The volume of the audio file. The topic Solving tasks for encoding sound information. Sound File Formats Examples

The lesson is devoted to the analysis of the task 9 EGE on computer science

9 Theme - "Coding of information, the volume and transmission of information" is characterized as the tasks of the base level of complexity, the execution time is about 5 minutes, the maximum score - 1

Coding text information

n. - Characters

i. - Number of bits per 1 character (encoding)

Coding graphic information

Consider some concepts and formulas necessary for the decision of the EGE on the computer science of this topic.

• Pixel- This is the smallest element of a bitmap image that has a certain color.
• Resolution- This is the number of pixels on an inch image size.
• Color depth - This is the number of bits required to encode pixel color.
• If the coding depth is i. bits per pixel, the code of each pixel is selected from 2 I. possible options, so you can use no more 2 I. Different colors.

Formula for finding the number of colors in the palette used:

• N. - Number of flowers
• i. - Color depth
• In the RGB color model (Red (R), Green (G), Blue (B)): R (0..255) G (0..255) B (0..255) -\u003e Receive 2 8 Options for each of three colors.
• R G B: 24 Bit \u003d 3 bytes - true Color mode (true color)

Find the formula of the memory for storing a raster image:

• I. - memory required for storage
• M. - Width of the image in pixels
• N. - Height of the image in pixels
• i. - color coding depth or permission

Or you can write the formula like this:

I \u003d n * i bits

• where N. - number of pixels (M * N) and i. - color coding depth (coding bit)

* To specify the amount of dedicated memory there are different notation ( V. or I.).

• You should also remember the conversion formulas:

1 MB \u003d 2 20 bytes \u003d 2 23 bits,
1 KB \u003d 2 10 byte \u003d 2 13 bits

Encoding audio information

We will get acquainted with the concepts and formulas necessary for solving tasks 9 EGE on computer science.

Example: with ƒ \u003d 8 kHz, coding depth 16 bits on the countdown and duration of the sound 128 S.. required:

✍ Solution:

I \u003d 8000 * 16 * 128 \u003d 16384000 bits
I \u003d 8000 * 16 * 128/8 \u003d 2 3 * 1000 * 2 4 * 2 7/2 3 \u003d 2 14/2 3 \u003d 2 11 \u003d
\u003d 2048000 byte

Determination of information transfer rate

• Communication channel always has limited throughput (information transfer rate), which depends on the properties of the equipment and the line itself (cable)

The amount of information transmitted I is calculated by the formula:

• I. - amount of information
• v. - communication channel bandwidth (measured in bits per second or similar units)
• t. - Transmission time

* Instead of setting speed V. Sometimes used Q.
* Instead of the designation of the message I. Sometimes used Q.

The data transfer rate is determined by the formula:

and measured B. bIT / S.

Decision of tasks 9 EGE on computer science

Subject: Image Coding

9_1: Ege in computer science 2017 Task 9 FIPI Option 1 (Krylov S.S., Churkina ie):

What is the minimum memory (in KB) need to be reserved so that you can save any raster image size 160 x 160. pixels provided that in the image can be used 256 Different colors?

✍ Solution:

• We use the formula for finding the volume:
• We will calculate every fact that in the formula, trying to bring the numbers to the degrees of the twos:
• M x n:

160 * 160 \u003d 20 * 2³ * 20 * 2³ \u003d 400 * 2 6 \u003d 25 * 2 4 * 2 6

Finding the depth of coding i.:

256 \u003d 2 8. 8 bits per pixel (from the formula number of colors \u003d 2 i)

Find volume:

I. \u003d 25 * 2 4 * 2 6 * 2 3 \u003d 25 * 2 13 - Total bit on all image

Transfer to Kable:

(25 * 2 13) / 2 13 \u003d 25 KB

Result:25

Detailed persisals of the task of 9 ege on computer science We offer to look in the video:

EGE on computer science task 9.2 (Source: option 11, K. Polekov):

Figure size 128 on the 256 pixels is in memory 24 KB (excluding compression). number of flowers In the picture palette.

✍ Solution:

• where M * N. - Total number of pixels. We will find this value using for the convenience of decendibility:

128 * 256 = 2 7 * 2 8 = 2 15

In the above formula i. - This is the depth of color from which the number of colors in the palette depends:

Number of colors \u003d 2 i

Find i. From the same formula:

i \u003d i / (m * n)

We take into account what 24 KB It is necessary to translate B. bits. We get:

2 3 * 3 * 2 10 * 2 3: i \u003d (2 3 * 3 * 2 10 * 2 3) / 2 15 \u003d 3 * 2 16/2 15 \u003d 6 bits

Now we find the number of colors in the palette:

2 6 = 64 Color options in color palette

Result:64

See Task Video Relationship:

Topic: Image Coding:

EGE on computer science task 9.3 (Source: 9.1 Option 24, K. Poles):

After converting raster 256-color Graphic file B. 4-color The format its size decreased on 18 KB. What was it the sizethe source file in Kablei?

✍ Solution:

• By the formula of the image file, we have:

where N. - Total number of pixels,
but i.

• i. Can be found, knowing the number of colors in the palette:

number of colors \u003d 2 i

Before conversion: i \u003d 8 (2 8 \u003d 256) after conversion: i \u003d 2 (2 2 \u003d 4)

Make a system of equations based on the available information, we will take x. Number of pixels (permission):

I \u003d x * 8 i - 18 \u003d x * 2

Express x. In the first equation:

x \u003d i / 8

I. (file size):

I - 18 \u003d i / 4 4i - i \u003d 72 3i \u003d 72 i \u003d 24

Result:24

A detailed analysis of 9 tasks of the exam. Look at the video:

Topic: Image Coding:

EGE on computer science task 9.4 (Source: 9.1 Option 28, K. Polyakov, S. Loginova):

The color image was digitized and saved as a file without the use of data compression. The size of the file received - 42 MB 2 times less and color coding depth increased in 4 times more compared to the original parameters. Data compression has not been produced. Specify file size in MBobtained during repeated digitization.

✍ Solution:

• By the formula of the image file, we have:

where N.
but i.

• In this kind of tasks it is necessary to take into account that a decrease in the resolution is 2 times, it implies a decrease in 2 times the pixels separately in width and height. Those. In general, N decreases 4 times!
• We will make a system of equations based on the existing information in which the first equation will correspond to the data before transformation of the file, and the second equation after:

42 \u003d n * i i \u003d n / 4 * 4i

Express i. In the first equation:

i \u003d 42 / n

Substitute to the second equation and find I. (file size):

\\ [I \u003d \\ FRAC (N) (4) * 4 * \\ FRAC (42) (N) \\]

After cuts, we get:

I \u003d. 42

Result:42

Topic: Image Coding:

EGE on computer science task 9.5 (Source: 9.1 Option 30, K. Polyakov, S. Loginova):

The image was digitized and saved as a raster file. The resulting file was transferred to cities on the communication channel for 72 seconds. Then the same image was digitized again with permission to 2 times larger and depth of color coding in 3 times less than the first time. Data compression has not been produced. The resulting file was transferred to city B., communication channel bandwidth with city B in 3 higher than the communication channel with the city of A.
B.?

✍ Solution:

• According to the file transfer rate formula, we have:

where I. - the size of the file, and t. - Time

• By the formula of the image file, we have:

where N. - Total number of pixels or permission,
but i. - color depth (number of bits highlighted by 1 pixel)

• For this task, it is necessary to clarify that the resolution actually has two beings (pixels in the width of * pixels in height). Therefore, with an increase in the resolution, both numbers will increase, i.e. N. will increase B. 4 Once instead of two times.
• Change the formula for obtaining a file for the city B.:

\\ [I \u003d \\ FRAC (2 * N * i) (3) \\]

• For the city a and b, replace the volume values \u200b\u200bin the formula for obtaining speed:

\\ [V \u003d \\ FRAC (N * i) (72) \\]

\\ [3 * v \u003d \\ FRAC (\\ FRAC (4 * N * i) (3)) (T) \\]

\\ [T * 3 * V \u003d \\ FRAC (4 * N * i) (3) \\]

• We substitute the value of the speed from the formula for the city and in the formula for the city b:

\\ [\\ FRAC (T * 3 * N * i) (72) \u003d \\ FRAC (4 * N * i) (3) \\]

• Express t.:

T \u003d 4 * 72 / (3 * 3) \u003d 32 Seconds

Result:32

Another way to solve see the video language:

Topic: Image Coding:

EGE on computer science task 9.6 (Source: Option 33, K. Polekov):

Camera makes photographs size 1024 x 768. pixels. The storage of one frame is given 900 KB.
Find the maximum possible number of flowers In the picture palette.

✍ Solution:

• The number of colors depends on the depth of the color coding, which is measured in the bits. For storage of frame, i.e. total pixels allocated 900 Krib. Transfer to bits:

900 KB \u003d 2 2 * 225 * 2 10 * 2 3 \u003d 225 * 2 15

Consider the total number of pixels (from the specified size):

1024 * 768 = 2 10 * 3 * 2 8

We define the amount of memory required for storing not the total number of pixels, but one pixel ([frame memory] / [Pixel])):

\\ [\\ FRAC (225 * 2 ^ (15)) (3 * 2 ^ (18)) \u003d \\ FRAC (75) (8) \\ APPROX 9 \\]

9 bits per 1 pixel

9 bits is i. - Depth of color coding. Number of colors \u003d 2 i:

2 9 = 512

Result:512

Watch a detailed video solution:

Topic: Image Coding:

9_8: DEVEROVIYS EGE 2018 Informatics:

Automatic camera produces raster images in size 640 × 480 pixels. In this case, the volume of the image file cannot exceed 320 Krib, data packaging is not performed.
What Maximum number of colors Can I use in the palette?

✍ Solution:

• By the formula of the image file, we have:

where N. i. - Depth of color coding (the number of bits allocated to 1 pixel)

• Let's see that we are already given from the formula:

I.\u003d 320 kb, N. \u003d 640 * 420 \u003d 307200 \u003d 75 * 2 12 total pixels, i. - ?

The number of colors in the image depends on the parameter i.which is unknown. Recall the formula:

number of colors \u003d 2 i

Since the color depth is measured in bits, then the volume is necessary to translate from kilobytes to the bits:

320 KB \u003d 320 * 2 10 * 2 3 bits \u003d 320 * 2 13 bits

Find i.:

\\ [i \u003d \\ FRAC (I) (N) \u003d \\ FRAC (320 * 2 ^ (13)) (75 * 2 ^ (12)) \\ APPROX 8.5 bits \\]

We find the number of colors:

2 i \u003d 2 8 \u003d 256

Result: 256

A detailed solution of this 9 tasks from the demo level of the 2018 year, look at the video:

9_21: : EGE on computer science task 9.21 (Source: K. Polekov, 9.1 Option 58):

For storage in the information system, documents are scanned with permission 300 ppi.. Image compression methods are not used. The average size of the scanned document is 5 MB. In order to save it, it was decided to switch to permission 150 PPI and color system containing 16 colors. The average size of a document scanned with modified parameters is 512 KB.

Determine number of flowers In the palette before optimization.

✍ Solution:

• By the formula of the image file, we have:

where N. - total number of pixels or permission, and i. - Depth of color coding (the number of bits allocated to 1 pixel).

• Since on the task we have permission expressed in pixels per inch, then actually this means:

I \u003d PPI 2 * N * i value

• Formula of the number of colors:

number of colors \u003d 2 i

• Let's see that we have already been given from the formula to the economical option and during the economical version:

Uneconom of: I.\u003d 5 MB \u003d 5 * 2 23 bits, N. - ?, i. -? 300 ppi. Economy option: I.\u003d 512 KB \u003d 2 9 * 2 13 bits \u003d 2 22 bits, N. - ?, i. \u003d 4 bits (2 4 \u003d 16) 150 ppi

Since in economical mode, we know all components of the formula, except for the resolution (N), we will find permission:

N \u003d i / (I * 150 * 150 ppi) n \u003d 2 22 / (4 * 22500)

We substitute all known values, including found N, in the formula for the uneconomic mode:

I \u003d n * 300 * 300 ppi * i 5 * 2 23 \u003d (2 22 * \u200b\u200b300 * 300 * i) / (22500 * 4);

Express i. and calculate its value:

i \u003d (5 * 2 23 * 22500 * 4) / (2 22 * \u200b\u200b300 * 300) \u003d 9000/900 \u003d 10 bits

By the formula for finding the number of colors in the palette, we have:

2 10 = 1024

Result: 1024

Subject: Sound Coding

9_7: EGE on computer science 2017 Task 9 FIPI Option 15 (Krylov S.S., Churkina ie):

In studios at four-channel ( quadro) S. sound recording 32 -bit permission to be 30 seconds was recorded a sound file. Data compression has not been produced. It is known that the file size turned out to be 7500 Krib.

With what sampling frequency (in kHz) was recorded?As an answer, specify only the number, you do not need to specify the units of measurement.

✍ Solution:

• By the formula of the audio file, we will get:

I \u003d β * T * ƒ * s

• From the task we have:

I.\u003d 7500 KB β \u003d 32 bits t.\u003d 30 seconds S.\u003d 4 channels

ƒ - The sampling frequency is unknown, express it from the formula:

\\ [ƒ \u003d \\ FRAC (I) (S * B * T) \u003d \\ FRAC (7500 * 2 ^ (10) * 2 ^ 2 bits) (2 ^ 7 * 30) Hz \u003d \\ FRAC (750 * 2 ^ 6 ) (1000) kHz \u003d 2 ^ 4 \u003d 16 \\]

2 4 = 16 kHz

Result: 16

For a more detailed analysis, we suggest see video solutions of this 9 task EGE on computer science:

Subject: Sound coding:

EGE on computer science task 9_9 (Source: 9.2 Option 36, K. Polekov):

The musical fragment was digitized and recorded as a file without the use of data compression. The resulting file was transferred to the city BUT via communication channel. Then the same musical fragment was digitized with resolution in 2 3 times less than the first time. Data compression has not been produced. The resulting file was transferred to the city B. per 15 seconds; Communication channel communication with city B. in 4 3 times higher than communication channel with city BUT.

How many seconds the file transfer to the city lasted A.? In response, write down only an integer, the unit of measurement is not necessary.

✍ Solution:

• To solve, it will be necessary for the formula for finding the transfer rate of the formula:
• Recall also the formula of the audio file:

I \u003d β * ƒ * T * s

where:
I. - Volume
β - coding depth
ƒ - sampling frequency
t. - Time
S. - Number of channels (if not specified, then mono)

• Drink separately, all data relating to the city B. (pro BUT Almost nothing is known):

city b: β - 2 times higher ƒ - 3 times less t. - 15 seconds, bandwidth (speed V.) - 4 times higher

Based on the previous paragraph, for the city, and get the inverse values:

cities: β b / 2 ƒ b * 3 I b / 2 V b / 4 t b / 2, t b * 3, t b * 4 -?

Let us give an explanation to the data:

as coding depth ( β ) For the city B. Above B. 2 times, then for the city BUT It will be lower in 2 times, respectively, and t. will decrease in 2 Once:

t \u003d T / 2

as sampling frequency (ƒ) For the city B. Less than B. 3 times, then for the city BUT It will be higher in 3 times; I. and t. It is changed in proportion, it means that, with an increase in the sampling frequency, not only volume will increase, but also time:

t \u003d T * 3

speed \u200b\u200b( V.) (bandwidth) for the city B. Above B. 4 times, then for the city BUT It will be 4 times lower; times the speed below, then the time is higher in 4 times ( t. and V. - inversely proportional dependence from the formula V \u003d I / T):

t \u003d T * 4

Thus, taking into account all the indicators, time for the city BUT changes like this:

\\ [T_A \u003d \\ FRAC (15) (2) * 3 * 4 \\]

90 seconds

Result: 90

For a detailed solution, see the video:

Subject: Sound coding:

EGE on computer science task 9.10 (Source: 9.2 Option 43, K. Polekov):

The musical fragment was recorded in the format of stereo ( two-channel recording), digitized and saved as a file without the use of data compression. The size of the file received - 30 MB. Then the same musical fragment was recorded re-in format monoand digitized with resolution 2 times higher and frequency of sampling in 1,5 times less than the first time. Data compression has not been produced.

Specify file size in MBRecord received. In response, write down only an integer, the unit of measurement is not necessary.

✍ Solution:

I \u003d β * ƒ * T * s

I. - Volume
β - coding depth
ƒ - sampling frequency
t. - Time
S. - Channels

• We discard separately, all data relating to the first state of the file, then the second state after the conversion:

1 Condition: S \u003d 2 channels i \u003d 30 MB 2 Condition: S \u003d 1 channel β \u003d 2 times higher ƒ \u003d 1.5 times lower I \u003d?

Since it was originally 2 Communication channel ( S.), and began to be used one communication channel, the file has decreased in 2 Once:

I \u003d i / 2

Coding depth ( β ) Increased B. 2 times, then the volume ( I.) will increase B. 2 times (proportional dependence):

I \u003d i * 2

Sampling frequency ( ƒ ) Reduced B. 1,5 times, it means volume ( I.) will also decrease in 1,5 Once:

I \u003d i / 1.5

Consider all changes in the volume of the transformed file:

I \u003d 30 MB / 2 * 2/15 \u003d 20 MB

Result: 20

See the video collection of this task:

Topic: Encoding audio files:

EGE on computer science task 9_11 (Source: 9.2 Option 72, K. Polekov):

The musical fragment was digitized and recorded as a file without the use of data compression. The resulting file was transferred to cities on the communication channel for 100 seconds. Then the same musical fragment was digitized by resolution 3 times higher and sampling frequency 4 times lessthan for the first time. Data compression has not been produced. The resulting file was transferred to city B. per 15 seconds.

How many times the speed (channel bandwidth) to the city B. More channel bandwidth to the city BUT ?

✍ Solution:

• Recall the formula of the audio file:

I \u003d β * ƒ * T * s

I. - Volume
β - coding depth
ƒ - sampling frequency
t. - Time

• Drink separately, all data regarding the file transmitted to the city BUTthen the transformed file transmitted to the city B.:

BUT: T \u003d 100 C. B: β \u003d 3 times higher ƒ \u003d 4 times lower than t \u003d 15 c.

✎ 1 solution method:

Data transfer rate (bandwidth) depends on file transfer time: the longer the time, the lower the speed. Those. How many times the transmission time will increase, speed and vice versa decrease in so many times.

From the previous item we see that if we calculate how many times the file transfer to the city will decrease or increase B. (Compared to the city a), then we will understand how many times the data transfer rate to the city will increase or decrease B. (inverse dependence).

Accordingly, imagine that the transformed file is transmitted to the city BUT. The file volume has changed in 3/4 times (coding depth (β) in 3 the above, the sampling rate (ƒ) in 4 times lower). Volume and time varies proportionally. So time will change in 3/4 Once:

T a for concessions. \u003d 100 seconds * 3/4 \u200b\u200b\u003d 75 seconds

Those. The transformed file would be transmitted to the city BUT 75 Seconds, and in the city B. 15 seconds. Calculate how many times the transfer time has decreased:

75 / 15 = 5

Once time transfer to the city B. decreased in 5 times, respectively, speed increased in 5 time.

Answer: 5

✎ 2 Decision method:

Drink out separately all data relating to the file transmitted to the city BUT: BUT: t a \u003d 100 c. V a \u003d i / 100

Because the increase or decrease in some times the resolution and sampling frequency leads to an appropriate increase in or reducing the file size (proportional dependence), we write the known data for the converted file transmitted to the city B.:

B: β \u003d 3 times higher ƒ \u003d 4 times lower than t \u003d 15 c. I b \u003d (3/4) * i v b \u003d ((3/4) * i) / 15

Now find the ratio V b to V A:

\\ [\\ FRAC (V_B) (V_A) \u003d \\ FRAC (3 / _4 * i) (15) * \\ FRAC (100) (I) \u003d \\ FRAC (3 / _4 * 100) (15) \u003d \\ FRAC (15 ) (3) \u003d 5 \\]

(((3/4) * i) / 15) * (100 / i) \u003d (3/4 * 100) / 15 \u003d 15/3 \u003d 5 S. - Number of channels

For the simplicity of calculations, we will not take into account the number of channels. Consider what data we have, and which of these must be translated into other units:

β \u003d 32 bits ƒ \u003d 32kc \u003d 32000Hz t \u003d 2 min \u003d 120 s

Substitute data in the formula; We take into account that the result must be obtained in MB, respectively, the product will be divided by 2 23 (2 3 (bytes) * 2 10 (KB) * 2 10 (MB)):

(32 * 32000 * 120) / 2 23 = =(2 5 * 2 7 * 250 * 120) / 2 23 = = (250*120) / 2 11 = = 30000 / 2 11 = = (2 4 * 1875) / 2 11 = = 1875 / 128 ~ 14,6

The result is the result of the value of the volume to multiply on 4 Taking into account the number of communication channels:

14,6 * 4 = 58,5

Nearest multiple 10. - this is 60 .

Result: 60

Watch a detailed solution:

Subject: Sound coding:

9_19: State graduation exam of the HBE 2018 (Infi Infi Infi Infi Infi, Task 7):

Produced two-channel (Stereo) digital sound recording. The signal value is fixed 48,000 times per second, to record each value used 32 bits. Recording lasts 5 minutesThe results are recorded in a file, data compression is not performed.

Which of the following values \u200b\u200bis closest to the size of the received file?

1) 14 MB
2) 28 MB
3) 55 MB
4) 110 MB

✍ Solution:

I \u003d β * ƒ * T * s

Substitute the existing values \u200b\u200bin the formula:

I \u003d 48000 * 32 * 300 * 2

Since the values \u200b\u200bare large, it is necessary 48000 and 300 express in degrees of two:

48000 | 2 24000 | 2 12000 | 2 6000 | 2 = 375 * 2 7 3000 | 2 1500 | 2 750 | 2 375 | 2 - no longer divided by 187.5 300 | 2 \u003d. 75 * 2 2 150 | 2 75 | 2 - no longer divisible 37.5

We get:

I \u003d 375 * 75 * 2 15

In the proposed answer options we see that the result is everywhere in MB. So it is necessary to divide the result obtained by 2 23 (2 3 * 2 10 * 2 10):

I \u003d 375 * 75 * 2 15/2 23 \u003d 28125/2 8

Find close to the number 28125 value to degree degree:

2 10 = 1024 1024 * 2 2048 * 2 4096 * 2 8192 * 2 16384 * 2 32768

We get:

2 10 * 2 5 = 2 15 = 32768 2 10 * 2 4 = 2 14 = 16384

Number 28125 lies between these meanings, it means that we take them:

2 15 / 2 8 = 2 7 = 128 2 14 / 2 8 = 2 6 = 64

Select the answer, the value in which is between the two these numbers: option 4 (110 MB)

Result: 4

Detailed decision of the HBE task 7 2018. Watch the video:

Subject: Sound coding:

9_20: Decision 9 of the tasks of the EGE on computer science (diagnostic version of the examination work of 2018, S.S. Krylov, D.M. Ushakov):

Produced two-channel (stereo) sound recording with the frequency of sampling 4 kHz and 64-bit resolution. Recording lasts 1 minuteThe results are recorded in a file, data compression is not performed.

Determine approximately the size of the resulting file (in MB). As an answer, specify the nearest number, multiple to the size of the file. 2 .

✍ Solution:

• By the formula of the audio file, we have:

I \u003d β * ƒ * T * s

I - volume β - coding depth \u003d 32 bits ƒ - sampling frequency \u003d 48000 Hz T - Time \u003d 5 min \u003d 300 s - number of channels \u003d 2

Substitute the existing values \u200b\u200bin the formula. For convenience, we will use de-degree degrees:

ƒ \u003d 4 kHz \u003d 4 * 1000 Hz ~ 2 2 * 2 10 B \u003d 64 bits \u003d 2 6/2 23 MB T \u003d 1 min \u003d 60 C \u003d 15 * 2 2 C s \u003d 2

We substitute the values \u200b\u200bin the formula of the audio file:

I \u003d 2 6 * 2 2 * 2 10 * 15 * 2 2 * 2 1/2 23 \u003d 15/4 ~ 3.75

The nearest whole, multiple two - this is the number 4

Result: 4

Task video collection:

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

Objectives: The use of theoretical knowledge in practice.
Tasks lesson:
Teach the principle of binary coding during sound digitization;
Introduce the concept of temporary sampling of sound;
Establish the relationship between sound coding quality, coding depth and sampling frequency;
Teach evaluating the information volume of the audio file;
Record sound using a computer, save it in sound files in WAV format, play.

During the classes:

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

The topic of our lesson "binary encoding of sound information". Today we will get acquainted with the concept of temporary sampling of sound, install experimental by the dependence between the quality of sound coding, coding depth and sampling frequency, learn how to evaluate the amount of audifiles, write sound using a computer, save it in sound files in WAV format and play.

II. Actualization of students' knowledge. Questions: (Answers to record in Blank №1)

1. List the types of information? (Number, textual, graphic, sound).
2. What key word can be chosen to the video. (information encoding).
3. What do the sound depth? (The sound depth or coding depth is the number of data bits on sound encoding).
4. What volume levels may have sound? (The sound may have different volume levels.

5. What is called sampling frequency? (Sampling frequency - the number of measurements of the input level per unit time (1 second).
6. What formula is calculated by the size of the digital monoadyphyle?
(A \u003d D * T * i).
Discretization frequency;
T- sounding or sound recording;
Register II.
7. What formula is calculated by the size of the digital stereo audio file?
A \u003d 2 * d * t * i

III. Solving tasks. Task number 1 (Semaakin. №88 p. 157, Task No. 1). Blank number 1.

Determine the amount of memory for the storage of a digital audio file, the sound time of which is two minutes at a sampling frequency of 44.1 kHz and expansion of 16 bits.

IV. Studying a new material.

Since the beginning of the 90s, personal computers have been able to work with sound information. Each computer having a sound card, a microphone and column, can record, save and play audio information.
With the help of special software (sound recordings), there are extensive opportunities for creating, editing and listening to sound files. Create speech recognition programs and, as a result, the ability to control a computer using a voice.
From the course of physics, you know that the sound is a mechanical wave with a continuously changing amplitude and frequency (Fig. 1). The higher the amplitude, the louder the sound, the less frequency, the lower the tone. Computer is a digital, therefore a continuous beep must be converted to a sequence of electrical pulses (zeros and units). For this, the plane on which the sound wave is graphically represented, is divided into horizontal and vertical lines (Fig. 2 and Fig. 3). Horizontal lines are volume levels, and vertical - the number of measurements for 1 second (one measurement per second is one hertz), or the discretization frequency (Hz). This method allows to replace the continuous dependence on the 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 depth of the sound - the number of bytes, used to encode one level. Usually 8 kHz and quantization level (code 8 bits length).
where N is the number of volume levels, and i - the depth of sound (bits)

Example: Blank number 3
Decision:
1) Coding with a frequency of 5 Hz - this means that the sound height measurements in 1 sec occurs. The depth of 4 bits means 16 volume levels are used.
"Round" the sound height values \u200b\u200bwill be up to the nearest lower level. (Coding result: 1000 1000 1001 O11O 0111)

2) To calculate the information volume of the encoded sound (a), a simple formula is used: A \u003d D * I * T, where: D is the sampling rate (Hz); i - the depth of sound (bit); T - Sound time (sec).
We get: a \u003d 5 Hz * 4 bits * 1 sec \u003d 20 bits.

V. Training independent work. Blank number 5.

Vi. Research task. Blank number 6.

Groups number 1-5. Establish the relationship between the quality of binary sound coding and the audio file information for sound information of various content (monologue, dialogic speech, poem, song); Dependence between file information and recording mode (mono, stereo).

Research work:

1) Fill out the form number 2.
2) Write results in the table obtained during the experiment.
3) Make a conclusion.

VII. Summing up work in groups
VIII. Mini Project Musical and Sound Opportunities.
Designations: Program: "A Christmas tree was born in the forest"
SCRN 7.
Line (20.0) - (300,180), 2, BF
For i \u003d l to 2000
X \u003d 280 * RND + 20 Y \u003d 180 * RND
C \u003d 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 "in the forest she grew up,"
Play "Caab-\u003e 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 "Winter and Summer Slim"
Play "C 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 Outcome lesson

one). Control of the level of learning software
1. At the sampling frequency of 8 kHz, the quality of the sampled sound signal corresponds to:

a) the quality of audio-CD sound;
b) quality of radio broadcasts;
c) medium quality.

2. In which format sound files are saved:

a) doc;
b) WAV;
c) BMP.

3. The quality of the coding of the continuous beep depends:

but) from the sampling frequency and coding depth;
b) from the depth of color and the resolution of the monitor;
c) from the International Coding Standard.

4. Two sound files are recorded with the same sampling frequency and coding depth. The information volume of the file recorded in stereo gear, more information on the file recorded in the monodemif:

a) 4 times;
b) the volumes are the same;
c) 2 times.

2). Assessment of knowledge and skills of students.
3). The word of the teacher.

Of course, the assessment of the quality of sound is largely subjective and depends on our perception. The computer, as well as a person, encodes sound information for the purpose of storing and subsequent playback. Think, and what is the difference between sound information stored in PC memory and in a person's memory? (Answer: a person has a sound coding process closely associated with emotions).
Thus, the computer stores the sound, and the man music !!!Music is a single language, on which the soul speaks with soul (Bertold Averbach). She can raise in heaven, awaken feelings, to sow the mind and instill fear. For each person music has its own. What emotions or association cause you "Moon Sonata"? ... Warm look of a loving person, gentle touch of the mother hand, and now it is possible that these charming sounds will remind you about the lesson of computer science. All this, you see, inaccessible to a digital binary code.

H. Homework Tasks number 89,91.92 pp 157.

The calculation of the information volume of the audio file can be performed according to the following formula (4):

V audio \u003d D * T * N channels * I / K compression, (4)

where V is the information volume of the audio file, measured in bytes, kilobytes, megabytes; D - the sampling frequency (number of points per second to describe audio records); T - audio file time; N channels - the number of audio file channels (stereo - 2 channels, system 5.1 - 6 channels); I - the depth of sound, which is measured in bits, K compression is the data compression ratio, it is equal to 1.

Calculation of the information of the animation

The calculation of the information volume of the animation can be performed according to the following formula (5):

V anim \u003d k * t * v. * I / K compression, (5)

where V anim is an information volume of a raster graphic image that is measured in bytes, kilobytes, megabytes; K - the number of pixels (points) in the image determined by the resolution of the media of the information (monitor screen, scanner, printer); T - animation time; v. - frequency of shift frames per second; I - the depth of color that is measured in the bits per pixel, K compression is the data compression ratio, it is equal to 1.

Calculation of the formation volume of the video file

The calculation of the video file of the video file can be performed according to the following formula (5):

V video \u003d V Anim + V Audio + V Sub, (5)

where v video is a video file information that is measured in bytes, kilobytes, megabytes; V anim is an animation information (video range), measured in bytes, kilobytes, megabytes; V Audio is an information field of the Audeo file, measured in bytes, kilobytes, megabytes (in the video clip can contain audio track files for several languages, then multiply the volume of the audio file by the number of language tracks); V Sub is the informational scope of subtitle files, measured in bytes, kilobytes, megabytes (if several subtitle files, then you need to fold the dimensions of each file).

Practical part

Parameters / Options
Frame frequency
Image size
Color Depth, Bit
Image compression ratio
Audio track
Number of languages
Sound Depth, Bit
Discretization frequency of audio stream, Hz
Audio track compression ratio
Number of subtitles, pcs.
Encoding text subtitles
The number of characters in the subtitle file, pcs.
Coefficient of text compression

When solving tasks, students relieve the following concepts:

Temporary discretization - the process at which, during the coding of a continuous sound signal, the sound wave is divided into separate small temporary sections, and a certain amount of amplitude is established for each such site. The more the amplitude of the signal, the louder the sound.

Sound Depth (coding depth) - The number of bit on sound encoding.

Volume levels (signal levels) - The sound may have different volume levels. The number of different volume levels are calculated by the formula N. = 2 I. Where I. - Sound depth.

Sampling frequency – the number of measurements of the input signal level per unit time (for 1 s). The greater the discretization frequency, the more accurate the binary coding procedure. Frequency is measured in Hertz (Hz). 1 Measurement for 1 second -1 Hz.

1000 measurements for 1 second 1 kHz. Denote the frequency of sampling letter D.. For encoding, choose one of three frequencies: 44.1 kHz, 22.05 kHz, 11.025 kHz.

It is believed that the frequency range that the person hears is from20 Hz to 20 kHz .

Quality of binary coding - The value that is determined by the depth of coding and the sampling frequency.

Audio adapter (sound fee) - a device that converts electrical oscillations of sound frequency in a numeric binary code when entering sound and back (from numerical code into electrical oscillations) when playing sound.

Audio adapter characteristics: Discretization frequency and the discharge of the register.).

The discharge of the register - the number of bits in the audio adapter register. The greater the discharge, the smaller the error of each individual transformation of the electrical current value and back. If the discharge is equal I. , when measuring the input signal can be obtained 2 I. = N. Different values.

Digital monoadylfile size ( A. ) It is measured by the formula:

A. = D. * T. * I. /8 where D. – discretization frequency (Hz), T. - sound time or sound recording, I. the discharge of the register (permission). According to this formula, the size is measured in bytes.

Digital stereo audio file size ( A. ) It is measured by the formula:

A. =2* D. * T. * I. /8 The signal is recorded for two columns, as the left and right sound channels are separately encoded.

Students are useful to issue table 1.showing how many MB will occupy a coded one minute of sound information with different sampling frequency:

Frequency Discretization, kHz

44,1

22,05

11,025

16 bits, stereo

10.1 MB

5.05 MB

2.52 MB

16 bits, mono

5.05 MB

2.52 MB

1.26 MB

8 bits, mono

2.52 MB

1.26 MB

630 KB

1. Digital file size

Level "3"

1. Determine the size (in bytes) of the digital audio file, the sound time of which is 10 seconds at a discretization frequency of 22.05 kHz and a resolution of 8 bits. The file compression is not subject. (, p. 156, Example 1)

Decision:

Formula for calculating the size(in bytes) Digital audio file: A. = D. * T. * I. /8.

To transfer to bytes, the value obtained must be divided into 8 bits.

22.05 kHz \u003d 22.05 * 1000 Hz \u003d 22050 Hz

A. = D. * T. * I. /8 = 22050 x 10 x 8/8 \u003d 220500 bytes.

Answer: File size 220500 byte.

2. Determine the amount of memory for storing the digital audio file, the sound time of which is two minutes at a sampling frequency of 44.1 kHz and a resolution of 16 bits. (, p. 157, №88)

Decision:

A. = D. * T. * I. /eight. - The amount of memory for storing the digital audio file.

44100 (Hz) x 120 (C) x 16 (BIT) / 8 (bit) \u003d 10584000 byte \u003d 10335,9375 KB \u003d 10.094 MB.

Answer: ≈ 10 MB

Level "4"

3. The user has a memory of 2.6 MB. It is necessary to record a digital audio file with a duration of a 1 minute sound. What should be the sampling frequency and the discharge? (, p. 157, №89)

Decision:

Formula for calculating the frequency of sampling and discharging:D.* I. \u003d A / t

(Memory capacity in bytes): (Sound time in seconds):

2, 6 MB \u003d 2726297.6 byte

D.* I. \u003d A / t \u003d 2726297.6 byte: 60 \u003d 45438.3 byte

D \u003d45438.3 bytes : I.

The discharge of the adapter can be 8 or 16 bits. (1 byte or 2 bytes). Therefore, the discretization frequency may beeither 45438.3 Hz \u003d 45.4 kHz ≈ 44.1 kHz. - Standard characteristic frequency of discretization, or 22719.15 Hz \u003d 22.7 kHz ≈ 22.05 kHz - Standard characteristic sampling rate

Answer:

4. The volume of free memory on the disk is 5.25 MB, the discharge of the sound card - 16. What is the duration of the digital audio file recorded with a discretization frequency of 22.05 kHz? (, p. 157, №90)

Decision:

Formula for calculating the duration of sound: T \u003d A / D / I

(Memory capacity in bytes): (sampling frequency in Hz): (Sound board bit in bytes):

5.25 MB \u003d 5505024 byte

5505024 byte: 22050 Hz: 2 bytes \u003d 124.8 sec
Answer: 124.8 seconds

5. One minute of recording of a digital audio file occupies 1.3 MB on a disk, the discharge of the sound card - 8. What frequency of sampling is the sound recorded? (, p. 157, №91)

Decision:

Formula for calculating sampling frequency: D. \u003d A / t / I.

(memory capacity in bytes): (recording time in seconds): (Sound card discharge in bytes)

1.3 MB \u003d 1363148.8 byte

1363148.8 Byte: 60: 1 \u003d 22719.1 Hz

Answer: 22.05 kHz

6. Two minutes of recording of a digital audio file occupy 5.1 MB on a disk. Discretization frequency - 22050 Hz. What is the discharge of an audio adapter? (, p. 157, №94)

Decision:

Formula for calculating bit: (memory capacity in bytes): (Sound time in seconds): (sampling rate):

5, 1 MB \u003d 5347737.6 byte

5347737.6 byte: 120 sec: 22050 Hz \u003d 2.02 bytes \u003d 16 bits

Answer: 16 bits

7. The volume of free memory on the disk is 0.01 GB, the discharge of the sound card - 16. What is the duration of the digital audio file recorded with a sampling frequency of 44100 Hz? (, p. 157, №95)

Decision:

Formula for calculating the duration of sound T \u003d A / D / I

(Memory capacity in bytes): (Frequency of sampling in Hz): (Sound board bit in bytes)

0.01 GB \u003d 10737418.24 byte

10737418.24 Byte: 44100: 2 \u003d 121.74 sec \u003d 2.03 min
Answer: 20.3 minutes

8. Evaluate the monoadyophile information of the sound of 1 min. If the "depth" of coding and the frequency of the sound signal is equal, respectively:
a) 16 bits and 8 kHz;
b) 16 bits and 24 kHz.

(, p. 76, №2.82)

Decision:

but).
16 bits x 8 000 \u003d 128000 bits \u003d 16000 byte \u003d 15,625 kb / s
15,625 KB / S X 60 C \u003d 937,5 KB

b).
1) The information volume of the sound file last time is 1 second is:
16 bits x 24 000 \u003d 384000 bits \u003d 48000 byte \u003d 46,875 KB / s
2) The information of the audio file with a duration of 1 minute is:
46,875 KBIT / C x 60 C \u003d 2812.5 KB \u003d 2.8 MB

Answer: a) 937.5 KB; b) 2.8 MB

Level "5"

Used Table 1.

9. What amount of memory is required to store a digital audio file with a high-quality sound recording provided that the sound time is 3 minutes? (, p. 157, №92)

Decision:

High sound quality is achieved at the sampling frequency of 44,1kHz and the discharge of an audio adapter equal to 16.
The formula for calculating the amount of memory: (recording time in seconds) x (the bit of sound card in bytes) x (sampling frequency):
180 С x 2 x 44100 Hz \u003d 15876000 byte \u003d 15.1 MB
Answer: 15.1 MB

10. The digital audio file contains a low-quality sound recording (the sound is grim and muted). What is the duration of the file of the file, if its volume is 650 kb? (, p. 157, №93)

Decision:

For gloomy and muted sound, the following parameters are characteristic: the sampling frequency is 11, 025 kHz, the audio adapter's discharge - 8 bits (see Table 1). Then T \u003d A / D / I. We translate volume to bytes: 650 kb \u003d 665600 byte

T \u003d 665600 bytes / 11025 Hz / 1 byte ≈60.4 s

Answer: The duration of the sound is 60.5 s

Decision:

The information of the audio file is duration of 1 second equal:
16 bits x 48 000 x 2 \u003d 1 536,000 bits \u003d 187.5 KB (multiplied by 2, as stereo).

Information volume of the audio file with a duration of 1 minute is:
187.5 KB / s x 60 s ≈ 11 MB

Answer: 11 MB

Answer: a) 940 KB; b) 2.8 MB.

12. Calculate the time of the monoadiophile sound, if with a 16-bit coding and sampling frequency of 32 kHz its volume is equal to:
a) 700 KB;
b) 6300 KB

(, p. 76, №2.84)

Decision:

but).
1) The information volume of the sound file last time is 1 second is:

700 KB: 62,5 KB / C \u003d 11.2 C

b).
1) The information volume of the sound file last time is 1 second is:
16 bits x 32 000 \u003d 512000 bits \u003d 64000 byte \u003d 62.5 kb / s
2) 700 KB mona audio file sounding time is:
6300 KB: 62.5 KB / C \u003d 100.8 C \u003d 1.68 min

Answer: a) 10 seconds; b) 1.5 min.

13. Calculate how many bytes of information occupies one second stereo recording on a CD (frequency 44032 Hz, 16 bits per value). How much does one minute take? What is the maximum disk capacity (counting the maximum duration of 80 minutes)? (, p. 34, Exercise №34)

Decision:

Formula for calculating memoryA. = D. * T. * I. :
(Recording time in seconds) * (Sound Circuit Blossomy) * (sampling frequency). 16 BIT -2 byte.
1) 1C x 2 x 44032 Hz \u003d 88064 bytes (1 second stereo recording on a CD)
2) 60c x 2 x 44032 Hz \u003d 5283840 byte (1 minute stereo maps on a CD)
3) 4800c x 2 x 44032 Hz \u003d 422707200 byte \u003d 412800 kb \u003d 403,125 MB (80 minutes)

Answer: 88064 bytes (1 second), 5283840 byte (1 minute), 403,125 MB (80 minutes)

2. Determining sound quality.

To determine the quality of the sound, it is necessary to find the sampling frequency and use Table No. 1

256 (2 8) signal intensity levels - quality of radio broadcast sound, using 65536 (2 16) signal intensity levels - audio-CD sound quality. The highest quality frequency corresponds to the music recorded on the CD. The magnitude of the analog signal is measured in this case 44 100 times per second.

Level "5"

13. Determine the quality of the sound (quality of radio broadcasting, average quality, audio-CD quality) if it is known that the volume of monoadiophile in the duration of the sound in 10 seconds. equal to:
a) 940 KB;
b) 157 KB.

(, p. 76, №2.83)

Decision:

but).
1) 940 KB \u003d 962560 byte \u003d 7700480 bits
2) 7700480 BIT: 10 sec \u003d 770048 bits / s
3) 770048 BIT / s: 16 bits \u003d 48128 Hz-particle sampling - close to the highest 44.1 kHz
Answer: Audio-CD quality

b).
1) 157 KB \u003d 160768 byte \u003d 1286144 bits
2) 1286144 BIT: 10 sec \u003d 128614.4 bits / s
3) 128614.4 bt / s: 16 bits \u003d 8038.4 Hz
Answer: Quality of Radio Translation

Answer: a) Quality CD; b) quality of radio broadcasts.

14. Determine the duration of the audio file, which will fit on a flexible diskette 3.5 ". Note that it is allocated to storing data on such a disket. There are 2847 sectors of 512 bytes.
a) with low sound quality: mono, 8 bits, 8 kHz;
b) with high quality sound: stereo, 16 bits, 48 \u200b\u200bkHz.

(, p. 77, №2.85)

Decision:

but).

8 bits x 8 000 \u003d 64,000 bits \u003d 8000 bytes \u003d 7.8 kb / s
3) The sounding time of the monoadiophile of 1423.5 KB is equal to:
1423,5 KB: 7.8 KBA / C \u003d 182.5 C ≈ 3 min

b).
1) The information volume of the diskette is equal to:
2847 sectors x 512 byte \u003d 1457664 byte \u003d 1423,5 kb
2) The information of the sound file with a duration of 1 second is:
16 bits x 48 000 x 2 \u003d 1 536,000 bits \u003d 192,000 byte \u003d 187.5 KB
3) 2,423,5 KB) stereoid appearance
1423,5 KB: 187.5 KB / C \u003d 7.6 C

Answer: a) 3 minutes; b) 7.6 seconds.

3. Binary sound coding.

When solving problems, uses the following theoretical material:

In order to encode the sound, analog signal shown in the figure,

the plane is divided into vertical and horizontal lines. Vertical partition - This is sampling analog signal (signal measurement frequency), horizontal partition -quantization by level. Those. The smaller the grid - the better the analog sound is approximated using numbers. Eight-bit quantization is used to digitize the usual speech (telephone conversation) and radiochanids on short waves. Sixteen - for digitizing music and VHF (ultra-short-wave) radio transmissions.

Level "3"

15. Analog beep was discretized first using 256 signal intensity levels (radio broadcast sound quality), and then using 65536 signal intensity levels (audio-CD sound quality). How many times the information volumes of digitized sound differ? (, p. 77, №2.86)

Decision:

The length of the analog signal code using 256 signal intensity levels is 8 bits, using 65536 Signal intensity levels are 16 bits. Since the length of the code of one signal has doubled, the information volumes of digitized sound differ by 2 times.

Answer: 2 times.

Level " 4 »

16. According to the Nyquist-Kotelnikov Theorem, in order for an analog signal to be accurately restored by its discrete representation (by its references), the sampling frequency should be at least twice the maximum sound frequency of this signal.

What should be the frequency of sampling of sound perceived by a person?

What should be more: speech sampling frequency or sampling rate of symphony orchestra?

Objective: to acquaint students with characteristics of hardware and software tools with sound. Activities: Attracting knowledge from the course of physics (or work with reference books). (, p. ??, Task 2)

Decision:

It is believed that the frequency range that the person hears is from 20 Hz to 20 kHz. Thus, according to the Nyquist Kotelnikov Theorem, in order for an analog signal to be accurately restored by its discrete representation (according to its references),the sampling frequency should be at least twice the maximum sound frequency of this signal. Maximum sound frequency that person hears -20 kHz, it means that the device rA and software should ensure the frequency of discretization of at least 40 kHz, or rather 44.1 kHz. Computer processing of the symphony orchestra sound suggests a higher sampling rate than speech processing, since the frequency range in the case of a symphony orchestra is significantly larger.

Answer: Not less than 40 kHz, the sampling frequency of the symphony orchestra is greater.

Level »5"

17. The figure shows the sound recorder 1 second speech. Encoding it in a binary digital code with a frequency of 10 Hz and a length of 3 bits code. (, p. ??, Task 1)

Decision:

Coding with a frequency of 10 Hz means that we must measure the height of the sound 10 times in a second. Choose equidestive moments of time:

Code length in 3 bits means 2 3 \u003d 8 quantization levels. That is, as a numeric code of the height of the sound at each selected point in time, we can set one of the following combinations: 000, 001, 010, 011, 100, 101, 110, 111. There are only 8 of them, therefore, the height of the sound can be measured by 8 " levels ":

"Rounding" the height of the sound will be up to the nearest lower level:

Using this encoding method, we will get the following result (spaces are delivered for the convenience of perception): 100 100 000 011 111 010 011 100 010 110.

Note.It is advisable to draw the attention of students on how inaccurately the code transmits the change in the amplitude. That is, the discretization frequency is 10 Hz and the level of quantization 2 3 (3 bits) are too small. Usually, the sampling rate of 8 kHz is chosen for sound (voices), i.e. 8000 times per second, and quantization level 2 8 (Code 8 bits long).

Answer: 100 100 000 011 111 010 011 100 010 110.

18. Explain why the quantization level refers, along with the sampling frequency, to the main characteristics of the sound representation in the computer.Objectives: consolidate an understanding of the concepts of the concepts of "the accuracy of presentation of data", the "measurement error", the "presentation error"; Repeat with students binary coding and code length. Type of activity: Working with definitions of concepts. (, p. ??, Task 3)

Decision:

In geometry, physics, technology is the concept of "measurement accuracy", closely associated with the concept of "measurement error". But there is also a concept"Presentation accuracy." For example, a person's growth can be said that he: a) about. 2 m, b) a little more than 1.7 m, c) is 1 m 72 cm, d) equals 1 m 71 cm 8 mm. That is, 1, 2, 3 or 4 digits can be used to designate the measured growth.
Also for binary coding. If you use only 2 bits to use the height of the sound at a specific point in time, then, even if the measurements were accurate, only 4 levels can be transferred: low (00), below the average (01), above average (10), high (11). If you use 1 byte, you can transfer 256 levels. Thanabove the level of quantization , or, what is the same thanmore bits are discharged to record the measured value, the more accurately transmitted this value.

Note. It should be noted that the measuring tool must also support the selected quantization level (length measured by a range with decimetary divisions, it makes no sense to represent with an accuracy of a millimeter).

Answer: The higher the level of quantization, the more accurate sound is transmitted.

Literature:

[ 1] Computer science. Task-workshop in 2 tons. / Ed. IG Semaakina, E.K. Henner: Volume 1. - Basic knowledge laboratory, 1999 - 304 p.: Il.

Workshop on computer science and information technology. Tutorial for general education institutions / N.D. Ugrinovich, L.L. Bosova, N.I. Mikhailova. - M.: Binom. Laboratory of Knowledge, 2002. 400 p.: Il.

Informatics at school: Annex to Informatics and Education magazine. №4 - 2003. - M.: Education and Informatics, 2003. - 96 p.: Il.

Kushnirenko A.G., Leonov A.G., epicthets MG and others. Information culture: odd information. Information models. 9-10 Class: Textbook for general educational institutions. - 2nd ed. - M.: Drop, 1996. - 208 C.: Il.

Gain A.G., Senokosov A.I. Informatics handbook for schoolchildren. - Yekaterinburg: "U-Factoria", 2003. - 346. C54-56.

Temporary sound sampling.

The sound is a sound wave with a continuously changing amplitude and frequency. The greater the amplitude of the signal, the more louder for a person, the greater the frequency of the signal, the higher the tone. In order for a computer to process the sound, a continuous beep should be turned into a sequence of electrical pulses (binary zeros and units).

In the process of encoding a continuous audio signal, its temporary discretion is made. The continuous sound wave is divided into separate small temporary sections, and for each such site a certain amount of amplitude is set.
Discretization is the conversion of continuous signals into a set of discrete values, each of which is assigned a certain binary code.

Thus, the continuous dependence of the amplitude of the signal from time A (T) is replaced by a discrete volume of volume levels. On the chart it looks like replacing a smooth curve to the sequence of "steps".

Each "step" is assigned the value of the volume of the sound volume, its code (1, 2, 3, and so on). The volume levels can be considered as a set of possible states, respectively, the greater the amount of volume levels will be highlighted during the encoding process, the more information will be the value of each level and the more high-quality will be the sound. Modern audio cards provide a 16-bit sound coding depth. The number of different signal levels (states with this coding) can be calculated by the formula:
N \u003d 2 16 \u003d 65356 [sound levels],
where i is the depth of coding.

Thus, modern sound cards can provide encoding 65536 signal levels. Each value of the amplitude of the audio signal is assigned a 16-bit code.

With binary encoding of a continuous beep, it is replaced by a sequence of discrete signal levels. The coding quality depends on the number of signal level measurements per unit of time, that is, sampling frequency. The more measurement is performed in 1 second (the greater the sampling frequency), the more precisely the binary coding procedure.

The quality of the binary sound encoding is determined by the coding depth and the sampling frequency.

The number of measurements per second may lie in the range from 8,000 to 96,000, that is, the sampling frequency of the analog sound signal can take values \u200b\u200bfrom 8 to 96 [kHz]. With a frequency of 8 [kHz], the quality of the discretized audio signal corresponds to the quality of radio broadcasts, and at a frequency of 96 [kHz] - the quality of the audio-CD sound. It should also be considered that both mono and stereo modes are possible.

Informational Sound File

To determine the volume of the audio file V of the ZF, multiply the amount of measurements K is via the coding depth (the number of bits per level) V 1ness:

V zf \u003d k change of * v 1ness

Where the number of measurements of K out depends on:

Task 1.

Homework

1 Determine the volume of the audio stereo file, with a sampling frequency (DD) [kHz], the sound time (Gg) [C] for (mm) -bit coding.

2 Determine the sound time in [C] sound mono file having a volume equal to (Hg) [KB], with a coding depth (mm) [bit] and sampling frequency (DD) [kHz].
Where (DD) - date of your birth, (mm) - the month of your birth, (GG) is the year of your birth.