

1Impulse Response

Time (sec)

Ampl

itude

Analytical Solutions (K. Ogata, Modern Control Engineering, 4th edition, Prentice Hall pp. 268-271), where L{}=Laplace transform L{x(t)=sin ωt } G(s) =k/(1+Ts) L{y(t)} 1/ Kc L{ xest(t)} where static calibration constant is Kc = G(j0) For X(s)= ω/(s2+ω2) and unit impulse input δ(s)=1 δ(s)=1 ω/(s2+ω2) k/(1+Ts) y(t) 1/ Kc L{ xest(t)} For k=5; T=0.01; results |x(t)|=1 x(t)=1 sin ωt G(s) =k/(1+Ts) G(jω) =k/(1+T jω)=5/(1+0.01 jω)= 5(1-j0.01ω)/(1+0.012ω2)= 5(1-j0.01ω)/(1+0.012ω2) |G(jω)|= 5/(1+0.012ω2) 1/2 Φ = tan -1 (-0.01ω) Kc = G(j0)=k=5 y(t)=1 |G(jω)| sin (ωt+ Φ) xest(t)= y(t)/ Kc =(|G(jω)|/ Kc) sin (ωt+ Φ)= (|G(jω)|/5) sin (ωt+ Φ)

|xest(t)|/|x(t)|= |G(jω)|/5 For ω = 10 results |G(jω)|= 5/(1+0.012ω2) 1/2 =5/(1+0.012102) 1/2 ≈5 Φ = tan -1 (-0.01ω) =-5.71≈0 y(t)=1 |G(jω)| sin (ωt+ Φ) ≈5 sin ωt xest(t)= y(t)/k ≈1 sin ωt=x(t) |xest(t)|/|x(t)| ≈1 Summary of results: ω [rad/s] f[Hz] Period [s] |G(jω)| Φ y(t) xest(t) |xest(t)|/|x(t)| 10 1.58 0.63 ≈5 ≈0 ≈5sin ωt ≈sin ωt ≈1 100 15.8 0.063 ≈3.5 ≈-45 3.5sin(ωt-45) 0.7sin(ωt-45) ≈0.7 1000 158 0.0063 ≈3.5 ≈-90 0.5sin(ωt-90) 0.1sin(ωt-45) ≈0.1 The result for ω = 100, gives |xest(t)|/|x(t)| ≈0.7 20 log |xest(t)|/|x(t)| = - 03 dB, i.e ω = 100 = ωb Consequently, for ω > ωb |xest(t)|/|x(t)| <0.7 ω > >ωb |xest(t)|/|x(t)| <<0.7 ω = 1000 |xest(t)|/|x(t)| ≈0.1 i.e the estimation is only 10% of the amplitude of the sensor input signal Dynamic estimation (calibration) can use inverse problem solution, increasing gains, in this case 1/0.7 for ω = 10 for and 1/0.1=10 for ω = 1000. (E. Doebelin, Measurement Systems, McGraw Hill, 1990, Ch. 10.5 Dynamic Compensation, pp. 804-808) Obviously these gains would increase with ω and can lead to various difficulties (overflow in numerical computations, over-amplification of noise high frequency-low amplitude components in the y(t) signal etc), to be studied as part of inverse problem theory, a topic of advanced mechatronics.

Bode diagram G-1(s) = (1+Ts)/ k L{x(t)=sin ωt } G(s) k/(1+Ts) L{y(t)} G-1(s) = (1+Ts)/ k 1/ Kc xest(t) MATLAB program for the dynamic compensator k=5; T=0.01; num=[T 1]; den=[0 k]; bode(num,den);grid;

-20

-10

0

10

20

30

Mag

nitu

de (d

B)

100

101

102

103

104

0

45

90

Phas

e (d

eg)

Bode Diagram

Frequency (rad/sec) The magnitude of the inverse dynamic compensator |G-1(s)| = |(1+Ts)/ k| shows 20 dB/decade increase beyond bandwidth cutoff frequency, ωb =100, indicating growing computational difficulties as ω>> ωb . In general, inverse dynamic compensator for increasing sensors bandwidth is subjet to various difficulties :

- computational difficulties as ω>> ωb due to increasing magnitude of the inverse dynamic compensator |G-1(s)| = |(1+Ts)/ k| for ω> ωb ; digital word length limitation can lead to overflow.

- high frequency noise in the sensor output is amplified by increasing magnitude of the inverse dynamic compensator |G-1(s)| = |(1+Ts)/ k| for ω> ωb reducing signal to noise ratio

- unmodelled dynamics and parametric uncertainty result in reduced effect of inverse dynamics compennsator

- non-minimum phase sytems have unstable inverse dynamics Some solutions to the above difficulties are :

- low pass filter for removing high frequency noise in the sensor output L{x(t)} G(s) k/(1+Ts) L{y(t)} Low Pass Filter G-1(s)=(1+Ts)/ k 1/ Kc xest(t)

- Modified Output Approach (MOA) applied to non-minimum phase sytems to avoid unstable inverse dynamics

L{x(t)} G(s)k/(1+Ts) L{y(t)} Low Pass Filter G MOA

-1(s)=(1+Ts)/ k 1/ Kc xest(t)

B) Second order Instruments (E. Doebelin, Measurement Systems, McGraw Hill, 1990, pp. 123-157) b) Second order transfer function (K. Ogata, Modern Control Engineering, Prentice Hall, 1997, Ch. 4-3, pp. 141-187) for example for a servo system (pp.142-146), or a mass-spring-damper system (Example 4-7, pp. 174-175). G(s)=k /(s2+2 ζ ωn s + ωn

2) where k=gain, ωn = undamped natural frequency, ζ= damping ratio An example could be a force f(t) transducer f(t) a force measuring spring based instrument d(t) position measurement potentiometer v(t) where f(t) [N] is the input force to measure d(t) [m] is the output displacement v(t) [V]is the output voltage of the potentiometer Assume the force measuring spring based instrument measured by a mass-spring-damper M-B-K system moving horizontally (such that gravity effect can be ignored in deriving motion equation) F(s)/ D(s)= 1/ (M s2+B s + K) where D(s) =L{d(t)} F(s) =L{f(t)} Assume the approximate transfer function of the position measurement potentiometer V(s)/D(s)= Kp where V(s)= L{v(t)} D(s)=L{d(t)} Kp [V/m] is the calibration constant of the potentiometer F(s) 1/ (M s2+B s + K) D(s) Kp V(s) The overall transfer function is V(s)/F(s)= Kp / (M s2+B s + K)= (Kp / M)/ (s2+( B / M) s +( K /M) ) or

G(s)=V(s)/F(s)= k /(s2+2 ζ ωn s + ωn2)

where ωn

2 = K /M 2 ζ / ωn = B / M k= Kp / M Time response of such second order instruments are significantly dependent on the value of ζ = B ωn / (2 M) ζ < 1 underdamped response ζ = 1 critically damped response ζ > 1 overdamped response MATLAB Simulations For k =1, ωn = 10 rad/s fn= 1.58 Hz period = 0.63 sec 1) Step response of G(s) = k /(s2+2 ζ ωn s + ωn

2) = k /(s2+b s + c) where b=2 ζ ωn c= ωn

2 Steady state of v(t) for unit step input f(t) is obtained using limit value theorem for unit step input F(s)=1/s for s tending towards zero V = lim s G(s) /s= lim G(s) =lim k /(s2+2 ζ ωn s + ωn

2)= 1/ωn2 =0.01

a) for ζ = 0, b=2 ζ ωn =0 MATLAB program for k=1; b=0; c=100; num=[ 0 0 k]; den=[1 b c]; step(num,den);grid Fig. v(t) plot shows an undamped oscillatory response, obviously not usefull in practical applications. Higher values for ζ are required.

0 10 20 30 40 50 600

0.002

0.004

0.006

0.008

0.01

0.012

0.014

0.016

0.018

0.02Step Response

Time (sec)

Ampl

itude

b) for ζ = 0.1, b= 2 ζ ωn =2 MATLAB program k=1; b=2; c=100; num=[ 0 0 k]; den=[1 b c]; step(num,den);grid The results show significant maximum overshoot of 70% and long 2% settling time of 4/(ζ ωn ) = 4/ (0.1 x 10)= 4 sec.

0 1 2 3 4 5 60

0.002

0.004

0.006

0.008

0.01

0.012

0.014

0.016

0.018Step Response

Time (sec)

Ampl

itude

c) for ζ = 0.6, b= 2 ζ ωn =12 MATLAB program k=1; b=12; c=100; num=[ 0 0 k]; den=[1 b c]; step(num,den);grid The results show significant reduced overshoot of 5% and reduced 2% settling time of 4/(ζ ωn ) = 4/ (0.6 x 10)= 1.7 sec.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.80

0.002

0.004

0.006

0.008

0.01

0.012Step Response

Time (sec)

Ampl

itude

d) for overdamped case, ζ = 1.2, b= 2 ζ ωn =24 MATLAB program k=1; b=24; c=100; num=[ 0 0 k]; den=[1 b c]; step(num,den);grid The results show no overshoot but sluggish response.

0 0.2 0.4 0.6 0.8 1 1.20

0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

0.01Step Response

Time (sec)

Ampl

itude

Present day practice is to provide underdamped response for ζ = B ωn / (2 M) in the range of 0.6-0.7, but selecting a damping coefficient of B = 2 M ζ / ωn (E. Doebelin, Measurement Systems, McGraw Hill, 1990, pp. 131). Bode diagram for such a case ζ = 0.6, b= 2 ζ ωn =12 is given by MATLAB program k=1; b=12; c=100; num=[ 0 0 k]; den=[1 b c]; bode(num,den); grid;

-120

-100

-80

-60

-40

-20M

agni

tude

(dB)

10-1

100

101

102

103

-180

-135

-90

-45

0

Phas

e (d

eg)

Bode Diagram

Frequency (rad/sec) Cutoff frequency, ωb ,for the above case of a second order transfer function G(s) = k /(s2+2 ζ ωn s + ωn

2) = k /(s2+b s + c) where k=1; b=12; c=100; gives G(jω)=1/( -ω2 +12jω + 100) its zero-frequency value is G(j0)=1/100=0.01 or 20 log|G(j0)|=- 40 dB Cutoff frequency, ωb ,defining the bandwidth, is obtained from the equation 20 log|G(jωb)|= 20 log|G(j0)| - 3 =-40 – 3 = -43 dB The Magnitude diagram gives ωb ≈10 rad/sec fb ≈10/(2π) ≈1.4 Hz Dynamic estimation (calibration) can use inverse problem solution, increasing gains, in this case Obviously these gains would increase with ω and can lead to various difficulties (overflow in numerical computations, over-amplification of noise high frequency-low amplitude components in the y(t) signal etc).

Bode diagram G-1(s) = (s2+b s + c) / k F(s) G(s) D(s) G-1(s) V(s) MATLAB program k=1; b=12; c=100; den =[ 0 0 k]; num =[1 b c]; bode(num,den); grid;

20

40

60

80

100

120

Mag

nitu

de (d

B)

10-1

100

101

102

103

0

45

90

135

180

Phas

e (d

eg)

Bode Diagram

Frequency (rad/sec) The magnitude of the inverse dynamic compensator |G-1(s)| = |(1+Ts)/ k| shows 40 dB/decade increase beyond bandwidth cutoff frequency, ωb =10 , indicating growing computational difficulties as ω>> ωb , even more significant the in the case of first order instruments. N-order instrument will have 20 N dB/decade increase beyond bandwidth cutoff frequency.

Some solutions to the above difficulties were outlined for the case of first oreder instruments:

- low pass filter for removing high frequency noise in the sensor output

- Modified Output Approach (MOA) applied to non-minimum phase sytems to avoid unstable inverse dynamics

In practice the increase beyond bandwidth cutoff frequency, ωb is normally limited up to a maximum useful frequency component ωuseful which might be achieved by inverse dynamic compensator approach without reachng unacceptable high magnitudes of the inverse dynamic compensator, making this approach advantageus in computer based instrumentation.

Sinusoidal Response of the second order sensor with ζ =0.6 For F(s) 1/ (M s2+B s + K) D(s) V(s) the overall transfer function is V(s)/F(s)= Kp / (M s2+B s + K)= (Kp / M)/ (s2+( B / M) s +( K /M) ) or G(s)=V(s)/F(s)= k /(s2+2 ζ ωn s + ωn

2) where ωn

2 = K /M 2 ζ ωn = B / M k= Kp / M Time response of such second order instruments are significantly dependent on the value of ζ = B ωn / (2 M) where ζ =0.6 MATLAB simulation of x(t)=sin ωt k/( s2+2 ζ ωn s + ωn

2) y(t) 1/Kc xest(t) where static calibration constant is Kc = G(j0) For X(s)= ω/(s2+ω2) and unit impulse input δ(s)=1 can be achieved with MATLAB instruction impulse (num, den) for δ(s)=1 ω/(s2+ω2) k/( s2+2 ζ ωn s + ωn

2) y(t) 1/ Kc xest(t) The transfer function for the g eneral MATLAB program is k ω/[(s2+ω2)( s2+2 ζ ωn s + ωn

2)] = k ω/(s4+ 2 ζ ωn s3 +(ωn2 +ω2)s2 +2 ζ ωn ω2 s + ωn

2ω2) = k ω /( s4 +as3+ (b +ω2)s2 +a ω2s + bω2) where a= 2 ζ ωn b= ωn

2 Let us assume k =1, ζ = 0.6 ωn = 10 rad/s such that a= 2 ζ ωn =12 b= ωn

2=100 or

a=12; b=100; num=[ 0 0 0 0 ω]; den=[1 a b+ω*ω a*ω*ω b*ω*ω]; impulse(num,den); MATLAB Simulations 1) ω=1 rad/s f= 0.158 Hz period = 6.3 sec MATLAB program is a=12; b=100; num=[ 0 0 0 0 1]; den=[1 a b+1 a b]; impulse(num,den);

0 50 100 150 200 250 300-0.015

-0.01

-0.005

0

0.005

0.01

0.015Impulse Response

Time (sec)

Ampl

itude

The result agrees to Bode diagram amplitude for ω=1of -40 dB or 0.01 2) ω=5 rad/s

a=12; b=100; num=[ 0 0 0 0 5]; den=[1 12 125 300 2500]; impulse(num,den);

0 10 20 30 40 50 60 70 80 90 100-0.015

-0.01

-0.005

0

0.005

0.01

0.015Impulse Response

Time (sec)

Ampl

itude

The result agrees to Bode diagram amplitude for ω=5of -40 dB or 0.01 Analytical Solutions (K. Ogata, Modern Control Engineering, 4th edition, Prentice Hall pp. 268-271), where L{}=Laplace transform L{x(t)=sin ωt } G(s)= k/( s2+2 ζ ωn s + ωn

2) L{y(t)} 1/ Kc L{ xest(t)} or for X(s)= ω/(s2+ω2) and unit impulse input δ(s)=1 δ(s)=1 ω/(s2+ω2) k/( s2+2 ζ ωn s + ωn

2) y(t) 1/ Kc L{ xest(t)} For |x(t)|=1 x(t)=1 sin ωt ζ = 0.6 ωn = 10 rad/s such that a= 2 ζ ωn =12 b= ωn

2=100 G(s) = k/( s2+12 s + 100) G(jω) =1/( -ω2+100+12 jω)= ( -ω2+100-12 jω)/ [( -ω2+100+12 jω) ( -ω2+100-12 jω)] = (100-ω2-12 jω)/( (100-ω2)2+122ω2) |G(jω)|= 1/((100-ω2)2+122ω2) 1/2

Φ = tan -1 (-12 ω/(100-ω2) ) y(t)=1 |G(jω)| sin (ωt+ Φ) |G(j0)|= 1/((100)2) 1/2 =1/100 Kc =G(j0)=1/100 xest(t)= y(t)/ Kc =(|G(jω)|/ Kc) sin (ωt+ Φ)= (|G(jω)|/(1/100)) sin (ωt+ Φ) |xest(t)|/|x(t)|= |G(jω)|/ Kc =|G(jω)|/|G(j0)| For ω = 10 results |G(jω)|= 1/((100-ω2)2+122ω2) 1/2 =1/((100-102)2+122102) 1/2 =1/120 Φ = tan -1(-12 ω/(100-ω2) )= tan -1 (-120/(100-102) ) = tan -1 (-120/0)=-900

y(t)=1 (|G(jω)| / Kc)sin (ωt+ Φ) ≈((1/120)/(1/100)) sin( ωt-90) |G(j10)|/ |G(j0)| = (1/120)/(1/100)=100/120 = |xest(t)|/|x(t)| ≈0.83 For ω = 10, Bode diagram gives appr -42 dB and -900 which agrees with the above result. Exact calculation of the cutoff frequency, ωb ,defining the bandwidth, obtained from the equation 20 log|G(jωb)|= 20 log|G(j0)| - 3 or log|G(j0)| - log|G(jωb)| = log(|G(j0)| / |G(jωb)| )=3/20 or |G(j0)| /|G(jωb)| = log-1 3/20 = 1.4125 or |G(jωb)| =|G(j0)| /1.4125 =0.0709 |G(j0)| which shows that, at cutoff frequency, ωb , the amplitude |G(jωb)| drops to 0.0709 of the |G(j0)| For the above second order instrument |G(jω)|= 1/((100-ω2)2+122ω2) 1/2 results |G(j0)|= 1/((100-02)2+12202) 1/2 = 1/100 The cutoff frequency, ωb can be obtained from |G(jωb)| =|G(j0)| /1.4125 =(1/100)/ 1.4125=1/141.25 For obtaining ωb the equation to solve is then 1/((100-ωb

2)2+122ωb2) 1/2 =1/141.25

or ((100-ωb

2)2+122ωb2) 1/2 =141.25

or (100-ωb

2)2+122ωb2 =141.252 =19952.6

or ωb

4 -200 ωb2 +10000-19952.6+ 144ωb

2 =0 or ωb

4 -56 ωb2 -9952.6 =0

The solution for ωb2 is

ωb2 =28±√(282+9952.6) =28±103.6 or 131.6 and – 75.6

Only the positive solution gives ωb =√131.6= 11.47 [rad/s] that can be recognized on the second order Bode diagram for the magnitude of -40-3 =-43 dB