The problem of finding the length of recordable audio
Audio is sampled 11,000 times per second, and each sampled value is recorded as 8-bit data.
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?
A 77 B 96 C 775 D 969
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.
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.
Since the capacity of the flash memory is 512 x 10 6 bytes, data of 512 x 10 6 ÷ 11000 = 46545.45 … seconds can be recorded.
Since the answer is obtained in minutes, 46545.45 … ÷ 60 = 775.75 … minutes.
Since 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.
correct answer hare
Problem finding the sampling interval
Question 4 (Fall 2016)
When 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?
A 15.6 B 46.8 C 125 D 128
Now the question is to find the sampling interval (how many seconds to sample).
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.
The way of thinking when calculating is shown below.
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 ÷ 8 = 8000 times.
Since 8000 samples were taken per second, the time interval is 1 ÷ 8000 = 0.000125 seconds.
Since the question is “how many microseconds?”, The correct answer is 125 microseconds, which is 0.000125 seconds in microseconds.
correct answer hare
Problem finding buffering time
Q31 (Fall 2014)
To 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?
A 50 B 100 C 150 D 250
Finally, let’s fix an issue with a slightly different coat color. Finding the buffering time when downloading digitized audio data is a problem.
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.
The idea when calculating is shown below, so please check each one carefully. Here, M = 1000 k.
A code rate of 192 kbits / s means that the amount of digitized data is 192 kbits per second.
The 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 ÷ 192 k = 19200 k ÷ 192 k = 100 seconds.
Since the communication speed is 128 kbit / s, data of 128 k x 100 = 12800 kbit can be transferred in 100 seconds.
However, 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.
Since the communication speed is 128 kbit / s, it takes 6400 k ÷ 128 k = 50 seconds to buffer the 6400 kbit data.
From the above, the buffering time is 50 seconds and option A is the correct answer.
