Data volume units
1tb=2^10gb=2^20mb=2^40kb=2^50b (bytes) =2^53bit (bit)
The digital rate in network technology
1tbit/s=10^3gbit/s=10^6mbit/s=10^9kbit/s=10^12bit/s
① transmit delay (s seconds) = Data frame length (bit bit) ÷ channel bandwidth (bit/s bits per second)
② propagation delay (s seconds) = Channel length (m m) ÷ propagation rate of electromagnetic waves on the channel (m/s meters per second)
The propagation rate of electromagnetic wave in free space is about 3*10^8m/s
The propagation rate of electromagnetic wave in copper cable is about 2.3*10^8m/s
The propagation rate of electromagnetic waves in light is about 2.0*10^8m/s
Example: 1000km long fiber line propagation delay of about 5ms (1000*10^3/(2*10^8) =5*10^-3) data transmission rate/data rate/bit rate is the number of bits transmitted per second (bit/s bits per second; bps)
Code Rate/Modulation rate/baud rate number of times per second (symbol/s symbol per second, also do band)
Bit rate (bit/s bit per second; also do bps) =
Baud rate (symbol/s symbol per second, also do band) *log2 (Power level (Bit/symbol bit per symbol))
bitrate = Baud rate * Number of bits corresponding to a single modulation state
I=S*LOG2 (N)
The width of the letter number of T-seconds, the baud rate =1/t; The maximum data transfer rate formula for a noise-free, limited-bandwidth channel is deduced:
Data transfer Rate c = 2bxlog2 (V) (bps)
C = data transfer rate, Unit bit/s (bps), B = bandwidth, Unit hz,v= signal encoding series
Any signal through a low-pass filter with a bandwidth of H, then sampled 2H per second can be fully reproduced the signal, the signal is divided into V-level
Shannon's rationale is given in the following formula:
C=BLOG2 (1+s/n| non-dimensional signal-to-noise ratio)
Where c is the available link speed and B is the bandwidth of the link
(s/n|db) =10LOG10 (s/n| non-dimensional signal-to-noise ratio)
∴ can also be c=blog2 (1+10^ ((s/n|db) ÷10) AAA bitstream 0 1 0 1 0 1 0 1 0 1
Unipolar nrz____|￣￣|____|￣￣|____|￣￣|____|￣￣|____|￣￣| does not return zero
Unipolar rzx____|￣|______|￣|______|￣|______|￣|_____|￣|__ Zeroing
Bipolar nrzxxxx|￣￣| xxxx|￣￣| xxxx|￣￣|xxxx|￣￣|xxxx|￣￣|
xxxxxxxx____| xxx|____| xxx|____| xxx|____| Xxx|____|xxx| does not return to zero
Bipolar Rzxxxx |￣| XXXXXX |￣| XXXXXX |￣| XXXXXX |￣| XXXXX |￣|
xxxxxxxx__|￣xx￣|__|￣xxx￣|__|￣xx￣|__|￣xx￣|__|￣xx￣| return to zero
Manchester x |￣￣↓xxx|￣￣↓xxxx |￣￣↓xxx |￣￣↓xxxx|￣￣↓
Xxxxxx__↑xxx|____↑xxx |____↑xxx|____↑xxx |____↑xxx |__ Ethernet data sent to Manchester code
DifferenceXxx |￣| Xxx |￣| x|-→| xx↑￣| xx|￣| XX |-→| xx↑￣| Manchester ↑x|-→| x↓-| Xxx| -| X |-→| x↓-| Xxx| -|xxx |-→ the remainder to the left-hand position of the information code to get the complete CRC code.
The "example" assumes that the generated polynomial used is g (X) =x3+x+1.
The original 4-bit message is 1010 and the encoded message is obtained.
Solution: 1, the resulting polynomial g (X) =x3+x+1 converted to the corresponding binary divisor 1011.
2, the problem generated polynomial has 4 bits (r+1) (Note: 4-bit generation polynomial calculation of the check code is 3 bits, R is the number of check code),
To shift the original message C (X) left 3 (R) bits into
1010 000
3, with the generation of a polynomial corresponding to the binary number of left 3 bits after the original message is modulo 2 (high alignment),
Equivalent to bitwise XOR OR:
1010000
1011
------------------then binary divisor aligns to the first nonzero
0001000
0001011
------------------
0000011
Get the remainder 011,
So the final code is: 1010 011 The contention period for Ethernet is set to 51.2μs
て= Contention period ÷2=25.6μs
T0 Transmit delay = data frame length (bit bit) ÷ transmit rate (bit/s bits per second)
Occupy channel time =t0+て;
Parameter a=t0÷て=t0÷25.6 (shortest) Frame length ÷ data rate >=2* maximum signal propagation delay
That is
(shortest) Frame length ÷ data transfer rate >=2* channel length (m m) ÷ propagation rate of electromagnetic waves on the channel (m/s meters per second)
Get the equal sign that the shortest x stations are connected to a Y-Mbit/s Ethernet hub
The bandwidth that each station can get y/x Mbit/s
X stations are connected to a ymbit/s Ethernet switch
Each station can get the bandwidth Y Mbit/s TCP datagram length x bytes, fixed header 20 bytes, the second network maximum data length is max, should be divided into several slices? Data field length, slice offset field, and MF,DF. "TCP datagram length x bytes = Data part + Fixed header"
Several pieces: (x-20)/(max-20) fetch quotient m (quotient m for the whole number only into the shed)
The following units are byte b
Total, data field length, slice offset field, and MF,DF.
Max; max-20; 0/8, 1,0
Max; max-20; (max-20)/8, 1,0
Max; max-20; (max-20)/8,1,0
......... 1,0
(x-20)% (max-20) +20; (x-20)% (max-20); (m-1) * (max-20)/8; 0,0
Up and down two a little difference, the next one should not copy the wrong bar. The original data sheet of the above shard removes the fixed header of 20.
The following UDP instead adds a 8-byte header.
Maximum Ethernet frame load max, IP data that can be hosted <=max-20; UDP datagram Length x+8 bytes, should be divided into several slices? Data field length, slice offset field, and MF,DF.
Several pieces: (x+8)/(max-20) fetch quotient m (quotient m for the whole number only into the shed)
The following units are byte b
Total, data field length, slice offset field, and MF,DF.
Max; max-20; 0/8; 1,0
Max; max-20; (max-20)/8; 1,0
Max; max-20; (max-20)/8; 1,0
......... 1,0
(x+8)% (max-20) +20; (x+8)% (max-20); (m-1) * (max-20)/8; 0,0
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