1. INFORMATION TRANSFER

2. Information transfer Information transfer is a physical process through which information is transferred in space. This process is characterized by the following components:  Source of information  Information receiver  The transmission medium (an information channel) Source of information Information receiver information channel

3. Message Message – a form of expression (presentation) of information, convenient for transmission over a distance. Messages intended for processing on a computer are called data.  Optical communications  Sound message  Documentary communications

4. A signal is a physical process that displays the message being transmitted. Display of the message is provided by change of any physical quantity characterizing process. This value is an information parameter of the signal.  instantaneous sound pressure value  electromagnetic field strength  the intensity of the electric field in the cable Signals

5. Kind of signals Continuous (analog) signal Discrete signal

6. Communication channel Scheme of information transfer using information technology SMEC М CLDMADRM RM – the recipient of the message SM is the source of the message. Registers (fixes) information on any carrier, resulting in a signal. Communication channel SMEC М CLDMADRM

7. Канал связи Scheme of information transfer using information technology SME УМЛСДМВ DRM D-message decoder E-message encoder enCOderDECoderCODEC Communication channel SMEC М CLDMADRM

8. Канал связи Scheme of information transfer using information technology SMECC ЛС AADRM A – device for allocation of compacted signal C-compacted signal. Facilitates the transmission of multiple signals on the same line of communication LS. CAMULTIPLEXER Communication channel SMEC М CLDMADRM

9. Канал связи Scheme of information transfer using information technology ИСКУМЛСДМВДКПС DM – demodulator. Selects the information signal from the carrier signal. M-signal modulator. Changes the information characteristics of the carrier signals, imposing on it an information signal. MOdulatorDEModulator. МО DEM Communication channel SMEC М CLDMADRM

10. Modulation Сигнал - носитель Информа - ционный сигнал Модули - рованный сигнал 0110 назад For example, amplitude

11. Канал связи Scheme of information transfer using information technology ИСКУМЛСДМВДКПС CL-communication line – the physical environment and the technical means in it, which are used to transmit a signal over a distance. Communication channel SMEC М CLDMADRM

12. Basic requirements for communication lines  implementation of communication on practically required distances;  suitability for the transmission of various types of messages;  protection from mutual influences and external interference, as well as from physical effects (atmospheric phenomena, corrosion, etc.).);  stability of line parameters, stability and reliability of communication; efficiency of communication system as a whole.

13. Types of communication lines  Cable and air communication lines based on metal conductors  Radio link  Fiber-optic communication lines

14. Tasks Amount of information transmitted Q Q = q∙t q – channel capacity (a synonym for line of communication) t – transmission time Basic tricks :  Convert all to the same units, such as bits and bps  Calculations with large numbers are carried out through powers of 2

15. Vasya has access to the Internet via a high-speed one-way radio channel, which provides a speed of 218 bits per second. Petya does not have high-speed Internet access, but it is possible to receive information from Vasya via a low-speed telephone channel with an average speed of 215 bits per second. Petya agreed with Vasya that he would download 5 MB of data for him via a high-speed channel and relay them to Petya via a low- speed channel. Vasya’s computer starts the retransmission of data before they are received on the first 512 KB of the data. What is the minimum possible period of time (in seconds), from the moment you start downloading the data, to the full receipt of Petya? Tasks

16. T = t 1 + t 2 t 1 – Vasya’s download time (the first fragment in 512K) t 2 – download time of the entire file, 5M Q 1 = 512K = 2 9 K = 2 9 ∙ 2 10 = 2 9 ∙ 2 10 ∙ 2 3 bit = 2 22 bit Q 2 =5М=5∙ 2 10 K =5∙ 2 10 ∙ 2 10 byte =5∙ 2 10 ∙ 2 10 ∙ 2 3 bit = 5∙ 2 23 bit t 1 = Q 1 /q 1 = 2 22 /2 18 = 2 4 = 16 s t 2 = Q 2 /q 2 = 5∙ 2 23 /2 15 = 5∙ 2 8 = 1280 s T = 1280+16 = 1296 s Tasks Vasya : q 1 = 2 18 bit/s Petya : q 2 = 2 15 bit/s