Information transfer rate. Data transfer rate

Internet channel bandwidth or, more simply put, , represents the maximum number of data received personal computer or transmitted to the Network for a certain unit of time.

Most often you can find a measurement of data transfer speed in kilobits/second (Kbps; Kbps) or in megabits (Mb/s; Mbps). File sizes are usually always specified in bytes, KB, MB and GB.

Since 1 byte is 8 bits, in practice this will mean that if the speed of your Internet connection is 100 Mbps, then the computer can receive or transmit no more than 12.5 Mb of information per second (100/8 = 12.5). This is simpler. can be explained in this way, if you want to download a video whose volume is 1.5 Gb, then it will take you only 2 minutes.

Naturally, the above calculations were made under ideal laboratory conditions. For example, the reality may be completely different:

Here we see three numbers:

1. Ping – this number means the time during which Network packets are transmitted. The smaller the value of this number, the better quality Internet connection (preferably the value should be less than 100ms).
2. Next comes the speed of receiving information (incoming). This is exactly the number that Internet providers offer when connecting (it is for this number of “Megabits” that you have to pay your hard-earned dollars/hryvnia/rubles, etc.).
3. The third number remains, indicating the information transfer speed (outgoing). Naturally, it will be less than the speed of receiving data, but providers are usually silent about this (although, in fact, a higher outgoing speed is rarely required).

What determines the speed of an Internet connection?

• The speed of the Internet connection depends on the tariff plan set by the provider.
• The speed is also affected by the technology of the information transmission channel and the load on the Network by other users. If the overall channel capacity is limited, then the more users are on the Internet and the more information they download, the more the speed drops, since there is less “free space” left.
• There is also a dependency on which you are accessing. For example, if at the time of loading the server can provide data to the user at a speed of less than 10 Mbit/sec, then even if you have a maximum connection tariff plan, you won't achieve more.

Factors that also affect Internet speed:

• When checking, the speed of the server you are accessing.
• Settings and Wi-Fi speed router, if you are connected to the local network through it.
• At the time of scanning, all programs and applications running on the computer.
• Firewalls and antiviruses that run in the background.
• Your settings operating system and the computer itself.

How to increase internet speed

If there is malware or unwanted software on your computer, then this may affect the speed of your Internet connection. Trojans, viruses, worms, etc. that got into the computer can take part of the channel bandwidth for their needs. To neutralize them, you must use anti-virus applications.

If you use Wi-Fi that is not password protected, then other users will usually connect to it and are not averse to using free traffic. Be sure to set a password to connect to Wi-Fi.

Parallel running programs also reduce speed. For example, simultaneous download managers, Internet messengers, automatic update operating system leads to an increase in processor load and therefore the speed of the Internet connection decreases.

These actions, in some cases, help increase Internet speed:

If you have a high Internet connection, but the speed leaves much to be desired, increase the port bandwidth. This is quite easy to do. Go to the "Control Panel" then to "System" and to the "Hardware" section, then click on "Device Manager". Find "Ports (COM or LPT)", then expand their contents and look for " Serial port(COM 1)".

After that, right-click and open “Properties”. After this, a window will open in which you need to go to the “Port parameters” column. Find the “Speed” parameter (bits per second) and click on the number 115200 – then OK! Congratulations! Now you have throughput speed port increased. Because the default speed is set to 9600 bps.

To increase speed, you can also try disabling the QoS packet scheduler: Run the gpedit.msc utility (Start - Run or Search - gpedit.msc). Next: Computer Configuration - Administrative Templates - Network - QoS Packet Scheduler - Limit reserved bandwidth - Enable - set to 0%. Click "Apply" and restart the computer.

General information

In most cases, information is transmitted sequentially in networks. Data bits are transmitted one by one over a communication channel, cable or wireless. Figure 1 shows the sequence of bits, transmitted by computer or any other digital circuit. This data signal is often called the original signal. The data is represented by two voltage levels, for example, a logical one corresponds to a voltage of +3 V, and a logical zero corresponds to +0.2 V. Other levels can be used. In the non-return-to-zero (NRZ) code format (Figure 1), the signal does not return to the neutral position after each bit, unlike the return-to-zero (RZ) format.

Bitrate

The data rate R is expressed in bits per second (bps or bps). The rate is a function of bit lifetime or bit time (T B) (Figure 1):

This speed is also called the channel width and is denoted by the letter C. If the bit time is 10 ns, then the data transfer rate is defined as

R = 1/10 × 10 - 9 = 100 million bps

This is usually written as 100 MB/s.

Service bits

Bitrate, as a rule, characterizes the actual data transfer speed. However, in most serial protocols, the data is only part of a more complex frame or packet that includes source address, destination address, error detection, and code correction bits, as well as other information or control bits. In a protocol frame, the data is called useful information(payload). Bits that are not data are called overhead. Sometimes the number of overhead bits can be significant - from 20% to 50%, depending on the total number of useful bits transmitted over the channel.

For example, an Ethernet protocol frame, depending on the amount of payload data, can have up to 1542 bytes or octets. The payload can be from 42 to 1500 octets. With the maximum number of useful octets, only 42/1542, or 2.7%, will be service octets. There would be more of them if there were fewer useful bytes. This ratio, also known as protocol efficiency, is usually expressed as a percentage of the amount of useful data from maximum size frame:

Protocol efficiency = payload/frame size = 1500/1542 = 0.9727 or 97.3%

As a rule, to show the true data transfer speed over the network, the actual line speed is increased by a factor depending on the amount of service information. In One Gigabit Ethernet, the actual line speed is 1.25 Gb/s, while the payload speed is 1 Gb/s. For 10-Gbit/s Ethernet these values ​​are 10.3125 Gb/s and 10 Gb/s, respectively. When assessing the speed of data transfer over a network, concepts such as throughput, payload rate, or effective data transfer rate can also be used.

Baud rate

The term "baud" comes from the name of the French engineer Emile Baudot, who invented the 5-bit teletype code. The baud rate expresses the number of signal or symbol changes in one second. A symbol is one of several changes in voltage, frequency, or phase.

The NRZ binary format has two symbols represented by voltage levels, one for each 0 or 1. In this case, the baud rate or symbol rate is the same as the bitrate. However, it is possible to have more than two symbols in a transmission interval, whereby several bits are allocated for each symbol. In this case, data over any communication channel can only be transmitted using modulation.

When the transmission medium cannot process the original signal, modulation comes to the fore. Of course we are talking about wireless networks. The original binary signals cannot be transmitted directly, they must be transferred to a radio carrier frequency. Some cable data protocols also use modulation to improve transmission speeds. This is called "broadband transmission".
Above: modulating signal, original signal

By using composite symbols, multiple bits can be transmitted in each symbol. For example, if the symbol rate is 4800 baud and each symbol consists of two bits, full speed data transfer will be 9600 bps. Typically the number of symbols is represented by some power of 2. If N is the number of bits in a symbol, then the number of symbols required will be S = 2N. So the total data transfer rate is:

R = baud rate × log 2 S = baud rate × 3.32 log 1 0 S

If the baud rate is 4800 and there are two bits per character, the number of characters is 22 = 4.

Then the bitrate is:

R = 4800 × 3.32log(4) = 4800 × 2 = 9600 bps

With one character per bit, as is the case with the NRZ binary format, the bit and baud rates are the same.

Multi-level modulation

High bitrate can be achieved by many modulation methods. For example, frequency shift keying (FSK) typically uses two different frequencies to represent logical 0s and 1s in each symbol interval. Here the bit rate is equal to the baud rate. But if each symbol represents two bits, then four frequencies (4FSK) are required. In 4FSK, the bit rate is twice the baud rate.

Another common example is phase shift keying (PSK). In binary PSK, each character represents 0 or 1. Binary 0 represents 0°, and binary 1 represents 180°. At one bit per character, the bit rate is equal to the baud rate. However, the bit-to-symbol ratio is easy to increase (see Table 1).

For example, in quadrature PSK there are two bits per symbol. Using this structure and two bits per baud, the bit rate is twice the baud rate. With three bits per baud, the modulation will be designated 8PSK, and eight different phase shifts will represent three bits. And with 16PSK, 16 phase shifts represent 4 bits.

One of the unique forms of multilevel modulation is quadrature amplitude modulation(QAM). To create symbols representing multiple bits, QAM uses a combination different levels amplitudes and phase shifts. For example, 16QAM encodes four bits per symbol. The symbols are a combination of different amplitude levels and phase shifts.

To visually display the amplitude and phase of the carrier for each value of the 4-bit code, a quadrature diagram is used, which also has the romantic name “signal constellation” (Figure 2). Each point corresponds to a certain carrier amplitude and phase shift. A total of 16 characters are encoded at four bits per character, resulting in a bitrate that is 4 times the baud rate.

Why several bits per baud?

By transmitting more than one bit per baud, you can send data from high speed through a narrower channel. It should be recalled that the maximum possible data transfer rate is determined by the bandwidth of the transmission channel.
If we consider the worst-case scenario of alternating zeros and ones in the data stream, then the maximum theoretical bit rate C for a given bandwidth B will be equal to:

Or bandwidth at maximum speed:

To transmit a signal at a speed of 1 Mb/s you need:

B = 1/2 = 0.5 MHz or 500 kHz

When using multi-level modulation with several bits per symbol, the maximum theoretical data rate will be:

Here N is the number of characters in the character interval:

log 2 N = 3.32 log10N

The bandwidth required to provide the desired speed at a given number of levels is calculated as follows:

For example, the bandwidth required to achieve a transfer rate of 1 Mb/s at two bits per symbol and four levels can be defined as:

log 2 N = 3.32 log 10 (4) = 2

B = 1/2(2) = 1/4 = 0.25 MHz

The number of symbols required to obtain the desired data rate in a fixed bandwidth can be calculated as:

3.32 log 10 N = C/2B

Log 10 N = C/2B = C/6.64B

N = log-1 (C/6.64B)

Using the previous example, the number of symbols required to transmit at 1 Mbps over a 250 kHz channel is determined as follows:

log 10 N = C/6.64B = 1/6.64(0.25) = 0.60

N = log-1 (0.602) = 4 characters

These calculations assume that there is no noise in the channel. To take into account noise, you need to apply the Shannon-Hartley theorem:

C = B log 2 (S/N + 1)

C - throughput channel in bits per second,
B is the channel bandwidth in hertz,
S/N - signal to noise ratio.

In decimal logarithm form:

C = 3.32B log 10 (S/N + 1)

What is maximum speed in a 0.25 MHz channel with an S/N ratio of 30 dB? 30 dB translates to 1000. Therefore, the maximum speed is:

C = 3.32B log 10 (S/N + 1) = 3.32(0.25) log 10 (1001) = 2.5 Mb/s

The Shannon-Hartley theorem does not specifically state that multilevel modulation must be used to achieve this theoretical result. Using the previous procedure, you can find out how many bits are required per character:

log 10 N = C/6.64B = 2.5/6.64(0.25) = 1.5

N = log-1 (1.5) = 32 characters

Using 32 characters implies five bits per character (25 = 32).

Baud rate measurement examples

Almost everything high speed connections use some form of broadband transmission. In Wi-Fi, modulation schemes with orthogonal multiplexing frequency division channels (OFDM) QPSK, 16QAM and 64QAM are used.

The same is true for WiMAX and technology cellular communications Long-Term Evolution (LTE) 4G. Transmission of analogue and digital television cable TV and high-speed Internet access systems are based on 16QAM and 64QAM, while satellite communications use QPSK and different versions QAM.

Modulation standards have recently been adopted for public safety land mobile radio systems speech information and data using 4FSK. This bandwidth narrowing technique is designed to reduce bandwidth from 25 kHz per channel to 12.5 kHz, and ultimately to 6.25 kHz. As a result, more channels for other radio stations can be placed in the same spectral range.

High-definition television in the United States uses a modulation method called eight-level vestigial sideband, or 8VSB. This method allocates three bits per symbol at 8 amplitude levels, which allows transmission of 10,800 thousand symbols per second. At 3 bits per symbol, the total speed would be 3 × 10,800,000 = 32.4 Mbps. Combined with the VSB method, which transmits only one full sideband and part of the other, high-definition video and audio data can be transmitted over television channel 6 MHz wide.

Key words:

· data transfer speed

bits per second

Data transfer rate – most important characteristic communication lines. After studying this paragraph, you will learn how to solve problems related to transmitting data over a network.

Units of measurement

Let's remember in what units speed is measured in situations already familiar to us. For a car, speed is the distance traveled per unit of time; speed is measured in kilometers per hour or meters per second. In liquid pumping problems, speed is measured in liters per minute (or per second, per hour).

It is not surprising that in data transmission problems we will refer to the amount of data transmitted over the network per unit of time (most often per second) as speed.

The amount of data can be measured in any units of information quantity: bits, bytes, KB, etc. But in practice, data transfer speed is most often measured in bits per second (bps).

In high-speed networks, the data exchange rate can be millions and billions of bits per second, so multiple units are used: 1 kbit/s (kilobits per second), 1 Mbit/s (megabits per second) and 1 Gbit/s (gigabits per second).

Please note that here the prefixes "kilo-", "mega-" and "giga-" denote (as in international system SI units) increase exactly a thousand, a million and a billion times. Recall that in traditional units of measurement amount of information“kilo-” means an increase of 1024 times, “mega-” - 1024 2 and “giga-” - 1024 3.

Tasks

Let the data transfer rate over some network be v bps This means that in one second it is transmitted v bits, and for t seconds - v×t bits

Problem 1. The data transfer rate over the communication line is 80 bit/s. How many bytes will be transferred in 5 minutes?

Solution. As you know, the amount of information is calculated by the formula I = v×t. IN in this case v= 80 bps and t= 5 min. But the speed is given in bits in second, and time is in minutes, so to get the correct answer you need to convert minutes to seconds:

t= 5 × 60 = 300 s

and only then perform the multiplication. First we get the amount of information in bits:

I= 80 bps × 300 s = 24000 bits

Then we convert it to bytes:

I= 24000: 8 bytes = 3000 bytes

Answer: 3000 bytes.

Problem 2. The data transfer rate over the communication line is 100 bit/s. How many seconds will it take to transfer a 125 byte file?

Solution. We know the data transfer rate ( v= 100 bit/s) and the amount of information ( I= 125 bytes). From the formula I = v×t we get

t= I: v.

But the speed is set to bits per second, and the amount of information – in bytes. Therefore, in order to “match” units of measurement, you must first convert the amount of information into bits (or speed into bytes per second!):

I= 125 × 8 bits = 1000 bits.

Now we find the transmission time:

t= 1000 : 100 = 10 s .

Answer: 10 seconds.

Problem 3. What is the average data transfer rate (in bits per second) if a 200 byte file was transferred in 16 seconds?

Solution. We know the amount of information ( I= 200 bytes) and data transfer time ( t= 16 s). From the formula I = v×t we get

v= I: t.

But the file size is set to bytes, and the transmission speed must be obtained in bits per second. Therefore, first we convert the amount of information into bits:

I= 200 × 8 bits = 1600 bits.

Now we find the average speed

v= 1600 : 16 = 100 bps .

Please note that we are talking about the average transfer speed, because it could change during data exchange.

Answer: 100 bps.

1. In what units is the data transfer rate measured? computer networks?

2. What do the prefixes “kilo-,” “mega-,” and “giga-” mean in units of data transfer speed? Why do you think these prefixes are not the same as in units of measuring the amount of information?

3. What formula is used to solve data transfer speed problems?

4. What do you think is the main reason mistakes in solving such problems?

1. How many bytes of information will be transmitted in 24 seconds over a communication line at a speed of 1500 bits per second?

2. How many bytes of information will be transmitted in 15 seconds over a communication line at a speed of 9600 bps?

3. How many bytes of information are transmitted in 16 seconds over a communication line at a speed of 256,000 bits per second?

4. How many seconds will it take to transfer a 5 KB file over a 1024 bps link?

5. How many seconds will it take to transfer an 800 byte file over a 200 bps link?

6. How many seconds will it take to transfer a 256 KB file over a communication line at 64 bytes per second?

7. A book containing 400 pages of text (each page contains 30 lines of 60 characters each), encoded in 8-bit encoding. How many seconds will it take to transmit this book over a communication line at a speed of 5 kbit/s?

8. How many bits per second are transmitted over a communication line if a 400 byte file was transmitted in 5 s?

9. How many bits per second are transmitted over a communication line if a 2 KB file was transferred in 8 s?

10. How many bytes per second are transferred over a communication line if a 100 KB file was transferred in 16 s?

Of course, instead of 0 and 1, you can use any two characters.

English word bit is an abbreviation for the expression binary digit, "binary digit".

There is another type of language, which includes Chinese, Korean, Japanese languages. They use hieroglyphs, each of which denotes separate word or concept.

English word pixel is an abbreviation for picture element, element of the picture.