## Second-order Sensor Measurement Systems

On: 06 Jun, 2017

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A sensor is of second order when it contains two energy-storing elements and one energy-dissipating element. Its i...

A sensor is of second order when it contains two energy-storing elements and one energy-dissipating element. Its input x(t) and output y(t) are related by a second-order linear differential equation of the form
a2 d2y(t) + a1 dy(t) + aoy(t) = x(t) (1.9)

The corresponding transfer function is
Y(s) k& (110
X(s) s2 + 2ws +
where k is the static sensitivity, is the damping coefficient, and w is the natural frequency for the sensor. In order to determine the dynamic behavi or, 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 a0, a1, and a2 are 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. = 1, or > I. 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 damping coefficient to archive the same – 3 dB attenuation at that frequency.

A-3 dB attenuation means

3 = 20 log a, a = 10 = 0.707

The relative magnitude for a second-order response is

We want to have a magnitude of 0.707 at wn. Therefore From the above equation we obtain 21/2/2 = 0.707.

The dynamic error and delay in a second-order system depends not only on the input waveform but also on &, and . Their expressions are much more involved than in first-order systems, and to analyze them several fact ors related to and are defined.
When the input is a unit step, if the system is overdamped ( > 1) or is critically damped ( 1), there is neither overshoot nor dynamic error in the response.
In an underdamped system, < 1, the dynamic error is zero, but the speed and the overshoot are related (Figure 1 .9a). In general, the faster the speed, the larger the overshoot. The rise time r is defined as the time spent to rise from 10% to 90% of the final output value, and it is given by
= arctan(—d/) 114
where = is the so-called attenuation, and Wd w(l 2)I/2 is the damped natural frequency.
The time elapsed to the first peak t, is
(1.15)
and the maximum overshoot M is
Mpexp(_)7r (1.16)
The me for the output to settle within a defined band around the final value t, or settling time, depends on the width of that band. For 0< < 0.9, for a ±2% band, t 4I, and it is minimal when 0.76; for a ±5% band
3/s, and it is minimal when = 0.68. The speed of response is optimal for between 0.5 and 0.8 [6].
Figure 1 .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
2R
ed = (1.17)
and the delay is 2/w.

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
= w(1 22)l/2 (1.18)
and the amplitude of that resonance at & Wd is Mr.
Mr
2(1 2)l/2 (1.19)
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 of the 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 mas-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 x0 of the mass M with respect to the armature fixed to the element undergoing an acceleration I, then the equation for equilibrium is
M(x1 I) Kx0 + Bx0 (1.20)
where K is the spring constant and B is the internal viscous friction. The Laplace transform of I is s2X(s), from which we obtain
Ms2X1(s) = X0(s)[K + Bs + Ms21 (1.21)

from which the transfer function is
X0(s) M KIM (1 22)
s2X1(s) K s2 + s(BIM) + KIM
Therefore k = MIK; BI[2(KM)”2], and w, (KIM)”2.

To consider also the acceleration of gravity when the axis of the accelero meter 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(t) would be defined as x0 + (Mg sin 0)/K, and its Laplace transform would be given by (1.22), with Y(s) replacing X0(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 X0(s)1X1(s). We find that for acceleration measurements w must be higher than the maximal frequency variation of the acceleration to be measured but that for the measurement of vibration displacement w1, must be lower than the frequency of the displacement.

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