Sensor Specifications: Analog-to-Digital Conversion Specifications: Sytem Noise:
ID: 2079032 • Letter: S
Question
Sensor Specifications:
Analog-to-Digital Conversion Specifications:
Sytem Noise:
- There is signifigant noise at 20kHz during transmission between the amplifier and the A/D converter -->
The filter's cut-off frequency needs to be set at 1kHz above the highest frequency component of signal of interest.The passband gain needs to be 1.
(4) Assume the sampled input to the A/D converter is 3.58 volts. Determine the binary output.
WHAT I'VE DONE-
The Analog to Digital Converter opted will be 8 bit converter.
Since for the input of VMIN = 0 V & VMAX = 5V, there are minimum 5 voltage levels as the analog input so minimum 5 bits are required for that converstion but the converters are in the form of 2N where N is any natural number.
So there may be 21, 22, 23 ................ and so on.
so 2, 4, 8 bits are the possible converters and 5 lies in 8 bit converters so 8 bit converter is suitable.
Nyquist Frequency = 2 * highest frequency of the frequency spectrum.
= 2 * 4 kHz
= 8 kHz
for the given question sampling frequency is 120 % os Nyquist frequency.
so, Sampling Frequency = 120% os 8 kHz
= (120/100) * 8 kHz
= 9.6 kHz
(1) Assume the sampled input to the A/D converter is 3.58 volts. Determine the binary output.
Sensor Amplifier Filter A/D Data StorageExplanation / Answer
if we want to take 8 bit A/D converters
and Vmax = 5v
then the resolution is 5v/255 = 0.01960784313v
=>that is 19.6 mv
if we gave 19.6 mv to the 8 bit A/D converter it gives the digital value= 0000 0001
now our problem is if we give 3.58v to the A/D converter what is the binary value
the decimal value = 3.58/19.6m = 182.65
so almost it is 183
now we convert the 183 to binary = 10110111
so the final answer is 10110111
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