Analog to Digital Converter (Estimation of Input Voltage)

AI Thread Summary
To estimate the input voltage for a 0-10 V, 10-bit A/D converter displaying the binary output 1010110111, the resolution is calculated as 9.77 mV per code. Converting the binary output to decimal gives a value of 683, which is then multiplied by the resolution to find the input voltage. The resulting input voltage is approximately 6.67 V, with a quantization error of ±4.88 mV due to the 1 LSB consideration. Visualizing the transfer function can aid in understanding the relationship between input and output voltages. This approach clarifies the estimation process for the input voltage based on the A/D converter's output.
GreenLRan
Messages
59
Reaction score
0

Homework Statement



A 0-10 V, 10-bit A/D converter displays an output in straight binary code of 1010110111. Estimate the input voltage to within 1 LSB (Least Significant Bit).

Homework Equations



Resolution = Efsr/2^M. (where M= 10 in this case)

Eout = X/2^M (where Eout is the output Voltage, X is the actual input binary number [I'm also not sure if this equation is valid... I found it under the digital to analog converter section rather than analog to digital])

The Attempt at a Solution



I calculated the resolution and got it to be 10 V/2^10 = 9.77 mV. I am unsure of what to do next (my book is terrible). I was thinking of converting the output binary (which i believe to be the output voltage) to base 10, then I didn't know if there was an equation for the input Voltage as a function of Output voltage.


I also thought about taking the quantization error to be 1/2 the resolution. the somehow getting the input voltage from that.

Honestly, I am lost. Any help would be great. Thanks!
 
Physics news on Phys.org
You're making this too hard.

You found the resolution, which is 9.77 mV per code. An input of 0 V to 9.77 mV would produce an output of 00 0000 0000. An input of 9.77 mV to 19.54 mV would produce an output of 00 0000 0001. Each additional step represents 9.77 mV.

Convert the 10-bit output code to decimal -- that's the number of codes above zero -- and multiply it by 9.77 mV. That's the boundary between the given 10-bit code and the next one up.

You might want to draw a picture of the transfer function (not all 2^10 codes of course) to help you visualize the behavior.

- Warren
 
Back
Top