The serial data rate is usually referred to as the bit rate. However, another commonly used unit is the baud rate. Although they are not the same thing, there are certain similarities between both units under certain circumstances. The article provides a clear explanation of the differences between these concepts.
general information
In most cases, information is transmitted sequentially in networks. Data bits are transmitted in turn over a communication channel, cable or wireless. Figure 1 shows a sequence of bits transmitted by a computer or some other digital circuit. Such a data signal is often referred to as the original. The data is represented by two voltage levels, for example +3 V for logic one and +0.2 V for logic zero. Other levels can be used. In the non-return-to-zero (NRZ) code format (Figure 1), the signal does not return to neutral 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 the bit lifetime or bit time (T B) (Figure 1):
This rate is also called the channel width and is denoted by the letter C. If the bit time is 10 ns, then the data rate is given by
R = 1/10 × 10 - 9 = 100 Mbps
This is usually written as 100 Mbps.
Service bits
Bitrate generally characterizes the actual data transfer rate. 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 bits. Sometimes the number of service 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 useful 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, there will be only 42/1542 service octets, or 2.7%. 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 payload 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 rate over the network, actual speed line increases by a factor depending on the amount of service information. On One Gigabit Ethernet, the actual line speed is 1.25 Gb/s, while the payload data rate is 1 Gb/s. For 10-Gbit/s Ethernet, these values are 10.3125 Gb/s and 10 Gb/s, respectively. When estimating the data rate of a network, concepts such as throughput, payload rate, or effective data 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 voltage, frequency, or phase changes.
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 bit rate. However, it is possible to have more than two symbols in a transmission interval, whereby several bits are assigned to each symbol. In this case, data on 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 frequency carrier. Some cable protocols also use modulation to increase the transmission speed. This is called "broadband transmission".
Above: modulating signal, original signal
Using composite characters, each can carry several bits. For example, if the symbol rate is 4800 baud and each symbol consists of two bits, the total data rate will be 9600 bps. Usually the number of characters is represented by some power of 2. If N is the number of bits in a character, then the number of required characters will be S = 2N. So the total data 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 symbol per bit, as in the case of the binary NRZ format, the bit and baud rates are the same.
Multilevel modulation
A high bit rate can be provided by many modulation methods. For example, in frequency shift keying (FSK), two different frequencies are typically used in each symbol interval to represent logical 0s and 1s. Here, the bit rate is equal to the baud rate. But if each character 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 symbol represents 0 or 1. Binary 0 corresponds to 0°, and binary 1 to 180°. With one bit per symbol, the bit rate is equal to the baud rate. However, the ratio of the number of bits and characters is easy to increase (see Table 1).
| Table 1. | Binary phase shift keying. | ||||||||||
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For example, quadrature PSK has two bits per symbol. With this structure and two bits per baud, the bit rate is twice the baud rate. With three bits per baud, the modulation would be 8PSK and eight different phase shifts would represent three bits. And at 16PSK, 16 phase shifts represent 4 bits.
One unique form of multilevel modulation is quadrature amplitude modulation (QAM). To create symbols representing multiple bits, QAM uses a combination of different amplitude levels and phase offsets. For example, 16QAM encodes four bits per symbol. The symbols are a combination of different amplitude levels and phase shifts.
For a visual display of 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 with four bits per character, resulting in a bit rate of 4 times the baud rate.
Why multiple 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 interleaving of zeros and ones in the data stream, then the maximum theoretical bit rate C in bits 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 multiple bits per symbol, the maximum theoretical data rate would be:
Here N is the number of characters in a character interval:
log 2 N = 3.32 log10N
The bandwidth required to provide the desired speed for a given number of levels is calculated as follows:
For example, the bandwidth required to achieve a transmission rate of 1 Mbps with 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 a rate of 1 Mbps over a 250 kHz channel is given by:
log 10 N = C/6.64B = 1/6.64(0.25) = 0.60
N = log-1 (0.602) = 4 symbols
These calculations assume that there is no noise in the channel. To account for noise, you need to apply the Shannon-Hartley theorem:
C = B log 2 (S/N + 1)
C - channel bandwidth in bits per second,
B - channel bandwidth in hertz,
S/N - signal-to-noise ratio.
In the form of a decimal logarithm:
C = 3.32B log 10 (S/N + 1)
What is the maximum speed on 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 Mbps
The Shannon-Hartley theorem does not specifically state that multilevel modulation must be applied 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 means five bits per character (25 = 32).
Baud Rate Measurement Examples
Almost all high speed connections use some form of broadband transmission. In Wi-Fi, orthogonal frequency division multiplexing (OFDM) modulation schemes use QPSK, 16QAM, and 64QAM.
The same is true for WiMAX and technology cellular communication Long-Term Evolution (LTE) 4G. The transmission of analogue and digital television signals in cable TV systems and high-speed Internet access is based on 16QAM and 64QAM, while satellite communications use QPSK and various versions of QAM.
For public safety land mobile radio systems, 4FSK voice and data modulation standards have recently been adopted. This bandwidth narrowing method is designed to reduce the bandwidth from 25 kHz per channel to 12.5 kHz, and eventually to 6.25 kHz. As a result, more channels for other radios can be placed in the same spectral range.
High-definition television in the US uses a modulation technique called eight-level vestigial sideband (8-level signaling with partially suppressed sideband), or 8VSB. This method allocates three bits per symbol at 8 amplitude levels, allowing 10,800 symbols per second to be transmitted. With 3 bits per symbol, the total speed will be 3 × 10,800,000 = 32.4 Mbps. In combination with the VSB method, which transmits only one full sideband and part of another, high definition video and audio data can be transmitted over a 6 MHz television channel.
Claims that his program is able to make the most of Ethernet resources. Due to its own network driver, its own TCP stack and work bypassing the kernel operating system it is indeed capable of approaching the physical limitations of the Ethernet standard.
Masscan scanner developer Robert Graham has published results that demonstrate the real-world performance of his program.
For the scanner, the number of packets sent per second is important. The Ethernet standard requires that there be a 12-byte "silence" period between packets, which determines the end of one packet and the beginning of the next. At the end of each packet, a CRC code (4 bytes) must also be transmitted to check the integrity of the transmission, and at the beginning of the packet, a mandatory preamble of 8 bytes. There is one more restriction - the minimum packet size is 60 bytes, this is an ancient restriction from the 80s, which does not make sense nowadays, but is kept for the sake of compatibility.
Given all the restrictions, then the packets must be at least 84 bytes. Thus, for a 1 Gbps network, we get a theoretical limit of 1,000,000,000/84*8 = 1,488,095 packets per second.
On a modern 10 Gigabit network, this number can be increased tenfold: 14,880,952 packets per second.
When scanning ports, we do not need to use all 60 bytes, 20 bytes for the IP header and 20 bytes for the TCP header are enough, 40 bytes in total. That is, the effective packet rate is 1488095 x 40 = 476 Mbps. In other words, even if we use the physical Ethernet resource at 100%, the provider or the program for measuring traffic on a gigabit channel will show a data transfer rate of 476 Mbps. Such a discrepancy is understandable, because during normal surfing packets of 40 bytes are not used, there packets are usually 500 bytes each, so the overhead from the service data can be ignored.
In practice, the scanner can ignore some Ethernet standards, such as reducing the pause between packets from 12 to 5 bytes, and the preamble from 8 to 4 bytes. The minimum packet size can be reduced from 84 bytes to 67 bytes. In this case, 1,865,671 packets per second can be transmitted over a gigabit channel, which increases the speed demonstrated in tests from 476 Mbps to 597 Mbps. True, it is possible backfire: a router in the path of your packets may drop some of them, which will reduce the real effective speed data transmission.
There are other problems as well. For unknown reasons, Linux is unable to overcome the milestone of 1.488 million packets per second on gigabit Ethernet. On the same system, but with a 10Gb link connected, Linux barely breaks the 2Mpps mark. In practice, the real speed in a Linux system is approximately 1.3 million packets per second on a gigabit link. Again, Robert Graham has no idea why this is.
Internet bandwidth or, more simply, Internet speed, represents the maximum number of data received personal computer or transferred to the Network for a certain unit of time.
Most often, you can meet the measurement of data transfer speed in kilobits / second (Kb / s; Kbps) or in megabits (Mb / s; Mbps). File sizes are usually always specified in bytes, KBytes, MBytes, and GBytes.
Since 1 byte is 8 bits, in practice this will mean that if your Internet connection speed is 100 Mbps, then the computer can receive or transmit no more than 12.5 Mb of information per second (100/8=12.5). It’s easier can be explained in this way, if you want to download a video, the volume of which is 1.5 Gb, then it will take you only 2 minutes.
Naturally, the above calculations are made under ideal laboratory conditions. For example, the reality may be quite different:
Here we see three numbers:
- Ping - this number means the time for which Network packets are transmitted. The lower the value of this number, the better quality Internet connection (it is desirable that the value be less than 100ms).
- Next comes the speed of obtaining information (incoming). It is this figure that Internet providers offer when connecting (it is precisely for this number of "Megabits" that you have to pay your hard-earned dollars / hryvnias / rubles, etc.).
- The third number remains, indicating the speed of information transfer (outgoing). It will naturally be less than the speed of receiving data, but providers are usually silent about this (although, in fact, a large outgoing speed is rarely required).
What determines the speed of the Internet connection
- The speed of the Internet connection depends on the tariff plan that the provider sets.
- The speed is also affected by the technology of the information transmission channel and the workload of the Network by other users. If the total bandwidth of the channel is limited, then the more users are on the Web and the more they download information, the more the speed drops, because there is less "free space".
- There is also a dependence on the download speed of the sites you access. For example, if at the time of loading the server can give the user data at a speed of less than 10 Mbps, then even if you have the maximum tariff plan you won't get more.
Factors that also affect internet speed:
- When checking, the speed of the server you are accessing.
- Setting and wifi speed router if you are connected through it to the local network.
- At the time of the scan, all programs and applications running on the computer.
- Firewalls and antiviruses that run in the background.
- Settings for your operating system and the computer itself.
How to increase internet speed
If there is malware or unwanted software on your computer, it can slow down 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 who are not averse to using free traffic will usually connect to it. Be sure to set a password to connect to Wi-Fi.
Reduce the speed and parallel running programs. For example, simultaneous download managers, Internet messengers, automatic OS updates lead to an increase in processor load and therefore the speed of the Internet connection decreases.
These actions, in some cases, help to increase the speed of the Internet:
If you have a high Internet connection, and the speed leaves much to be desired, increase the port bandwidth. To do this is quite simple. Go to "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 that, a window will open in which you need to go to the "Port Settings" column. Find the "Speed" parameter (bits per second) and click on the number 115200 - then OK! Congratulations! Now you have increased the throughput of the port. Since the speed is set to 9600 bps by default.
To increase the 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 your computer.
The exchange of information is carried out through channels of information transmission.
Information transmission channels can use various physical principles. So, when people communicate directly, information is transmitted using sound waves, and when talking on the phone - using electrical signals that propagate through communication lines.
Link- technical means that allow data transmission at a distance.
Computers can exchange information using communication channels of various physical nature: cable, fiber optic, radio channels, etc.
Information transfer rate (information flow rate) - the amount of information transmitted per unit of time.
The general information transmission scheme includes an information sender, an information transmission channel, and an information recipient.
The main characteristic of information transmission channels is their throughput.
Channel capacity - the maximum rate of information transfer over the communication channel per unit of time.
The bandwidth of a channel is equal to the amount of information that can be transmitted over it per unit of time.
The amount of transmitted information \(V\) is calculated by the formula:
where \(q\) is the bandwidth of the link (in bits per second or similar units), and \(t \) - transmission time.
Bandwidth is usually measured in bits per second (bps) and multiples of Kbps and Mbps.
However, sometimes a byte per second (byte / s) and multiples of it are used as units Kbyte / s and Mbyte / s.
Relations between units bandwidth information transmission channels are the same as between units of measurement of the amount of information:
1 byte = 2 3 bits = 8 bits; 1 kbits = 2 10 bits = 1024 bits; 1 Mbps = 2 10 Kbps = 1024 Kbps; 1 Gbps = 2 10 Mbps = 1024 Mbps.
Example:
How many seconds would it take a modem transmitting messages at \(28,800 \)bps to transmit \(100\) pages of text in \(30\) lines of \(60\) characters each, assuming that each character encoded by \(1\) bytes?
Solution. Let's calculate the file size in bits V = 100 ⋅ 30 ⋅ 60 ⋅ 8 bits = 1440000 bits.
Message transfer rate \(q = 28 800 \)bps.
The time is t = V q = 1440000 28800 = 50 seconds.
Let's consider a more complex problem.
Example:
Device \(A\) transmits information to device \(C\) through device \(B\) according to the following rules:
1. Information is transmitted in packets of \(200\) bytes.
2. The device \(B\) can simultaneously receive information from the device \(A\) and transmit previously received information to the device \(C\).
3. The device \(B\) can send the next packet to the device \(C\) only after it has completely received this packet from the device \(A\).
4. The device \(B\) has an unlimited buffer, in which it can store packets received from the device \(A\), but not yet transmitted to the device \(C\).
The bandwidth between \(A\) and \(B\) is \(100\) bytes per second.
The bandwidth between \(B\) and \(C\) is \(50\) bytes per second.
Three packets of information were sent. In how many seconds will \(C\) finish receiving all information from \(A\)?
Solution. Since the rate of information reception by the device \(B\) is greater than the rate of its transmission to the device C, the transmission time will consist of two stages.
Everyone has repeatedly heard about networks of the second, third and fourth generation mobile communications. Some may have already read about the networks of the future - the fifth generation. But questions - what does G, E, 3G, H, 3G +, 4G or LTE mean on a smartphone screen and what is faster among this are still of concern to many people. We will answer them.
These icons indicate the type of connection your smartphone, tablet or modem has to the mobile network.
1. G(GPRS - General Packet Radio Services): The slowest and most obsolete packet data connection option. First standard mobile internet, performed by an add-on over GSM (after a CSD connection up to 9.6 kbps). The maximum speed of the GPRS channel is 171.2 kbps. At the same time, the real one, as a rule, is an order of magnitude lower, and the Internet here is not always functional in principle.
2. E(EDGE or EGPRS - Enhanced Data rates for GSM Evolution): Faster add-on over 2G and 2.5G. Technology of digital data transmission. The speed of EDGE is about 3 times higher than GPRS: up to 474.6 kbps. However, she also belongs to the second generation wireless communication and is outdated. The real speed of EDGE is usually kept in the region of 150-200 kbps and directly depends on the location of the subscriber - that is, the workload base station in a specific area.
3. 3 G(Third Generation - third generation). Here, not only data transfer is possible over the network, but also “voices”. The quality of voice transmission in 3G networks (if both interlocutors are within their range) can be an order of magnitude higher than in 2G (GSM). Internet speed in 3G is also much higher, and its quality, as a rule, is already quite sufficient for comfortable work on mobile devices and even desktop computers via USB modems. At the same time, your current position may affect the data transfer rate, incl. whether you are in one place or moving in transport:
- Stay still: typically up to 2 Mbps
- Drive at speeds up to 3 km/h: up to 384 kbps
- Travel at speeds up to 120 km/h: up to 144 kbps.
4. 3,5 G.3G+,h,H+(HSPDA - High-Speed Downlink Packet Access): The next high-speed packet data add-on is already over 3G. In this case, the data transfer rate is very close to 4G and in H mode it is up to 42 Mbps. AT real life mobile internet in this mode average works for mobile operators at speeds of 3-12 Mbps (sometimes higher). For those who do not understand: it is very fast and quite enough to watch online video in not too high quality (resolution) or download heavy files with a stable connection.
Also in 3G there was a video call function:
5. 4G, LTE(Long-Term Evolution - long-term development, the fourth generation of mobile Internet). This technology used only for data transmission (not for "voice"). The maximum download speed here is up to 326 Mbps, upload - 172.8 Mbps. The real values are again an order of magnitude lower than the declared ones, but they still amount to tens of megabits per second (in practice, often comparable to mode H; in Moscow, usually 10-50 Mbps). At the same time, faster PING and the technology itself make 4G the most preferred standard for mobile Internet in modems. Smartphones and tablets in 4G (LTE) networks hold a battery charge longer than in 3G.
6. LTE-A(LTE Advanced - LTE upgrade). The peak data transfer rate here is up to 1 Gbps. In reality, the Internet is capable of operating at speeds up to 300 Mbps (5 times faster than conventional LTE).
7. VoLTE(Voice over LTE - voice over LTE, as an additional development of technology): a technology for transmitting voice calls over LTE networks based on IP Multimedia Subsystem (IMS). The connection speed is up to 5 times faster compared to 2G/3G, and the quality of the conversation itself and voice transmission is even higher and cleaner.
8. 5 G(fifth generation of cellular communications based on IMT-2020). The standard of the future is still under development and testing. The data transfer rate in the commercial version of the networks is promised to be up to 30 times higher than LTE: the maximum data transfer can be up to 10 Gb / s.
Of course, you can use any of the above technologies if your equipment supports it. Also, its work depends on the capabilities of the mobile operator itself at a particular location of the subscriber and his tariff plan.
