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Here, we have one half of a sine wave being represented by 20 samples. The samples are further digitized by a 8 bit DAC. Also, the sine wave is offset by half of the max amplitude. Let the max amplitude be 1. Thus, the sine wave will be of the form 0.5 + sin(wt). Since the frequency is not provided, we assume frequency to be 1 Hz. Thus, the decimal, binary and the hex sample are noted here:
n | Decimal | Binary | Hex |
1 | 0.5 | 0 1 0 0 1 1 0 0 | 32 |
2 | 0.6564 | 0 1 0 0 0 0 1 0 | 42 |
3 | 0.809 | 1 0 0 0 1 0 1 0 | 51 |
4 | 0.95 | 1 1 1 1 1 0 1 0 | 5F |
5 | 1.0878 | 1 0 1 1 0 1 1 0 | 6D |
6 | 1.2071 | 1 0 0 1 1 1 1 0 | 79 |
7 | 1.3090 | 1 1 0 0 0 0 0 1 | 83 |
8 | 1.3910 | 1 1 0 1 0 0 0 1 | 8B |
9 | 1.4511 | 1 0 0 0 1 0 0 1 | 91 |
10 | 1.4877 | 1 0 1 0 1 0 0 1 | 95 |
11 | 1.5 | 0 1 1 0 1 0 0 1 | 96 |
12 | 1.4877 | 1 0 1 0 1 0 0 1 | 95 |
13 | 1.4511 | 1 0 0 0 1 0 0 1 | 91 |
14 | 1.3910 | 1 1 0 1 0 0 0 1 | 8B |
15 | 1.3090 | 1 1 0 0 0 0 0 1 | 83 |
16 | 1.2071 | 1 0 0 1 1 1 1 0 | 79 |
17 | 1.0878 | 1 0 1 1 0 1 1 0 | 6D |
18 | 0.9540 | 1 1 1 1 1 0 1 0 | 5F |
19 | 0.8090 | 1 0 0 0 1 0 1 0 | 51 |
20 0.6564 0 1 0 0 0 0 1 0 42
Note: There was an error in the table with 20 rows, hence had to put the last row outside the table.
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