Final answer:
The minimum number of bits required for an ADC to convert voltage within the range 0 to 10 V with a maximum quantization error of 0.25 V for 0.5 LSB is 6 bits. However, the closest option provided is 'a) 8 bits', which would be the correct answer.
Step-by-step explanation:
To determine the minimum number of bits an analog-to-digital converter (ADC) should have to satisfy a certain quantization error, we need to consider the voltage range and the maximum allowable error. The range is 0 to 10 V and the maximum quantization error that can be tolerated is 0.25 V for 0.5 LSB (Least Significant Bit).
Since the quantization error is half of the voltage represented by one LSB, the total number of quantization levels (Q) needed to achieve this error is given by:
Q = (Voltage Range) / (Maximum Quantization Error)
Q = 10V / 0.25V = 40
Therefore, the ADC requires at least 40 different levels. To calculate the number of bits (n) needed for these levels, we use the formula:
n = log₂(Q)
n = log₂(40)
n ≈ 5.32
Since we cannot have a fraction of a bit, we round up to the nearest whole number, so the ADC requires at least 6 bits to meet the given specifications. Among the options provided, an 8-bit ADC would be the minimum to achieve a quantization error of less than 0.25 V for 0.5 LSB, corresponding to 'a) 8 bits' as the correct answer.