f*********t
发帖数: 37 | 1 To make things simple, assuming this is a static wireless channel. There is
no relative movement between tranmitter and receiver.
Intuitively you can understand that receiver receives the same transmitted signal from multiple
dirrections because of the wireless channel. This is multipath. If all those
signals from different dirrections arrive at the receiver at almost the
same time comparing with the time duration of one transmission symbol. Then
the channel can be modelled as a single tap y(n)= |
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r**********r
发帖数: 13 | 2 LTE中每个symbol前有一个Cyclic Prefix (CP, 4.7 us). CP is just a repeat of
the end of the previous symbol, and is used to give a "settling time" to
allow for delay spread in propagation channel.
Does this meant that if the multipath delay is longer than 4.7 us, the LTE
system will suffer from significant inter symbol interference?
If no, what am I missing here?
谢谢了. |
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z*****n
发帖数: 7639 | 3 if the multipath delay is longer than 4.7us,
the rake receiver will simply ignore it.
The assumption is that all the most significant
arrivals are within this time interval. |
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S***i
发帖数: 50 | 4 You are not missing any thing.
If the multipath delay is longer than period of CP, ISI will be introduced.
The length of CP is chosen to balance the performance, minimizing ISI and
efficiency. |
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s********k
发帖数: 6180 | 5 主要是用于做multiuser detection，理论上也是对multipath很有效啊 |
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z*****n
发帖数: 7639 | 6 怎么个有效了？
所有高symbol rate的调制方式都有multipath问题。 |
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s********k
发帖数: 6180 | 7 sorry,突然有个问题晕了，multipath fading应该是针对carrier frequency，不是针
对baseband signal吧？ |
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z*****n
发帖数: 7639 | 8 man，从物理角度来考虑，multipath当然是个物理现象，
作用于 channel 中的信号，不过经过解调后它也影响baseband啊。 |
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z*****n
发帖数: 7639 | 9 你这20M是比特率？
假设802.11b，（802.11a/g都是OFDM，不是DSSS，
受multipath影响很小）载波2.4GHz，chip rate = 11Mcps,
两个path，一个line of sight 40m，一个reflected by wall
70m，then 它们的到达时间差是 30/300000000 = 0.1us，
差不多超过一个chip的时间。如果是常规spread code，
和延迟到达的信号就不是正交了，会出现解码错误。
是1 |
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m**t
发帖数: 1292 | 10 in CDMA system, there is less
problem for multipath due to the high chip rate and low correlaation
between successive chips. Equalization is not required in the case that
multipath components are delayed more than a chip duration...
Rake receiver provides diversity using the multipath components, the
result seems can have the effect to reduce the multipath interference too, maybe
use Rake receiver in a different way? like a pilot aided rake demodulation or sth?
i am a bit confused on this, c |
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t***o
发帖数: 4265 | 11 1. 我认为 multipath channel 与 Rayleigh channel 不是一回事，虽然在书里和
paper里它们经常关联出现。
Rayleigh channel 是time varying 的。但一个multipath channel 可能time varying
也可能 time nonvarying, 比如在室内，两个antennas 放在两个角落，位置不变，室
内的环境也没有变化，这样任何时候信号总是通过同样的multipath channel 从一个
antenna 到达另一个antenna的。这样的channel用一个FIR filter描述。
Time-varying multipath channel 很容易理解了，并且其变化服从Rayleigh
distribution也好理解。当然也可以用FIR描述，但filter coefficients，甚至filter
长度都可能是time varying。
我的这个理解有错不？
2.
看了几篇MIMO的paper，如Alamouti(1998), MRT(Lo, 1999),其中的channel 都是
Ray |
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s*****o
发帖数: 1540 | 12 给我站内发信,告诉我你能审那一篇的编号．要求
１：PHD already or almost obtained in CS or EE。为保证审稿质量，必须是相关专
业的。不可能给其它无关专业博士审理，见谅。
2: let me know your edas id. (If you don't have an edas, I assume you are
not working in these fields).
Send me the number of the paper to my mitbbs mailbox.
Two conferences are regular quality conference.
------------ conference 1 due Aug 23 -----------
A) Performance of Multipath Congestion Control, Wireless Networks, "linked
increases" algorithm (MPTCP-LIA), "opportunistic linked increases... 阅读全帖 |
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p**y
发帖数: 10 | 13 Equalization IS a big deal in CDMA systems because
CDMA signal is inheritantly wideband and have to
deal with multipath problem.
For gaochao's problem, it depends on your problem.
If you are working on some practical system like
IS-95 system, in which long-code is used, no
equalization, rake receiver is used to perform a
maximum ratio combining to combat multipath.
In academia/research CDMA literature, people did
lots of things about short-code CDMA systems. In
this case, a lot of thing become m |
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i*****y
发帖数: 47 | 14 ICC是他們的共同平台
39. Grace Xingxin Gao, Lu Mingquan, and Feng Zhenming, \Asymmetric Hexagonal
QAM
Based OFDM System," IEEE International Conference on Communications,
Circuits and
Systems and West Sino Expositions 2002, Xi'an, China, June 2002.
40. Grace Xingxin Gao, Lu Mingquan, and Feng Zhenming, \Optimal Wavelet
Packet Based
Multicarrier Modulation over Multipath Wireless Channels," IEEE
International Conference
on Communications, Circuits and Systems and West Sino Expositions 2002, Xi'
an, China,... 阅读全帖 |
|
发帖数: 1 | 15 GPS需要相对论是谣言，这种谣言一到具体，很容易不攻自破
GPS 不是广义相对论的实验证据
王令隽
有人宣称GPS全球定位系统必须进行相对论修正，否则就不干活。许多人也信了。
最近和孙南生教授通信，他推荐了一篇业内专家R Li 的不同意见：
作者：R Li
链接：http://www.zhihu.com/question/24796597/answer/45911638
来源：知乎
摘要如下：
GPS不是相对论的应用
相对论改正只是卫星钟误差中的一个影响很小的系统误差改正项。
GPS系统有两个用途
主要用途， to drop 5 bombs in the same hole – 定位用途（这也可以看出设计这
个系统时军方目标精度是米级）相对论影响对目标精度的影响可忽略不计。
–主要用途中衍生出来的高精度测绘级定位用差分观测法消掉卫星钟误差，不需考虑相
对论影响（直接被绕过了）。
GPS定位是从未知点观测观测已知点从而确定未知点的坐标方位（专业说法叫resection
后方交会），观测量是未知点，你手持的GPS到已知坐 标的卫星的距离（卫星发射基于
卫星钟时钟信号，其实是有长周... 阅读全帖 |
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c*****d
发帖数: 6045 | 16 传统企业的IT部门，在Minneapolis MN，有on-call的要求
要求熟悉Linux v5/v6，Solaris v10
熟悉DNS, NIS, NFS, SAMBA, LDAP, Symantec Cluster Services，Storage
Foundation，SAN environment; multipathing, Symantec NetBackupExec
email c*********[email protected] if you are interested |
|
c*****d
发帖数: 6045 | 17 传统行业的IT部门，在Maple Grove, MN
有on-call的要求
要求熟悉Linux v5/v6，Solaris v10
熟悉DNS, NIS, NFS, SAMBA, LDAP, Symantec Cluster Services，Storage
Foundation，SAN environment; multipathing, Symantec NetBackupExec
email c*********[email protected] if you are interested |
|
c*****d
发帖数: 6045 | 18 传统行业的IT部门，在Minneapolis MN，有on-call的要求
要求熟悉Linux v5/v6，Solaris v10
熟悉DNS, NIS, NFS, SAMBA, LDAP, Symantec Cluster Services，Storage
Foundation，SAN environment; multipathing, Symantec NetBackupExec
email c*********[email protected] if you are interested |
|
s*****g
发帖数: 1055 | 19 I believe he/she is saying a L2 "routing" protocol very similar to IS-IS,
such that L2 switching path can also have native ECMP (or multipath, if ECMP
is a L3 term). This is not new, not every vendor likes this idea. |
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t*******r
发帖数: 3271 | 20 来自主题: EmergingNetworking版 - nnd ATT 沙特电信
Announcement bits (4): 0-KRT 3-KRT 5-Resolve tree 1
6-Resolve tree 2
AS path: 6453 39386 25019 I Unrecognized Attributes: 39
bytes
AS path: Attr flags e0 code 80: 00 00 fd 88 40 01 01 02
40 02 04 02 01 5b a0 c0 11 04 02 01 fc da 80 04 04 00 00 00 01 40 05 04
00 00 00 64
Accepted Multipath |
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L******t
发帖数: 1985 | 21 The main reason to use L2 DCN is because of VM mobility. Now with VXLAN, VM
mobility is supported as well, but this time over L3 DCN. In L3 DCN,
multipath is no longer a problem. And VXLAN can also fix the MAC scalability
issue in similar way as FabricPath. And VXLAN is even better because it
improves vlan scalability.
So in my understanding VXLAN makes FabricPath/TRILL much less appealing than
before. |
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l*p
发帖数: 74 | 22 呵呵，讨论深化了，先仍一砖头
1。就原题"为什么受机没有电视那么宽的带宽",是因为一个是
一对多一条信道， 另一个是多对多多个信道，如果大家
只从基站受一路信号，那手机带宽也回很宽。XDMA不就是
相仿设法增加单个拥护的等效带宽吗？
2。注意是multipath引发 time delay spread ,thus
cause ISI，
time delay spread/coherence bandwidth 和dopper 无关
dopper印发 time selective fading/coherence time
我想这是两个感念
3。并不是有个doppler，就cannot trasmite high speed date
in high speed mobile communications" 了。香农公式中
并没有速度那一项。给定带宽，whatever vehical speed
and how large the Fd,不断增加eb/ |
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s**o
发帖数: 29 | 23 基本上，在无线通信中有两大类四种small-scale fading
1. Time dispersion due to "multipath delay spread"
a) flat fading
b) frequency selective fading
2. Frequency despersion due to "Doppler spread"
a) fast fading
b) slow fading
以上这两大类fading基本上是independent of one another
关于flat fading，通俗讲，就是传输信号的带宽小于信道的相干带宽，所以，信号
经过衰落信道后，在receiver处，信号的spectral characteristics依旧不变。
其对应的则是frequency selective fading，其信号的带宽大于信道的相干带宽，
所以，在RX处，接收信号在不同的频率其衰落特性也不同，即spectral characteristics
变了。
具体的，可以参考 Rappaport的"wireless com |
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a*******e
发帖数: 346 | 24 h = rayleighchan()
....
plot(h)
Thanks |
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w****j
发帖数: 237 | 25 根据EM的reciprocal theory,passive天线的接受增益和发射增益肯定是一样的。
至于通讯距离,google一下link budget就可以了，基本取决于：发射EIRP=Pt*Gt; 接受
的Gr/Te,传播的Path loss,然后就是调制方式的BER vs. C/No了。当然，multipath的
就不再讨论范围了。 |
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R****a
发帖数: 199 | 26 Mostly, use an outdoor antenna is a best solution if you have a roof to
place that.
MIMO might not be easy since the multipath or obstruction is a big issue
that not easily handle by MIMO if I understands correctly. |
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f*********t
发帖数: 37 | 27 Thanks a lot for sharing those questions, Those answers are really good.
Maybe there are a little more to say for the cyclic prefix.
According to the answer leave the cyclic prefix part to be bland, ISI can
also be avoided. In real multipath environment, we cannot leave the cyclic
prefix to be blank.
I don't think this is the reason that they didn't give you onsite. Qualcomm
is very BT. I just want to dicuss if there are any better answers for those
questions. |
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f*********t
发帖数: 37 | 28 You already answered it in other question. The multipath effects on a symbol
with CP is just like a circular convolution. That's the reason you say it
will be a multiplication on frequency domain, isn't it?
I still dont know why you think orthogonal property requires infinity in time domain?
Is there any reason?
"Q: just now. You said that convolution in time domain is corresponding to
multiplication in frequency
domain. This is for continuous time. How about the property for discrete
time signa |
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w*******i
发帖数: 525 | 29 google了一下，还真有一些文件，不过没找到好好解释的。 |
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g******s
发帖数: 410 | 30 Now I'm a little confused. Is there a rake receiver in OFDM systems? Can you
please give more details? Thanks! |
|
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s********k
发帖数: 6180 | 32 比如2.4G的载频，20M的wifi信号，假设用DSSS，那么这和symbolrate是算1/2.4G还是1
/100M? |
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s********k
发帖数: 6180 | 33 我糊涂不在这个地方，而是为什么不用2.4G的信号算？毕竟最后在空间中传播的是2.4G
信号，2.4G信号也有symbol吧,原来一直也是按你这样算 |
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z*****n
发帖数: 7639 | 34 不管是啥频率的信号，propagation延迟只跟距离有关，
而延迟产生的影响只跟一个symbol的duration有关。
802.11b的symbol rate是11Msps。
我觉得你对symbol的理解有偏差。
a symbol is a signal element, within its duration
it doesn't change form.
对于基带信号，比如unipolar NRZL，一个symbol 就是
信号电平维持在+V 或者0V的时间。对于调制信号，
比如QPSK，一个symbol就是信号相位维持在某个
constellation point的时间。跟载波频率没啥关系，
跟基带信号的symbol rate一致。
4G |
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s********k
发帖数: 6180 | 35 那么载频的2.4G在时域上表现出来的特征是什么呢？发现高了半天Baseband没有对这个
问题想清楚。一个BPSK的基带调制到载频2.4G上，时域上理论上就是加了个复数，实际
上是什么回事？ |
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z*****n
发帖数: 7639 | 36 “时域上理论上就是加了复数”，in real situation
a signal is ALWAYS a function of time can be captured
by oscilloscope.
BPSK signal can be expressed by
s(t) = A \cos(2*pi*fc*t + b_k*pi)
where fc is the carrier freq.
b_k is baseband signal (unipolar NRZ) exprssed as
b_k(t) = 1 or 0, for t \in (kT, (k+1)T],
T=1/R as R is the bit rate.
so when b_k=1, s(t) has a phase of pi, when b_k=0, phase 0. |
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