C******e
发帖数: 1850 | 1 假设我有一个数据文件，文件内容是完全随机的数据，现在我以每秒100个的速度把数
据读出来，并打算把它们通过调制的方法传送出去，我有两个问题：
1. 我的数据速率是100符号/秒，如果用PSK调制，那么载波频率至少应该是多少呢？
2. 当我按100符号/秒的速度读数据的时候，这个数据流的带宽是多少？按照奈奎斯特
定理，似乎应该是小于50赫兹，但是前面说过，这些数据是纯随机数据，难道不应该是
个宽带信号么？
请板上高手指教。 | k*******d
发帖数: 1340 | 2 1, 载波频率大大与带宽就行，1K以上都行吧
2， 我觉得应该还是100Hz，原因是Nyquist准则是在基带上说的，基带上的带宽是50Hz，但是调制
以后负频率部分变到正频率了还要占50Hz，这里用的是PSK，正负频率不对称，还是要传双边带信号，
所以是
100Hz（这一点我不确定，希望高人能够指点），纯随机二进制数据和宽带信号没啥关系啊，
还是窄带的 | C******e
发帖数: 1850 | 3 为什么随机数是窄带信号呢？不是可以把它们看作白噪声吗？另外，传统的MODEM好像
是8KHZ的采样频率，而载波只有1.65KHZ，那么我们用MODEM下载数据文件的时候，难道
这些数据文件中的数据带宽都小于1.65KHZ？感觉不太对啊。有什么环节限制了信号源
的带宽吗？ | V******i
发帖数: 306 | 4 PSK的带宽用Carson's rule估计。
【在 C******e 的大作中提到】
: 假设我有一个数据文件，文件内容是完全随机的数据，现在我以每秒100个的速度把数
: 据读出来，并打算把它们通过调制的方法传送出去，我有两个问题：
: 1. 我的数据速率是100符号/秒，如果用PSK调制，那么载波频率至少应该是多少呢？
: 2. 当我按100符号/秒的速度读数据的时候，这个数据流的带宽是多少？按照奈奎斯特
: 定理，似乎应该是小于50赫兹，但是前面说过，这些数据是纯随机数据，难道不应该是
: 个宽带信号么？
: 请板上高手指教。
| k*******d
发帖数: 1340 | 5 ?? 我知道FSK带宽用Carson's rule估计，PSK也用吗？
【在 V******i 的大作中提到】
: PSK的带宽用Carson's rule估计。
| f*********t
发帖数: 37 | 6 I am not 高人.
You cannot know the modulated signal bandwidth just from the symbol rate.
This really depends on your modulation scheme. Like in GSM system, symbol
rate is 270.08333kHZ. The baseband bandwidth of the signal is only 100kHz.
When your symbol rate is higher than two times of the bandwidth of a
baseband signal, then ISI cannot be avoided. We need equalization to detect
signal.
note: GSM channel is usually known to be 200KHz, this number is passband
bandwidth after modulated on radio fre
【在 k*******d 的大作中提到】
: 1, 载波频率大大与带宽就行，1K以上都行吧
: 2， 我觉得应该还是100Hz，原因是Nyquist准则是在基带上说的，基带上的带宽是50Hz，但是调制
: 以后负频率部分变到正频率了还要占50Hz，这里用的是PSK，正负频率不对称，还是要传双边带信号，
: 所以是
: 100Hz（这一点我不确定，希望高人能够指点），纯随机二进制数据和宽带信号没啥关系啊，
: 还是窄带的
| f*********t
发帖数: 37 | 7 8KHZ sampling rate in digital modem is the sampling rate for voice. Those
8kHz samples will be encoded to 64Kbps data stream. 64kbps data stream will
be compressed, channel encoded and modulated before transmision. The 64kbps
audio signal can be then transmitted in rate 4.7kbps to 12kbps. The higher
rate the better quality of the voice.
传统的MODEM I think you mean the old modem that supports 4.8k -9.6 kbps. I
think they use 16QAM modulation and symbol rate is 1.2khz-2.4khz. This
signal will be mod
【在 C******e 的大作中提到】
: 为什么随机数是窄带信号呢？不是可以把它们看作白噪声吗？另外，传统的MODEM好像
: 是8KHZ的采样频率，而载波只有1.65KHZ，那么我们用MODEM下载数据文件的时候，难道
: 这些数据文件中的数据带宽都小于1.65KHZ？感觉不太对啊。有什么环节限制了信号源
: 的带宽吗？
| C******e
发帖数: 1850 | 8 isn't the same modem used for data transfer as well. for example the dial up
modem for internet service. in data transfer, the baseband signal, for
example a data file containing random data, could have a wider bandwidth
than the carrier frequency, how do we do to make it band limited to the
carrier freq?
will
64kbps
baseband
telephone
higher
【在 f*********t 的大作中提到】
: 8KHZ sampling rate in digital modem is the sampling rate for voice. Those
: 8kHz samples will be encoded to 64Kbps data stream. 64kbps data stream will
: be compressed, channel encoded and modulated before transmision. The 64kbps
: audio signal can be then transmitted in rate 4.7kbps to 12kbps. The higher
: rate the better quality of the voice.
: 传统的MODEM I think you mean the old modem that supports 4.8k -9.6 kbps. I
: think they use 16QAM modulation and symbol rate is 1.2khz-2.4khz. This
: signal will be mod
| f*********t
发帖数: 37 | 9 I don't know about this :). Yeah you maybe right, there maybe some data
loads on this system. I say it is 'old' just because most people use DSL (
ADSL) modem for data service now. The 1.6k carrier frequency system is what
they call POTS (Plain old telephone service or Post Office Telephone System)
. This system takes 300 to 3400 Hz to transmit.
I don't know how they transmit 64kbps non-compressed speech source in this
band. To make system to be band limited is not hard. Simplest way is to put
a
【在 C******e 的大作中提到】
: isn't the same modem used for data transfer as well. for example the dial up
: modem for internet service. in data transfer, the baseband signal, for
: example a data file containing random data, could have a wider bandwidth
: than the carrier frequency, how do we do to make it band limited to the
: carrier freq?
:
: will
: 64kbps
: baseband
: telephone
| t****t
发帖数: 6806 | 10 you guys are messing up basic concepts of signal and system, and probably
you'd better go back and refresh yourself. for example:
* a data file itself, doesn't have any "bandwidth"; it all depends on the
modulation.
* random data is definitely NOT white noise; go check what white noise is.
for one thing, the band-unlimited white noise also has unlimited variance.
* it is true that any implementable data has unlimited frequency (since it's
time-limited), but for usually when people say "bandwidth
【在 C******e 的大作中提到】
: isn't the same modem used for data transfer as well. for example the dial up
: modem for internet service. in data transfer, the baseband signal, for
: example a data file containing random data, could have a wider bandwidth
: than the carrier frequency, how do we do to make it band limited to the
: carrier freq?
:
: will
: 64kbps
: baseband
: telephone
| | | t****t
发帖数: 6806 | 11 GSM data rate is 270.833kbps, not symbol rate. 1 symbol can have multiple
bits.
if symbol rate is 270k, then bandwidth will be at least half of it. that's
nyquist theorem.
detect
【在 f*********t 的大作中提到】
: I am not 高人.
: You cannot know the modulated signal bandwidth just from the symbol rate.
: This really depends on your modulation scheme. Like in GSM system, symbol
: rate is 270.08333kHZ. The baseband bandwidth of the signal is only 100kHz.
: When your symbol rate is higher than two times of the bandwidth of a
: baseband signal, then ISI cannot be avoided. We need equalization to detect
: signal.
: note: GSM channel is usually known to be 200KHz, this number is passband
: bandwidth after modulated on radio fre
| z*****n
发帖数: 7639 | 12 Well, does the OP's 100"symbol"/sec mean bitps or byteps?
I take bit here.
For PSK (and I assume 2PSK here) applied, as long as
the receiver can detect the phase correctly, the carrier
frequency can be the same as the symbol rate, which means
in each symbol interval if the carrier can complete one
2*pi cycle, the detector can figure the data out.
50Hz，但是调制
要传双边带信号，
关系啊，
【在 k*******d 的大作中提到】
: 1, 载波频率大大与带宽就行，1K以上都行吧
: 2， 我觉得应该还是100Hz，原因是Nyquist准则是在基带上说的，基带上的带宽是50Hz，但是调制
: 以后负频率部分变到正频率了还要占50Hz，这里用的是PSK，正负频率不对称，还是要传双边带信号，
: 所以是
: 100Hz（这一点我不确定，希望高人能够指点），纯随机二进制数据和宽带信号没啥关系啊，
: 还是窄带的
| C******e
发帖数: 1850 | 13 thank you for correcting us. could you further help with the answer to
my original question? If I want to send a data stream at the symbol rate
of 1000 symbols per second, what is the minimum carrier frequency I
should use? Could it be anything or there is a constraining
relationship?
probably
the
is.
variance.
(since it's
-
【在 t****t 的大作中提到】
: you guys are messing up basic concepts of signal and system, and probably
: you'd better go back and refresh yourself. for example:
: * a data file itself, doesn't have any "bandwidth"; it all depends on the
: modulation.
: * random data is definitely NOT white noise; go check what white noise is.
: for one thing, the band-unlimited white noise also has unlimited variance.
: * it is true that any implementable data has unlimited frequency (since it's
: time-limited), but for usually when people say "bandwidth
| V******i
发帖数: 306 | 14 Man, these are basic questions of the undergraduate communication course.
You can easily find the answer in any communication textbook. If you don't
have these basic concepts, you won't understand the answers given here
either. For example, if you do not know Carson's rule, there is no way you
can understand the bandwidth of PSK or FSK. You also messed up the concepts
of data rate, symbol rate, sampling rate, etc. How can anyone really teach
these concepts on BBS...
【在 C******e 的大作中提到】
: thank you for correcting us. could you further help with the answer to
: my original question? If I want to send a data stream at the symbol rate
: of 1000 symbols per second, what is the minimum carrier frequency I
: should use? Could it be anything or there is a constraining
: relationship?
:
: probably
: the
: is.
: variance.
| C******e
发帖数: 1850 | 15 I don't understand why this has to do anything with Carson's rule. Let's
use ASK for example to rule out the Carson argument. If a data file is
read out at R symbols/second, what is the requirement of the ASK carrier
frequency?
course.
don't
you
concepts
teach
【在 V******i 的大作中提到】
: Man, these are basic questions of the undergraduate communication course.
: You can easily find the answer in any communication textbook. If you don't
: have these basic concepts, you won't understand the answers given here
: either. For example, if you do not know Carson's rule, there is no way you
: can understand the bandwidth of PSK or FSK. You also messed up the concepts
: of data rate, symbol rate, sampling rate, etc. How can anyone really teach
: these concepts on BBS...
| V******i
发帖数: 306 | 16 The file symbols do not have to be equivalent to the transmitted symbols.
Carson's rule is used to estimate the bandwidth of PSK and FSK. If you
assume transmitted symbols must be the symbols you read from the file, then
your transmitted symbol rate is R. If you assume M-ary ASK, then your
transmission bandwidth must be at least R/2. But, this is a toy question, no
one really do this in practice.
【在 C******e 的大作中提到】
: I don't understand why this has to do anything with Carson's rule. Let's
: use ASK for example to rule out the Carson argument. If a data file is
: read out at R symbols/second, what is the requirement of the ASK carrier
: frequency?
:
: course.
: don't
: you
: concepts
: teach
| f*********t
发帖数: 37 | 17 "if symbol rate is 270k, then bandwidth will be at least half of it. that's
nyquist theorem. "
Are you sure Nyquist said that?
To my understand, Nyquist said if the bandwidth is less than half of the
symbol rate then you cannot achieve ISI (inter symbol interference) free
transmission. ISI is just a kind of distortions you can correct at receivers
. This means you still can transmit the data with this rate, you just have to deal with the ISI at receivers.
If you really want to know what will be
【在 t****t 的大作中提到】
: GSM data rate is 270.833kbps, not symbol rate. 1 symbol can have multiple
: bits.
: if symbol rate is 270k, then bandwidth will be at least half of it. that's
: nyquist theorem.
:
: detect
| t****t
发帖数: 6806 | 18 ok, i am not familiar with GSM. but i checked it just now, seems the
bandwidth is 200K, instead of 100K.
as for nyquist, if the bandwidth is W Hz, then you can reconstruct the
waveform with >2W/s sample rate, right? of course you can transmit more than
that much symbols and filter the extra band out, but that's still
equivalent to transmit 2W symbol/s, with a different symbol table.
s
receivers
have to deal with the ISI at receivers.
look at the channel capacity theorem instead of Nyquist theor
【在 f*********t 的大作中提到】
: "if symbol rate is 270k, then bandwidth will be at least half of it. that's
: nyquist theorem. "
: Are you sure Nyquist said that?
: To my understand, Nyquist said if the bandwidth is less than half of the
: symbol rate then you cannot achieve ISI (inter symbol interference) free
: transmission. ISI is just a kind of distortions you can correct at receivers
: . This means you still can transmit the data with this rate, you just have to deal with the ISI at receivers.
: If you really want to know what will be
| f*********t
发帖数: 37 | 19 look at the notes in my previous post, I said 100kHZ is the baseband
bandwidth. After up convert to carrier frequency, then it becomes 200kHz. I
believe in Nyquist theorem, the bandwidth is baseband bandwidth.
"as for nyquist, if the bandwidth is W Hz, then you can reconstruct the
waveform with >2W/s sample rate, right"
If this is how you get your conclusion, then we are not talking about the
same thing. In digital communication, we don't really reconstruct the whole
tranmitted signal. All we do
【在 t****t 的大作中提到】
: ok, i am not familiar with GSM. but i checked it just now, seems the
: bandwidth is 200K, instead of 100K.
: as for nyquist, if the bandwidth is W Hz, then you can reconstruct the
: waveform with >2W/s sample rate, right? of course you can transmit more than
: that much symbols and filter the extra band out, but that's still
: equivalent to transmit 2W symbol/s, with a different symbol table.
:
: s
: receivers
: have to deal with the ISI at receivers.
| t****t
发帖数: 6806 | 20 i am not saying that you need to reconstruct the waveform. but for a W Hz
bandwidth signal, you only need (>) 2W/s sample; that's all the information
it contains. of course, the new (resampled) samples is not the transmitted
sample (e.g. MSK), you probably need much more bits/sample to demod it. but
in the end, if you transmit a 100K sym/s BPSK or BMSK signal (=100Kbps) and
limit the band by filtering it to 30KHz, when you demod it, you only need
60K sampling rate. any higher sampling rate is th
【在 f*********t 的大作中提到】
: look at the notes in my previous post, I said 100kHZ is the baseband
: bandwidth. After up convert to carrier frequency, then it becomes 200kHz. I
: believe in Nyquist theorem, the bandwidth is baseband bandwidth.
: "as for nyquist, if the bandwidth is W Hz, then you can reconstruct the
: waveform with >2W/s sample rate, right"
: If this is how you get your conclusion, then we are not talking about the
: same thing. In digital communication, we don't really reconstruct the whole
: tranmitted signal. All we do
| | | C******e
发帖数: 1850 | 21 ok, so with a bandwidth of R/2, the carrier must be at least R/2 Hz as
well to be able to transmit the whole band. But in v.90 modem, R is 8000
symbols/sec (or 64kbps with 8-bit/symbol), however, the carrier frequency
is only 1650Hz. How could this happen? What is missing here?
symbols.
then
question, no
【在 V******i 的大作中提到】
: The file symbols do not have to be equivalent to the transmitted symbols.
: Carson's rule is used to estimate the bandwidth of PSK and FSK. If you
: assume transmitted symbols must be the symbols you read from the file, then
: your transmitted symbol rate is R. If you assume M-ary ASK, then your
: transmission bandwidth must be at least R/2. But, this is a toy question, no
: one really do this in practice.
| V******i
发帖数: 306 | 22 You just talked about a toy problem and messed up the concepts, I do not
think you can understand the detail of V90 modem. The short answer is, with
advanced modulation and coding schemes, the transmitted symbol rate of the
modem is much less than the "symbol rate" they presented to you. For example
if you group 4 symbols together as one transmitted symbol, then your
transmitted symbol rate is only 1/4 of the "symbol rate" you want to support
. Again, knowing that these concepts of "symbol rates
【在 C******e 的大作中提到】
: ok, so with a bandwidth of R/2, the carrier must be at least R/2 Hz as
: well to be able to transmit the whole band. But in v.90 modem, R is 8000
: symbols/sec (or 64kbps with 8-bit/symbol), however, the carrier frequency
: is only 1650Hz. How could this happen? What is missing here?
:
: symbols.
: then
: question, no
| f*********t
发帖数: 37 | 23 You got the simplest way to explain this :).
with
example
support
【在 V******i 的大作中提到】
: You just talked about a toy problem and messed up the concepts, I do not
: think you can understand the detail of V90 modem. The short answer is, with
: advanced modulation and coding schemes, the transmitted symbol rate of the
: modem is much less than the "symbol rate" they presented to you. For example
: if you group 4 symbols together as one transmitted symbol, then your
: transmitted symbol rate is only 1/4 of the "symbol rate" you want to support
: . Again, knowing that these concepts of "symbol rates
| f*********t
发帖数: 37 | 24 I agree with you at this point. Symbol rate is not signal bandwidth. To
reconstruct signal, we need sampling rate at least twice of the signal
bandwidth. In real system, using just twice of the signal bandwidth sampling
rate may casue a lot of design issues and sampling rate is usually higher
than this.
Bandwidth here means single side baseband bandwidth.
information
but
and
(
【在 t****t 的大作中提到】
: i am not saying that you need to reconstruct the waveform. but for a W Hz
: bandwidth signal, you only need (>) 2W/s sample; that's all the information
: it contains. of course, the new (resampled) samples is not the transmitted
: sample (e.g. MSK), you probably need much more bits/sample to demod it. but
: in the end, if you transmit a 100K sym/s BPSK or BMSK signal (=100Kbps) and
: limit the band by filtering it to 30KHz, when you demod it, you only need
: 60K sampling rate. any higher sampling rate is th
| k*******d
发帖数: 1340 | 25 Yes, I think 楼主 is discussing a toy question. I understand that in
practice even if you have ISI you can still recover the signal by
equalization. 我想GSM里面的GMSK其实Gaussian滤波就会造成一些ISI. I think 楼主
的问题是没有ISI的情况下，最小带宽需要多少，如果要回答楼主的问题的话，假如
symbol rate是100Hz,我认为最小带宽在基带上是50Hz，调制以后则是100Hz。
sampling
【在 f*********t 的大作中提到】
: I agree with you at this point. Symbol rate is not signal bandwidth. To
: reconstruct signal, we need sampling rate at least twice of the signal
: bandwidth. In real system, using just twice of the signal bandwidth sampling
: rate may casue a lot of design issues and sampling rate is usually higher
: than this.
: Bandwidth here means single side baseband bandwidth.
:
: information
: but
: and
| t****t
发帖数: 6806 | 26 where did you get the idea that "R/2 bandwidth need at least R/2 carrier"?
and where did you get the idea that V.90 is using 1.65K carrier?
【在 C******e 的大作中提到】
: ok, so with a bandwidth of R/2, the carrier must be at least R/2 Hz as
: well to be able to transmit the whole band. But in v.90 modem, R is 8000
: symbols/sec (or 64kbps with 8-bit/symbol), however, the carrier frequency
: is only 1650Hz. How could this happen? What is missing here?
:
: symbols.
: then
: question, no
|
|
