Signal Coding Lab In this lab, we will develop the digital signal code for a very basic, notional codec. A specialized application has determined the proper sampling frequency, 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. 3.5V +3.0V +2.5V +2.0V +1.5V +1.0V +0.5V 0.0V -0.5V -1.0V -1.5V -2.0V -2.5V -3.0V -3.5V 15msec 10msec When the blank lines on the left of the figure are filled in with your digital code, the graph will be similar to (but not the same as) graphs in figure 3-20 of the text. 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. 1. (4 pts) How many different voltage levels must be represented? a. 15 levels 2. (6 pts) How many bits are needed to represent at least this many levels? What about 17 levels? Show your work. -3.5V to 3.5V represents a range of 7 volts. Each voltage level is represented in 0.5-volt increments. Therefore, the number of levels can be calculated as (7 volts / 0.5 volts) + 1 = 15 levels. To represent 15 levels, we need 4 bits (2^4 = 16, which is greater than 15). For 17 levels, we would need 5 bits (2^5 = 32, which is greater than 17). 3. (10 pts) Fill in the below table -3.0V = -2.5V = -2.0V = -1.5V = -1.0V = -0.5V = 0.0V = +0.5V = +1.0V = +1.5V = +2.0V = +2.5V = +3.0V =
Signal Coding Lab In this lab, we will develop the digital signal code for a very basic, notional codec. A specialized application has determined the proper sampling frequency, 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. 3.5V +3.0V +2.5V +2.0V +1.5V +1.0V +0.5V 0.0V -0.5V -1.0V -1.5V -2.0V -2.5V -3.0V -3.5V 15msec 10msec When the blank lines on the left of the figure are filled in with your digital code, the graph will be similar to (but not the same as) graphs in figure 3-20 of the text. 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. 1. (4 pts) How many different voltage levels must be represented? a. 15 levels 2. (6 pts) How many bits are needed to represent at least this many levels? What about 17 levels? Show your work. -3.5V to 3.5V represents a range of 7 volts. Each voltage level is represented in 0.5-volt increments. Therefore, the number of levels can be calculated as (7 volts / 0.5 volts) + 1 = 15 levels. To represent 15 levels, we need 4 bits (2^4 = 16, which is greater than 15). For 17 levels, we would need 5 bits (2^5 = 32, which is greater than 17). 3. (10 pts) Fill in the below table -3.0V = -2.5V = -2.0V = -1.5V = -1.0V = -0.5V = 0.0V = +0.5V = +1.0V = +1.5V = +2.0V = +2.5V = +3.0V =
The Correct Answer and Explanation is:
1. How many different voltage levels must be represented?
Answer: 15 levels
2. How many bits are needed to represent at least this many levels? What about 17 levels? Show your work.
The range of voltage levels from -3.5V to +3.5V with 0.5V increments includes: Number of levels=(+3.5−(−3.5))0.5+1=70.5+1=14+1=15\text{Number of levels} = \frac{(+3.5 – (-3.5))}{0.5} + 1 = \frac{7}{0.5} + 1 = 14 + 1 = 15
To represent 15 levels, we need 4 bits:
- 23=82^3 = 8 → Not enough
- 24=162^4 = 16 → Sufficient
If we had 17 levels:
- 25=322^5 = 32 → Minimum bits needed: 5
3. Fill in the table with digital codes (minimum 4-bit code)
| Voltage Level | Binary Code |
|---|---|
| -3.5V | 0000 |
| -3.0V | 0001 |
| -2.5V | 0010 |
| -2.0V | 0011 |
| -1.5V | 0100 |
| -1.0V | 0101 |
| -0.5V | 0110 |
| 0.0V | 0111 |
| +0.5V | 1000 |
| +1.0V | 1001 |
| +1.5V | 1010 |
| +2.0V | 1011 |
| +2.5V | 1100 |
| +3.0V | 1101 |
| +3.5V | 1110 |
Explanation
In digital signal processing, analog signals must be quantized into discrete voltage levels for digital encoding. This lab focuses on converting analog voltage values between -3.5V and +3.5V, incremented by 0.5V, into a digital binary code.
The range from -3.5V to +3.5V is 7 volts. Dividing this range by the step size of 0.5V gives us 14 intervals, but we must include both endpoints, resulting in 15 distinct voltage levels. To represent these 15 levels digitally, we need enough binary combinations. A binary system using n bits can represent up to 2n2^n unique values. For 15 levels, 4 bits are sufficient since 24=162^4 = 16. While 3 bits only allow for 8 levels, 4 bits offer 16 combinations—more than enough.
The coding scheme starts at -3.5V with binary 0000 and increments by one for every 0.5V increase, ending at +3.5V with 1110. The code 1111 remains unused in this scheme but offers flexibility for potential expansion (e.g., error codes or future voltage ranges).
For example, if a sensor reads -2.0V, it is encoded as 0011. This encoding ensures efficient digital transmission and storage while preserving the analog signal’s resolution to within 0.5V, which meets the application’s sensitivity requirement.
Using the fewest bits reduces data size and transmission time, which is especially critical in bandwidth-limited systems. However, care must be taken to handle the unused code (1111) to avoid misinterpretation.
This process is foundational in digital communication systems, such as audio/video encoding, analog-to-digital conversion, and data compression.
