Hello,

overall, your code is not bad, it just needs to be improved. I mean: 

- you didn't create a signal, but symbols. In order to simulate a single carrier signal, you just have to oversample (zero pading) and filter (root raised cosine for instance) the symbols.

- I don't understand why you multiply your channel by sqrt(P/2). If it is for a normalization matter, then it is wrong. You may remove it. However, your 3 taps channel is OK, the Rayleigh-ness of the channel is intrinsic to the function randn. In further  works, just take into account the delays and the gains of the taps. But for a first test your code is fine.  

- Then you can use the conv function to filter the signal by the channel.

- The SNR is usually defined as the power of the received signal over the noise variance. So for a given SNR : assess the power of the received signal P = sum(abs(x).^2)/length(x), then create a noise vector of power 1 : N1 = sqrt(0.5)*(randn(1,L)+1i*randn(1,L)), and finally create the noise vector with expected power: 

N = sqrt(P_N)*N1 with P_N = P*10^(-SNR/10)

- Finally, the signal is the sum of the transmitted signal over the channel plus noise N.

Regards

A Rayleigh flat-fading channel is described, in part, by the equations in the Matlab listing. The factor of 1/sqrt(2) is there so that the expected value of is P. Normally, one would set P = 1 for a lossless channel -- all of the signal gets through on the average, its just redistributed in time.

What is missing from this code is the second-order statistics of the channel. Physical channels do not change in an uncorrelated manner from sample to sample -- there is temporal correlation described by the power spectral density of the fading. If one believes that communications link has no memory such as that of error correction coding, interleaving, and decoding or that memory is short compared to the decorrelation time of the channel, then the time averaged error rate will approach the ensemble average error rate, and the latter is what this code is set up to do, compute the ensemble error rate in a slow fading channel.

Thank you for answering my question.

A channel is said to flat-fading if symbol time Ts is larger than maximum excess delay time Tm. How do you know the channel is flat-fading from this code ?

The h in my code is impulse response, right ?

I have seen following code,

h=(1/sqrt(2))*((randn(1,L))+j*(randn(1,L)));

r=abs(h).*s+n;     % s is transmitted signal (length L), n is WGN

What is different meaning between above code and my code?

When fading is obvious, I have to set the expected value P larger than 1, right?

what is effect in the missing second-order statistics?

I wrote the code to simulate a OFDM signal through Rayleigh channel. To calculating bit error rate, I use ZF equalizer to compensate channel effect in receiver ( received signal divided by fft(h) in frequency-domain ). 

If you don't mind, Would you please explain it or give a example in matlab code?

It's clearer to me. Because my English is basic, I worry to misunderstand your meaning. Thank you.

Hello,

overall, your code is not bad, it just needs to be improved. I mean: 

- you didn't create a signal, but symbols. In order to simulate a single carrier signal, you just have to oversample (zero pading) and filter (root raised cosine for instance) the symbols.

- I don't understand why you multiply your channel by sqrt(P/2). If it is for a normalization matter, then it is wrong. You may remove it. However, your 3 taps channel is OK, the Rayleigh-ness of the channel is intrinsic to the function randn. In further  works, just take into account the delays and the gains of the taps. But for a first test your code is fine.  

- Then you can use the conv function to filter the signal by the channel.

- The SNR is usually defined as the power of the received signal over the noise variance. So for a given SNR : assess the power of the received signal P = sum(abs(x).^2)/length(x), then create a noise vector of power 1 : N1 = sqrt(0.5)*(randn(1,L)+1i*randn(1,L)), and finally create the noise vector with expected power: 

N = sqrt(P_N)*N1 with P_N = P*10^(-SNR/10)

- Finally, the signal is the sum of the transmitted signal over the channel plus noise N.

Regards

I'll answer some of your questions here. But the general impulse response function h(time,delay) is quite complicated, as can be seen from the ACIRF (Antenna/Channel Impulse Response Function) code description. (You can find some documentation from a Google search on ACIRF). The time dependence is described by a Doppler frequency spectrum that describes how the "impulse ringing of the channel" varies with time. The Fourier transform of the CIRF with delay gives the frequency spectrum of the channel across the signal bandwidth. The frequency and Doppler spectra are not the same; together they describe the second-order statistics of the channel.

Most of the time one says "flat fading" the channel delay is very small compared to the duration of a channel symbol, and the channel does not create intersymbol interference. I described your channel as flat fading, in a non-conventional manner, because you do not include any tapering of the CIRF power with delay. Each of your delay paths has equal average power, much like thermal noise, so its frequency spectrum across the signal bandwidth is flat (as I understand it, h(1) has delay 0, h(2) has delay Ts, and h(3) had delay 2Ts in your formulation and each has average power P). This is a very non-physical channel; there is no tapering of CIRF power with delay, the channel is producing equal power delays over 2 symbol periods, and the average power of the CIRF is 3P.

The other application of the CIRF that you describe involving abs(h) is incorrect. The correct physical application is the convolution that you use. Taking the absolute value of the CIRF eliminates the very real phase fluctuations of the channel that are what causes the most problem for coherent phase demodulation such used in a standard QPSK receiver -- essentially the phase fluctuations break the phase-lock loop tracking.

The average power of a CIRF P cannot be larger than 1 as that implies that the channel somehow amplifies the average signal power -- physical nonsense for a passive channel. (Instantaneously the CIRF power can and does exceed 1, but on the average

I referred to a code and wrote my code. I don't sure whether the factor sqrt(P/2) is necessary.

Why the SNR is negative in this formulation ( P_N = P*10^( - SNR / 10 )  ) ?

 Is the x received signal or transmitted signal? If x is defined as received, how do i determine the received signal before adding noise.

To define a impulse response, I have to first set delay, power of path, second-order statistics and son on.

h=sqrt(P/2)*(randn(1,3)+1i*randn(1,3)) is not a impulse response, but a channel ( a non-strict flat-fading channel ) . Is that right ?

I found some information about CIRF. I thank I have to spend some time on it.

Thank you.

(1) The channel is generated as Rayleigh fading process; Let me call it h = x + i y where x and y are generated using the Matlab function randn that generates zero-mean, unity variance, normally distributed random samples. As each call to randn generates independent, uncorrelated samples, the mean power of h is = + . But the variance of x and y is unity, so the mean power of h as written is 2. Therefore I normalize h by 1/sqrt(2) so that the mean power of h as I have written it is 1.

(2) As I have discussed above, h as you have it represents a sampled version of the impulse response function with three taps, separated by the symbol period, each with power P. This is physical nonsense, but you can simulate nonsense if you want.

(3) The author of this code is setting the noise power spectral density N0 (the noise floor) below the signal symbol energy Es by the SNR. Presumably, one gets the SNR from the link margin equation, something that you should know already. As you do not seem to have a specified value for Es, set it to 1 (Es = 1). Then the equation N0=Es/10^(SNR_dB/10) sets N0 below the signal. Now you can compute performance parametric with SNR_dB. Suggested values: SNR_dB are (0:2:20).

(3) I contend that h as you generate it is a sampled version of the CIRF. The delay taps are separated by one symbol when you apply conv(QPSK , h). Each tap has equal power P. A better definition of P might be P = [0.9  0.09  0.009] which applies a power taper -- longer delays have less relative power. Still ad hoc, but more realistic.

(3) For a passive channel, the sum of the power in the CIRF should be 1. My definition has total mean power slightly less that one, meaning you need more taps to capture all of the delayed signal energy propagation through the channel. This is a realistic situation -- you would size an equalizer to capture some fraction (

I appreciate your detail.

As you discussed, h is a  sampled version of the impulse response function. Is the concept similar to a way that a continuous-time signal is represented as discrete signal by sampling process?

Is signal symbol energy Es and abs(1+1i) ( the distance from QPSK symbol to origin in the constellation diagram) connected ?

I tried to follow your explains and finished a matlab code (attached file). Would you please check it ?

I will spend some time to study.

Thanks all people's help.Thank you so much.

(1) The concept of a sampled h is similar to, but different from, the concept of a sampled continuous-time signal. For the latter, one relies on band-limiting and the Nyquist sampling theorem to say that two samples per modulation period is sufficient for the sampled signal to represent all of the information contained in the continuous signal.

(2) But the CIRF is a band-unlimited process in the time of arrival or delay domain and a stationary, correlated process in the frequency domain. That is, one can sample it as finely as one wants in the delay domain and each of the samples are uncorrelated. The number of samples per modulation symbol depends on the application -- e.g, demodulation may have a different requirement than that required for a delay tracking loop for example. A rule of thumb is that at least 4 samples per demodulation period are required for delay tracking, but if you assume perfect delay tracking then as few as one sample may be required for demodulation, as you have assumed. A few of my reports/articles listed in ResearchGate address this issue in detail.

(3) Ideal QPSK demodulation performance, i.e., performance not affected by quantization or tracking loop errors for example, is determined only by the SNR, not Es or N0 separately. By setting Es to 1 and N0 below Es by the SNR, we are tacitly assuming that an automatic gain control (AGC) circuit that does this scaling. Yes, the distance of the modulation voltage in the I-Q plan is connected to Es, but demodulation performance depends on that distance compared to the RMS distance of the noise samples alone.

(4) I have given you enough information to simulate this process in a physically realistic manner. It is now your responsibility to ensure your code is correct.