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Alamouti STBC

by Krishna Sankar on October 16, 2008

In the recent past, we have discussed three receive diversity schemes – Selection combining, Equal Gain Combining and Maximal Ratio Combining. All the three approaches used the antenna array at the receiver to improve the demodulation performance, albeit with different levels of complexity. Time to move on to a transmit diversity scheme where the information is spread across multiple antennas at the transmitter. In this post, lets discuss a popular transmit diversity scheme called Alamouti Space Time Block Coding (STBC). For the discussion, we will assume that the channel is a flat fading Rayleigh multipath channel and the modulation is BPSK.

Alamouti STBC

A simple Space Time Code, suggested by Mr. Siavash M Alamouti in his landmark October 1998 paper – A Simple Transmit Diversity Technique for Wireless Communication, offers a simple method for achieving spatial diversity with two transmit antennas. The scheme is as follows:

1. Consider that we have a transmission sequence, for example $\{x_1, x_2, x_3, \ldots, x_n \}$

2. In normal transmission, we will be sending $x_1$ in the first time slot, $x_2$ in the second time slot, $x_3$ and so on.

3. However, Alamouti suggested that we group the symbols into groups of two. In the first time slot, send $x_1$ and $x_2$ from the first and second antenna. In second time slot send $-x_2^*$ and $x_1^*$ from the first and second antenna. In the third time slot send $x_3$ and $x_4$ from the first and second antenna.In fourth time slot, send $-x_4^*$ and $x_3^*$ from the first and second antenna and so on.

4. Notice that though we are grouping two symbols, we still need two time slots to send two symbols. Hence, there is no change in the data rate.

5. This forms the simple explanation of the transmission scheme with Alamouti Space Time Block coding.

Figure: 2-Transmit, 1-Receive Alamouti STBC coding

Other Assumptions

1. The channel is flat fading – In simple terms, it means that the multipath channel has only one tap. So, the convolution operation reduces to a simple multiplication. For a more rigorous discussion on flat fading and frequency selective fading, may I urge you to review Chapter 15.3 Signal Time-Spreading from [DIGITAL COMMUNICATIONS: SKLAR]

2. The channel experience by each transmit antenna is independent from the channel experienced by other transmit antennas.

3. For the $i^{th}$ transmit antenna, each transmitted symbol gets multiplied by a randomly varying complex number $h_i$ . As the channel under consideration is a Rayleigh channel, the real and imaginary parts of $h_i$ are Gaussian distributed having mean $\mu_{h_i}=0$ and variance $\sigma^2_{h_i}=\frac{1}{2}$ .

4. The channel experienced between each transmit to the receive antenna is randomly varying in time. However, the channel is assumed to remain constant over two time slots.

5. On the receive antenna, the noise $n$ has the Gaussian probability density function with

$p(n) = \frac{1}{\sqrt{2\pi\sigma^2}}e^{\frac{-(n-\mu)^2}{2\sigma^2}$ with $\mu=0$ and $\sigma^2 = \frac{N_0}{2}$ .

7. The channel $h_i$ is known at the receiver.

Receiver with Alamouti STBC

In the first time slot, the received signal is,

$y_1 =h_1x_1 + h_2x_2 + n_1 = [h_1\ h_2] \left[\begin{eqnarray}x_1 \\ x_2 \end{eqnarray}\right]+n_1$ .

In the second time slot, the received signal is,

$y_2 =-h_1x_2^* + h_2x_1^* + n_2 = [h_1\ h_2] \left[\begin{eqnarray}-x_2^* \\ x_1^*\end{eqnarray}\right]+n_2$ .

where

$y_1$ , $y_2$ is the received symbol on the first and second time slot respectively,
$h_1$ is the channel from $1^{st}$ transmit antenna to receive antenna,
$h_2$ is the channel from $2^{nd}$ transmit antenna to receive antenna,
$x_1$ , $x_2$ are the transmitted symbols and
$n_1,\ n_2$ is the noise on $1^{st}, 2^{nd}$ time slots.

Since the two noise terms are independent and identically distributed,

$E\left{\left[\begin{eqnarray}n_1\\n_2^*\end{eqnarray}\right]\left[\begin{eqnarray}n_1^*\ n_2\end{eqnarray}\right]\right}=\left[\begin{eqnarray}|n_1|^2\ \ \ \ 0 \\0\ \ \ \ |n_2|^2\end{eqnarray}\right]$ .

For convenience, the above equation can be represented in matrix notation as follows:

$\left[\begin{eqnarray}y_1 \\ y_2^*\end{eqnarray}\right] = \underbrace{\left[\begin{eqnarray}\ h_1\ \ \ h_2 \\ h_2^*\ -h_1^*\end{enarray}\right]}\left[\begin{eqnarray}x_1 \\ x_2 \end{eqnarray}\right]+\left[\begin{eqnarray}n_1\\n_2^* \end{eqnarray}\right]$ .

Let us define $\mathbf{H}= \left[\begin{eqnarray}\ h_1\ \ \ h_2 \\ h_2^*\ -h_1^*\end{enarray}\right]$ . To solve for $\left[\begin{eqnarray}x_1 \\ x_2 \end{eqnarray}\right]$ , we know that we need to find the inverse of $\mathbf{H}$ .

We know, for a general m x n matrix, the pseudo inverse is defined as,

$\mathbf{H^+}=(H^HH)^{-1}H^H$ .

The term,

$(H^HH) = \left[\begin{eqnarray}\ h_1^*\ \ \ h_2 \\ h_2^*\ -h_1\end{eqnarray}\right]\left[\begin{eqnarray}\ h_1\ \ \ h_2 \\ h_2^*\ -h_1^*\end{eqnarray}\right]=\left[\begin{eqnarray}|h_1|^2+|h_2|^2\ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \\ 0\ \ \ \ \ \ \ \ \ \ \ \ \|h_1|^2+|h_2|^2\end{eqnarray}\right]$ . Since this is a digonal matrix, the inverse is just the inverse of the diagonal elements, i.e

$(H^HH)^{-1} = \left[\begin{eqnarray}\frac{1}{|h_1|^2+|h_2|^2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \\ 0\ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{\|h_1|^2+|h_2|^2}\end{eqnarray}\right]$ .

The estimate of the transmitted symbol is,

$\begin{eqnarray}\hat{\left[\begin{eqnarray}x_1 \\ x_2\end{eqnarray}\right]} & = &(H^HH)^{-1}H^H\left[\begin{eqnarray}y_1 \\ y_2^* \end{eqnarray}\right]\\<br /> & = & (H^HH)^{-1}H^H\left(H\left[\begin{eqnarray}x_1 \\ x_2 \end{eqnarray}\right]+\left[\begin{eqnarray}n_1\\n_2^* \end{eqnarray}\right]\right)<br /> \\&=&\left[\begin{eqnarray}x_1 \\ x_2 \end{eqnarray}\right] + (H^HH)^{-1}H^H\left[\begin{eqnarray}n_1\\n_2^* \end{eqnarray}\right]\\\end{eqnarray}$ .

If you compare the above equation with the estimated symbol following equalization in Maximal Ratio Combining, you can see that the equations are identical.

BER with Almouti STBC

Since the estimate of the transmitted symbol with the Alamouti STBC scheme is identical to that obtained from MRC, the BER with above described Alamouti scheme should be same as that for MRC. However, there is a small catch.

With Alamouti STBC, we are transmitting from two antenna’s. Hence the total transmit power in the Alamouti scheme is twice that of that used in MRC. To make the comparison fair, we need to make the total trannsmit power from two antennas in STBC case to be equal to that of power transmitted from a single antenna in the MRC case. With this scaling, we can see that BER performance of 2Tx, 1Rx Alamouti STBC case has a roughly 3dB poorer performance that 1Tx, 2Rx MRC case.

From the post on Maximal Ratio Combining, the bit error rate for BPSK modulation in Rayleigh channel with 1 transmit, 2 receive case is,

$P_{e,MRC} =p_{MRC}^2\left[1+2(1-p_{MRC})\right]$ , where

$p_{MRC}=\frac{1}{2}-\frac{1}{2}\left(1+\frac{1}{E_b/N_0}\right)^{-1/2}$ .

With Alamouti 2 transmit antenna, 1 receive antenna STBC case,

$p_{STBC}=\frac{1}{2}-\frac{1}{2}\left(1+\frac{2}{E_b/N_0}\right)^{-1/2}$ and Bit Error Rate is

$P_{e,STBC} =p_{STBC}^2\left[1+2(1-p_{STBC})\right]$ .

Key points

The fact that $(H^HH)$ is a diagonal matrix ensured the following:
1. There is no cross talk between $x_1$ , $x_2$ after the equalizer.

2. The noise term is still white.

$E\left{H^H\left[\begin{eqnarray}n_1\\n_2^*\end{eqnarray}\right]\left[\begin{eqnarray}n_1^*\ n_2\end{eqnarray}\right]H\right}=H^H\left[\begin{eqnarray}|n_1|^2\ \ \ \ 0 \\0\ \ \ \ |n_2|^2\end{eqnarray}\right]H=\left[\begin{eqnarray}|n_1|^2\ \ \ \ 0 \\0\ \ \ \ |n_2|^2\end{eqnarray}\right]\left[\begin{eqnarray}{|h_1|^2+|h_2|^2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \\ 0\ \ \ \ \ \ \ \ \ \ \ \ {\|h_1|^2+|h_2|^2}\end{eqnarray}\right]$ .

Simulation Model

The Matlab/Octave script performs the following

(a) Generate random binary sequence of +1’s and -1’s.

(b) Group them into pair of two symbols

(d) Equalize the received symbols

(e) Perform hard decision decoding and count the bit errors

(f) Repeat for multiple values of $\frac{E_b}{N_0}$ and plot the simulation and theoretical results.

Click here to download Matlab/Octave script for simulating BER for 2 transmit, 1 receive Alamouti STBC coding for BPSK modulation in Rayleigh fading channel

Figure: BER plot for BPSK in Rayleigh channel with 2 Transmit and 1 Receive Alamouti STBC

Observations

Compared to the BER plot for nTx=1, nRx=2 Maximal ratio combining, we can see the Alamouti Space Time Block Coding has around 3dB poorer performance.

Reference

[DIG-COMM-BARRY-LEE-MESSERSCHMITT] Digital Communication: Third Edition, by John R. Barry, Edward A. Lee, David G. Messerschmitt

A Simple Transmit Diversity Technique for Wireless Communication Siavash M Alamouti, IEEE Journal on selected areas in Communication, Vol 16, No, 8, October 1998

Tagged as: Alamouti, BPSK, Rayleigh, STBC

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{ 1 trackback }

MIMO with Zero Forcing equalizer: October 24, 2008 at 5:53 am

{ 130 comments… read them below or add one }

1 Alvina October 17, 2008 at 1:54 pm

well explained.

2 Krishna Pillai October 17, 2008 at 10:18 pm

@Avlina: Thanks, glad it helped.

3 Honghai October 17, 2008 at 10:55 pm

Excellent! Hope you can put more posts on MIMO.

4 Mercy October 19, 2008 at 7:13 pm

thanks for t heinfo… I cant for posts on MIMO.

5 Mercy October 19, 2008 at 7:16 pm

thanks for the info…. It has been quite helpful. I would like information pertaining to MIMO and associated channel capacity

6 Krishna Pillai October 24, 2008 at 6:45 pm

@Honghai , @Mercy: Hope the recent post on 2×2 MIMO with Zero forcing equalisation (for BPSK in Rayleigh channel) helps.
URI: http://www.dsplog.com/2008/10/24/mimo-zero-forcing/

7 ila October 29, 2008 at 9:19 pm

hi krishna,
need help to explain:

hEq(:,[2:2:end]) = kron(ones(1,N/2),[1;-1]).*flipud(reshape(h,2,N/2)); % [h1 h2 ... ; h2 -h1 ...]
hEq(1,:) = conj(hEq(1,:)); % [h1* h2* ... ; h2 -h1 .... ]

why is it not the same as in the derivation??
H= [h1 h2; h2* -h1*]

8 ila October 29, 2008 at 9:54 pm

hi krishna.
sorry. pls ignore my earlier question. i think there is typo mistake in the notes:

should be:
H* = [h1* h2* ... ; h2 -h1* .... ]

in notes written:
H* = [h1* h2 ...; h2* -h1]

btw, i tried to implement this for QPSK (modify from your entry on BER QPSK in Rayleigh – but seems the curve is very weird.. forever error rate ~0.5). is there a different equalization method?

btw, thanks for all the update! they’re all very helpful. without this, i dont think i can start with my assignment at all

9 Krishna Pillai October 31, 2008 at 6:02 am

@ila: Can you please point to the line in the code where the error in the note is present?
Also, kindly do note that in the simulation model, to speed up simulation time, I have tried to use matrix operations as far as possible. For doing so, I may have played around with matrix dimensions (which might not be very intuitive). Hopefully the comments help.

Similar equalization should hold good for QPSK too. Try for a no-noise case, see whether you can recover the QPSK symbols accurately.

10 Alvina November 3, 2008 at 9:05 pm

Hi krishna

can u tell plz why you h

11 Alvina November 3, 2008 at 9:16 pm

hi Krishna

can you plz tell me why you have taken the conjugate of estimated received signal X. (after this sentences)
The estimate of the transmitted symbol is,

12 Alvina November 4, 2008 at 12:55 am

can u plz tell me why you have taken tne conjugate (* sign) over noise symbol n2 and n1 (after the sentences)
Since the two noise terms are independent and identically distributed,

13 Krishna Pillai November 5, 2008 at 5:43 am

@alvina: If you look at the system model equation, we took the conjugate of the second equation to enable us conveniently represent both the equation in matrix form.

So, when we solve for that equation, what we estimate is the conjugate of what we have transmitted.

For the noise term, we are finding the average noise power, which is the the E{n*n^H}.

14 farie November 19, 2008 at 5:45 pm

hey Krishna

Do you have Alamouti 2×1, 2×2 STBC in 4QAM and 16QAM codes? I am having problems in combining the modulated 4qam and 16qam signals with channels.

15 Krishna Pillai November 21, 2008 at 5:46 am

@farie: No, I do not. However, may I suggest that debugging 4QAM, 16QAM cases should be reasonably easy. I think as a first step, you can try with
(a) Make all the channel coefficients unity
(b) No noise.

Once you are able to receive perfectly what was transmitted, then move on to add channel and noise.

Hope this helps.

16 Paul January 27, 2009 at 10:15 am

Hi Krishna,

Could you explain what E{.} expectation operater for me please. What is it for? why do we have to find E{.}

Thank you

17 Krishna Pillai January 28, 2009 at 6:01 am: @Paul: The E{.} operator find the mean of a set of observations. The E{.} operator is used to show that mean of |n1}^2 corresponds to noise variance. The mean of n1*n2 = 0 as the noise is uncorrelated.

Reply

18 Paul January 28, 2009 at 8:01 pm

Thank you so much Krishna

19 lia January 29, 2009 at 3:33 pm

hi,

i was informed that equalization like ZF and MMSE is only used when you expect the received signal is not orthogonal.

so, in the case of for Alamouti 2×2 MIMO (orthogonal), how do I modify the equalization matrix into 2 receiver for BPSK?
what’s the method to expand this to QPSK&QAM?

e.g. in the given Matlab code:
rx_Mod = kron(reshape(rx,2,N/2),ones(1,2));
rx_Mod(2,:) = conj(rx_Mod(2,:));
% forming the equalization matrix
hEq = zeros(2,N);
hEq(:,[1:2:end])=reshape(h,2,N/2);
hEq(:,[2:2:end]) = kron(ones(1,N/2),[1;-1]).*flipud(reshape(h,2,N/2));
hEq(1,:) = conj(hEq(1,:)); hEqPower = sum(hEq.*conj(hEq),1);

please help

20 Krishna Pillai January 30, 2009 at 5:52 am: @lia: Yes, when there is cross-coupling ZF and MMSE wont be optimal.
However, in the presence of Alamouti coding, the coding structure ensures that there is no cross-coupling. So ZF is optimal even in 2Tx 1Rx Alamouti case.

I would think that, to extend to 2Tx, 2Rx case one needs to write the received symbol equations for the second antenna in addition. The information from multiple rx antennas can be combined the MRC way (i think). I need to write the equations to confirm. Can you please try that. Please share your results.

Good luck.

Reply

21 Rocco January 30, 2009 at 3:25 pm

Hello Krishna,

your explanation is very clear and really helped me a lot.

Thank you

22 Krishna Pillai January 31, 2009 at 6:40 am: @Rocco: Thanks

Reply

23 giri February 5, 2009 at 9:06 pm

Can you explain what is IRC. In LTE there is mention of IRC.

24 Krishna Pillai February 10, 2009 at 7:48 pm: @giri: Sorry. Studying LTE spec is in my to-do list for a while. Hope to do it soon.

Reply

25 Suchitra February 9, 2009 at 1:02 pm

hi krishna…
please suggest me code for Alamouti with OFDM…

26 Krishna Pillai February 10, 2009 at 8:09 pm: @Suchitra: Please refer to the post on
http://www.dsplog.com/2008/10/16/alamouti-stbc/

Reply
27 Krishna Pillai February 10, 2009 at 8:09 pm: @Suchitra: Please refer to the post on
http://www.dsplog.com/2008/10/16/alamouti-stbc/

Reply

28 lia February 10, 2009 at 5:30 am

Hi Krishna & others,

I’m afraid I’m still struggling with Alamouti 2×2.

Can someone pls help to show/point where I’m doing wrong in my current simulation (getting extremely high BER):
1) input data modified to Alamouti scheme
2) generate 4 paths (divided into 2 sets to ensure the channel remain constant for 2 timeslot at least at each antenna)
3) 2 independent AWGN
4a) combine path1&2 with signal & add noise 1.
4b) combine path 3&4 with signal & add noise2.
5) perform 2 separate channel equalization
to get 2 estimated transmitted data
6) perform MRC based on original channel in (2)
7) count error

please help.. running out of ideas

29 Jose C February 11, 2009 at 12:09 am: First you should try to duplicate the process (2×1) of equalization for the second receiving antenna and retrieving data using only the second antenna, you should obtain the same BER.

Then you can use MRC to obtain BER of 2×2 correctly, note that as stated before the equations for MRC and Alamouti equalization are the same so when after equalizating Alamouti at the second (or duplicated process) antenna applying MRC is reduced to add both of “yHat”

Reply

30 Arun February 12, 2009 at 11:14 am

hi..can we combine OFDM with Alamouti forming STF code..i am not getting pls can u suggest code if possible

31 Krishna Pillai February 19, 2009 at 6:01 am: @Yes we can. However, I have not written posts on it. Just picked up a URL from googling
http://www.mk.tu-berlin.de/mitarbeiter/tub/lehrbeauftragte/intro_stc-sfc-coding

Reply

32 Murali February 27, 2009 at 9:39 am

Hai..i am doing my M.Tech project on COSTBC can u send matlab code for flat rayleigh fading channel with QPSK modulation or suggest some useful sites regarding my project

33 Krishna Pillai March 5, 2009 at 4:18 am: @Murali: Hmm… I have not worked on COSTBC.
For BPSK modulation in flat Rayleigh channel, you may look up @
http://www.dsplog.com/2008/08/10/ber-bpsk-rayleigh-channel/
For QPSK in AWGN, you may look up @
http://www.dsplog.com/2007/11/06/symbol-error-rate-for-4-qam/

Hope this helps.

Reply

34 Murali February 27, 2009 at 9:42 am

Hai krishna sir can u please send matlab programm for QOSTBC based on flat rayleigh fading channel with QPSK modulation

35 Krishna Pillai March 5, 2009 at 4:18 am: @Murali: Sorry, I have not worked on QOSTBC

Reply

36 DRAM March 12, 2009 at 10:44 pm

Krishna: At first, good post.
Now, few questions: You wrote “For convineance, the above equation can be represented in matrix notation as follows:”..and this is followed by a matrix notation of the system response. I find that in such a matrix, the functionality of X* has been moved to H*. Although we are allowed to do such operations freely, in practise it seems to suggest that the channel reposnse during the second time slot is the complext conjugate of its own response in the first time slot – right? Don’t you think over a short interval of time such a generalization is rather too extreme?
In my view, transmitting X* during second time slot is going to be totally independant of channel reposnse during its transmission. So, convertibility between X* to H* is not going to be simple over such short time periods. May be over the long it’s possible…

Please feel free to critiqe my understanding

Thanks for you time,
DRAM

37 Krishna Pillai March 21, 2009 at 7:26 am: @DRAM: For STBC, we assume that the channel is same over two time slots. Given that we make the assumption, its just for mathematical notional convineance, we formulate the channel in matrix form by using H*. So, the fact that we are using H* in matrix notation does not mean that the channel H in first time slot became H* in the second time slot. Hope this helps.

Reply

38 SHMAA April 1, 2009 at 11:12 am

great job krishna i hope you continue you’r
devolopment of alamouti……..

39 Krishna Pillai April 4, 2009 at 4:43 pm: @SHMAA: Thanks. I wrote a follow up post on 2 transmit 2 receive alamouti STBC @
http://www.dsplog.com/2009/03/15/alamouti-stbc-2-receive-antenna/

Reply

40 hannah April 6, 2009 at 5:51 pm

hi krishna,
Thanks for your excellent work.
I am doing a work on MIMO-OFDM with Alamouti 2×1 code.Would you like to send me a matlab code for flat rayleigh fading channel with QPSK modulation?Furthermore,would you model it using the simulink toolbox and send me a *.mdl file?
Hoping your replay.
Thanks very much.

41 Krishna Pillai April 11, 2009 at 6:17 am

@hannah: The Matlab model provided in this post can be easily adapted to QPSK case. Good luck in your project.

42 abubaker April 11, 2009 at 5:33 pm

hi krishna..do we need to divide the snr by 2 in the equation of bpsk to get qpsk curve in theoretical case or is it the same..or is it like only for high snr case only..im bit confused abot that can u pls explain it

43 Krishna Pillai April 16, 2009 at 5:22 am: @abubaker: Well, the symbol error rate for QPSK is around 3dB poorer than that for BPSK. However the BER performance is the same. Please refer to the following post for bit more details
http://www.dsplog.com/2008/07/08/compare-bpsk-qpsk-4pam-16qam-16psk-64qam-32psk/

Reply

44 abubaker April 7, 2009 at 4:58 pm

hi..can u tel me what is the equation for probablity of error for correlated raylegh flat fading channel (with correlation factor )for mrc (1*2) and alamouti (2*1) using qpsk ..

45 Krishna Pillai April 11, 2009 at 6:35 am

@abubaker: Sorry, I have not studied the effect of channel correlation in MRC and alamouti. From a quick googling, found that the link http://www.comm.utoronto.ca/~rsadve/Notes/DiversityReceive.pdf discuss about correlation for MRC.

Please also do reply to the comment, if you find more pertinent literature.

46 abubaker April 11, 2009 at 5:30 pm

hi krishna thanks for your reply..this is link..http://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=04109656..ieee paper with title “Symbol Error Probability for Space–Time Block Codes over Spatially Correlated Rayleigh Fading Channels” thsis gives the equations for alamouti scheme for all psks annd qams…

47 Krishna Pillai April 12, 2009 at 7:02 am

@abubaker: Thanks

48 abubaker April 12, 2009 at 4:41 pm: one more thing the equation u have given the above link its wrong..they have small printing mistake so its better to look in this text book .R. Janaswamy, Radiowave Propagation and Smart Antennas for Wireless Communications.
Kluwer Academic Publishers, 2000.

49 gly April 8, 2009 at 1:41 pm

Hi,Krishna Pillai,
would you help me Model the STBC-OFDM or DSTBC-OFDM system using the simulink toobox? Email:115473189@qq.com
Thank you!

50 Krishna Pillai April 11, 2009 at 7:02 am: @gly: Sorry, I do not have Simulink

Reply

51 UP April 11, 2009 at 3:59 am

Hi, Instead of hard decision coding, can u actually implement a Maximum likehood decoder for M-PSK encoding ? How do you think could that be implemented ?

52 Krishna Pillai April 11, 2009 at 7:54 am: @UP: Well, the brute force ML method is to find the minimum of error between received transmit constellation and receieved constellation. However, I think there is no advantge in doing the brute force ML as the hard decision decoding is also optimal, agree?

Reply

53 UP April 12, 2009 at 5:26 am

I was implementing the code with QPSK. Would everything other than the encoding (given below) be the same ?

ip = (2*(rand(1,N)>0.5)-1) + j*(2*(rand(1,N)>0.5)-1);

The decoder would also be modified accordingly.

I am not getting any gain over 1TX using QPSK. Pl. help! Thanks!

54 UP April 12, 2009 at 9:10 am

Hi ,

I’m here with one more doubt. I have a project on transmitter diversity. Can the alamouti scheme be applied to the UPLINK for CDMA systems– (where noise interference would be due to multiple users) ? Or it has to be at the base station ? I am having this conceptual doubt …

Sorry to have asked so many questions at one go..
Would be really grateful if you could help! Thanks!

55 Krishna Pillai April 16, 2009 at 5:33 am: @UP: Well, theoretically there is nothing stopping us from using Alamouti STBC in either uplink or downlink. In the past, I recall reading that Alamouti coding is done at the Base station (which has more computational power and more transmit atennas) and the subcriber stations (have the capability for doing demodulation).

Reply

56 Krishna Pillai April 16, 2009 at 5:29 am

@UP: Well, thats surprising. Are you scaling the transmit power accordingly?

57 UP April 17, 2009 at 4:18 am

Hi , Could you please tell me by what factor should I scale the transmit power ?

58 Krishna Pillai April 18, 2009 at 8:09 am: @UP: If you have two transmit antennas, it should be ensured that the sum power of the signals coming from both antennas should be equal to transmit power in the case of single antenna.

Reply

59 abubaker April 12, 2009 at 5:01 pm

hi krishna..do we need to divide the snr by 2 in the equation of bpsk to get qpsk curve in theoretical case or is it the same..or is it like only for high snr case only..for finding the probablity of error for diversity schemes ..im bit confused about that can u pls tel that

60 sara April 12, 2009 at 10:15 pm

hi krishna!!
i am working on cooperative communication systems…… can u help me in this regard???? if u have simulated cooperative mimo enviornment kindly info me.

the point that confuses me is that how we decide at the destination node to use which type of equalizer i-e LMS, DFE etc etc.
secondly do we combine signals before equalization or after that

regards

61 Krishna Pillai April 16, 2009 at 5:37 am: @sara: Sorry, I have not worked on co-operative communications. Anyhow, trying to answer the queries:
1/. Typically, the type of equalizer which is selected depends on the complexity considerations and the performance gain
2/, I think equalizer takes care of the job to effectively combine the signal resulting in minimal BER.

Reply

62 R@juelo April 16, 2009 at 7:56 pm

I already read the receive diversity blog, and I need to simulate alamouti 2X1 with channel correlation, how can I implement in the code because I only find the theoretical formula, I hope you can help me with this issue.
Regards

63 Krishna Pillai April 18, 2009 at 8:14 am: @R@juelo: Sorry, I have not tried modeling the channel correlation case.

Reply

64 UP April 26, 2009 at 12:43 am

@ Krishna and @Abubaker.

Can u please tell me how to modify alamouti to QPSK. I am gettin simulated BER 6 db higher than BPSK case. and also confused about theoritical values. Pl help!!!

65 Krishna Pillai April 30, 2009 at 5:21 am: @UP: Recalling the results from QPSK results in AWGN, the BER vs Eb/N0 curve is identical for both QPSK and BPSK. So, I do not see the reason for 6dB difference.

Reply

66 gly May 17, 2009 at 5:56 pm

Hi,Krishna,would you help me coding a program about Maximum Likelihood Detection for Alamouti-STBC system? Under QPSK modulation condition,My program didn’t get perfect result,So I’d like to see your help.Thanks!

67 Krishna Pillai May 20, 2009 at 5:32 am: @gly: Sorry, due to time considerations, I would not have time to help you in the coding for QPSK systems. However you may use the post on Symbol Error Rate for QPSK in AWGN as a reference
http://www.dsplog.com/2007/11/06/symbol-error-rate-for-4-qam/

I would think that the Zero forcing equalizer discussed in this post is optimal (and gives the same performance as ML) for Alamouti STBC system. Hence, I think it should be reasonably simple to extend this post to QPSK systems.

Reply

68 SAEED May 18, 2009 at 10:50 am

plz krishna;
give me another one matlab code to find ber/snr for 1tx-1rx & 2tx-1rx(alamouti)

69 Krishna Pillai May 20, 2009 at 5:43 am: @SAEED: The 1 transmit 1 receive case is discussed in the post
http://www.dsplog.com/2008/08/10/ber-bpsk-rayleigh-channel/

Reply

70 fof May 20, 2009 at 5:08 pm

plz help me…..i need a code mor alamouti stbc for BPSK (likelihood detection)
using only for loops or while loops…..without using reshape,rand functions……so can you help me please??
thannx alot…

71 fof May 22, 2009 at 1:19 am

can you help me where’s my error??

N = 10^6; % number of bits or symbols
Eb_N0_dB = [0:25]; % multiple Eb/N0 values

for ii = 1:length(Eb_N0_dB)

% Transmitter
ip = rand(1,N)>0.5; % generating 0,1 with equal probability
s = 2*ip-1; % BPSK modulation 0 -> -1; 1 -> 0

h = 1/sqrt(2)*[randn(1,2) + j*randn(1,2)]; % Rayleigh channel

n = 1/sqrt(2)*[randn(1,2) + j*randn(1,2)]; % white gaussian noise, 0dB variance

S_c = [s(1) -conj(s(2)); s(2) conj(s(1))];
% Channel and noise Noise addition
y = sum(hMod.*S_c,1) + 10^(-Eb_N0_dB(ii)/20)*n;

S_S(1) = conj(h(1))*y(1)+h(2)*conj(y(2));
S_S(2) =conj(h(2))*y1-h(1)*conj(y(2));
%Reciever
mpower=h*conj(h); %equalization
H1 = S_S(1)/mpower;
H2 = S_S(2)/mpower;
decoded_S=real(y)>0;
% counting the errors
nErr(ii) = size(find([ip- decoded_S]),2);
PER=nErr(ii)/Eb_N0_dB;
plot(PER,N);
hold

end
simBer = nErr/N; % simulated ber
EbN0Lin = 10.^(Eb_N0_dB/10);
theoryBer_nRx1 = 0.5.*(1-1*(1+1./EbN0Lin).^(-0.5));

p = 1/2 – 1/2*(1+1./EbN0Lin).^(-1/2);
theoryBerMRC_nRx2 = p.^2.*(1+2*(1-p));

pAlamouti = 1/2 – 1/2*(1+2./EbN0Lin).^(-1/2);
theoryBerAlamouti_nTx2_nRx1 = pAlamouti.^2.*(1+2*(1-pAlamouti));

72 Krishna Pillai May 22, 2009 at 5:30 am

@fof: Sorry. But, what is the error which you are observing?

73 fof May 22, 2009 at 1:06 pm: y = sum(hMod.*S_c,1) + 10^(-Eb_N0_dB(ii)/20)*n;
i also tried…
y=h*s+10^(-Eb_N0_dB);
but there’s also error….

Reply

74 Phan Minh Hoang May 23, 2009 at 6:12 am

Hi Krishna Pillai

first of all thank you for your post. it’s clearly and understandable

I’ve just studied a bit about transmit diversity scheme of Mr Alamouti. I just have some general questions.
1. Can you tell me why Alamouti scheme is introduced? in fact it is 3dB poorer than MRC.
2. If we use this scheme we’ll need double power in order to get the similiar BER curve like MRC. I just wonder why?
3. Could you suggest me a recevier which is with lower complexity than the MLSE

I know that i don’t have much time but hope you enjoy answering and discussing some questions. thanks in advance

Phan

75 Krishna Pillai May 31, 2009 at 8:06 pm: @Phan Minh Hoang: My replies
1. Though it is 3dB poorer than MRC, Alamouti scheme provided a way where we can put the complexity of two transmit antennas at the base station side. If you recall, when Alamouti scheme came about circa 1998, there were not too many takers where a mobile device can have multiple receive antennas

2. Thats because the math plays out that way.

3. For MIMO communications, we can use equalizers like ZF, MMSE, successive interference cancellation etc. Please check http://www.dsplog.com/tag/mimo

Hope this helps.

Reply

76 chadi June 20, 2009 at 4:07 pm

dear
We haveMultiple InputMultiple Output (MIMO) digital communication system consisting of two transmit
antennas (NT = 2) and one receive antenna (NR = 1).
Alamouti space-time (ST) block code is defined by folowing NT × L code matrix
X = x1 −x
2
x2 x
1
where L is a codeword length.
• implement a digital modulator involving Alamouti STBC
• modulated signal passes frequency-flatMIMO fading channel with AWGN
• construct a decoder for given communication system
• compare the performance of given system with the reference (scalar, SISO) system — display BER
(or SER) curves in the same figure
thanks, chadi_lb@hotmail.com

77 Krishna Pillai June 21, 2009 at 12:58 pm: @chadi: Good luck for your assignment.

Reply

78 pofi July 2, 2009 at 5:49 pm

but why do we use alamoutis STBC? what ist actually the advantage of if?

79 Khattak July 6, 2009 at 8:43 pm

Exellent theory and simulation i found kit useful.

80 Krishna Pillai July 15, 2009 at 4:36 am: @Khattak: Glad to hear.

Reply

81 mohanad July 21, 2009 at 9:24 pm

Hi krishna
I have seen your code and it was great. But I want to ask u about the simulation of MIMO channel in three dimensions …. The vertical coordinate is the fading envelope, and the two horizontal coordinates are the time and distance…. I have simulated the fading with the time by using jakes mode but I don’t know how i can simulate the fading with distance for 2*2 real world MIMO channel. At least give me an equation of fading with distance.
Thank u in advance…
This is my fading simulation with time:-
clear
t_sample = 0.001;
t = (0:8/t_sample)*t_sample;
fc = 220.5625*10^6;
N = 100;
c = 2.998*10^8;
v = 45*1000/3600;
lambda = c/fc;
fd = v/lambda;
f = linspace(-1/t_sample*1/2,1/t_sample*1/2,16000);
Eo = 1;
sigma = 1;
Cn = 1/(2*pi);
alpha = rand(1,N).*2*pi;
phi = rand(1,N).*2*pi;
Tc = zeros(1,length(t));
Ts = zeros(1,length(t));
for n = 1:N; % loop to superimpose the N signals
Tc_n = Cn*cos(fd*2*pi*t*cos(alpha(n))+phi(n));
Ts_n = Cn*sin(fd*2*pi*t*cos(alpha(n))+phi(n));
Tc = Tc+Tc_n;
Ts = Ts+Ts_n;
end
Env = (Tc.^2 + Ts.^2).^0.5;
plot(t(1:4/t_sample),10*log10(Env(1:4/t_sample)))
title(’Received Signal Envelope’);
xlabel(’time (s)’)
ylabel(’Magnitude of E Field (dB)’)

82 Krishna Pillai July 24, 2009 at 4:08 am: @mohanad: Am not sure, I fully understood your question nor the code which you have posted. Typically, more the distance, higher will be the RMS delay spread of the channel.
One good reference for MIMO channel model is defined by the High Throughput study group for 802.11n standards development –
Tgn Channel Models, Vinko Erceg et al. The document provides a good overview of MIMO channel modeling – including the effect of antenna correlation (based on antenna spacing). effect of fluorescent lights, doppler, indoor multipath characteristics.

Hope this helps.

Reply

83 mohanad July 24, 2009 at 5:38 pm

Thank u krishna for your help…..
If my viewpoint didn’t clear, can u please read the pages (6-7) of this book (MIMO Wireless Communications.2007) in this web site :-www.4shared.com/file/112621705/f9ab90f6/MIMO_Wireless_Communications2007.html
It will give u a clear overview of what I had pointed to
Thank u again dear Krishna

84 Krishna Pillai July 28, 2009 at 4:28 am: @mohanad: I looked at Fig 1.3 in the reference which you pointed. As you said,
a) the vertical co-ordinate ‘dB’ corresponds to attenuation of the channel.
b) the horizontal co-ordinate ‘tones’ corresponds to frequency i.e attenuation vs frequency and
c) the horizontal co-ordinate ’samples’ corresponds to many realization of the channel i.e attenuations vs time

Reply

85 Mahdi July 29, 2009 at 6:18 pm

Dear Krishna
Good work…
Execuse me , I have a one question …is there a Threshold in Alamouti scheme in detection side ?
thank you

86 Krishna Sankar July 30, 2009 at 5:39 am: @Mahdi: Did you mean threshold for determining whether the received symbol corresponds to 1 or 0? Yes, there is.

Reply

87 Umesh Bhaskar August 5, 2009 at 4:05 pm

A very useful post which dispelled my misconception – that when STBC is used , the minimum number of receive antennas should be 2. Thanks

88 Krishna Sankar August 10, 2009 at 5:47 am: @umesh: Glad to help

Reply
89 WiMAX February 17, 2010 at 3:35 am: Are you sure about that?

Reply

90 surbhi August 13, 2009 at 12:13 pm

please tell me when simulating the almouti scheme….
one matlab keyword using is “kron” that leads to kronecker tensor product. but in the literature it is mentioned that kronecker algebra is being used for almouti scheme.
please tell me what is significance of using “kron” when coding for STBC.

91 Krishna Sankar August 14, 2009 at 5:08 am: @surbhi: In the Matlab code, I used kron() to do all the operations as matrix operations and to avoid for-loops. Using kron() does not have any theoretical significance in understanding Alamouti STBC coding.

Reply

92 sreenu August 24, 2009 at 4:41 pm

can you help me please,
i am doing project based on MIMO-OFDM for reduction of peak to average power ratio. in this case i am using CI-OFDM with SFBC instead of STBC.
can you suggest any better technique than CI-OFDM, and how to write matlab code for CI-OFDM with SFBC.
thank you sir.

93 Krishna Sankar August 30, 2009 at 10:52 am

@sreenu: what is CI-OFDM?

94 sreenu September 12, 2009 at 8:31 pm

Carrier Interferometry Orthogonal Frequency Division Multiplexing
(in CI OFDM , puts each bit/symbol on all carriers simultaneously)

95 Krishna Sankar September 14, 2009 at 5:29 am

@sreenu: Whats the advantage in having such a scheme? and where is this used?

96 sreenu September 22, 2009 at 4:02 pm: this CI-OFDM scheme provide frequency diversity benefit and also reduces the peak to average power ratio(for preventing non linear distortion of High Power Amplifier).
First this(CI) scheme used at MC-CDMA. Now this(CI) used at OFDM.
this block used between SFBC encoder and OFDM block.

97 mohammad August 26, 2009 at 11:06 am

hi Krishna
My theses in MIMO in WIMMAX ,plz give me an idea for simulation.
Best regards.

98 Krishna Sankar September 7, 2009 at 4:58 am: @mohammad: Which blocks are you planning to simulate? Good luck.

Reply

99 mak_m August 29, 2009 at 4:30 am

thanks very for providing a very gud understanding of AL.STBC
krishna can u plz tel me ..
can u use the matlab builtin command: rayleighchan(ts,fd)
same for bpsk modulation command :psk.modem..will it work correctly becoz in ur all codes u have made ur own qpsk n bpsk modulators…plz help me ..n one more thing to ask if m using mimo ofdm then can i considered channel as flat fading channel as u did in it..bcoz of ofdm each subcarrier will suffer flat fading instead of frequency selective fading..plz do comment whether m right or wrong
many thanks in advance
waiting for ur reply

100 Krishna Sankar September 7, 2009 at 5:18 am: @mak_m: My replies:
1/ Sorry, I do not have the function rayleshchan, psk.modem etc. Hence I built my own.
2/ In most cases, you can consider it as flat fading for each subcarrier.

Reply

101 sam September 1, 2009 at 10:38 am

hello sir,

Till now u have worked on ALMOUTI full rate schemes(2 tx* 2 rx). will u extend this work on fractional rate(e.g. 3 tx – 4 rx or any other combination like e.g 4 tx * 3rx). if u have any idea of implementing STBC with antenna selection{ a technique which is used to select specific antenna’s (let 2& 4) out of given four antenna’s at receiver end }. if u have any matlab program for antenna selection plz send to me

102 Krishna Sankar September 7, 2009 at 5:36 am: @sam: I have discussed 2 transmit – 1 receive alamouti and 2 transmit – 2 receive alamouti codes. In future, I might consider discussing other structures which you have proposed.

I have one post on Selection diversity @
http://www.dsplog.com/2008/09/06/receiver-diversity-selection-diversity/

Reply

103 Sivam September 4, 2009 at 6:59 am

Hi sir,
I wanted to use your alamouti BPSK to change it to QPSK to verify that the results will be the same. However, i couldnt get the correct curve.

Do i need to change any of the codes in the alamouti calculation portion?
Or do i leave it as it is, and change only the ip and s variable?
Which is…..

% Alamouti STBC
sCode = zeros(2,N);
sCode(:,1:2:end) = (1/sqrt(2))*reshape(s,2,N/2); % [x1 x2 ...]
sCode(:,2:2:end) = (1/sqrt(2))*(kron(ones(1,N/2),[-1;1]).*flipud(reshape(conj(s),2,N/2))); % [-x2* x1* ....]

h = 1/sqrt(2)*[randn(1,N) + j*randn(1,N)]; % Rayleigh channel
hMod = kron(reshape(h,2,N/2),ones(1,2)); % repeating the same channel for two symbols

n = 1/sqrt(2)*[randn(1,N) + j*randn(1,N)]; % white gaussian noise, 0dB variance

% Channel and noise Noise addition
y = sum(hMod.*sCode,1) + 10^(-Eb_N0_dB(ii)/20)*n;

% Receiver
yMod = kron(reshape(y,2,N/2),ones(1,2)); % [y1 y1 ... ; y2 y2 ...]
yMod(2,:) = conj(yMod(2,:)); % [y1 y1 ... ; y2* y2*...]

% forming the equalization matrix
hEq = zeros(2,N);
hEq(:,[1:2:end]) = reshape(h,2,N/2); % [h1 0 ... ; h2 0...]
hEq(:,[2:2:end]) = kron(ones(1,N/2),[1;-1]).*flipud(reshape(h,2,N/2)); % [h1 h2 ... ; h2 -h1 ...]
hEq(1,:) = conj(hEq(1,:)); % [h1* h2* ... ; h2 -h1 .... ]
hEqPower = sum(hEq.*conj(hEq),1);

yHat = sum(hEq.*yMod,1)./hEqPower; % [h1*y1 + h2y2*, h2*y1 -h1y2*, ... ]
yHat(2:2:end) = conj(yHat(2:2:end));

104 Krishna Sankar September 9, 2009 at 5:37 am: @Sivam: I guess changing only the variable s and the receiver ipHat should suffice.

Reply

105 Sivam September 5, 2009 at 12:46 pm

How do we explain this?

pAlamouti = 1/2 – 1/2*(1+2./EbN0Lin).^(-1/2);
theoryBerAlamouti_nTx2_nRx1 = pAlamouti.^2.*(1+2*(1-pAlamouti));

Why in theory we can model alamouti using this formula??
How is it being done?

Thanks so much!

106 Krishna Sankar September 9, 2009 at 5:42 am: @Sivam: The equation is same as the theoretical equation for BER for Maximal Ratio Combining, but poorer by 3dB. You can find the post on Maximal Ratio Combining @
http://www.dsplog.com/2008/09/28/maximal-ratio-combining/

Reply

107 surender September 15, 2009 at 6:12 am

you had done this alamouti stbc under BPSK modulation……..Is it possible to implement it with 16QAM modulation…….If so how to proceed along with this model.Plz help me on this…………….

108 Krishna Sankar September 18, 2009 at 5:37 am: @surender: I have discussed couple of posts on 16QAM modulation
http://www.dsplog.com/2007/12/09/symbol-error-rate-for-16-qam/
http://www.dsplog.com/2008/06/05/16qam-bit-error-gray-mapping/
Hope you will be able to adapt those to Alamouti case.

Reply

109 Aditya October 4, 2009 at 9:54 am

Hi Krishna,

Thanks for sharing the information. The website and all its articles are very informative….

110 Krishna Sankar October 8, 2009 at 5:21 am: @Aditya: Thanks

Reply

111 balu October 4, 2009 at 2:24 pm

hi krishna,
do u have matlab code for qrd-m algorithm in mimo-ofdm systems , if so plz help me.
thank u.

112 Krishna Sankar October 8, 2009 at 5:21 am: @balu: Sorry, no.

Reply

113 Rory November 15, 2009 at 8:02 pm

The prob. dens. function of your flat fading channel looks strange. I used dfittool with the data input abs(h) to create the PDF. It looks like a rayleigh pdf with sigma = 1 but multiplied by sqrt(2). Is there any particular reason for this? I’m using jakes model for flat fading and the pdf i get is the same as;
x = 0:0.1:3;
p = raylpdf(x,1);
plot(x,p)

Thanks for the code, it’s a great help with checking my own.

114 Krishna Sankar December 3, 2009 at 5:42 am: @Rory: I have created the Rayleigh channel with gaussian distributed real and imaginary parts. Both real and imaginary parts are divided by 1/sqrt(2) such that the total variance is unity. Agree?

Reply

115 joel November 16, 2009 at 9:24 pm

Hello Krishna,
After the sentence “The estimate of the transmitted symbol is,” you multiplicate the the channel matrix inverse and the [y1 y2*] . The typo mistake in my opinion is that y2* depends on x2 and not x2* as you have written. So the estimation vector is [x1 x2] and not [x1 x2*]. Agree ? Apart from that you have forgotten the conjugation for x1 in the formula after the sentence “In the second time slot, the received signal is,”
The rest is a good work, thank you !

116 Krishna Sankar December 6, 2009 at 3:38 pm: @joel: Thanks for the close review. I corrected both the typos.

Reply

117 nikunj December 4, 2009 at 12:17 pm

hi krishna,
ur code is excllent for BPSK for 2X1 .
i have modified it for MPSK and MQAM to calculate BER.
its working for 4,8 psk but doesnt give proper curve for 16psk.
is it necessary to demodulate signa at receiver side??
i have used c=modem.pskdemod(M,(360/M)) for demodulation after ML decoding.
is it correct or not???

118 Krishna Sankar December 7, 2009 at 5:12 am: @nikunj: Well, as I do not have pskdemod() function, I cannot comment. You might want to have a look @
http://www.dsplog.com/2008/03/18/symbol-error-rate-for-16psk/

Reply

119 KimMyungsong December 14, 2009 at 12:06 pm

Dear krishna, lots of engineers are ask for the MIMO 2×1 with QPSK. And I changed your source, but the result failed. Can you help me to check the source? Thansk.
clear

Eb_N0_dB = [0:25]; % multiple Eb/N0 values

nd = 6; % number of symbols
ml = 2; % modulation levels
para=256; % Number of parallel channel to transmit (points)
N=para*nd; % number of bits or symbols

for ii = 1:length(Eb_N0_dB)

% Transmitter
% ip = rand(1,N)>0.5; % generating 0,1 with equal probability
ip = rand(1,para*nd*ml)>0.5; % generating 0,1 with equal probability

% s = 2*ip-1; % BPSK modulation 0 -> -1; 1 -> 1

% QPSK modulation**********************************************************
paradata = reshape(ip,para,nd*ml);
[ich,qch]=qpskmod(paradata,para,nd,ml);
kmod=1/sqrt(2);
s=(ich+qch.*i).*kmod;
s = reshape(s,1,para*nd);

% IFFT*********************************************************************
% s=ifft(s);

% IFFT*********************************************************************
% sCode=ifft(sCode);

h = 1/sqrt(2)*[randn(1,N) + j*randn(1,N)]; % Rayleigh channel
hMod = kron(reshape(h,2,N/2),ones(1,2)); % repeating the same channel for two symbols

n = 1/sqrt(2)*[randn(1,N) + j*randn(1,N)]; % white gaussian noise, 0dB variance

% Channel and noise Noise addition
y = sum(hMod.*sCode,1) + 10^(-Eb_N0_dB(ii)/20)*n;

% Receiver
yMod = kron(reshape(y,2,N/2),ones(1,2)); % [y1 y1 ... ; y2 y2 ...]
yMod(2,:) = conj(yMod(2,:)); % [y1 y1 ... ; y2* y2*...]

% FFT**********************************************************************
% yMod = fft(yMod);

yHat = sum(hEq.*yMod,1)./hEqPower; % [h1*y1 + h2y2*, h2*y1 -h1y2*, ... ]
yHat(2:2:end) = conj(yHat(2:2:end));

% receiver – hard decision decoding
% ipHat = real(yHat)>0;

% Hard decision decoding***************************************************
yHat = reshape(yHat,para,nd);
S_re = real(yHat)/kmod;
S_im = imag(yHat)/kmod;
[demodata]=qpskdemod(S_re,S_im,para,nd,ml);
ipHat = reshape(demodata,1,ml*N);

% FFT**********************************************************************
% ipHat = fft(ipHat);

% counting the errors
nErr(ii) = size(find([ip- ipHat]),2);

end

simBer = nErr/(ml*N); % simulated ber
EbN0Lin = 10.^(Eb_N0_dB/10);
theoryBer_nRx1 = 0.5.*(1-1*(1+1./EbN0Lin).^(-0.5));

p = 1/2 – 1/2*(1+1./EbN0Lin).^(-1/2);
theoryBerMRC_nRx2 = p.^2.*(1+2*(1-p));

pAlamouti = 1/2 – 1/2*(1+2./EbN0Lin).^(-1/2);
theoryBerAlamouti_nTx2_nRx1 = pAlamouti.^2.*(1+2*(1-pAlamouti));

close all
figure
semilogy(Eb_N0_dB,theoryBer_nRx1,’bp-’,'LineWidth’,2);
hold on
semilogy(Eb_N0_dB,theoryBerMRC_nRx2,’kd-’,'LineWidth’,2);
semilogy(Eb_N0_dB,theoryBerAlamouti_nTx2_nRx1,’c+-’,'LineWidth’,2);
semilogy(Eb_N0_dB,simBer,’mo-’,'LineWidth’,2);
axis([0 25 10^-5 0.5])
grid on
legend(’theory (nTx=1,nRx=1)’, ‘theory (nTx=1,nRx=2, MRC)’, ‘theory (nTx=2, nRx=1, Alamouti)’, ’sim (nTx=2, nRx=1, Alamouti)’);
xlabel(’Eb/No, dB’);
ylabel(’Bit Error Rate’);
title(’BER for BPSK modulation with Alamouti STBC (Rayleigh channel)’);

===============================================
qpskmod.m
% Program 3-9
% qpskmod.m
%
% Function to perform QPSK modulation
%
% Programmed by H.Harada
%

function [iout,qout]=qpskmod(paradata,para,nd,ml)

%****************** variables *************************
% paradata : input data (para-by-nd matrix)
% iout utput Ich data
% qout utput Qch data
% para : Number of paralell channels
% nd : Number of data
% ml : Number of modulation levels
% (QPSK ->2 16QAM -> 4)
% *****************************************************

m2=ml./2;

paradata2=paradata.*2-1;
count2=0;

for jj=1:nd

isi = zeros(para,1);
isq = zeros(para,1);

for ii = 1 : m2
isi = isi + 2.^( m2 – ii ) .* paradata2((1:para),ii+count2);
isq = isq + 2.^( m2 – ii ) .* paradata2((1:para),m2+ii+count2);
end

iout((1:para),jj)=isi;
qout((1:para),jj)=isq;

count2=count2+ml;

end

%******************** end of file ***************************

qpskdemod.m
% Program 3-10
% qpskdemod.m
%
% Function to perform QPSK demodulation
%
% programmed by H.Harada
%

function [demodata]=qpskdemod(idata,qdata,para,nd,ml)

%****************** variables *************************
% idata :input Ich data
% qdata :input Qch data
% demodata: demodulated data (para-by-nd matrix)
% para : Number of paralell channels
% nd : Number of data
% ml : Number of modulation levels
% (QPSK ->2 16QAM -> 4)
% *****************************************************

demodata=zeros(para,ml*nd);
demodata((1:para),(1:ml:ml*nd-1))=idata((1:para),(1:nd))>=0;
demodata((1:para),(2:ml:ml*nd))=qdata((1:para),(1:nd))>=0;

%******************** end of file ***************************

120 Krishna Sankar December 22, 2009 at 5:43 am

@KimMyungsong: For QPSK demod, instead of qpskdemod(), you can try something as simple as
ipHat1 = real(yHat)>0
ipHat2 = imag(yHat)>0
Make sure that we do the ungrouping of each constellation symbol into two bits correctly.

121 KimMyungsong January 27, 2010 at 7:34 am

I already solved, thanks a lot.

Just need to delete below line.
%yHat(2:2:end) = conj(yHat(2:2:end));

122 Krishna Sankar January 28, 2010 at 5:32 am: @KimMyungsong: Glad to hear

Reply

123 Abrar December 16, 2009 at 2:29 am

Hi
My question is that if we compare MIMO system with single antenna system, how we make fair comparison by setting same power for both case.
for example, consider two antenna at transmitter like 2 X 1 alamouti scheme, if power=1 is used for single antenna system, so it will be half poer for each antenna to make fair comparison. Am I wright.

Thnaks

124 Krishna Sankar December 22, 2009 at 5:59 am: @Abrar: Yes, you are right

Reply

125 wap January 2, 2010 at 6:31 pm

hi sir………..

hi krishna…………..

im try youre source programs……….if i use stbc with 4,6,or 8 tx n tx (tx,rx > 2) where must change in the source??

126 Alex January 8, 2010 at 12:58 pm

can you help me please, i read a lot of your articles, that helped me a lot.
Hi, i have to finish my thesis by comparing some Ber curves. I am not so good in matlab, can you please show me how to modify your matlab file without rayleigh channel. (MISO channel without OFDM and only with AWGN channel, ) My modulation is QPSK but i think the ber curver are the same as for BPSK.
I have to be ready for 12 genn
If it helps, here are other Ber graphics: (with/without Rayleigh, MIMO)
http://zdarova.it/tesi/7genn/differenza%20tra%208%20curve.fig

127 manoj sharma February 9, 2010 at 7:04 am

hi
i am working on channel estimation.problem i am facing is how to write a code for higher order modulations at ofdm transmitter i.e.8 qam and above.how to write a code for a matrix relaed to received signal data pl guide

128 Krishna Pillai April 16, 2009 at 5:34 am

@abubaker: Can you please point to the error in the equation. If there is indeed an error, I will correct it.
I referred Digital Communications by Barry, Lee, Messerchmit.

129 fof May 19, 2009 at 1:09 am

i need it to be with Maximum Likelihood Detection for Alamouti-STBC system Under QPSK modulation condition….not with hard decision….so can you help me please…

130 Krishna Sankar October 1, 2009 at 5:08 am

@sreenu: Thanks. Can you plz point me to articles on the net or in print.

Alamouti STBC

Alamouti STBC

Other Assumptions

Receiver with Alamouti STBC

BER with Almouti STBC

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Simulation Model

Observations

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