C******e
发帖数: 1850 | 1 If a communications receiver has a DAC sampling at frequency fs, what is
the maximum bit rate can this receiver receive in theory? My answer is
fs/2. The reason is as follows. Suppose the transmitted symbol has N
bits(e.g. N=2 for QPSK), then there are 2^N constellation states. That
means the receiver must have at least a resolution of 2^N states per
sample, therefore, the symbol rate is limited by fs/(2^N), with this
symbol rate and N bits per symbol, the bit rate is N*fs/(2^N). The maximum
of | f*********t
发帖数: 37 | 2 "That means the receiver must have at least a resolution of 2^N states per
sample, therefore, the symbol rate is limited by fs/(2^N)" Don't understand
how you get this conclusion.
You didn't say anything about channel condition. If the channel is a perfect
flat without any noise, each sample can be any one of the 2^N
constellations and N can be as large as you want.
【在 C******e 的大作中提到】
: If a communications receiver has a DAC sampling at frequency fs, what is
: the maximum bit rate can this receiver receive in theory? My answer is
: fs/2. The reason is as follows. Suppose the transmitted symbol has N
: bits(e.g. N=2 for QPSK), then there are 2^N constellation states. That
: means the receiver must have at least a resolution of 2^N states per
: sample, therefore, the symbol rate is limited by fs/(2^N), with this
: symbol rate and N bits per symbol, the bit rate is N*fs/(2^N). The maximum
: of
| C******e
发帖数: 1850 | 3 But if you sample a sine wave of frequency f0 with a clock fs, you would
only get at most fs/f0 different sample values, right?
understand
perfect
【在 f*********t 的大作中提到】
: "That means the receiver must have at least a resolution of 2^N states per
: sample, therefore, the symbol rate is limited by fs/(2^N)" Don't understand
: how you get this conclusion.
: You didn't say anything about channel condition. If the channel is a perfect
: flat without any noise, each sample can be any one of the 2^N
: constellations and N can be as large as you want.
| a****l
发帖数: 8211 | 4 If a communication transceiver has a DAC sampling at a frequency fs>0, my
answer is that the maximum bit rate is Infinite. Reason: it depends on how
accurate the DAC is.
【在 C******e 的大作中提到】
: If a communications receiver has a DAC sampling at frequency fs, what is
: the maximum bit rate can this receiver receive in theory? My answer is
: fs/2. The reason is as follows. Suppose the transmitted symbol has N
: bits(e.g. N=2 for QPSK), then there are 2^N constellation states. That
: means the receiver must have at least a resolution of 2^N states per
: sample, therefore, the symbol rate is limited by fs/(2^N), with this
: symbol rate and N bits per symbol, the bit rate is N*fs/(2^N). The maximum
: of
| z*****n
发帖数: 7639 | 5 If the input signal is baseband, such as NRZ or NRZI, the maximum
bit rate can be achieved is f_s: thinking of the worst case, which
the inputing bit pattern is 10101010... at f_s b/s, your sampler
takes sample at middle of each bit, you will get it.
However, this demands strict bitwise synchronization, as you will
find that when the sampling freq. drifts away from input signal
clock, you will eventually get error reception.
In practice, we often use over-sampling plus digital phase
loop-lock (D
【在 C******e 的大作中提到】
: If a communications receiver has a DAC sampling at frequency fs, what is
: the maximum bit rate can this receiver receive in theory? My answer is
: fs/2. The reason is as follows. Suppose the transmitted symbol has N
: bits(e.g. N=2 for QPSK), then there are 2^N constellation states. That
: means the receiver must have at least a resolution of 2^N states per
: sample, therefore, the symbol rate is limited by fs/(2^N), with this
: symbol rate and N bits per symbol, the bit rate is N*fs/(2^N). The maximum
: of
| V******i
发帖数: 306 | 6 fs smapling rate means your maximum rate of (independent) symbols is fs/2
symbol/sec. But each symbol can carry more than one bit.
【在 C******e 的大作中提到】
: If a communications receiver has a DAC sampling at frequency fs, what is
: the maximum bit rate can this receiver receive in theory? My answer is
: fs/2. The reason is as follows. Suppose the transmitted symbol has N
: bits(e.g. N=2 for QPSK), then there are 2^N constellation states. That
: means the receiver must have at least a resolution of 2^N states per
: sample, therefore, the symbol rate is limited by fs/(2^N), with this
: symbol rate and N bits per symbol, the bit rate is N*fs/(2^N). The maximum
: of
| w*******d
发帖数: 3714 | 7 DAC的精度只要足够高，哪怕你每秒钟只测量一次电压，你还是可以获得无限的信息。v
=0.3020394828374987059283498723876597198234
当然前提是DAC精度，和采样频率无关。所以你的问题缺条件。。。 | z*****n
发帖数: 7639 | 8 only in the case the signal is noise free.
... so what you said also 缺条件。。。
。v
【在 w*******d 的大作中提到】
: DAC的精度只要足够高，哪怕你每秒钟只测量一次电压，你还是可以获得无限的信息。v
: =0.3020394828374987059283498723876597198234
: 当然前提是DAC精度，和采样频率无关。所以你的问题缺条件。。。
| a****l
发帖数: 8211 | 9 neither of these two things are present in the original post...
【在 z*****n 的大作中提到】
: only in the case the signal is noise free.
: ... so what you said also 缺条件。。。
:
: 。v
| z*****n
发帖数: 7639 | 10 haha, you are right.
【在 a****l 的大作中提到】
: neither of these two things are present in the original post...
| C******e
发帖数: 1850 | 11 Sorry I missed one important item in my previous post, which is the carrier
frequency of the MPSK. So please allow me to restate the problem as follows.
If a communications transmitter and receiver both use a sampling rate of Fs
at the DAC and ADC, and use MPSK modulation, we know that the maximum symbol
rate possible is Fs/2. However, suppose the carrier of the M-PSK is
generated in the digital domain with the same sampling rate of Fs, there is
a hard limit on the number of bits per symbol. Let
【在 V******i 的大作中提到】
: fs smapling rate means your maximum rate of (independent) symbols is fs/2
: symbol/sec. But each symbol can carry more than one bit.
| f*********t
发帖数: 37 | 12 "we know that the maximum symbol rate possible is Fs/2" I am not sure where
you get this.
If you mean Nyquist theorem. There are two versions.
1. For band limited signal, sampling rate should be higher than two times of
baseband signal bandwidth to recover the signal.
2. Symbol rate should be less than two times of baseband signal bandwidth to
avoid ISI.
Neither of them is what you said.
For the later part, I think you missed something in your system. you should
have an analog demodulator before
【在 C******e 的大作中提到】
: Sorry I missed one important item in my previous post, which is the carrier
: frequency of the MPSK. So please allow me to restate the problem as follows.
: If a communications transmitter and receiver both use a sampling rate of Fs
: at the DAC and ADC, and use MPSK modulation, we know that the maximum symbol
: rate possible is Fs/2. However, suppose the carrier of the M-PSK is
: generated in the digital domain with the same sampling rate of Fs, there is
: a hard limit on the number of bits per symbol. Let
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