![\"Sample](\"https:\/\/www.mixcorner.com\/IMG\/png\/sinusoide2.png?1620726079\")
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Audio is sampled 11,000 times per second, and each sampled value is recorded as 8-bit data.<\/p>\n
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At this time, what is the maximum length of audio that can be recorded in flash memory with a capacity of 512 x 10 6 bytes?<\/p>\n
A 77 B 96 C 775 D 969<\/p>\n
Next is the question of finding the length of audio that can be recorded. With the knowledge acquired so far, you can easily find a calculation method. The way of thinking when calculating is shown below.<\/p>\n
Since 11000 samples are taken per second and each one is recorded as 8 bits = 1 byte of data, the amount of data per second is 11000 x 1 = 11000 bytes.
\nSince the capacity of the flash memory is 512 x 10 6 bytes, data of 512 x 10 6 \u00f7 11000 = 46545.45 … seconds can be recorded.
\nSince the answer is obtained in minutes, 46545.45 … \u00f7 60 = 775.75 … minutes.
\nSince the question says “maximum number of minutes”, the maximum is 775 minutes rounding down the fraction of 775.75 …, and option c is the correct answer.<\/p>\n
correct answer hare<\/p>\n
Problem finding the sampling interval
\nQuestion 4 (Fall 2016)
\nWhen the audio was sampled using the PCM method, converted to 8-bit digital data and transferred in real time without compression, the transfer rate was 64,000 bits \/ sec. How many microseconds is the sampling interval at this time?<\/p>\n
A 15.6 B 46.8 C 125 D 128<\/p>\n
Now the question is to find the sampling interval (how many seconds to sample).<\/p>\n
The “transfer rate” is involved, but since the audio was transferred in real time, 64000 bits \/ second is the same as the encoded capacity per second. Once you know that, you will be able to find a calculation method with the knowledge you have gained so far.<\/p>\n
The way of thinking when calculating is shown below.<\/p>\n
Since there are 64000 bits per second and the size of the encoded data is 8 bits, the number of samples taken per second is 64000 \u00f7 8 = 8000 times.
\nSince 8000 samples were taken per second, the time interval is 1 \u00f7 8000 = 0.000125 seconds.
\nSince the question is “how many microseconds?”, The correct answer is 125 microseconds, which is 0.000125 seconds in microseconds.<\/p>\n
correct answer hare<\/p>\n
Problem finding buffering time
\nQ31 (Fall 2014)
\nTo play 2.4 Mbytes of audio data at 192 kbit \/ s encoding speed without interruption during downloading using a network with 128 kbit \/ s communication speed, buffering the data before the start of playback How many seconds do you need in less?<\/p>\n
A 50 B 100 C 150 D 250<\/p>\n
Finally, let’s fix an issue with a slightly different coat color. Finding the buffering time when downloading digitized audio data is a problem.<\/p>\n
Buffering is downloading some data before starting to play. This allows you to play the audio without interruptions, even if the download speed is slow.<\/p>\n
The idea when calculating is shown below, so please check each one carefully. Here, M = 1000 k.<\/p>\n
A code rate of 192 kbits \/ s means that the amount of digitized data is 192 kbits per second.
\nThe total amount of data is 2.4 Mbytes = 2.4 M x 8 = 19.2 M bits, which is expressed in seconds as 19.2 M \u00f7 192 k = 19200 k \u00f7 192 k = 100 seconds.
\nSince the communication speed is 128 kbit \/ s, data of 128 k x 100 = 12800 kbit can be transferred in 100 seconds.
\nHowever, since the total data capacity is 19.2 Mbit = 19200 kbit, the difference of 19200 kbit – 12800 kbit = 6400 kbit of data must be buffered beforehand.
\nSince the communication speed is 128 kbit \/ s, it takes 6400 k \u00f7 128 k = 50 seconds to buffer the 6400 kbit data.
\nFrom the above, the buffering time is 50 seconds and option A is the correct answer.<\/p>\n","protected":false},"excerpt":{"rendered":"