Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Transcript

4.1 Remainder 147 2 73 2 1 36 2 1 18 2 0 92 0 42 1 22 0 12 0 0 1 4.9 To find the decimal value for 10010001, we must first subtract 1 from the LSB thus giving: 10010000 Next, we invert the 1's and 0's as follows: 01101111 Finally, we evaluate the decimal: N10 027 126 125 024 123 122 121 120 111 N10 111 4.23 Quantization error is computed using equation 4.1. Vru Vrl 2N For 8 bits this becomes 8 (8) input res error 0.5 0.0313V and this is 0.42% of the 7.5V input. 28 For 12 bits this becomes 8 (8) input res error 0.5 0.00195V and this is 0.026% of the 7.5V input. 212 For 16 bits this becomes 8 (8) input res error 0.5 0.000122 V and this is 0.0016% if the 7.5V input. 216 input res error 0.5 4.27 Since the signal from the transducer varies between 15mV (0.015V) and the A/D converter input range is 10V, we can select a gain of 100 which will yield an input of 1.5V. A gain of 100 is chosen such that the amplified signal is not saturated (i.e. greater than the input range). The quantization error from Eq. 4.1 is as follows: V V Quantization Error 0.5 ru N rl Volts 2 10 10 0.5 212 2.44 10 3 Volts The transducer voltage is 3.75mV but after a gain of 100 it becomes 0.375V. Thus, the quantization error as a percent reading is as follows: 2.44 10 3Volts 100 0.651% 0.375Volts If the transducer output were attenuated by a factor of 2/3 (to 10 mV) the gain could be set to 1000 without saturating the A/D converter. The 3.75 mv output would then become 3.7510-3(2/3)1000 = 2.5 V at the input to the A/D converter and the resolution error would be reduced to 0.098%. CHAPTER 5: 5.3 V(t) = 25t 0 t 1 T = 1sec o = 2f = 2(1/T) = 2(1/1) = 2 rad/sec By Eq. 5.5. 2 T bn f (t ) sin n o tdt T 0 Thus, 2 1 b1 25t sin( 2t )dt 1 0 = 2(-3.9789) = -7.9578 2 1 25t sin( 2 2t )dt 1 0 = 2(-1.9894) = -3.9788 b2 ao is the average, Eq. 5.3 1 T a o f t cos n o tdt T 0 Thus, 25 0 1 2 1 1 25t cos(0)dt 1 0 = 12.5 a0 and an 2 T f t cos no tdt T 0 2 1 25t cos(2t )dt 1 0 =0 a1 2 1 25t cos(2 2t )dt 1 0 =0 a2 5.16 The function has frequencies of 5 and 20 Hz. The minimum sampling rate would then be 40 Hz to avoid aliasing. Sampling at 30 Hz would not produce aliases for the 5 Hz signal but would for the 20 Hz signal. The alias frequency can be evaluated using the folding diagram in Section 5.1. fN = 30/2 = 15 Hz. f/fN = 20/15 = 1.3333. From the folding diagram, fa/fN = 0.666. Thus fa = 0.66615 = 10 Hz. This is the difference between the sampling rate and the signal frequency. 5.19 The maximum frequency in the signal of problem 5.11 is 20 Hz. The sampling rate should exceed twice this value, or 40 Hz. 5.26 Use Appendix A-3. fN = 1500/2 = 750 Hz. f/fN = 1000/750 = 1.333. From Figure A.1, fa/fN = 0.666. Thus fa = 0.666*750 = 500 Hz. 147 2 73 2 36 2 18 2 9 2 4 2 2 2 1 2 0 11001001 From the remainders, we get the 8-bit number as follows: N2 = 1 0 0 1 0 0 1 1 4.9 To find the decimal value for 10010001, we must first subtract 1 from the LSB thus giving: 10010000 Next, we invert the 1's and 0's as follows: 01101111 Finally, we evaluate the decimal: 4.23 Quantization error is computed using equation 4.1. For 8 bits this becomes and this is 0.42% o For 12 bits this becomes For 16 bits this becomes and this is 0.0016% if the 7.5V input. V err 4.27 is chosen such that the amplified signal is not saturated (i.e. greater than the input range). The quantization error from Eq. 4.1 is as follows: ts The transducer voltage is 3.75mV but after a gain of 100 it becomes 0.375V. Thus, the quantization error as a percent reading is as follows: to 1000 without saturating the A/D converter. The 3.75 mv output would then become -3 on error