Measurement noise and signal processing

 

5.5.2 Signal attenuation

One method of attenuating signals by analogue means is to use a potentiometer connected in a voltage-dividing circuit, as shown in Figure 5.11. For the potentiometer slider positioned a distance of x along the resistance element of total length L, the voltage level of the processed signal V₀ is related to the voltage level of the raw signal V_i by the expression:

Unfortunately, the potentiometer is unsuitable as a signal attenuator when it is followed by devices or circuits with a relatively low impedance, since these load the potentiometer circuit and distort the input–output V0/Vi relationship above. In such cases, an operational amplifier is used as an attenuator instead. This is connected in the same way as the amplifier shown in Figure 5.9, except that R₁ is chosen to be greater than R₂. Equation 5.3 still holds and therefore, if R₁ is chosen to be 10 MΩ and R2 as 1 MΩ, an attenuation factor of ten is achieved (gain = 0.1). Use of an operational amplifier as an attenuating device is a more expensive solution than using a potentiometer, but, apart from being relatively unaffected by the circuit that is connected to its output, it has further advantages in terms of its small size and low power consumption.

5.5.3 Differential amplification

Figure 5.12 shows a common amplifier configuration that is used to amplify the small difference that may exist between two voltage signals V_A and V_B. These may represent, for example, the pressures either side of an obstruction device put in a pipe to measure the volume flow rate of fluid flowing through it (see Chapter 16). The output voltage V₀ is given by:

                                                         V₀ = R₃/R₁ (V_B - V_A)

A differential amplifier is also very useful for removing common mode noise voltages. Suppose V_A and V_B in Figure 5.12 are signal wires such that V_A = +Vs volts and V_B = 0 volts. Let us assume that the measurement circuit has been corrupted by a common mode noise voltage Vn such that the voltages on the +Vs and 0 V signal wires become (Vs + Vn) and (Vn). The inputs to the amplifier V1 and V2 and the output V0 can then be written as:

If the resistance values are chosen carefully such that R4/R2 D R3/R1, then equation (5.4) simplifies to:
                                                      V0 = - R_3/R₁ (Vs)

i.e. the noise voltage Vn has been removed.