The basic unit of information transmission speed. Units of information measurement. Information transfer speed. Communication channel capacity. Signal transmission methods
Before reading, it is recommended that you familiarize yourself with the Firefox profile structure.
%USERPROFILE%- path to profile Windows user(not to be confused with the Firefox profile). Typically located in the C:\Documents and Settings\username folder. %AppData%- path to the Windows user profile subfolder that contains application data. Typically located in the %USERPROFILE%\Application Data folder. Firefox Profiles are stored inside the %AppData%\Mozilla\Firefox folder.Q: What happens when launching Firefox?
A: When Firefox starts, it looks for the %AppData%\Mozilla\Firefox folder, where the profiles.ini file is located, containing information about all available profiles: pointer to active profile, profile names and relative ways to them.
Q: Why can't I just copy the %AppData%\Mozilla\Firefox folder?
A: This is possible, but only if the paths to %AppData% match
(Method 2). Unfortunately, in currently many paths
the transferred profile worked, you need to correct all absolute paths in
all profile files. If such an operation is relatively correct according to
towards text formats, then replacing strings in binary files Maybe
damage their structure.
Classification of operations with profiles
Backup
Cloning:
Using cloning, you can create a duplicate profile and conduct any experiments on it without affecting the working profile.
To create a clone profile, follow these simple steps:
- Create a new profile
- Delete its contents and copy the cloned profile into it
- Using any batch file search and text replacement program, replace the name string of the profile being cloned (for example, uhxh6u8f.default) to the clone name (for example, daf9u3bb.cloned). Advanced Find & Replace, EmEditor and others are suitable as such a program.
Transfer:
Method 1 - If the %AppData% paths on both computers are the same
Just copy the %AppData%\Mozilla\Firefox folder from one
computer to another. If the target computer already has such a folder
exists (and you don’t care about its contents - existing ones on the machine
profiles), clear it before copying.
Method 2 - If %AppData% paths do not match
Method 2a. Symbolic links or reparse points
Possible on a file system that supports symbolic links (reparse points): this is NTFS 5.x and file systems Linux.
- Copy the Mozilla folder to its old location (for example, if it was C:\Documents and Settings\name\Application Data\Mozilla\Firefox on the original machine, copy it there).
- Delete the %AppData%\Mozilla\Firefox folder on the new computer if it exists.
- Create a symbolic link
(reparse point) from the folder copied in step 1 to the folder deleted in step 2
(in Windows this is done using the NTFS Links or Junction utility).
Method 2b. Modification of profiles.ini
Possible if the drive letters on which the Documents and Settings folder is located match, but the Windows user names do not match.
- Copy the Mozilla\Firefox folder to its old location (for example, if it was C:\Documents and Settings\name\Application Data\Mozilla\ on the original machine, copy it there).
- Copy the profiles.ini file to the %AppData%\Mozilla\Firefox folder
- Open it with any text editor and correct the line containing the path to the profile to this:
Method 3 - If it is not possible to create new folders and paths
Copy the profile over the existing one (after clearing it
any batch search and string replacement program (Advanced Find &
Replace, EmEditor, etc.) replace absolute path old profile on
This method is not the most correct of all possible, therefore
100% identical operation of the transferred profile is not guaranteed, in
difference from methods 1 and 2. But this is the only possible way transfer
profile between Windows and Linux.
After migrating the profile, delete the XUL .MFL file!
What is Internet speed and in what units is the Internet connection speed measured: bits or maybe bytes?
Internet speed represents the maximum amount of data received personal computer(PC) or transmitted to the Network for a certain unit of time. If you look at the change in data transfer speed, you can most often find it in kilobits/second (Kb/sec; Kbit/sec), or in megabits (Mb/sec; Mbit/sec). The size of any files is always indicated in bytes, KB, MB and GB.
We all know that 1 byte is 8 bits. If the speed of your Internet connection is 100 Mbit/s, then following the calculations (100/8=12.5), we can conclude that the computer can transmit or receive no more than 12.5 in one second MB of information. If the size of the file you want to download is 1.5 GB, then the download process will take you no more than two minutes.
What does the speed of the Internet connection depend on?
First of all, the speed of your Internet connection depends on your tariff plan that was set for you internet provider. The speed is also affected by the technology of the information transmission channel and the congestion of the Network by other users. If you limit the total bandwidth of the channel, then the more users are on the Internet and the more files and information they download, the more the connection speed drops, since there is less " free space" online.
Secondly, there is a dependence on the loading speed of the sites you are on. For example, if at the time of loading the server can provide the user with data at a speed of at least 10 Mbit/sec, then even if you are connected to the maximum tariff plan, you can't expect more.
Factors that influence Internet speed:
1. When checking, the speed of the server you are accessing.
2. Speed and settings of your Wi-Fi router.
3. All programs and applications running on the computer at the time of verification.
4. Also firewalls and antiviruses running in the background.
5. Your settings operating system(OS) and the computer itself.
How can you increase the speed of your Internet connection?
1. Malicious or unwanted software, it can primarily affect the reduction in Internet connection speed.
2. Viruses, worms and Trojans, which randomly got into your computer, they can take away part of the channel bandwidth. To solve this problem you need to use antivirus programs that will fight against infection of your PC.
3. When you use Wi-Fi that is not password protected You are also putting yourself at risk since other users usually connect to it. Therefore on WiFi you need to set a password.
4. Parallel running programs also reduce the speed of the Internet connection, because they lead to an increase in processor load, so the speed decreases sharply.
Some actions are capable increase internet speed connections, for example:
1. Increasing port throughput speed. This is if you have connected high internet connection, and the speed dropped sharply. Go to the "Start" menu, then "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 this you need to click right click mouse and open "Properties". After this, a window will open in which you need to go to the “Port parameters” column. After the window has opened, click the “Speed” parameter (bits per second) and click on the number 115200 - then OK! After all these actions throughput speed port increased. Because the silent speed is set to 9600 bps.
2. You can also try disabling the QoS packet scheduler to increase speed. To do this, you need to run the gpedit.msc utility. In Start, search for gpedit.msc. Next, you need to click “Computer Configuration” after “Administrative Templates”. Then go to “Network”, then “QoS Packet Scheduler”. Next you need to “Limit reserved bandwidth” then “Enable” and set it to 0%. Click "Apply" and restart the computer.
3.Reboot your router. Restarting your modem or router can resolve many connection problems. Unplug, wait 30 seconds and turn it back on.
These actions, in some cases, will help you increase your speed.
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 - +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 data transfer rate of 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. Certainly, we're talking about O 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).
| Table 1. | Binary phase shift keying. | ||||||||||
|
|||||||||||
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. IN total The 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 is the channel capacity 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, orthogonal frequency division multiplexing (OFDM) modulation schemes use QPSK, 16QAM, and 64QAM.
The same is true for WiMAX and technology cellular communications Long-Term Evolution (LTE) 4G. Transmission of analogue and digital television in cable TV and high-speed Internet access systems is based on 16QAM and 64QAM, while in 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.
A television high definition in the US 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.
