An analog-to-digital converter (ADC) samples an analog voltageand computes a dig

Question

An analog-to-digital converter (ADC) samples an analog voltageand computes a digital value representing that voltage. Assume anADC accepts voltages between -1.0 V and +1.0 V and hte digitalvalues are 16-bit, two's complement fractions. What is therepresetation of the following voltages: A. 0.875 V B. -0.875 V C. -0.1 V An analog-to-digital converter (ADC) samples an analog voltageand computes a digital value representing that voltage. Assume anADC accepts voltages between -1.0 V and +1.0 V and hte digitalvalues are 16-bit, two's complement fractions. What is therepresetation of the following voltages: A. 0.875 V B. -0.875 V C. -0.1 V

Explanation / Answer

One way of converting a decimal into its binary is to doubleit and drop any 1 that carries out. If the 1 carries out,that bit is a 1; if no overflow, it is a zero.

A) .875 * 2 = 1.75 bit = 1, remove theinteger part    .750 * 2 = 1.50 bit =1, remove the integer part    .500 * 2 = 1.00 bit =1, remove the integer part    0 *2 = 0 for all lower orderbits.
So, answer is 111. Adding 0 sign bit and the rest of thebits gives 0111 0000 0000 0000

B) -.875: Use the above answer but convert tonegative using 2's complement
Answer is 1000 1111 1111 1111 + 1 in the lowestbit Answer is 1001 0000 0000 0000
C) -0.1: First find +0.1, convert later:
.1 * 2 = .2, bit = 0 .2 * 2 = .4, bit = 0 .4 * 2 = .8, bit = 0 .8 * 2 = 1.6, bit = 1, remove integer .6 * 2 = 1.2, bit = 1, remove integer .2 * 2 = .4, bit =0
At this point we can see that we have a pattern that willrepeat. Here's the positive answer:
0000 1100 1100 1101 Rounded up at 16 bits, asbit 17 is a 1.
Make negative with 2's complement
1111 0011 0011 0010 + 1 in the lowest bit gives
answer is 1111 0011 0011 0011
A) .875 * 2 = 1.75 bit = 1, remove theinteger part    .750 * 2 = 1.50 bit =1, remove the integer part    .500 * 2 = 1.00 bit =1, remove the integer part    0 *2 = 0 for all lower orderbits.
So, answer is 111. Adding 0 sign bit and the rest of thebits gives 0111 0000 0000 0000

B) -.875: Use the above answer but convert tonegative using 2's complement
Answer is 1000 1111 1111 1111 + 1 in the lowestbit Answer is 1001 0000 0000 0000
C) -0.1: First find +0.1, convert later:
.1 * 2 = .2, bit = 0 .2 * 2 = .4, bit = 0 .4 * 2 = .8, bit = 0 .8 * 2 = 1.6, bit = 1, remove integer .6 * 2 = 1.2, bit = 1, remove integer .2 * 2 = .4, bit =0
At this point we can see that we have a pattern that willrepeat. Here's the positive answer:
0000 1100 1100 1101 Rounded up at 16 bits, asbit 17 is a 1.
Make negative with 2's complement
1111 0011 0011 0010 + 1 in the lowest bit gives
answer is 1111 0011 0011 0011

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