Information encoding. Tasks for encoding audio information

Task 2. Determine the size (in bytes) of a digital audio file whose playing time is 10 seconds at a sampling rate of 22.05 kHz and a resolution of 8 bits.

Task 3. The amount of free memory on the disk is 5.25 MB, the bit depth of the sound card is 16. What is the duration of the sound of a digital audio file recorded with a sampling frequency of 22.05 kHz?

Task 4. Determine the information volume stereo audio 1 second sound file with high sound quality (16 bits, 48 ​​kHz).

Task 5. Determine the amount of storage space for a digital audio file that has a two-minute playing time at a sampling rate of 44.1 kHz and a resolution of 16 bits.

Task 6. One minute of recording a digital audio file occupies 1.3 MB on the disk, the bit depth of the sound card is 8. With what sampling rate was the sound recorded?

Task 7. How much storage space is required to store a high quality digital audio file with a 3 minute playback time?

Task 8. The digital audio file contains a low quality audio recording (sound is dark and muffled). What is the duration of the sound of a file if its volume is 650 KB?

Task 9. Two minutes of digital audio recording takes up 5.05 MB of disk space. Sampling frequency - 22 050 Hz. What is the bitness of the audio adapter?

Task 10. The amount of free memory on the disk is 0.1 GB, the bit depth of the sound card is 16. What is the duration of the sound of a digital audio file recorded with a sampling rate of 44 100 Hz?

• Task 11. Estimate the information volume of a stereo audio file with a sound duration of 1 second (1 minute) with high sound quality (16 bit; 48 kHz)
• Task 12. Calculate the playing time of a mono audio file if, with 16-bit encoding and a sampling rate of 32 kHz, its volume is equal to:
• A) 700 Kbytes;
• B) 6300 KB.

Information encoding

One of the main lines in the teaching of computer science is the coding of information. This topic corresponds to the main goals of the discipline "Informatics and ICT": the formation of ideas about information, ideas about the measurements of the amount of information located in various types perception.

Depending on the way of perception, tasks for encoding information are divided into three types: encoding images, text and sound. Any of the listed types of tasks is always present in the GIA, the Unified State Examination and Olympiads. Therefore, it is most important to understand this area of ​​computer science.

Graphic arts

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Example . Calculate the amount of video memory required to store a bitmap that occupies the entire screen of a monitor with a resolution of 640x480 pixels, if a palette of 65536 colors is used.

Decision. The color depth can be found using the formula N = 65536 = 2I, which means I = 16 bits. Therefore, to store one image pixel, 16/8 = 2 bytes of video memory are required. Thus, to store the entire image, Vgr = W*H*I = 640*480*2 = 614400 bytes = 600 Kb is required.

Sound

Alphabet" href="/text/category/alfavit/" rel="bookmark">alphabet) as various possible equiprobable states (events). Then each text symbol of any alphabet (letter, number, punctuation mark, etc.) can assign a code - an integer, which means that their information capacity depends on their number in the alphabet ( power alphabet), the more their number, the large quantity information is carried by one symbol. For example, in the Russian alphabet (without the letter “ё”) there are 5 bits of information per character (32=2I)..

Example. Determine the information volume of the proverb in bits "Small spool, but expensive" if the power of the alphabet is 16 bits (without quotes).

Decision: You just need to count the number of characters in the sentence, not including quotes, but including spaces. (At the same time, assuming that there is one space between words, and there are no spaces before punctuation marks). Multiply the resulting value by I = 4 bits. That is, Vt \u003d 22 * ​​4 \u003d 44 bits.

These sections are a complement and generalization of each other, therefore, when teaching them, one should observe the correspondence in the theoretical presentation and in solving problems.

1. Sound is a sound wave with a continuously changing amplitude and frequency. For a person, the sound is louder, the greater the amplitude of the signal, and the higher the tone, the greater the frequency of the signal. A continuous signal does not carry information, so it must be turned into a sequence of binary zeros and ones - a binary (digital) code.

2. The sound is digitized by a special device on the sound card. It is called an analog-to-digital converter (ADC). The reverse process - the reproduction of the encoded sound is performed using a digital-to-analog converter (DAC). Let's take a closer look at these processes.

In the process of coding continuous sound signal it is discretized in time, or, as they say, "temporal discretization". The sound wave is divided into separate small time sections and a certain amplitude value is set for each section. This method is called PCM pulse-amplitude modulation ( Pulse Code Modulation ). Thus, a smooth curve is replaced by a sequence of "steps". Each "step" is assigned a sound volume value (1, 2, 3, etc.). The more "steps", the more volume levels will be allocated in the coding process, and the more information will be carried by the value of each level and the sound will be of better quality.

2. Characteristics of the digitized sound.

Audio quality depends on two characteristics - audio encoding depth and sampling rate. Let's consider these characteristics.

Audio encoding depth ( I ) is the number of bits used to encode different signal levels or states. Then the total number of such states or levels ( N ) can be calculated using the formula:

N=2I.

Modern sound cards provide 16-bit audio encoding depth, and then the total number of different levels will be:

N = 216 = 65536.

The sampling frequency (M) is the number of measurements of the audio signal level per unit of time. This characteristic indicates the sound quality and the accuracy of the binary encoding procedure. It is measured in hertz (Hz). One measurement in one second corresponds to a frequency of 1 Hz, 1000 measurements in one second - 1 kilohertz (kHz). The sampling frequency of the audio signal can take values ​​from 8 to 48 kHz. At a frequency of 8 kHz, the quality of the sampled audio signal corresponds to the quality of a radio broadcast, and at a frequency of 48 kHz, the sound quality of an audio signal CD.

High quality sound is achieved with a sampling rate of 44.1 kHz and an audio encoding depth of 16 bits. The dark, muffled sound is characterized by the following parameters: sampling rate - 11 kHz, encoding depth - 8 bits.

In order to find the volume of sound information, you must use the following formula: V =M*I*t , where M is the sampling frequency (in Hz), I - encoding depth (in bits), t — playing time (in seconds).

Example

Audio is played for 10 seconds at a sampling rate of 22.05 kHz and an audio depth of 8 bits. Determine its size (in bytes). Decision:

M \u003d 22.05 * 1000 \u003d 220500 Hz; I = 8/8=1 byte;

T = 10 seconds; V = 220500*10*1= 220500 bytes.

№1

Determine the amount of memory to store a mono audio file that has a playback time of five minutes at a sampling rate of 44 kHz and a 16-bit encoding depth.

Decision:

V=MIt = 44000Hz * 16 * 5 = 3520000 bits = 430 KB.

Answer: 430 KB.

№2

What should be the sampling frequency and encoding depth for "recording audio information lasting 2 minutes, if the user has a memory of 5.1 MB.

Decision:

M*I=V/t;

M*I = 5.1 *1024*1024*8/2/60 = 356515 ( Hz * bit).

Option 1

356515 (Hz * bits) = 22.05 kHz * 16 bits.

Answer: 22.05 kHz and 16 bits.

Option 2

356515 (Hz * bits) = 44.1 kHz * 8 bits.

Answer: 44.1 kHz and 8 bits.

№3

The amount of free disk space is 5.25 MB, the encoding depth is 8. Sound information is recorded with a sampling rate of 44.1 kHz. What is the duration of such information?