Signal Processing - Introduction Signal Processing - - PowerPoint PPT Presentation

▶
signal processing introduction
SMART_READER_LITE
LIVE PREVIEW

Signal Processing - Introduction Signal Processing - - PowerPoint PPT Presentation

Jul 08, 2023 556 likes •896 views

Instrumentation (and Signal Processing Process Control) Fall 1393 Bonab University Signal Processing - Introduction Signal Processing Analogue/digital filters: extensively used in sensor signal processing Pre-processing

slide-1
SLIDE 1

Instrumentation (and

Process Control)

Fall 1393 Bonab University

Signal Processing

slide-2
SLIDE 2

Signal Processing - Introduction

  • Analogue/digital filters: extensively used in sensor signal processing
  • Pre-processing
  • Post-processing
  • Analog filters: very common in dealing with: Aliasing phenomenon (acquisition)
  • Distortion: signal reconstructed from samples

is different from the original continuous signal

  • Digital filters: generally used to post-process

acquired signals + other sophisticated digital signal-processing techniques (say, Fast Fourier Transform to perform spectral analysis)

  • Analogue filter: analysis-synthesis
  • Passive components
  • Op-Amps
  • Digital: moving average (MA) and

autoregressive moving average (ARMA)

2

Signal Processing

slide-3
SLIDE 3

Signal Processing - Introduction

  • Aliasing problem

3

Signal Processing

slide-4
SLIDE 4

Analog Filters

  • Two main reasons:
  • Buffer and reduce the impedance of sensors for interface with data acquisition
  • Eliminate high-frequency noise to prevent aliasing in ADC
  • Passive Analog
  • Few simple electronic components (resistors and capacitors)
  • Basic low-pass filter (Fig) used to remove (or attenuate) high-frequency noise
  • Vo,1 may be very close to Vi,1, say 98% of this value, but Vo,2 may be about 70%
  • f Vi,2 and so on…
  • Why? low-pass filter attenuates each signal according to its frequency

4

Signal Processing

slide-5
SLIDE 5

Analog Filters

  • precise amount of attenuation?
  • frequency response (Bode plot)
  • Equation?
  • Kirchoff’s current law:
  • Laplace transform

τ = RC, time-constant: time it takes for the filter to respond to a step input function by reaching 63%

  • 1

τ= ω𝐷 : Corner frequency

  • Assuming sinusoidal inputs: ignoring the transient:

5

Signal Processing

_

Cutoff frequency 20 db /decade Pass Band Stop Band

slide-6
SLIDE 6

Analog Filters

  • Input-output relationship:
  • More standard form:
  • Higher frequencies (>> corner frequency of 10 r/s, Fig)  high attenuation
  • Active Analog (Using Op-Amps)
  • Simple RC can be easily implemented
  • However since it is entirely passive:
  • Draw current from the input
  • And “load” the circuit connected to the output

6

Signal Processing

_ _

_

_

slide-7
SLIDE 7

Active Analog Filter (Using Op-Amps)

  • Op-amps eliminate this problem:
  • Current drawn from the input :very small (because op-amps internal resistances, of the
  • rder of 10 MO)
  • Likewise, as active devices, op-amps supply current to drive their output  minimize the

impact on any output circuit, such as DAQ card

  • So, Op-Amps & R & C = Active filter
  • Current Summation at inverting input:

 Very similar to Passive (gain): Negative means min 180o lag

  • K & τ : 2 DoF to adjust gain, corner-f independently

7

Signal Processing

_ _ _

_ _ _ _

_ k τ

slide-8
SLIDE 8

Active Analog Filter (in frequency domain)

  • The ratio of output to input voltage:
  • Bode: the same, with 20 log(k) shift up
  • Making the filter (usually):
  • Resistor (fixed/variable)
  • Capacitor (ceramic plate)
  • Op-Amp (say, LM741)
  • Designing the filter:
  • Choose corner-F (Better: measure)
  • 2πf= 1/RC (1-equation, 2-un-known)
  • Choose a C (µF-pF)
  • Find R  standardize
  • Choose another C if (not 1k-1MΩ)
  • Keep in mind: tolerances

8

Signal Processing

_ _

slide-9
SLIDE 9

Other Active Signal Processing Circuits

  • Simple/complex tasks:
  • e.g. Adding two signals (say, for adjusting D.C offset)
  • Set vr = - D.C. offset
  • Generally:
  • Need: remove high-frequency noise in the signal before sampling it with an A/D
  • Usual color coding in the circuit:
  • Red: positive supply
  • Black: ground
  • Blue: signal

9

Signal Processing

_

slide-10
SLIDE 10

Digital Filters

  • Uses discrete data points sampled at regular intervals (from sensor output)
  • e.g. accelerometer (vibration in a beam)
  • Digital filters rely:
  • not only on the current value of the measured variable
  • but also on its past values (raw/filtered form)
  • Input Averaging Filter

Extend to more complex filter Called Moving Average (MA):

  • Filter with Memory
  • Large α  Current value of input has more weight
  • Small α  Past filtered signal has more weight (normally, α <1)

10

Signal Processing

Digital _

_ _ _ _

_

_

slide-11
SLIDE 11

Example (filter with memory):

11

Signal Processing

Simple averaging filter:

  • No single α is the best
  • But α=0.5 may be better
  • More similar to input pattern
  • To fit the application:
  • Fine-tune it
  • There are mathematical methods
  • Choose it based on frequency response (as in analog)
  • More effective Digital Filters : (higher order)
  • Combination of MA
  • And Autoregressive MA (ARMA)
slide-12
SLIDE 12

Variable Conversion Elements

  • Often: measurement sensors’ outputs = Voltage signal

 measured by: voltage indicating instruments

  • However, many cases: output ≠ voltage
  • translational displacements
  • changes in various electrical parameters:
  • Resistance
  • Inductance
  • Capacitance
  • Current
  • Variations in the phase / frequency of an a.c. electrical signal
  • So, how convert output of sensors with Non-Voltage form?
  • Variable conversion elements
  • Particularly important: bridge circuits

12

Variable Conversion

slide-13
SLIDE 13

Bridge Circuits

  • Output: voltage level that changes as the measured physical quantity changes
  • Accurate method of measuring:
  • Resistance
  • Inductance
  • Capacitance
  • Enable: detection of very small changes (about nominal value)
  • So, immense importance in measurement
  • Many transducers measuring physical quantities: output
  • Expressed as a change in
  • Resistance, inductance, or capacitance
  • Example: a displacement-measuring strain gauge
  • Bridge excitation:
  • DC: for resistance
  • AC: for inductance, or capacitance
  • Bridge type:
  • Null type: calibration purposes
  • Deflection types: used in closed loop automatic control

13

Variable Conversion

slide-14
SLIDE 14

Null-Type d.c. Bridge (Wheatstone Bridge)

  • 4 arms of the bridge:
  • Unknown resistance Ru
  • 2 equal value resistors R2 and R3
  • variable resistor Rv (usually a decade resistance box)
  • A d.c. voltage Vi : applied across the points AC
  • Resistance Rv is varied  voltage across points BD = zero
  • Null point is measured with a high sensitivity galvanometer
  • Normally (high impedance voltage-measuring instrument):
  • Im = 0  I1 = I3 & I2 = I4
  • At the null-point:  if R2=R3  Ru = Rv

14

Variable Conversion

slide-15
SLIDE 15

Deflection-Type d.c. Bridge

  • Null-type: tedious to use, but highly accurate
  • Main difference: variable resistance Rv is replaced by
  • fixed resistance R1
  • (same value as the nominal value of unknown resistance Ru)
  • Relationship:
  • Simplify: high impedance voltage measurement
  • Nonlinear:
  • Easier to use (less accurate)  preferred in general

15

Variable Conversion

_

slide-16
SLIDE 16

Deflection-Type d.c. Bridge

  • Example:
  • A certain type of pressure transducer (range 0–10 bar)
  • Consists of a diaphragm with a strain gauge
  • The strain gauge has a nominal resistance of 120 O (one arm)
  • Other 3 arms each having a resistance of 120 O
  • Instrument’s input impedance can be assumed infinity
  • To limit heating effects, gauge current < 30 mA
  • Maximum permissible bridge excitation voltage?
  • Sensitivity of the strain gauge = 338 mO/bar
  • maximum bridge excitation voltage is used
  • bridge output voltage when measuring a pressure of 10 bar?
  • Solution:
  • I1 = current flowing in path ADC:
  • Balance: Ru = 120  Vi = 7.2
  • Pressure=10bar: Resistance change = 3.38 O  Ru = 123.38  Vo = 50mv

16

Variable Conversion

slide-17
SLIDE 17

Sensitivity of the bridge

  • How to deal with the non-linear relationship?
  • Special case: δ Ru << nominal value of Ru
  • New voltage:

 This is Bridge Sensitivity

  • Such approximation (linearization) is valid for transducers such as strain gauges
  • However, many instruments (say, resistance thermometers)
  • are inherently linear themselves (at least over a limited measurement range)
  • exhibit large changes
  • Other actions to improve linearity:
  • A common solution:
  • Make: R2, R3 > 10x R1, Ru (nominal)
  • The effect can be seen in an example

17

Variable Conversion

_ _ _

slide-18
SLIDE 18

Sensitivity of the bridge – (non-linearity)

  • Example:
  • a platinum resistance thermometer:
  • Range of 0 – 50oC
  • Resistance at 0oC = 500 O
  • Resistance varies with temperature at the rate of 4 O/oC
  • Over this range of measurement, the output characteristic

itself is nearly perfectly linear

  • Assuming:
  • Non-linear:
  • From 0 – 25: Vo change = 0.455-0 = 0.455
  • From 25 – 50: Vo change = 0.833-0.455 = 0.378 Now if R2,R3

18

Variable Conversion

_ _ And Vi=26.1

0.424 0.833

0.424 0.409

slide-19
SLIDE 19

Sensitivity of the bridge – (non-linearity)

  • In increasing the values of R2 and R3
  • Also necessary to increase the excitation from 10 to 26.1
  • To obtain the same output levels
  • In practical applications: Vi set at maximum
  • Consistent with limitation (heating)
  • maximize the measurement sensitivity (V0/dRu relationship)
  • If smart system is used, the importance
  • f this non-linearity is decreased

19

Variable Conversion

slide-20
SLIDE 20

Case: current drawn by measuring instrument is not negligible

  • For various reasons, it is not always possible to meet these conditions:
  • impedance of the instrument measuring the bridge output voltage is sufficiently large
  • Current drawn by it to be negligible
  • Wherever the measurement current is not negligible, an alternative

relationship between bridge input and bridge output must be derived

  • To take the current into account
  • Thevenin Theorem:
  • Replacing voltage source Vi by zero internal resistance:

20

Variable Conversion

slide-21
SLIDE 21

Case: current drawn by measuring instrument is not negligible

21

Variable Conversion

_

slide-22
SLIDE 22

Case: current drawn by measuring instrument is not negligible

22

Variable Conversion

_

Substituting Eo, RDB: Simplifying:

_

Example:

The bridge is used to measure the value of the unknown resistance Ru of a strain gauge of nominal value 500 O. The output voltage measured across points DB in the bridge is measured by a voltmeter. Calculate the measurement sensitivity in volts per ohm change in Ru if: (a) resistance Rm of the measuring instrument is neglected (b) account is taken of the value of Rm

slide-23
SLIDE 23

Case: current drawn by measuring instrument is not negligible

  • Solution:

a) For Ru = 500 O, Vm = 0. To determine sensitivity:

  • Calculate Vmfor Ru = 501 O
  • So, if resistance of the measuring circuit is

neglected, the measurement sensitivity is 5.00 mV per ohm change in Ru b) So, if 10Ko is considered, sensitivity decrease to 4.76 mv per ohm change in Ru

23

Variable Conversion

_ _ _

slide-24
SLIDE 24

Error Analysis

  • Contribution of component value tolerances to total measurement system

accuracy limits = ?

  • Null-Type:
  • maximum measurement error:
  • 1. Find Ru, each parameter at tolerance limit, producing Max Ru
  • 2. Find Ru, each parameter at tolerance limit, producing Min Ru
  • 3. Error band = the span (Max-Min)
  • Example:
  • Rv is a decade resistance box with a specified inaccuracy ±0.2% and R2 = R3 = 500 O ±0.1%
  • If the value of Rv at the null position is 520.4 O, determine the error band for Ru expressed as a

percentage of its nominal value

  • Solution:
  • with Rv = 520.4 O + 0.2% = 521.44 O, R3 5000 O + 0.1% = 5005 O, R2 = 5000 O - 0.1% = 4995 O, we

get:

24

Variable Conversion

slide-25
SLIDE 25

Error Analysis

  • with Rv = 520.4 O - 0.2% = 519.36 O, R3 5000 O - 0.1% = 4995 O, R2 = 5000 O + 0.1% = 5005

O, we get:

  • So, the error band for Ru is ±0.4%
  • Although the maximum error in any one component is ±0.2%, the possible error in the measured

value of Ru is ±0.4% !!

  • If magnitude of error is not acceptable  to overcome the introduction of error by

component value tolerances:

  • One method: is the introduction of apex balancing:
  • One way: placing an additional variable resistor R5 at junction C between resistances R2

and R3 and applying excitation voltage Vi to the wiper

25

Variable Conversion

_

slide-26
SLIDE 26

Error Analysis

  • For calibration purposes, Ru and Rv are replaced by two equal resistances whose values are

known accurately, and R5 is varied until output voltage V0 is zero. At this point, if the portions of resistance on either side of the wiper on R5 are R6 and R7 (such that R5 = R6 + R7), we can write:

  • We have thus eliminated any source of error due to tolerance in

the value of R2 and R3, and the error in the measured value

  • f Ru depends only on the accuracy of one component, the

decade resistance box Rv

  • Example:
  • A potentiometer R5 is put into the apex of the bridge shown

in Figure to balance the circuit. The bridge components have the following values: Ru = 500 O, Rv = 500 O, R2 = 515 O, R3 = 480 O, R5 = 100 O.

26

Variable Conversion

slide-27
SLIDE 27

Error Analysis

  • Determine the required value of resistances R6 and R7 of the parts
  • f the potentiometer track either side of the slider in order to

balance the bridge and compensate for the unequal values of R2 and R3

  • Solution:
  • For balance, R2 + R7 = R3 + R6 ; hence, 515 + R7 = 480 + R6
  • Also, because R6 and R7 are the two parts of potentiometer track R5
  • whose resistance is 100 O: R6 + R7 = 100
  • thus 515 + R7 = 480 + (100 - R7); that is, 2R7 = 580 - 515 = 65
  • Thus, R7 = 32.5  R6 = 100 - 32.5 = 67.5 O

27

Variable Conversion

slide-28
SLIDE 28

a.c. Bridges

  • Bridges with a.c. excitation are used to measure unknown impedances:
  • Capacitances
  • Inductances
  • Both types exist:
  • Null-type
  • Deflection-type
  • Null-type impedance bridge
  • null point can be detected conveniently by monitoring

the output with a pair of headphones connected via an operational amplifier across points BD

  • Less expensive than galvanometer

28

Variable Conversion

slide-29
SLIDE 29

Null-type impedance bridge

  • If Zu is capacitive, that is:
  • Then Zv must consist of a variable capacitance box, which is readily available
  • If Zu is inductive:
  • Note:
  • inductive impedance has a resistive term in it because it is impossible to realize a pure inductor
  • An inductor coil always has a resistive component, although this is made as small as possible

by designing the coil to have a high Q factor (Q factor is the ratio inductance/resistance)

  • Therefore, Zv must consist of a variable resistance box and a variable inductance box
  • However, the latter is not readily available because it is difficult and hence expensive to

manufacture a set of fixed value inductors to make up a variable inductance

  • box. For this reason, an alternative kind of null-type bridge circuit, known as the Maxwell

bridge

29

Variable Conversion

slide-30
SLIDE 30

Maxwell bridge

  • The requirement for a variable inductance box

is avoided by introducing instead a second variable resistance

  • The circuit requires 1-standard fixed-value

capacitor, 2-variable resistance boxes, and 1- standard fixed-value resistor 

30

Variable Conversion

slide-31
SLIDE 31

Maxwell bridge

  • Substituting:
  • Taking Real and Imaginary parts:
  • Can be used to calculate Q:
  • If a constant frequency ω is used, Q almost= R1
  • Thus, the Maxwell bridge can be used to measure the Q value of a coil directly

31

Variable Conversion

slide-32
SLIDE 32

Example

  • Example:
  • In the Maxwell bridge shown in Figure 9.8, let the fixed-value bridge components have the

following values: R3 = 5 O ; C = 1 mF .

  • Calculate the value of the unknown impedance (Lu, Ru) if R1 = 159 O and R2 = 10 O at

balance

  • Solution:
  • Substituting values into the relations developed in:
  • What is Q at 50Hz?

32

Variable Conversion