## Second-order Measurement Systems in dynamic characteristic

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A sensor is of second order when it contains two energy-storing elements and one energy-dissipating element. Its input x(r) and output y(t) are related by a second-order linear differential equation of the form where k is the static sensitivity, (...

A sensor is of second order when it contains two energy-storing elements and one energy-dissipating element. Its input x(r) and output y(t) are related by a second-order linear differential equation of the form where k is the static sensitivity, ( is the damping coefficient, and ro, is the natural frequency for the sensor. In order to determine the dynamic behavior, two coefficients are necessary while a single one determines the static behavior. The expressions for a general second-order system are

Notice that these three parameters are related and that a modification in one of them may result in a change in another one. Only ao, at, and a2are

independent. In [2] the detailed procedure is given to obtain the expression of the output as a function of the input when this input is one of the simple test waveforms. Table 1.4 gives the results. Figure 1.9 shows the graphical characteristics. Note that the system behavior differs for 0 < ( < 1, {: l, or

( > l. The initial transient has been omitted for the sinusoidal input.

Example: In a measurement system, a first-order sensor is replaced by a second-order sensor with the same natural (corner) frequency. Calculate the

From the above equation we obtain ( : 2tt2l2 : 0.707. The dynamic error and delay in a second-order system depends not only on the input waveform but also on ar, and (. Their expressions are much more involved than in first-order systems, and to analyze them several factors related to c'rn and ( are defined. When the input is a unit step, if the system is overdamped (( > l) or is critically damped (( : 1), there is neither overshoot nor dynamic error in the response. In an underdamped system, ( < l, the dynamic error is zero, but the speed and the overshoot are related (Figure l.9a).In general, the faster the speed, the larger the overshoot. The rise time r. is defined as the time spent to rise from lDVo to 90% of the final output value, and it is given by

where 6 : (r, is the so-called attenuation, and a.r6 : arn(l - (2)l/2 is the damped natural frequency. The time elapsed to the first peak ro is

The Time for the output to settle within a defined band around the final value 1., or settling time, depends on the width of that band. For 0 < ( < O.g,

for a'+2Vo band, l, = 416, and it is minimal when ( : 0.76;for a -+5Voband ,. = 3/6, and it is minimal when ( : 0.68. The speed of response is optimal

for ( between 0.5 and 0.8 t6l. Figure l.9a suggests that underdamped sensors are useless because they exhibit a large overshoot. But, in practice, the input will never be a perfect step and the sensor behavior may be acceptable. That is the case with piezoelectric sensors (Section 6.2), for example.

When the input is a ramp, the dynamic error is

To describe the frequency response of a second-order system where 0 (( < 0.707, we note that the frequency of resonance is the same as the damped natural frequency

A simple example of a second-order sensor described by a transfer-function like equation (1.10) is a thermometer covered for protection. In this case

we must add the heat capacity and thermal resistance of the covering to the heat capacity ofthe sensing element and heat conduction resistance from the medium where it is placed. The system is an overdamped one. Examples of underdamped systems are the masB-spring systems used to measure displacement, velocity, and acceleration in vibratory movements or in long-range missiles. They are also the heart of seismographs. Using the notation of Figure 1.10, if we measure the displacement xo of the mass M with respect to the armature fixed to the element undergoing an acceleration ii, then the equation for equilibrium is

Therefore k : MIK; (: Bll2(KM)r/21, and 6n: (KlM)|t2. To consider also the acceleration of gravity when the axis of the accelerometer forms an angle 0 with respect to the horizontal, the term Mgsin 0 should be included in the right-hand member of (1.20). Then the output y(l) would be defined as .ro + (Mgsin 0)lK, and its Laplace transform would be given by (1.22), with Y(s) replacing Xo(s).

If instead of the input acceleration we want to obtain the displacement, we would multiply both sides of (1.22) by s2 to obtain X"(s)/&(s). We find that for acceleration measurements rrrn must be higher than the maximal frequency variation of the acceleration to be measured but that for the measurement of vibration displacement ar, must be lower than the frequency of the displacement.

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