vb.net barcode reader code Figure 613 Equivalentcircuit representation of Figure 612 in Software

Make QR-Code in Software Figure 613 Equivalentcircuit representation of Figure 612

Figure 613 Equivalentcircuit representation of Figure 612
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Frequency Response and System Concepts
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The Wheatstone bridge circuit of Examples 210 and Focus on Measurements: Wheatstone Bridge in 2 is used in a number of instrumentation applications, including the measurement of force (see Example 213, describing the strain gauge bridge) Figure 614 depicts the appearance of the bridge circuit When undesired noise and interference are present in a measurement, it is often appropriate to use a low-pass lter to reduce the effect of the noise The capacitor that is connected to the output terminals of the bridge in Figure 614 constitutes an effective and simple low-pass lter, in conjunction with the bridge resistance Assume that the average resistance of each leg of the bridge is 350 (a standard value for strain gauges) and that we desire to measure a sinusoidal force at a frequency of 30 Hz From prior measurements, it has been determined that a lter with a cutoff frequency of 300 Hz is suf cient to reduce the effects of noise Choose a capacitor that matches this ltering requirement
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Filter capacitor c Vnoise + _ RT R1 Va
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R3 + a C b Vb R4 Wheatstone bridge equivalent circuit
+ VT _
Vout _
d Vout = Va _ Vb
Figure 614 Wheatstone bridge with equivalent circuit and simple capacitive lter
Solution:
By evaluating the Th venin equivalent circuit for the Wheatstone bridge, e calculating the desired value for the lter capacitor becomes relatively simple, as illustrated at the bottom of Figure 614 The Th venin resistance e for the bridge circuit may be computed by short-circuiting the two voltage sources and removing the capacitor placed across the load terminals: RT = R1 R2 + R3 R4 = 350 350 + 350 350 = 350
Since the required cutoff frequency is 300 Hz, the capacitor value can be computed from the expression 0 = or C= 1 1 = 151 F = RT 0 350 2 300 1 RT C = 2 300
Part I
Circuits
The frequency response of the bridge circuit is of the same form as equation 616: 1 Vout (j ) = VT 1 + j CRT This response can be evaluated at the frequency of 30 Hz to verify that the attenuation and phase shift at the desired signal frequency are minimal: 1 Vout (j = j 2 30) = VT 1 + j 2 30 151 10 6 350 = 09951 57 Figure 615 depicts the appearance of a 30-Hz sinusoidal signal before and after the addition of the capacitor to the circuit
10 5 Volts 0 Noisy sinusoidal voltage
_5 _10 0 008 016 t (s) 024 032
10 5 Volts 0
Filtered noisy sinusoidal voltage
_5 _10 0 008 016 t (s) 024 032
Figure 615 Un ltered and ltered bridge output
Focus on Computer-Aided Tools An EWB simulation of this circuit may
be found in the accompanying CD-ROM Much more complex low-pass lters than the simple RC combinations shown so far can be designed by using appropriate combinations of various circuit elements The synthesis of such advanced lter networks is beyond the scope of this book; however, we shall discuss the practical implementation of some commonly used lters in s 12 and 15, in connection with the discussion of the operational ampli er The next two sections extend the basic ideas introduced in the preceding pages to high- and band-pass lters that is, to lters that emphasize the higher frequencies or a band of frequencies, respectively High-Pass Filters Just as you can construct a simple lter that preserves low frequencies and attenuates higher frequencies, you can easily construct a high-pass lter that passes
6
Frequency Response and System Concepts
RC high-pass filter The circuit preserves higher frequencies while attenuating the frequencies below the cutoff frequency, v0 = 1/RC C + Vi _ R + Vo _
mainly those frequencies above a certain cutoff frequency The analysis of a simple high-pass lter can be conducted by analogy with the preceding discussion of the low-pass lter Consider the circuit shown in Figure 616 The frequency response for the high-pass lter, H (j ) = Vo (j ) Vi
may be obtained by noting that Vo (j ) = Vi (j ) R j CR = Vi (j ) R + 1/j C 1 + j CR (624)
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