Encoding information. Amount of information

Basic Concepts

The sampling frequency (f) determines the number of samples stored in 1 second;

1 Hz (one hertz) is one count per second,

and 8 kHz is 8000 samples per second

Encoding depth (b) is the number of bits required to encode 1 volume level

Playing time (t)

Memory capacity for data storage of 1 channel (mono)

I=f b t

(to store information about sound lasting t seconds, encoded with a sampling frequency f Hz and a coding depth b bits are required I memory bit)

At two-channel recording (stereo) The amount of memory required to store data for one channel is multiplied by 2

I=f b t 2

Units I - bits, b - bits, f - Hertz, t - seconds Sampling frequency 44.1 kHz, 22.05 kHz, 11.025 kHz

Encoding of audio information

Basic theoretical principles

Time sampling of sound. In order for a computer to process sound, the continuous audio signal must be converted into digital discrete form using time sampling. A continuous sound wave is divided into separate small temporary sections, and for each such section a certain value of sound intensity is set.

Thus, the continuous dependence of sound volume on time A(t) is replaced by a discrete sequence of loudness levels. On the graph, this looks like replacing a smooth curve with a sequence of “steps”.

Sampling frequency. A microphone connected to the sound card is used to record analog audio and convert it to digital form. The quality of the resulting digital sound depends on the number of measurements of the sound volume level per unit time, i.e. sampling rates. The greater the number of measurements made in 1 second (the higher the sampling frequency), the more accurate the “ladder” of the digital sound signal repeats the analog signal curve.

Audio sampling rate is the number of sound volume measurements in one second, measured in hertz (Hz). Let's denote the sampling rate by the letter f.

Audio sampling rates can range from 8,000 to 48,000 audio volume measurements per second. For encoding, choose one of three frequencies: 44.1 KHz, 22.05 KHz, 11.025 KHz.

Audio encoding depth. Each “step” is assigned a specific sound volume level. Sound loudness levels can be thought of as a set of possible states N, which require a certain amount of information to encode b , which is called the audio coding depth

Audio coding depth is the amount of information needed to encode discrete loudness levels digital audio.

If the encoding depth is known, then the number of digital sound volume levels can be calculated using the formula N = 2 b. Let the audio encoding depth be 16 bits, then the number of audio volume levels is equal to:

N=2 b = 2 16 = 65,536.

During the encoding process, each sound volume level is assigned its own 16-bit binary code, the lowest sound level will correspond to the code 0000000000000000, and the highest - 1111111111111111.

Digitized sound quality. The higher the frequency and sampling depth of the sound, the higher the quality of the digitized sound. Lowest quality digitized audio corresponding to quality telephone communication, obtained at a sampling rate of 8000 times per second, a sampling depth of 8 bits and recording one audio track (mono mode). The highest quality digitized audio, corresponding to audio CD quality, is achieved with a sampling rate of 48,000 times per second, a sampling depth of 16 bits and recording of two audio tracks (stereo mode).

It must be remembered that the higher the quality of digital sound, the greater the information volume of the sound file.

Tasks for self-study.

1. Calculate the volume of a 10-second monophonic audio file with 16-bit encoding and a sampling rate of 44.1 kHz. (861 KB)

2. Two-channel (stereo) sound recording is performed with a sampling frequency of 48 kHz and 24-bit resolution. The recording lasts 1 minute, its results are written to a file, data compression is not performed. Which of the following numbers is closest to the size of the resulting file, expressed in megabytes?

1)0,3 2) 4 3) 16 4) 132

3. Single-channel (mono) sound recording is made with a sampling frequency of 11 kHz and a coding depth of 24 bits. The recording lasts 7 minutes, its results are written to a file, data compression is not performed. Which of the following numbers is closest to the size of the resulting file, expressed in megabytes?

1) 11 2) 13 3) 15 4) 22

4. Two-channel (stereo) sound recording is made with a sampling frequency of 11 kHz and a coding depth of 16 bits. The recording lasts 6 minutes, its results are written to a file, data compression is not performed. Which of the following numbers is closest to the size of the resulting file, expressed in megabytes?

1) 11 2) 12 3) 13 4) 15

Option 1

Laboratory work

“Coding and processing of audio information”

Goals:

educational
educational –
developing –

Progress:

Decide

File name

f-sampling frequency

k - sound depth

t- playing time

File type

44.1 kHz

16 bit

1 min

stereo

1.wav

8 kHz

8 bit

1 min

mono

2.wav

16 kHz

16 bit

1 min

stereo

3.wav

24 kHz

16 bit

1 min

mono

4.wav

32 kHz

16 bit

1 min

stereo

for tasks 7-9

5.wav

Show the partially completed table to your teacher.

Run sound editor Audacity .

Trim sound of the file offered to you up to 1 minute, highlighting the desired period of time, execute the command Edit - Trim along the edges.

Convert wav .

In a sound editor Audacity For example

Compare

Hand over report to the teacher for verification.

Option 2

Laboratory work

"Coding of audio information"

Goals:

educational- ensure the formation and use by students of knowledge about encoding audio information using a computer, as well as skills in processing it using applied software software;
educational – cultivate attentiveness, accuracy, independence;
developing – skills in using application software; ability to solve information problems.

Hardware and software requirements: headphones, sound files for students, sound editor Audacity, Sound Recording program OC Windows.

Progress:

Decide tasks from the table below.

Find the volume of sound information using the formula V = f *k *t, where

f - sampling frequency, k - sound depth, t - playing time

Present the solution to the problems in the form of a table.

In the “Calculated volume of sound file” column, write down the answers to the solved problems yourself. Give the answer in megabytes.

File name

f-sampling frequency

k - sound depth

t- playing time

File type

Estimated audio file size

Actual audio file size

44.1 kHz

16 bit

45 s

stereo

1.wav

8 kHz

8 bit

45 s

stereo

2.wav

1 1.025 kHz

16 bit

45 s

mono

3.wav

24 kHz

Launch the sound editorAudacity .

Trim sound of the file offered to you up to 45 seconds, selecting the desired period of time, execute the command Edit - Trim along the edges.

Convert the file offered to you into a file with the extension wav . Save this file with the same name.

In a sound editor Audacity create effects for the sound file offered to you. For example, make the last 10 seconds of the file fade out

Split the stereo track and then delete one of the tracks. Convert this file from stereo to mono. Save this file with a new name and wav extension.

Compare file volumes. Fill in the table with data.

Hand over report to the teacher for verification.

Basic Concepts

The sampling frequency (f) determines the number of samples stored in 1 second;

1 Hz (one hertz) is one count per second,

and 8 kHz is 8000 samples per second

Encoding depth (b) is the number of bits required to encode 1 volume level

Playing time (t)

Memory capacity for data storage of 1 channel (mono)

I=f b t

(to store information about sound lasting t seconds, encoded with a sampling frequency f Hz and a coding depth b bits are required I memory bit)

At two-channel recording (stereo) The amount of memory required to store data for one channel is multiplied by 2

I=f b t 2

Units I - bits, b - bits, f - Hertz, t - seconds Sampling frequency 44.1 kHz, 22.05 kHz, 11.025 kHz

Encoding of audio information

Basic theoretical principles

Sampling frequency. A microphone connected to the sound card is used to record analog audio and convert it to digital form. The quality of the resulting digital sound depends on the number of measurements of the sound volume level per unit time, i.e. sampling rates. The more measurements are taken in 1 second (the higher the sampling frequency), the more accurately the “ladder” of the digital audio signal follows the curve of the analog signal.

Audio sampling rate is the number of sound volume measurements in one second, measured in hertz (Hz). Let's denote the sampling rate by the letter f.

Audio sampling rates can range from 8,000 to 48,000 audio volume measurements per second. For encoding, choose one of three frequencies: 44.1 KHz, 22.05 KHz, 11.025 KHz.

Audio coding depth is the amount of information needed to encode discrete volume levels of digital audio.

N=2 b = 2 16 = 65,536.

During the encoding process, each sound volume level is assigned its own 16-bit binary code, the lowest sound level will correspond to the code 0000000000000000, and the highest - 1111111111111111.

Digitized sound quality. The higher the frequency and sampling depth of the sound, the higher the quality of the digitized sound. The lowest quality of digitized sound, corresponding to the quality of telephone communication, is obtained with a sampling rate of 8000 times per second, a sampling depth of 8 bits and recording of one audio track (mono mode). The highest quality digitized audio, corresponding to audio CD quality, is achieved with a sampling rate of 48,000 times per second, a sampling depth of 16 bits and recording of two audio tracks (stereo mode).

It must be remembered that the higher the quality of digital sound, the greater the information volume of the sound file.

Tasks for self-study.

1. Calculate the volume of a 10-second monophonic audio file with 16-bit encoding and a sampling rate of 44.1 kHz. (861 KB)

1)0,3 2) 4 3) 16 4) 132

1) 11 2) 13 3) 15 4) 22

1) 11 2) 12 3) 13 4) 15

1. General information

Complexity: basic.

Approximate solution time (for those who will do part 2): 2 minutes

Subject: Creation and processing of graphic and multimedia information

Subtopic: Digital audio recording

What is checked: Ability to evaluate quantitative characteristics sound recording process.

Brief theoretical information: Because the this type the task is new in the KIM Unified State Examination, we will give it (without justification yet, justification below) mathematical model recording process:

N = k F * L * T* (1)

N– file size (in bits) containing the sound recording;
k- number of recording channels (for example, 1 – mono, 2 – stereo, 4 – quad, etc.);
F– sampling frequency (in hertz), i.e. the number of sound amplitude values recorded in one second;
L– resolution, i.e. the number of bits used to store each measured value;
T– duration of the sound fragment (in seconds).

What might the task look like? For example, like this: The values of all required parameters of the sound recording process are specified, except one. You need to estimate the value of the remaining parameter, for example, the file size or the duration of the audio fragment.

Example condition:

Possible answers:

1) 0.2 MB

2. Example assignment

2.1. The task.

Problem 2012-A8-1.

Single-channel (mono) sound recording is performed with a sampling frequency of 16 kHz and 24-bit resolution. The recording lasts 1 minute, its results are written to a file, data compression is not performed. Which of the following values is closest to the size of the resulting file?

1) 0.2 MB 2) 2 MB 3) 3 MB 4) 4 MB

2.2. Solution.

We reduce the initial data to the dimension of bits-seconds-hertz and carry out calculations using formula (1):

Given:

k= 1, because single-channel (mono) sound recording;

F= 16 kHz = 16,000 Hz;

T= 1 min = 60 s.

FindN

Substitute the value of the known parameters into formula (1)

N=1*16000 *24*60 =(16 *1000) * (8*3) * (4*15)=

= 2 4 *(2 3 *125) *(2 3 *3)*) *(2 2 * 15) = 2 12 *5625 (bits)=

= 2 12 *5625 bits = (2 12 *5625)/2 3 bytes = 2 9 *5625 bytes =

= (2 9 *5625)/ 2 20 MB = 5625/2 11 MB = 5625/2048 MB.

Number 5625/2048 is between the numbers 2 and 3. Moreover, it is closer to 3 than to 2, because 3 * 2048 – 5625 < 1000; 5625 - 2 * 2048 > 1000.

Correct answer: No. 3 (3 MB)

Comment. Another solution idea is given in paragraph 3.3

3. Tips for teachers and students

3.1 What knowledge/abilities/skills does the student need to solve this problem?

1) You should not “memorize” formula (1). A student who represents the essence of the digital audio recording process must be able to formulate it independently.

2) You must be able to write down parameter values in the required dimensions, as well as basic arithmetic skills, incl. operating with powers of two.

A. Strong students.

1. Most likely, they will solve this problem anyway.

2. You can give students the task of testing formula (1) in practice by recording sound from a microphone into a file. Please note that this applies only if the recorded information is not compressed (WAV (PCM) format uncompressed). If compressed audio formats are used (WMA, MP3), then the size of the resulting file will, for obvious reasons, be significantly less than the calculated one. To experiment with digital audio recording, you can use the freely available audio editor Audacity (http://audacity.sourceforge.net/).

3. It is advisable to emphasize the conceptual commonality of the raster representation of sound and image, which are varieties of the same process of approximate representation of a continuous signal - a sequence of short discrete signals, i.e. digitization based on sampling. When bitmap Two-dimensional sampling of brightness in space is performed, and in the case of sound, one-dimensional sampling in time. In both cases, increasing the sampling rate (number of pixels or audio samples) and/or increasing the number of bits to represent one sample (color or sound bit depth) leads to an increase in the quality of digitization, while simultaneously increasing the size of the file with the digital representation. Hence the need for data compression.

4. It is advisable to mention alternative ways sound digitization – recording “parts” of instruments in MIDI format. Here it is appropriate to draw an analogy with raster and vector representation of images.

B. Not so strong students.

1. It is necessary to ensure the assimilation of relation (1). It is recommended to give tasks like “How will the file size change if the sound recording time is increased/decreased by p once? ",

“How many times can you increase/decrease the recording duration if maximum size file enlarge/reduce in p once? ", "How will the file size change if the number of bits for recording one value is increased/decreased in p once?" etc.

2. It is necessary to make sure that students are fluent in using dimensions, know that there are 23 bits in MB, etc.

3. It is necessary to make sure that students are sufficiently arithmetic literate and are fluent in mental calculation with powers of two (multiplication, division, isolating factors representing 2n).

4. Come up with your own approaches and try them.

3.3. Useful trick.

Powers of two often arise in problems like this. Multiplying and dividing powers is easier than arbitrary numbers: multiplying and dividing powers comes down to adding and subtracting exponents.

Note that the numbers 1000 and 1024 differ by less than 3%, the numbers 60 and 64 differ by less than 7%. Therefore, you can do this. Carry out calculations by replacing 1000 with 1024 = 2 10 and 60 with 64 = 2 6, taking advantage of operations with powers. The answer closest to the resulting number will be the one you are looking for. You can then double-check yourself by making accurate calculations. But we can take into account that the total calculation error in our approximation does not exceed 10%. Indeed, 60*1000 = 60000; 64*1024=65536;

60000 > 0.9 * 65536 = 58982.4

Thus, the correct result of multiplications according to formula (1) is slightly more than 90% of the obtained approximate result. If taking into account the error does not change the result, there is no doubt about the answer.

Example. (ege.yandex.ru, option 1).

Two-channel (stereo) sound recording is performed with a sampling frequency of 16 kHz and 32-bit resolution. The recording lasts 12 minutes, its results are written to a file, and no data compression is performed. Which of the following values is closest to the size of the resulting file?

1) 30 MB 2) 60 MB 3) 75 MB 4) 90 MB

Solution. The record size in bits is

2*16*1000*32*12*60

Taking into account the replacement of 1000 by 1024=2 10 and 60 by 64=2 6 we get:

2 1 *2 4 *2 10 *2 5 *3*2 2 *2 6 =3*2 28

As you know, 1 MB = 2 20 bytes = 2 23 bits. Therefore 3*2 28 bits = 3*32 = 96 MB. Reducing this number by 10%, we get 86.4 MB. In both cases, the closest value is 90 MB.

Correct answer: 4

1. Read the problem statement. Express the unknown parameter in terms of the known ones. Special attention Pay attention to the dimension of the known parameters. It should be bits-seconds-hertz (recall that 1 Hz = s -1). If necessary, reduce the parameter values to the required dimension, just as is done in physics problems.

2. Carry out calculations, trying to identify powers of two.

3. Please note that the condition requires you to select the most appropriate answer, so high accuracy of calculations to decimal places is not required. As soon as it becomes clear which of the answer options is closest to the calculated value, the calculations should be stopped. If the discrepancy with all answer options is very large (by several times or by an order of magnitude), then the calculations must be double-checked.

4. Problems for independent solution

4.1. Clones of problem 2012-A8-1.

Below are four more options for task 2012-A8-1.

A) Single-channel (mono) sound recording is performed with a sampling frequency of 32 kHz and 24-bit resolution. The recording lasts 15 seconds, its results are written to a file, and no data compression is performed. Which of the following values is closest to the size of the resulting file?

B) Two-channel (stereo) sound recording is made with a sampling frequency of 32 kHz and 24-bit resolution. The recording lasts 30 seconds, its results are written to a file, and no data compression is performed. Which of the following values is closest to the size of the resulting file?

1) 1.5 MB 2) 3 MB 3) 6 MB 4) 12 MB

C) Single-channel (mono) sound recording is performed with a sampling frequency of 16 kHz and 32-bit resolution. The recording lasts 2 minutes, its results are written to a file, data compression is not performed. Which of the following values is closest to the size of the resulting file?

D) Single-channel (mono) sound recording is performed with a sampling frequency of 16 kHz and 32-bit resolution. The recording lasts 4 minutes, its results are written to a file, data compression is not performed. Which of the following values is closest to the size of the resulting file?

1) 2 MB 2) 4 MB 3) 8 MB 4) 16 MB

Right answers:

A:1; B:3; AT 3; G:4.

4.2. Problem 2012-A8-2 (reverse to the previous one).

A) Single-channel (mono) sound recording is performed with a sampling frequency of 16 kHz and 24-bit resolution. The results are written to a file whose size cannot exceed 8 MB; data is not compressed. Which of the following values is closest to the maximum possible duration of a recorded sound fragment?

B) Two-channel (stereo) sound recording is performed with a sampling frequency of 16 kHz and 24-bit resolution. The results are written to a file whose size cannot exceed 8 MB; data is not compressed. Which of the following values is closest to the maximum possible duration of a recorded sound fragment?

1) 1 minute 2) 30 seconds 3) 3 minutes 4) 90 seconds

C) Single-channel (mono) sound recording is performed with a sampling frequency of 48 kHz and 8-bit resolution. The results are written to a file whose size cannot exceed 2.5 MB; data is not compressed. Which of the following values is closest to the maximum possible duration of a recorded sound fragment?

1) 1 minute 2) 30 seconds 3) 3 minutes 4) 90 seconds

D) Single-channel (mono) sound recording is performed with a sampling frequency of 48 kHz and 16-bit resolution. The results are written to a file whose size cannot exceed 5 MB; data compression is not performed. Which of the following values is closest to the maximum possible duration of a recorded sound fragment?

1) 1 minute 2) 30 seconds 3) 3 minutes 4) 90 seconds

Right answers:

A:3; B: 4 ; IN 1; G:1.

5.Addition. Some information about digital audio recording.

The propagation of sound in air can be considered as the propagation of pressure fluctuations. The microphone converts pressure fluctuations into vibrations electric current. This is an analog continuous signal. The sound card provides sampling input signal from the microphone. This is done as follows: a continuous signal is replaced by a sequence of values measured with a certain accuracy.

Analog signal graph:

Discrete representation of the same signal (41 measured values):

Discrete representation of the same signal (161 measured values, more high frequency sampling):

It can be seen that the higher the sampling frequency, the higher the quality of the approximate (discrete) signal. In addition to the sampling frequency, the quality of the digitized signal is affected by the number of binary bits allocated to record each signal value. The more bits allocated for each value, the more accurately the signal can be digitized.

An example of a 2-bit representation of the same signal (only 4 possible levels of signal magnitude can be numbered with two bits):

Now you can write down a dependency for the size of a file with digitized audio

file_size = (number_of_values_captured_per_1_second)*

*(number_of_binary_bits_for_recording_one_value)*

*(number_seconds_of_recording).

Taking into account the possibility of simultaneous recording of sound from several microphones (stereo, quad recording, etc.), which is done to enhance realism during playback, we obtain formula (1).

When audio is played back, digital values are converted to analog. Electrical vibrations transmitted to the speakers are converted back into air pressure fluctuations.

Knowledge is made up of small
grains of daily experience.
DI. Pisarev

Goals: Application of theoretical knowledge in practice.
Lesson objectives:
Teach the principle binary coding when digitizing sound;
Introduce the concept of time sampling of sound;
Establish the relationship between audio encoding quality, encoding depth and sampling frequency;
Learn to evaluate the information volume of an audio file;
Record sound using a computer, save it in audio files in WAV format, and play it back.

During the classes:

I. Organizational moment 1. Music is playing
2. Teacher's words:

The topic of our lesson is “Binary coding of audio information.” Today we will get acquainted with the concept of time sampling of sound, establish experimentally the relationship between sound encoding quality, encoding depth and sampling frequency, learn how to estimate the size of audio files, record sound using a computer, save it in sound files in WAV format and play it back.

II. Updating students' knowledge. Questions: (write answers in form No. 1)

1. List the types of existence of information? (numeric, text, graphic, sound).
2. Which keyword Can you match it to the video? (information coding).
3. What is called the depth of sound? (sound depth or encoding depth - the number of bits of information per audio encoding).
4. What volume levels can the sound have? (sound may have different volume levels.

5. What is the sampling rate? (Sampling frequency is the number of measurements of the input signal level per unit of time (per 1 second).
6. What is the formula for calculating the size of a digital mono audio file?
(A=D*T*I).
D - sampling frequency;
T is the time of sound playing or recording;
I - register bit size.
7. What is the formula for calculating the size of a digital stereo audio file?
A=2*D*T*I

III. Problem solving. Problem No. 1 (Semakin. No. 88 p. 157, problem book No. 1). Form No. 1.

Determine the amount of memory to store a digital audio file, the playing time of which is two minutes at a sampling frequency of 44.1 kHz and 16-bit extension.

IV. Learning new material.

Since the early 90s personal computers got the opportunity to work with audio information. Every computer that has a sound card, microphone and speakers can record, save and play audio information.
With the help of special software(sound recording editors) opens up wide possibilities for creating, editing and listening to sound files. Speech recognition programs are being created and, as a result, it becomes possible to control a computer using voice.
From your physics course, you know that sound is a mechanical wave with continuously changing amplitude and frequency (Fig. 1). The higher the amplitude, the louder the sound; the lower the frequency, the lower the tone. A computer is a digital device, so a continuous audio signal must be converted into a sequence of electrical impulses (zeros and ones). To do this, the plane on which the sound wave is graphically represented is divided into horizontal and vertical lines(Fig. 2 and Fig. 3). The horizontal lines are the volume levels, and the vertical lines are the number of measurements per second (one measurement per second is one hertz), or the sampling frequency (Hz). This method allows you to replace a continuous dependence with a discrete sequence of volume levels, each of which is assigned a value in binary code (Fig. 4).


Fig.1	Fig.2	Fig.3	Fig.4

The number of volume levels depends on the sound depth - the number of bytes used to encode one level. Typically 8 kHz and quantization level (8 bit code length).
, where N is the number of volume levels, and I is the sound depth (bits)

Example: Form No. 3
Solution:
1) encoding with a frequency of 5 Hz - this means that sound pitch measurements occur in 1 second. 4 bit depth means 16 volume levels are used.
We will “round” the pitch values to the nearest lower level. (Coding result: 1000 1000 1001 O11O 0111)

2) To calculate the information volume of encoded sound (A), a simple formula is used: A = D * i * T, where: D is the sampling frequency (Hz); i - sound depth (bits); T - playing time (sec).
We get: A = 5 Hz * 4 bits * 1 sec = 20 bits.

V. Educational independent work. Form No. 5

VI. Research assignment. Form No. 6

Groups No. 1-5. Establish a relationship between the quality of binary audio coding and the information volume of the audio file for audio information of various contents (monologue speech, dialogic speech, poem, song); the relationship between the information volume of the file and the recording mode (mono, stereo).

Progress of research work:

1) Fill out form No. 2.
2) Write down the results obtained during the experiment in a table.
3) Draw a conclusion.

VII. Summing up group work
VIII. Mini project Musical and sound capabilities.
Legend: Program: “A Christmas tree was born in the forest”
SCRN 7
LINE (20,0)-(300,180),2,BF
FOR I=l TO 2000
X=280*RND+20 Y=180*RND
C=16*RND
PSET(X,Y),C
NEXT I
SLEEP 1
LINE (150,140)-(170,160),6,BF
PSET(110,140)
LINE-(210,140), 10
LINE-(160,110),10
LINE-(110,140),10
PAINT (160,120), 10,10
LOCATE 24.10
PRINT “A Christmas tree was born in the forest”
PLAY “ms+80 02 18 caajafcc”
PSET (120,110)
LINE-(200,110),10
LINE-(160,85),10
LINE-(120,110),10
PAINT (160,90),10,10
LOCATE 24.10
PRINT "She grew up in the forest"
PLAY "caab->dc4"
PSET (130.85)
LINE-(190.85),10
LINE-(160,65),10
LINE-(130.85), 10
PAINT (160,70),10,10
LOCATE 24.10
PRINT “THE SLIMMER IN WINTER AND SUMMER”
PLAY "with PSET (140.65)
LINE-(180.65), 10
LINE -(160.50), 10
LINE - PAINT (160.60), 10.10
LOCATE 24.10
PRINT "GREEN WAS"
PLAY "caajofu"
SLEEP
STOP
IX Lesson summary

1). Monitoring the level of mastery of program material
1. At a sampling rate of 8 kHz, the quality of the sampled audio signal corresponds to:

radio broadcast quality;

2. In what format are audio files saved:

3. The quality of encoding a continuous audio signal depends on:

on sampling frequency and encoding depth;

4. Two audio files are recorded at the same sampling rate and encoding depth. The information volume of a file recorded in stereo mode is greater than the information volume of a file recorded in mono mode:

2). Assessment of students' knowledge and skills.
3). Teacher's word.

Of course, assessing sound quality is largely subjective and depends on our perception. A computer, just like a person, encodes sound information for the purpose of storage and subsequent playback. Think about it, what is the difference between audio information stored in PC memory and in human memory? (Answer: in humans, the process of encoding sound is closely related to emotions).
Thus, the computer stores the sound, and the person stores the music!!! Music is the only language in which soul speaks to soul (Berthold Auerbach). It can lift into the skies, awaken the senses, bind the mind and instill fear. Each person has their own music. What emotions or associations does “Moonlight Sonata” evoke in you?... The warm gaze of a loving person, the gentle touch of a mother’s hand, and now it is possible that these enchanting sounds will remind you of your computer science lesson. All this, you see, is inaccessible to digital binary code.

X. Homework Problems No. 89,91,92 page 157.

N = k * F * L * T (1)

N = k F * L * T* (1)