In this lab, we will develop the digital signal code for a very basic, notional codec. A specialized application has been developed which needs to quantize analog signals varying between -3.5 volts and +3.5 volts (see figure). The proper sampling frequency has been determined, and the sensitivity requires that each sample be within 0.5 volts of the true analog signal. Your job is to provide a digital (binary) code, using the minimum number of bits necessary, to represent each voltage, in 0.5-volt increments, from -3.5 volts to 3.5 volts. Turn-in Requirements: To complete the assignment, upload a Word (.docx) or Adobe (.pdf) file with answers to the 2 questions below, followed by a table or list indicating your digital code representation for each voltage level as your submission to the M03 Lab assignment in Blackboard. 2. (6 pts) How many bits are needed to represent at least this many levels? What about 17 levels? Show your work. a. 4 bits are needed to represent 15 levels.
The Correct Answer and Explanation is :
To represent the analog signal ranging from -3.5 volts to +3.5 volts in increments of 0.5 volts, we first need to calculate how many distinct voltage levels we have. The range from -3.5V to 3.5V in 0.5V steps includes the following voltage levels:
-3.5, -3.0, -2.5, -2.0, -1.5, -1.0, -0.5, 0.0, 0.5, 1.0, 1.5, 2.0, 2.5, 3.0, 3.5.
These are 15 distinct levels. Each voltage level must be encoded digitally, and to do this, we need to determine the number of bits required to represent these 15 levels.
The number of bits required to represent a given number of levels (L) is determined using the formula:
[
2^n \geq L
]
where (n) is the number of bits, and (L) is the number of levels. In our case, (L = 15). We need to find the smallest (n) such that (2^n \geq 15).
[
2^3 = 8 \quad \text{(not enough levels)}
]
[
2^4 = 16 \quad \text{(sufficient levels)}
]
Thus, 4 bits are sufficient to represent 15 levels because (2^4 = 16), which is enough to cover the 15 levels.
Next, for 17 levels, we use the same approach:
[
2^n \geq 17
]
[
2^4 = 16 \quad \text{(not sufficient)}
]
[
2^5 = 32 \quad \text{(sufficient)}
]
Therefore, 5 bits are needed to represent at least 17 levels, as (2^5 = 32), which comfortably covers the requirement of 17 levels.
Explanation
The number of bits required to represent a specific range of values is based on how many distinct values (levels) exist within that range. In our case, the voltage levels range from -3.5V to 3.5V, giving us 15 total levels. Since binary numbers double the number of representable values with each additional bit (i.e., 1 bit can represent 2 levels, 2 bits can represent 4 levels, etc.), we calculated that 4 bits are sufficient for 15 levels. If we expand the range to 17 levels, 5 bits are needed because 4 bits only represent up to 16 values, and 5 bits are required to represent more than 16 levels.