Instrumentasi & Pengukuran: • Karakteristik Statik • Karakteristik Dinamik

INSTRUMENTASI & PENGUKURAN • Karakteristik Statik • Karakteristik Dinamik

Karakteristik Statik • • • • • •

Kalibrasi statik Akurasi Presisi Bias Ketidakpastian Sensitivitas

• • • • • •

Threshold Resolusi Histerisis Offset Jangkah Linearitas

Karakteristik Dinamik • Karakteristik alat ukur terhadap sebuah perubahan dari satu pengukuran ke pengukuran lain – Respon kecepatan: kecepatan mengikuti dan memberikan tanggapan atas besaran ukur yang ditangkap sebagai sebuah sinyal masukan – Fidelity: kecepatan menunjukkan nilai pengukuran baru dengan tepat saat terjadi perubahan nilai objek ukur

Karakteristik Dinamik • Model matematis sistem pengukuran – Persamaan fungsi alih – Persamaan ruang-keadaan

• Instrumen orde nol • Instrumen orde satu • Instrumen orde dua

Model Matematis Sistem Masukan (x) Besaran yang diukur

Alat Pengukuran

Keluaran (y) Besaran yang terukur

• Hubungan masukan dan keluaran alat atau instrumen pengukuran dinyatakan sebagai fungsi persamaan diferensial dalam domain waktu

• Bentuk persamaan differensial umum dalam domain waktu: 𝑑𝑛 𝑦 𝑑𝑛−1 𝑦 𝑑 𝑑𝑚 𝑑 𝑚−1 𝑎𝑛 𝑛 + 𝑎𝑛_1 𝑛−1 + ⋯ + 𝑎1 𝑦 + 𝑎0 𝑦 = 𝑏𝑚 𝑚 𝑥 + 𝑏𝑚_1 𝑑𝑑𝑡 𝑥 + ⋯ 𝑏1 𝑥 𝑚−1 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 + 𝑏0 𝑥

untuk D =

𝑑 , 𝑑𝑡

maka fungsi alih:

𝑦 𝑏𝑚 𝐷𝑚 + 𝑏𝑚_1 𝐷𝑚−1 + ⋯ + 𝑏1 𝐷 + 𝑏0 𝐷 = 𝑥 𝑎𝑛 𝐷𝑛 + 𝑎𝑛_1 𝐷𝑛−1 + ⋯ + 𝑎1 𝐷 + 𝑎0

• Instrumen orde nol 𝑦 𝑏0 𝐷 = = 𝑲𝒑 𝑥 𝑎0

• Instrumen orde satu 𝑲𝒑 𝑦 𝑏0 𝐷 = = 𝑥 𝑎1 𝐷 + 𝑎0 𝜏𝐷 + 1

– –

𝒃𝟎 𝑲𝒑 = 𝒂𝟎 𝒂𝟏 𝝉𝒑 = 𝒂𝟎

: Penguatan statis (static gain) : Tetapan waktu (time constant)

• Instrumen orde dua 𝒅 𝒚 𝒅𝟐 𝒚 𝐛. 𝐟 𝐭 = 𝒂𝟎 𝒚 + 𝒂𝟏 + 𝒂𝟐 𝟐 𝒅𝒕 𝒅𝒕

𝐛. 𝐟 𝐭 = 𝒂𝟎 𝒚 + 𝒂𝟏

– –

𝒃 𝑲𝒑 = 𝒂𝟎 𝒂𝟏 𝝉𝒑 = 𝒂𝟎

constant)

𝒅 𝒚

𝒅𝒕 𝒅 𝒚 𝑲𝒑 . 𝐟 𝐭 = 𝒚 + 𝝉𝒑 𝒅𝒕

: Penguatan statis (static gain) : Tetapan waktu (time

Sistem Orde Nol

𝒃

𝑲𝒑 = 𝒂𝟎 : Penguatan statis (static gain) 𝟎

Semua tetapan kecuali a0 dan b0 menjadi nol. 𝑦 𝑏0 𝐷 = = 𝑲𝒑 𝑥 𝑎0

y (t )  Kx(t )

a0 y(t )  b0 x(t )

Karakteristiknya bergantung dari static sensitivity, Kp dan tetap tidak berubah (tidak bergantung dari frekuensi masukkan): karakteristik dinamis ideal Untuk K = static sensitivity = b0/a0

x m

Vr

+ x=0

y=V -

x V  Vr  here, K  Vr / xm xm Untuk 0  x  xm dan Vr tegangan referensi

Potentiometer linear dipakai sebagai sensor posisi merupakan sensor orde nol

Sistem Orde Satu Semua a dan b selain a1, a0 and b0 bernilai nol. a1

dy (t )  a0  b0 x(t ) dt



dy (t )  y (t )  Kx(t ) dt

y K ( D)  x D  1

Untuk K = b0/a0 : static sensitivity  = a1/a0 : konstanta waktu sistem (dimension of time)

Sistem Orde Satu: Step Response Misalkan, untuk t < 0, y = y0 , pada waktu = 0 nilai masukan, x naik tajam sehingga bernilai A. demikian pula untuk t > 0 0 t  0 x(t )  AU (t )   A t  0



dy (t )  y (t )  KAU (t ) dt

y(t )  Cet /  KA

Solution lengkap: 2

U(t)

yocf Transient response

yopi Steady state response

1

Applying the initial condition, diperoleh C = y0-KA, sehingga: 0 -1

0

1

2 Time, t

3

4

5

y (t )  KA  ( y0  KA)e

t /

Sistem Orde Satu: Step Response Error pembagian didefinisika sebagai:

y (t )  KA y (t )  y ()   e t /  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 /  KA  y0

.4 .2

y (t )  KA t / e y (0)  KA

.6 .4

0.368

.2

0.0

0.0

0

1

2

3 t/

4

5

0

1

3

2 t/

Non-dimensional step response of first-order instrument

4

5

Penentuan Konstanta Waktu y (t )  KA t  e t / ln em  2.3 log em   y (0)  KA  1

y (t )  KA  e t /  y (0)  KA

0.368

Error fraction, em

em 

.1

Slope = -1/

.01

.001 0

1

2

3 t

4

5

Sistem Orde Satu: Ramp Response Misalkan pada keadaan awal, y dan x = 0, pada waktu = 0, nilai masukan mulai berubah dengan laju tetap, sehingga didapat qis

Maka Solusi lengkap:

0 t  0 x(t )   qist t  0 dy (t )   y (t )  KqistU (t ) dt y (t )  Cet /  Kqis (t   ) Transient response

y (t )  Kqis (e t /  t   )

Memasukan kondisi awal: Kesalahan pengukuran

Steady state response

em  x(t ) 

y (t )  qise t /  qis K Transient error

Steady state error

Sistem Orde Satu: Ramp Response 10

Output signal, y/K

8 6 Steady state time lag = 

4 Steady state error = qis

2 0 0

2

4

6

8

10

t/

Non-dimensional ramp response of first-order instrument

Sistem Orde Satu: Frequency Response x(t )  A sin t Dari respon sistem orde satu untuk masukan sinus: diperoleh dy   y  KA sin t D 1y(t )  KAsin t

dt

Solusi lengkapnya :

y (t )  Cet / 

KA 1  ( ) 2

Transient response

sin t  tan 1  

Steady state response

=

Frequency response

Jika yang dicari hanya steady state response , persamaan umumnya: y(t )  Cet /  B( ) sin t   ( )

B( ) 

KA

1  ( ) 

2 1/ 2

 ( )   tan 1 

Untuk B() = amplitudo steady state response; () = pergeseran fasa

First-Order Instrument: Frequency Response M ( )  1

The amplitude ratio M ( ) 

B 1  KA 1   2 1/ 2



( ) 2  1



The phase angle is

Dynamic error

1.2

 ( )   tan 1 ( )

0

.8

-2

-3 dB

0.707

.6

-4

.4

-6 -8 -10

.2

Cutoff frequency

-20 0.0 .01

.1

1 

10

100

-20 Phase shift, ()

0 Decibels (dB)

Amplitude ratio

-10 1.0

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

.1

Frequency response of the first order system

1

10

100



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

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

Dynamic error, () = 1- M() 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()  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

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  is 0.2 s and static sensitivity K Superposition concept: For  = 2 rad/s: B(2 rad/s ) 

y(t)/K

For  = 20 rad/s: B(20 rad/s ) 

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)(0.24K ) sin( 20t  76o ) y(t )  0.93K sin( 2t  21.8o )  0.072K sin( 20t  76o )

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

Qo (i ) K  M ( )  Qi (i )  2 2  1

  M ( )  M (0) 1  Dynamic error  100%    1 100% 2 2 M (0)    1  From the condition |Dynamic error| < 5%, it implies that 0.95 

1

  1 2 2

 1.05

But for the first order system, the term 1 /  2 2  1can not be greater than 1 so that the constrain becomes 1

0.95 

Solve this inequality give the range

  1 2 2

1

0    0.33

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 give  = -9.33o and = -18.19o at 50 and 100 Hz respectively

0.33  0.52 ms 2 100 Hz

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

a2

d y(t ) dy (t )  a  a0 y(t )  b0 x(t ) 1 2 dt dt

 D 2 2   2  D  1 y(t )  Kx(t )  n n 

1 d 2 y(t ) 2 dy(t )   y(t )  Kx(t ) 2 2 n dt n dt The essential parameters K



b0 a0

a1 2 a0 a2

n 

a0 a2

= the static sensitivity = the damping ratio, dimensionless

= the natural angular frequency

Second-Order Systems Consider the characteristic equation 1 2 2 D  D 1  0 2 n n This quadratic equation has two roots:

S1, 2  n  n  2 1 Depending on the value of , three forms of complementary solutions are possible     2 1  t   n

    2 1  t   n

Overdamped ( > 1):

yoc (t )  C1e

Critically damped ( = 1):

yoc (t )  C1e nt  C2tent

Underdamped (< 1): :

yoc (t )  Ce nt sin n 1   2 t  



 C2 e



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

Case 2 Overdamped ( > 1):





S1, 2   n  n  2  1

S1, 2      2 1 n

   j d

Case 3 Critically damped ( = 1):

yt

Ae

S1, 2  n

 t

t

sin( d t   )

yt

 1

 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; forces=(mass)(acceleration)

dxo d 2 xo fi  B  K s xo  M 2 dt dt ( MD 2  BD  K s ) xo  fi the second-order model:

K

1 Ks

m/N

Ks rad/s M B  2 Ks M

n 

Second-order Systems: Step Response  D 2 2   2  D  1 y(t )  KAU (t )   n n 

1 d 2 y 2 dy   y  KAU (t ) For a step input x(t) 2 2 n dt n dt

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

The complete solution:

Overdamped ( > 1):

   2  1    y(t )  e 2 KA 2  1

Critically damped ( = 1):

y (t )  (1  nt )e nt  1 KA

Underdamped (< 1): :

y(t ) e  nt  sin 1   2 nt    1 KA 1  2



 2 1 nt 



   2 1 2  1 2



e

    2 1  t   n



1

  sin 1 1   2



Second-order Instrument: Step Response Ringing frequency:

Output signal, y(t)/KA

2.0

1.5

Td 

2 d

d  n 1   2

=0

Ringing frequency:

0.25

Rise time decreases  with but increases ringing

0.5

1.0

Optimum settling time can be obtained from  ~ 0.7

.5

1.0

Practical systems use 0.6<  <0.8

2.0

0.0 0

2

4

6

8

10

nt

Non-dimensional step response of second-order instrument

Dynamic Characteristics 1.4 overshoot

Output signal, y(t)/KA

1.2 100%  5%

1.0 .8 .6 .4

settling time

.2

rise time

0.0 0

5

10

15

Time, t (s) Typical response of the 2nd order system

20

Second-order System: Ramp Response For a ramp input x(t )  qistU (t )

1 d 2 y 2 dy   y  KqistU (t ) 2 2 n dt n dt

 nt   nt 1  e (1  )  2 

qo 2q  qist  is K n

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

Overdamped:

2qis  2 2  1  2  2  1    y (t )  qist  1 e 2 K n  4   1 

Critically damped:

Underdamped:

 2 2  1  2  2  1 4   1 2

e

 2 1  n t 

    2 1  t   n

   

2q  t y(t )   qist  is 1  (1  n )ent  K n  1 





 2qis  y(t ) e  nt 2  qist  sin 1   nt    1  2 K n  2 1   

2 1   2   tan 2 2  1 1

Second-order Instrument: Step Response Steady state error =

Output signal, y(t)/K

10 8

Steady state 2  time lag =

Ramp input

n

6 4

 = 0.3 1.0 2.0

2

0.6

0 0

2

4

2qis n

6

8

Time, t (s)

Typical ramp response of second-order instrument

10

Second-order Instrument: Frequency Response The response of a second-order to a sinusoidal input of the form x(t) = Asint KA y(t )  yoc (t )  sin t   ( ) 1/ 2 2 2 2 1   / n   2 / n 







where  ( )   tan 1

2  / n  n / 

The steady state response of a second-order to a sinusoidal input ysteady (t )  B( ) sin t   ( )

B( ) 

KA

1  /     2 /    2 2

n

2

1/ 2

 ( )   tan 1

n

2  / n  n / 

Where B() = amplitude of the steady state response and () = phase shift M ( ) 

B 1  KA 1   /  2 2  2 /  2 1/ 2 n n







Second-order Instrument: Frequency Response The phase angle

The amplitude ratio M ( ) 

1

 ( )   tan 1

1  /     2 /    2 2

2

n

1/ 2

n

2  / n  n / 

0 6

0.3

1.5

3 0.5

1.0

0 -3 -6 -10 -15

1.0

.5 2.0

0.0 .01

.1

1

10

100

0 = 0.1

-20 Phase shift, 

 = 0.1

Decibel (dB)

Amplitude ratio

2.0

0.3

-40

0.5

-60 -80

1.0 2.0

-100 -120 -140 -160 -180 .01

.1

n

Magnitude and Phase plot of second-order Instrument

1 n

10

100

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

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

Maximum overshoot:

M p  exp  / 1   2

Peak time:

tp 



 d

Resonance  r   n 1  2 2 frequency: 1 Resonance M r  2 1   2 amplitude:

Td

overshoot



1.2 1.0

o

tr 

Output signal, q (t)/Kqis

arctan( d /  ) d

Rise time:

1.4

.8 peak time

.6 .4

settling time

.2

rise time

0.0 0

where  = n ,  d   n 1   2 , and   arcsin( 1   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 10 to 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.

Dead time: the length of time from the application of a step change at the input of the sensor until the output begins to change