問88
頻出度★★☆
音声信号をディジタル化する。図の時刻 1 から時刻 5 のタイミングで標本化を行い，4 段階に量子化（標本点に最も近い段階を選択）を行った。その後 2 ビットで符号化を行った。結果は“11 01 00 10 11”であった。同じ手法でディジタル化を行うと“01 00 10 11 01”となる音声信号を示す図はどれか。
平成22年度 春期 ITパスポート試験 問88
テクノロジ系基礎理論 › 情報に関する理論
This question's explanation and your answers so far.
解説
問 88
Incorrectア
Reading the sample points of Figure ア in order: time 1 is 01, time 2 is also 01 (the small peak around time 1.5 does not fall on a sampling time, so it does not appear in the code), time 3 is a trough of 00, time 4 is 10, and time 5 is 11, giving the bit string 01 01 00 10 11. Compared with the target 01 00 10 11 01, this is a waveform whose trough and peak are each shifted one time unit later: the second group is 01 rather than 00, and the fifth group is 11 rather than 01, so it does not match.
- イ
正解 The correct answer is イ. First, confirm how to read the figures by using the example figure. In the example figure the waveform is near 11 at time 1 (the peak is just before time 1), is at 01 at time 2, reaches a trough of 00 at time 3, is at 10 at time 4, and rises back to 11 at time 5; lining these up two bits at a time gives exactly 11 01 00 10 11, as stated in the question. In other words, what you should look at is not the shape of the peaks and troughs themselves, but only the five points from time 1 to time 5, and which level on the vertical axis the waveform is closest to at each of them. Reading Figure イ with this in mind: at time 1 there is a small peak sitting exactly on the 01 gridline, at time 2 the curve has fallen to a trough of 00, at time 3 it passes the 10 gridline while rising sharply, at time 4 the peak around time 3.7 is still continuing so the value is 11, and at time 5 the curve comes back exactly to 01. Encoding these in order gives 01 00 10 11 01, which matches the required bit string exactly.
Incorrectウ
The sample points of Figure ウ are: 01 at time 1 (the curve is slightly below the 01 gridline, but still above the midpoint between 00 and 01, so the nearest level is 01), a trough of 00 at time 2, 10 at time 3, and 11 at time 4, which is the same as the target 01 00 10 11 up to this point. However, the fall after time 4 is gentle, and by time 5 the curve comes back down only as far as the 10 gridline. The bit string is 01 00 10 11 10, and only the last group differs from the target value of 01. Because the first half matches perfectly, it is easy to choose this option without reading through to the end, which makes it the most confusing distractor.
Incorrectエ
Figure エ is the only one of the four options drawn as a polyline of straight segments. Its sample points are 01 at time 1, 00 at time 2, and 10 at time 3, matching the target so far; but the apex around time 3.2 only slightly exceeds the midpoint between 10 and 11 and never reaches 11, and the curve turns downward from there, so time 4 is still 10. Time 5 is 01. The bit string is 01 00 10 10 01, and the fourth group is 10 rather than 11, so it does not match the target.
ここがポイント
- Digitizing sound (A/D conversion) has three stages: sampling, then quantization, then encoding. Sampling means reading the amplitude at fixed times, quantization means rounding that value to the nearest level, and encoding means representing the level as a string of bits. In problems that ask for the bit string from a figure, you only need to read the values at the sampling times in order; the key point is not to be misled by the shape of peaks and troughs that lie between the sampling times.
- The relationship between the number of quantization levels and the number of bits in the code is 2 raised to the power of (the number of bits) = the number of levels. This question uses 4 levels = 2 to the power of 2, so each sample takes 2 bits, and 5 samples make a bit string of 10 bits in total. Also remember the trade-off: increasing the number of levels reduces the quantization error, but it increases the number of bits (the amount of data) required.
用語