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    GENERALIZED PERFORMANCECHARACTERISTICS OF

    INSTRUMENTS

    Lecture 8Instructor : Dr Alivelu M Parimi

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    Example

    2

    IQ scores of result are normally distributed to mean of 100 and standard deviation of 13.

    Find (i) Fraction of students having an IQ > 133

    (ii) Fraction of students having an IQ > 90

    (iii) IQ value exceeded by upper quartile (from 75% to 100%)

    Solution

    (i) IQ >133

    z =13

    100-133= 2.54, Area corresponding to z = 2.54 is 0.4945, students having IQ 133

    are 0.5 + 0.4945 =0.9945, i.e. 99.45% have IQ 133

    Fraction of students having IQ > 133 are ( 0.50.4945 = .0055) . i.e. 0.55% will have IQ >

    133

    (ii) For IQ = 90, z =13

    100-90=0.77 Area from 0 to 0.77 = 0.2799 which is same as

    area for z = -0.77 to 0.

    Total area 0.5 + 0.2794 = 0.7794, i.e. 77.94% will have IQ score greater than 90

    (iii) Upper quartile implies area > 0.75

    from Table 3.2, find z corresponding to area of 0.25 which is z

    xx = 0.6745

    So xx = 0.6745 x 13 = 8.76,

    x = 108.76, hence upper quartile will be for students having IQ > 108.76

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    Gaussian distribution

    3

    Table 3.2: Values of Area under Gaussuain Distribution Curve

    Table list areas under the

    curve between zero andvarious values of z

    This table will have areas

    from 0 to maximum 0.5

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    DYNAMIC PERFORMANCE

    CHARACTERISTICS Instruments rarely respond instantaneously to changes in the

    measured variables.

    Instead, they exhibit slowness or sluggishness due to factors

    such as mass, thermal capacitance, fluid capacitance or

    electrical capacitance. In addition, pure delay in time is often encountered where the

    instrument waits for some reaction to take place.

    Such industrial instruments are nearly always used for

    measuring quantities that fluctuate with time. Therefore the dynamic and transient behavior of the

    instrument is as important as the static behavior.

    5

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    DYNAMIC PERFORMANCE

    CHARACTERISTICS The parameters characterizing the dynamic characteristics of

    an instrument are:

    Speed of response: It is the rapidity with which an instrument

    responds to changes in the measured quantity.

    Fidelity: It is the degree to which an instrument indicates, thechanges in the measured variable without dynamic

    error(faithful reproduction)

    Lag: It is the retardation delay in the response of an

    instrument to changes in the measured variable. Dynamic error: It is the difference between the true value of a

    quantity changing with time and the value indicated by the

    instrument . 6

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    DYNAMIC PERFORMANCE

    CHARACTERISTICS When measurement problems are concerned with rapidly

    varying quantities, the dynamic relations between the input

    and output are generally defined by the use of differential

    equations. For most measuring instruments, generalized

    relation between any particular input x and output y can bewritten as:

    7

    ann

    n

    dt

    yd + an-1

    1-n

    1-n

    dt

    yd . .+ a1

    dt

    dy+ a0y = bn

    n

    n

    dt

    xd + bn-1

    1-n

    1-n

    dt

    xd..+ b1

    dt

    dx+ b0x

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    DYNAMIC PERFORMANCE

    CHARACTERISTICS Types of Standard Inputs

    Once response of an instrument to certain standard inputs is

    known, its response to more complicated inputs can be

    determined using mathematical techniques.

    Step input:System is subjected to an instantaneous and finitechange in measured variable.

    8

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    DYNAMIC PERFORMANCE

    CHARACTERISTICS Linear/Ramp input: System is subjected to a measured

    variable which is changing linearly with time. Like

    thermometer is put in a bath whose temperature is rising

    linearly.

    9

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    DYNAMIC PERFORMANCE

    CHARACTERISTICS Pulse/Impulse input:System is subjected to step magnitude

    for small time and then is brought back to its initial state.

    When this small time is tending to zero, pulse becomes

    impulse, which is a theoretical concept not realizable

    practically as it implies in no time system is subjected to largechange. If the automobile running on road suddenly hits a pot

    hole on the road, then the force experienced by shock

    absorbing system may be modeled by impulse function

    10

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    DYNAMIC PERFORMANCE

    CHARACTERISTICS Sinusoidal input: System is subjected to measured variable,

    the magnitude of which changes in accordance with a

    sinusoidal function of constant amplitude.

    11

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    DYNAMIC PERFORMANCE

    CHARACTERISTICS Order of a System

    The order of the system relates to the number of energy

    storage elements in the measurement system.

    Zero order,

    First order,

    Second order instrument

    12

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    DYNAMIC PERFORMANCE

    CHARACTERISTICS Order of a System: examples

    For example, if you use a mercury-in-glass thermometer to

    measure temperature there is a single energy store (the heat

    capacity of the mercury), thermometer acts predominantly

    like a Firstorder system. RC and RL circuits are examples of firstorder system as only

    electrostatic or electromagnetic energy is involved.

    RLC circuit is an example of secondorder system as both

    electromagnetic and electrostatic energies are involved. Accelerometers and seismographs having mass-spring-damper

    arrangement have two energy storesthe spring (elastic

    potential energy) and the mass (kinetic energy).

    Thus, these devices respond like Secondorder systems.

    13

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    DYNAMIC PERFORMANCE

    CHARACTERISTICS Zero Order instruments

    14

    ann

    n

    dt

    yd + an-1

    1-n

    1-n

    dt

    yd . .+ a1

    dt

    dy+ a0y = bn

    n

    n

    dt

    xd + bn-1

    1-n

    1-n

    dt

    xd..+ b1

    dt

    dx+ b0x

    y =0

    0

    a

    bx

    Since y = Kx is algebraic, it is clear that no matter how x

    might vary with time, the instrument output (readings )follows it perfectly with no distortion or time lag of any

    sort. Thus, the zero order system represents ideal or

    perfect dynamic performance and is thus a standard

    against which less perfect instruments may be compared.

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    DYNAMIC PERFORMANCE

    CHARACTERISTICS A practical instrument of a zero order type is the

    displacement potentiometer shown in Figure, where a strip of

    resistance material is excited with a voltage and provided with

    a sliding contact. If the resistance is disturbed linearly along

    length L, we may write

    15

    e0 =L

    xiEb= K xi

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    DYNAMIC PERFORMANCE

    CHARACTERISTICS An ideal potentiometer acts like a zero order instrument.

    Output reflects input faithfully without any time delay.

    Ideal Operational amplifier used as an inverter is a zero order

    system with a gain of (1.).

    A wire strain gauge in which the change in the electrical

    resistance of the wire is proportional to the strain in the wire

    is also an example of zero order system.

    16

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    DYNAMIC PERFORMANCE

    CHARACTERISTICS Step response of a zero order instrument

    In a zero order instrument, y (t) = Kx (t), we have y (t) = 0 for t

    < 0, and y = K for t 0. Therefore the response to the unit step

    function is a step function with height K

    17

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    Dynamic performance characteristics:

    First Order instruments

    18

    The governing equation for the First order system is

    a1dt

    dy+ a0 y = b0 x

    Any instrument which follows this equation is by definition a first order instrument.

    The equation can be rewritten as

    0

    1

    a

    a

    dt

    dy+ y =

    0

    0

    a

    bx

    =0

    1

    a

    a, where is the time constant.

    K =0

    0

    a

    b, where K is static sensitivity/gain.

    Kxydt

    dy

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    Dynamic performance characteristics:

    First Order instruments

    Example

    Time constant of measurement system is determined by thephysical characteristics of the system.

    A measurement system with a small time constant will

    respond quicker to changes than a system with a large timeconstant.

    The time constant depends on the size of the energy storageelement as well as how quickly the energy can get in/out ofthat storage element.

    For example, mercury in glass thermometer will respondquicker when plunged into water than it does in air (fasterheat transfer with water).

    A thermometer with a large bead will respond slower than athermometer with a small bead

    19

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    Dynamic performance characteristics:

    First Order instruments

    Step response of first order instruments

    Assuming system is in equilibrium at t = 0 and then a step

    input of magnitude A is applied.

    20

    Kxydtdy

    y(t) = AK (1e- t/)

    Steady state value ( as t ) of y(t) = AK = yss

    Error = em= inputoutput =

    /t/t Ae)e1(AAK/)t(y)t(x

    Steady state error ( as t ) = em,ss= A

    So /

    ,

    t

    ssm

    m

    e

    e

    e

    X(s)Y(s) =

    )s(1K

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    Dynamic performance characteristics:

    First Order instruments

    21

    Nondimensional Response of First Order

    System to Step Input

    Non Dimensional Error Curve of First Order

    System

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    Dynamic performance characteristics:

    First Order instruments

    Ramp response of first order instruments

    Measuring instruments may be placed in an environment

    whose temperature, pressure is increasing or decreasing in

    linear (ramp) fashion.

    22

    X(s)

    Y(s)=

    )s(1

    K

    X(s) =2s

    M

    KM

    y(t)= -+ t + e

    -t/

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    Dynamic performance characteristics:

    First Order instruments

    23

    K

    )t(y= M ( t - + e

    -t/)

    At larger values of t, the exponential term will tend to zero.

    y = KM ( t - ); Input = Mt, implying there is a steady state lag of

    Measurement error emis given by:

    em = InputOutput = x(t)y(t)/K = MtM ( t - + e-t/

    )

    em = M- Me-t/

    = M( 1- e-t/

    )

    Steady state error = em,ss( emas t ) = M, Transient error = Me-t/

    For em,ssto be small, M and should be small.

    ssm,

    m

    ee = 1- e

    -t/

    At steady state, measurement error is equal to the steady state error.

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    Dynamic performance characteristics:

    First Order instruments

    Impulse response of the first order system

    A unit impulse function is defined as a signal which has zero

    values at all times except at t = 0, where its magnitude is

    infinite. In real, practical systems, it is not possible to produce

    a perfect impulse to serve as input for testing. Therefore, abrief pulse is used as an approximation of an impulse.

    Provided that the pulse is short compared to the impulse

    response, the result will be near enough to the true,

    theoretical, impulse response

    24

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    Dynamic performance characteristics:

    First Order instruments

    25

    For 0 < t < T:

    ( D + 1) y = K x =T

    KA

    Since, up until time T, this is no different from a step input of sizeT

    A. For initial condition y

    = 0 at t = 0, the complete solution is

    y =T

    KA ( 1-e

    -t/)

    at t = T, y =T

    KA ( 1-e

    -T/) (A)

    For t > T, x = 0

    Therefore, (D+1) y = Kx = 0

    y(t) = Ce-t/

    Therefore, y(t) = Ce-t/

    (B)

    At t = T, y(t) = Ce-T/

    Y(t) = Ce

    -t/

    = T

    KA

    /

    -T/e-1Te

    e

    -t/

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    Dynamic performance characteristics:

    First Order instruments

    26

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    Dynamic performance characteristics:

    First Order instruments

    Response of FOS to a sinusoidal input function

    The response of a system to sinusoidal input is known as the

    frequency response.

    If the input is sinusoidal, the steady state output is also

    sinusoidal.

    The output frequency equals the input frequency and the

    output phase lags behind the input phase.

    It is important to study the sinusoidal response as many

    natural phenomena are sinusoidal in character like vibration ofguitar string, currents in oscillating circuit. In generation and

    transmission of electric power and communication also

    sinusoidal signals are used. 27

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    Dynamic performance characteristics:

    First Order instruments

    28

    For example: if x = xmsin t then

    (t) =)1(

    )tsin(Kx22

    m

    On comparing the input with the output, it is observed that

    1. Frequency does not change in first order system.

    2. There is attenuation in amplitude by a factor)(1

    K

    22

    .

    3. Output lags behind the input by tan-1

    (- ).

    The frequency response has two elements which can be plotted graphically

    (i) The ratio of output amplitude to input amplitude plotted against frequency.

    (ii) The phase lag between output and input as a function of frequency.

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    Dynamic performance characteristics:

    First Order instruments

    29

    Frequency response is often specified by itsbandwidth: this is the range of frequency over

    which variation in magnitude is less than 70%

    (3dB).

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    Problem

    A certain thermometer has time constant of 15 sec and an

    initial temperature of 200 C. It is suddenly exposed to

    temperature of 1000C. Determine the time it would take to

    reach 90% of step size.

    30

    Solution

    = 15 sec. Step size = final valueinitial value = 10020 = 800C

    90 % of 800

    = 720C

    Therefore, total rise = 200C + 72

    0C = 92

    0C

    Y (t) = AK (1-e

    -t/

    ) + 20

    (assuming K=1)

    A = 800C

    92 = 80 (1-e-t/

    ) + 20 therefore t = 34.54 sec.

    The thermometer will take 34.54 seconds to reach 90 % of the step size.

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    Problem

    A mercury thermometer is kept in a bath of 400C and bath is

    heated such that its temperature increases linearly at the rate

    of 50C/min. Calculate the thermometer reading after 5

    minutes when is (i) 10 sec (ii) 2 min.

    31

    (i) = 10 sec.

    After 5 minute, actual temperature becomes 40 + 5*5 = 650C. But due to error thermometer

    indicates different value.

    K

    )t(Y= M (t+ e

    t/)

    t

    = sec30

    min5

    = 30 .

    Transient term is 10 e30

    0 and can be neglected

    Steady state error = M= 5min

    0C

    * min60

    10= 0.83

    0C

    Thermometer reading = 650.83 = 64.170C

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    Solution contd

    32

    (ii) = 2 min,2

    5t

    = 2.5

    et/

    = e2.5

    = 0.165

    Y (t) = M (t+ e2..5) + 40

    Y (t) = M ( 52 + e2.5) + 40 = 55.820C

    Error at t = 5 min is 6555.82 = 9.180C

    Steady state error = M= 5x 2 = 100C

    Note: Smaller is the time constant lesser is the steady state error.