Chapter 5 SENSORS, ACTUATORS, AND OTHER CONTROL SYSTEM COMPONENTS 5.1
A continuous voltage signal is to be converted into its digital counterpart using an analog-to-digital converter. The maximum voltage range is ± 30 V. The ADC has a 12-bit capacity. Determine: (a) number of quantization levels, (b) resolution, (c) the spacing of each quantization level, and the quantization error for this ADC. Solution : Number of quantization levels = 2 12 = 4096
R ADC ADC =
60 4096
= 0.01465 volts
−1
Quantization error = ± (0.01465)/2 =
5.2
A voltage signal with a range of zero to 115 V. is to be converted by means of an ADC. Determine the minimum number of bits required to obtain a quantization error of (a) ± 5 V maximum, (b) ± 1 V maximum, (c) ± 0.1 V maximum. Solution : (a) ± 5 volts max = ±
5.0 =
0.5(115 ) 2
n
−1
0.5(115 ) 2
n
−1
1 2
R ADC
2n
−1
n 1n(2) = 1n(576)
1 Range
n 2 2 − 1 2n = 12.5
=
1 Range
n 2 2 − 1 2n = 58.5
n = 4.069/0.693 = 5.87 → Use n = 6
(c) ± 0.1 volt max = ±
0.5(115 )
=
(2n-1) = 0.5(115)/1 = 57.5
,
n 1n(2) = 1n(58.5)
0.1 =
2
R ADC
n = 2.526/0.693 = 3.64 → Use n = 4
(b) ± 1 volt max = ± 1.0 =
1
(2n-1) = 0.5(115)/5 = 11.5,
,
n 1n(2) = 1n(12.5)
5.3
0.00732 volts
,
1 2
R ADC
=
1 Range
n 2 2 − 1
(2n-1) = 0.5(115)/0.1 = 575.0
2n = 576.0
n = 6.356/0.693 = 9.17 → Use n = 10
A digital-to digital-to-analo -analog g converter converter uses uses a referenc referencee voltage voltage of 120 120 V dc and has eight eight binary binary digit digit precision precision.. In one of the sampling instants, the data contained in the binary register = 01010101. If a zero-order hold is used to generate the output signal, determine the voltage level of that signal. Solution : Vo = 120{0.5(0) + 0.25(1) + 0.125(0) + 0.0625 (1) + 0.03125(0) + 0.015625(1) + 0.007812(0) +0.003906(1)} Vo = 39.84 volts
5.4
A DAC DAC uses uses a reference reference voltage voltage of 80 80 V and has 6-bit precision. precision. In four successive successive sampling sampling periods, periods, each each 1 second long, the binary data contained in the output register were 100000, 011111, 011101, and 011010. Determine the equation for the voltage as a function of time between sampling instants 3 and 4 using (a) a zero-order hold, and (b) a first-order hold. Solution : First sampling instant: 100000, V o = 80(0.5) = 40.0 volts Second sampling instant: 011111, Vo = 80(0.25 + 0.125 + 0.0625 + 0.03125 + 0.015625) = 38.75 volts Third sampling instant: 011101, Vo = 80(0.25 + 0.125 + 0.0625 + 0.015625) = 36.25 volts Fourth sampling instant: 011001, Vo = 80(0.25 + 0.125 + 0.015625) = 31.25 volts (a) Zero order hold: V(t) = 36.25 between instants 3 and 4
(b) First First order hold: hold: V(t) V(t) = 36.25 + a t between instants instants 3 and and 4 a = (36.25 - 38.75)/1 38.75)/1 = -2.5 -2.5 V(t) = 36.25 - 2.5t
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5.5
In Problem 5.4, suppose that a second order hold were to be used to generate the output signal. The equation for the second-order hold is E(t) = E0 + α t + β t2
(5.8)
where E0 = starting voltage at the beginning of the time interval. (a) For the binary data given in Problem 5.4, determine the values of α and β that would be used in the equation for the time interval between sampling instants 3 and 4. (b) Compare the first-order and second-order holds in anticipating the voltage at the 4th instant. Solution : t = 0: V(t) = 36.25 = 36.25 + a(0) + b(0) t = -1: V(t) = 38.75 = 36.25 + a(-1) + b(1) t = -2: V(t) = 40.0 = 36.25 + a(-2) + b(4) Simultaneous solution yields a = -3.125 and b = -.625 V(t) = 36.25 - 3.125t - .625t 2 At the fourth instant, the second order hold yields V(t) = 36.25 - 3.125(1) - .625(1) = 32.5 volts At the fourth instant, the first order hold yields V(t) = 36.25 - 2.5(1) = 33.75 volts The actual voltage level at the fourth instant is 32.5 volts. Hence, the second order hold more acurately projects the voltage.
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