barcode reader vb.net codeproject Integrating a Square Wave in Software

Generation QR Code ISO/IEC18004 in Software Integrating a Square Wave

EXAMPLE 126 Integrating a Square Wave
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vS(t) A
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Determine the output voltage for the integrator circuit of Figure 1231 if the input is a square wave of amplitude A and period T
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Known Quantities: Feedback and source impedances; input waveform characteristics
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Operational Ampli ers
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Find: vout (t) Schematics, Diagrams, Circuits, and Given Data: T = 10 ms; CF = 1 F; RS =
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10 k vout (0) = 0 integrator: vout (t) = = 1 RF CS 1 RF CS
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t
Assumptions: Assume ideal op-amp The square wave starts at t = 0 and therefore Analysis: Following equation 1267, we write the expression for the output of the
vS (t )dt =
1 R F CS
0
vS (t )dt +
vS (t )dt
vout (0) +
vS (t )dt
Next, we note that we can integrate the square wave in a piecewise fashion by observing that vS (t) = A for 0 t < T /2 and vS (t) = A for T /2 t < T We consider the rst half of the waveform: vout (t) = 1 RF CS
vout (0) +
vS (t )dt T 2
= 100 0 +
= 100At vout (t) = vout T 2
0 t < 1 RF CS
vS (t )dt = 100A
T /2
T 100 2
( A) dt
T /2
vout(t) 0 T 2 50 AT T t
= 100A
T T + 100A t 2 2
= 100A (T t)
T t <T 2
Since the waveform is periodic, the above result will repeat with period T , as shown in Figure 1232
Comments: The integral of a square wave is thus a triangular wave This is a useful fact
to remember Note that the effect of the initial condition is very important, since it determines the starting point of the triangular wave
FOCUS ON MEASUREMENTS
Charge Ampli ers
One of the most common families of transducers for the measurement of force, pressure, and acceleration is that of piezoelectric transducers These transducers contain a piezoelectric crystal that generates an electric charge in response to deformation Thus, if a force is applied to the crystal (leading to a displacement), a charge is generated within the crystal If the external force generates a displacement xi , then the transducer will generate a charge q according to the expression q = Kp xi Figure 1233 depicts the basic structure of the piezoelectric transducer, and a simple circuit model The model consists of a current source in parallel with a capacitor, where the current source represents the rate of change of the
Part II
Electronics
Electrodes + Leads i= dq dt C vo Circuit model
Crystal Piezoelectric transducer
Figure 1233 Piezoelectric transducer
charge generated in response to an external force and the capacitance is a consequence of the structure of the transducer, which consists of a piezoelectric crystal (eg, quartz or Rochelle salt) sandwiched between conducting electrodes (in effect, this is a parallel-plate capacitor) Although it is possible, in principle, to employ a conventional voltage ampli er to amplify the transducer output voltage, vo , given by vo = 1 C i dt = 1 C Kp xi q dq dt = = C dt C
it is often advantageous to use a charge ampli er The charge ampli er is essentially an integrator circuit, as shown in Figure 1234, characterized by an extremely high input impedance3 The high impedance is essential; otherwise, the charge generated by the transducer would leak to ground through the input resistance of the ampli er
CF iF (t) i= dq dt i in 0 C iC 0 + Transducer + vout(t)
Figure 1234 Charge ampli er
Because of the high input impedance, the input current into the ampli er is negligible; further, because of the high open-loop gain of the ampli er, the inverting-terminal voltage is essentially at ground potential Thus, the voltage across the transducer is effectively zero As a consequence, to satisfy KCL, the feedback current, iF (t), must be equal and opposite to the transducer current, i: iF (t) = i
3 Special
op-amps are employed to achieve extremely high input impedance, through FET input circuits
12
Operational Ampli ers
and since vout (t) = 1 CF iF (t) dt
it follows that the output voltage is proportional to the charge generated by the transducer, and therefore to the displacement: vout (t) = 1 CF i dt = 1 CF Kp xi dq q = dt = dt CF CF
Since the displacement is caused by an external force or pressure, this sensing principle is widely adopted in the measurement of force and pressure
The Ideal Differentiator Using an argument similar to that employed for the integrator, we can derive a result for the ideal differentiator circuit of Figure 1235 The relationship between input and output is obtained by observing that iS (t) = CS
+ + vout(t)
RF CS vS(t)
iF (t)
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