s*******e
发帖数: 7 | 1 我们用的无线网卡给出一个速度(say 54Mbps). 这是哪一个Layer的速度啊? Physical,
MAC or LLC?
谢谢大家了. |
t**o
发帖数: 1030 | 2 DLL
去维基搜802.11
Physical,
【在 s*******e 的大作中提到】
: 我们用的无线网卡给出一个速度(say 54Mbps). 这是哪一个Layer的速度啊? Physical,
: MAC or LLC?
: 谢谢大家了.
|
z*****n
发帖数: 7639 | 3 physical
Physical,
【在 s*******e 的大作中提到】
: 我们用的无线网卡给出一个速度(say 54Mbps). 这是哪一个Layer的速度啊? Physical,
: MAC or LLC?
: 谢谢大家了.
|
s*******e
发帖数: 7 | |
X*****r
发帖数: 2521 | 5 什么是DLL
【在 t**o 的大作中提到】
: DLL
: 去维基搜802.11
:
: Physical,
|
z*****n
发帖数: 7639 | 6 i think he means data link layer
【在 X*****r 的大作中提到】
: 什么是DLL
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X*****r
发帖数: 2521 | 7 oh
then I think you are right
【在 z*****n 的大作中提到】
: i think he means data link layer
|
z*****n
发帖数: 7639 | 8 If you have a clear understanding of "bit rate" and
"throughput", this should not be a question...
For 802.11a, it has 52 subcarriers, and among them 48 are
dedicated for data. The symbol rate is fixed as 250ksps.
with 64-QAM modulation and 3/4 coding rate (best channel condition). It can
achieve
250k x 48 x log2(64) x 3/4 = 54000kbps.
【在 s*******e 的大作中提到】
: 到底哪个对啊?....CONFUSING..
|
s*******e
发帖数: 7 | 9 Good point. This tells me that it is the physical layer throughput. If I
leave my mouse cursor on the Window wireless connection symbol, it shows me
a speed of 54 Mbps.
How can it measure the physical layer throughput? I mean it can only "count"
the data at MAC layer or up.
Thanks.
can
【在 z*****n 的大作中提到】
: If you have a clear understanding of "bit rate" and
: "throughput", this should not be a question...
: For 802.11a, it has 52 subcarriers, and among them 48 are
: dedicated for data. The symbol rate is fixed as 250ksps.
: with 64-QAM modulation and 3/4 coding rate (best channel condition). It can
: achieve
: 250k x 48 x log2(64) x 3/4 = 54000kbps.
|
z*****n
发帖数: 7639 | 10 Well, this is a bit complex to explain...
For physical layer, we usually (I think it should be always)
say "bit rate" of the transmission --- how fast a transmitter
can achieve when IT IS TRANSMITTING. The information you
saw was this value.
However, for upper layers, I think the right word is to use
"throughput" --- how many bits of data can be delivered in
one second or over a long-enough time. This value is usually
different from the physical bit rate, or more exactly, less
than that, because
【在 s*******e 的大作中提到】
: Good point. This tells me that it is the physical layer throughput. If I
: leave my mouse cursor on the Window wireless connection symbol, it shows me
: a speed of 54 Mbps.
: How can it measure the physical layer throughput? I mean it can only "count"
: the data at MAC layer or up.
: Thanks.
:
: can
|
|
|
w******t
发帖数: 241 | 11 在实际中难道48个频段全部要用到？还是一个节点用其中的一个20M的频段？
can
【在 z*****n 的大作中提到】
: If you have a clear understanding of "bit rate" and
: "throughput", this should not be a question...
: For 802.11a, it has 52 subcarriers, and among them 48 are
: dedicated for data. The symbol rate is fixed as 250ksps.
: with 64-QAM modulation and 3/4 coding rate (best channel condition). It can
: achieve
: 250k x 48 x log2(64) x 3/4 = 54000kbps.
|
d****i
发帖数: 1038 | 12 I think this calculation just demonstrated that 54k is dll rate, not
physical. It is eqivalent to say fast ethernet is 100M, which is also dll
rate.
can
【在 z*****n 的大作中提到】
: If you have a clear understanding of "bit rate" and
: "throughput", this should not be a question...
: For 802.11a, it has 52 subcarriers, and among them 48 are
: dedicated for data. The symbol rate is fixed as 250ksps.
: with 64-QAM modulation and 3/4 coding rate (best channel condition). It can
: achieve
: 250k x 48 x log2(64) x 3/4 = 54000kbps.
|
c****n
发帖数: 21367 | 13 would the coding redundancy be counted as throughput?
say, shall we multiply 3/4 to the actual data arrival
rate for 802.11a at 54Mbps mode?
thanks!
【在 z*****n 的大作中提到】
: Well, this is a bit complex to explain...
: For physical layer, we usually (I think it should be always)
: say "bit rate" of the transmission --- how fast a transmitter
: can achieve when IT IS TRANSMITTING. The information you
: saw was this value.
: However, for upper layers, I think the right word is to use
: "throughput" --- how many bits of data can be delivered in
: one second or over a long-enough time. This value is usually
: different from the physical bit rate, or more exactly, less
: than that, because
|
c****n
发帖数: 21367 | 14 hopping hopping...
【在 w******t 的大作中提到】
: 在实际中难道48个频段全部要用到？还是一个节点用其中的一个20M的频段？
:
: can
|
d****i
发帖数: 1038 | 15 that is why I said 54M is layer 2 rate, not layer 1.
at layer 2, you should multiply 3/4.
【在 c****n 的大作中提到】
: would the coding redundancy be counted as throughput?
: say, shall we multiply 3/4 to the actual data arrival
: rate for 802.11a at 54Mbps mode?
: thanks!
|
w******t
发帖数: 241 | 16 11没有用frequency hopping吧，bluetooth才用的
【在 c****n 的大作中提到】
: hopping hopping...
|
z*****n
发帖数: 7639 | 17 Com'on, 100M is physical speed.
【在 d****i 的大作中提到】
: I think this calculation just demonstrated that 54k is dll rate, not
: physical. It is eqivalent to say fast ethernet is 100M, which is also dll
: rate.
:
: can
|
z*****n
发帖数: 7639 | 18 channel coding is regarded as physical layer issue,
so this is not counted as the physical layer bit rate.
Just like 100M FDDI, the bit rate is 100Mbps exactly,
but due to the 4B5B channel coding, the actual baud rate/
symbol rate is 125Mbaudps.
【在 c****n 的大作中提到】
: would the coding redundancy be counted as throughput?
: say, shall we multiply 3/4 to the actual data arrival
: rate for 802.11a at 54Mbps mode?
: thanks!
|
z*****n
发帖数: 7639 | 19 Channel coding is a physical layer function, please.
【在 d****i 的大作中提到】
: that is why I said 54M is layer 2 rate, not layer 1.
: at layer 2, you should multiply 3/4.
|
z*****n
发帖数: 7639 | 20 Of course!
The bandwidth of one subcarrier is 312.5khz, total
bandwidth of one channel is 312.5*52 = 16.250MHz,
keep channel spacing 20MHz, 802.11g is totally
down-compatible with 802.11b in frequency band
allocations.
【在 w******t 的大作中提到】
: 在实际中难道48个频段全部要用到？还是一个节点用其中的一个20M的频段？
:
: can
|
|
|
p*****g
发帖数: 996 | 21 re~:)
【在 z*****n 的大作中提到】
: Channel coding is a physical layer function, please.
|
w******t
发帖数: 241 | 22 你的意思是52个信道总共才占用了20M带宽？就是说11的频段分配总共就20M？不是20*
52这么多的频段吗？
【在 z*****n 的大作中提到】
: Of course!
: The bandwidth of one subcarrier is 312.5khz, total
: bandwidth of one channel is 312.5*52 = 16.250MHz,
: keep channel spacing 20MHz, 802.11g is totally
: down-compatible with 802.11b in frequency band
: allocations.
|
z*****n
发帖数: 7639 | 23 What I meant was "subcarriers", not channels.
【在 w******t 的大作中提到】
: 你的意思是52个信道总共才占用了20M带宽？就是说11的频段分配总共就20M？不是20*
: 52这么多的频段吗？
|
w******t
发帖数: 241 | 24 那802.11究竟占用了多少的带宽？channel在11里面指的就是载波的意思吧
【在 z*****n 的大作中提到】
: What I meant was "subcarriers", not channels.
|
z*****n
发帖数: 7639 | 25 In 802.11b one channel is defined as 5Mhz bandwidth, and
working at 2400-2483Mhz ISM band.
But due to the modulation sideband, a 20Mhz seperation is
necessary. So at most 3 channels can be used at the
same place.
802.11g uses OFDM, and one channel is 20MHz, just
back-compatible with 11b.
【在 w******t 的大作中提到】
: 那802.11究竟占用了多少的带宽？channel在11里面指的就是载波的意思吧
|
w******t
发帖数: 241 | 26 你的意思是11g里面的20Mchannel被划分为52个sub channel还是20M就是一个channel，
然后总共52*20M?似乎是前者？
另外11b用DSSS，这个20M算的是spread之前的还是之后的？
【在 z*****n 的大作中提到】
: In 802.11b one channel is defined as 5Mhz bandwidth, and
: working at 2400-2483Mhz ISM band.
: But due to the modulation sideband, a 20Mhz seperation is
: necessary. So at most 3 channels can be used at the
: same place.
: 802.11g uses OFDM, and one channel is 20MHz, just
: back-compatible with 11b.
|
z*****n
发帖数: 7639 | 27 see my previous post, there are 52 "subcarriers" in
one "channel" of 802.11g.
802.11a/g use OFDM, I think you didn't get this point.
【在 w******t 的大作中提到】
: 你的意思是11g里面的20Mchannel被划分为52个sub channel还是20M就是一个channel，
: 然后总共52*20M?似乎是前者？
: 另外11b用DSSS，这个20M算的是spread之前的还是之后的？
|
d****i
发帖数: 1038 | 28 sorry, my bad. I was confused. damn...
You are right. it is physical layer rate, more accurately, physical layer
net rate.
data link layer uses this rate to transmit data, i.e., it is the service rate provided by the
physical layer.
data link layer has other overhead such as ARQ, which will make the actual data link layer rate much lower than the claimed
physical rate...
【在 z*****n 的大作中提到】
: see my previous post, there are 52 "subcarriers" in
: one "channel" of 802.11g.
: 802.11a/g use OFDM, I think you didn't get this point.
|
F*M
发帖数: 1861 | 29 The 54 Mbps is the physical layer information bit rate.
Here is the computation procedure:
Physical layer modulation symbol rate = symbol rate per sub-carrier x number
of carriers = 250kbps x 48 = 12 Msps
Physical layer coded symbol rate (with binary codes) = Moulation symbol rate
x modulation rate = 12 Msps x log2(64) = 72 Mcps
Physical layer information bit rate = 72 Mcps x (3/4) = 54 Mbps.
The DLL throughput would be less due to the need to strip off additioinal
overhead (~10%).
【在 d****i 的大作中提到】
: that is why I said 54M is layer 2 rate, not layer 1.
: at layer 2, you should multiply 3/4.
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