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MIMO with Zero Forcing equalizer

by Krishna Sankar on October 24, 2008

We had discussed three Single Input Multiple Output (SIMO also known as receive diversity) schemes – Selection combining, Equal Gain Combining, Maximal Ratio Combining and a Multiple Input Single Output (MISO, also known as transmit diversity) scheme – Alamouti 2×1 STBC. Let us now discuss the case where there a multiple transmit antennas and multiple receive antennas resulting in the formation of a Multiple Input Multiple Output (MIMO) channel. In this post, we will restrict our discussion to a 2 transmit 2 receive antenna case (resulting in a 2×2 MIMO channel). We will assume that the channel is a flat fading Rayleigh multipath channel and the modulation is BPSK.

2×2 MIMO channel

In a 2×2 MIMO channel, probable usage of the available 2 transmit antennas can be 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, as we now have 2 transmit antennas, we may 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 $x_3$ and $x_4$ from the first and second antenna, send $x_5$ and $x_6$ in the third time slot and so on.

4. Notice that as we are grouping two symbols and sending them in one time slot, we need only $\frac{n}{2}$ time slots to complete the transmission – data rate is doubled !

5. This forms the simple explanation of a probable MIMO transmission scheme with 2 transmit antennas and 2 receive antennas.

Having said this, some of you will wonder – the two transmitted symbols interfered with each other. Can we ever separate the two out? The rest of the post attempts to answer this question.

Figure: 2 Transmit 2 Receive (2×2) MIMO channel

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

4. The channel experienced between each transmit to the receive antenna is independent and randomly varying in time.

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 $h_{j,i}$ is known at the receiver.

Zero forcing (ZF) equalizer for 2×2 MIMO channel

Let us now try to understand the math for extracting the two symbols which interfered with each other. In the first time slot, the received signal on the first receive antenna is,

$y_1 =h_{1,1}x_1 + h_{1,2}x_2 + n_1 = [h_{1,1}\ h_{1,2}] \left[\begin{eqnarray}x_1 \\ x_2 \end{eqnarray}\right]+n_1$ .

The received signal on the second receive antenna is,

$y_2 = h_{2,1}x_1 + h_{2,2}x_2 + n_2 = [h_{2,1}\ h_{2,2}] \left[\begin{eqnarray}x_1 \\ x_2\end{eqnarray}\right]+n_2$ .

where

$y_1$ , $y_2$ are the received symbol on the first and second antenna respectively,

$h_{1,1}$ is the channel from $1^{st}$ transmit antenna to $1^{st}$ receive antenna,

$h_{1,2}$ is the channel from $2^{nd}$ transmit antenna to $1^{st}$ receive antenna,

$h_{2,1}$ is the channel from $1^{st}$ transmit antenna to $2^{nd}$ receive antenna,

$h_{2,2}$ is the channel from $2^{nd}$ transmit antenna to $2^{nd}$ receive antenna,

$x_1$ , $x_2$ are the transmitted symbols and

$n_1,\ n_2$ is the noise on $1^{st}, 2^{nd}$ receive antennas.

We assume that the receiver knows $h_{1,1}$ , $h_{1,2}$ , $h_{2,1}$ and $h_{2,2}$ . The receiver also knows $y_1$ and $y_2$ . The unknown s are $x_1$ and $x_2$ . Two equations and two unknowns. Can we solve it? Answer is YES.

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

$\begin{eqnarray}\left[\begin{eqnarray}y_1 \\ y_2\end{eqnarray}\right] & = & {\left[\begin{array}{cc}h_{1,1}& h_{1,2} \\h_{2,1}&h_{2,2}\end{array}\right]}\left[\begin{eqnarray}x_1 \\ x_2 \end{eqnarray}\right]+\left[\begin{eqnarray}n_1\\n_2 \end{eqnarray}\right]\end{eqnarray}$ .

Equivalently,

$\mathbf{y} = \mathbf{H}\mathbf{x} + \mathbf{n}$

To solve for $\mathbf{x}$ , we know that we need to find a matrix $\mathbf{W}$ which satisfies $\mathbf{WH=I}$ . The Zero Forcing (ZF) linear detector for meeting this constraint is given by,

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

This matrix is also known as the pseudo inverse for a general m x n matrix.

The term,

$\mathbf{H^HH} = \left[\begin{array}{cc}\ h_{1,1}^* & h_{2,1}^* \\h_{1,2}^*& h_{2,2}^*\end{array}\right]\left[\begin{array}{cc}\ h_{1,1} & h_{1,2} \\h_{2,1} & h_{2,2}\end{array}\right]=\left[\begin{array}{cc}|h_{1,1}|^2+|h_{2,1}|^2 & h_{1,1}^*h_{1,2}+h_{2,1}^*h_{2,2} \\ h_{1,2}^*h_{1,1}+h_{2,2}^*h_{2,1}& |h_{1,2}|^2+|h_{2,2}|^2\end{array}\right]$ .

BER with ZF equalizer with 2×2 MIMO

Note that the off diagonal terms in the matrix $\mathbf{H^HH}$ are not zero (Recall: The off diagonal terms where zero in Alamouti 2×1 STBC case). Because the off diagonal terms are not zero, the zero forcing equalizer tries to null out the interfering terms when performing the equalization, i.e when solving for $x_1$ the interference from $x_2$ is tried to be nulled and vice versa. While doing so, there can be amplification of noise. Hence Zero Forcing equalizer is not the best possible equalizer to do the job. However, it is simple and reasonably easy to implement.

Further, it can be seen that, following zero forcing equalization, the channel for symbol transmitted from each spatial dimension (space is antenna) is a like a 1×1 Rayleigh fading channel (Refer Section 3.3 of [WIRELESS-TSE, VISWANATH]). Hence the BER for 2×2 MIMO channel in Rayleigh fading with Zero Forcing equalization is same as the BER derived for a 1×1 channel in Rayleigh fading.

For BPSK modulation in Rayleigh fading channel, the bit error rate is derived as,

$P_{b}=\frac{1}{2}\left(1-\sqrt{\frac{(E_b/N_0)}{(E_b/N_0) +1}}\right)$ .

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 and send two symbols in one time slot

(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 plot for 2×2 MIMO channel with ZF equalizer (BPSK modulation in Rayleigh channel)

Figure: BER plot for 2×2 MIMO channel with ZF equalizer (BPSK modulation in Rayleigh channel)

Summary

1. As expected, the simulated results with a 2×2 MIMO system using BPSK modulation in Rayleigh channel is showing matching results as obtained in for a 1×1 system for BPSK modulation in Rayleigh channel.

2. As noted in Section 3.3 of [WIRELESS-TSE, VISWANATH], the Zero Forcing equalizer is not the best possible way to equalize the received symbol. The zero forcing equalizer helps us to achieve the data rate gain, but NOT take advantage of diversity gain (as we have two receive antennas).

3. We might not be able to achieve the two fold data rate improvement in all channel conditions. It can so happen that channels are correlated (the coefficients are almost the same). Hence we might not be able to solve for the two unknown transmitted symbols even if we have two received symbols.

4. It is claimed that there can be receiver structures which enables us to have both diversity gain and data rate gain. In future posts, the attempt will be to discuss receiver structures which hopefully enables us to find out approaches which will help us to keep the data rate gain, but still move from the 1×1 curve to 1×2 MRC curve.

References

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

[WIRELESS-TSE, VISWANATH] Fundamentals of Wireless Communication, David Tse, Pramod Viswanath

Tagged as: BPSK, MIMO, Rayleigh, ZF

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{ 3 trackbacks }

MIMO with MMSE equalizer: November 2, 2008 at 8:35 pm
MIMO with Zero Forcing Successive Interference Cancellation equalizer: November 9, 2008 at 6:43 pm
MIMO with ZF SIC and optimal ordering: November 29, 2008 at 1:34 pm

{ 85 comments… read them below or add one }

1 lvs November 23, 2008 at 9:25 pm

A DSP blog Amazing, the internet does have it all!

A long time ago I used to work in DSP and communications. Since then this world has completely changed. I am looking forward to catching up by reading your blog.

You seem to have put a lot of effort, even have equations in here.

I have started out with this article so I dont know if I have missed some things…but you could make it more easier and interesting for readers like me through real life examples and putting more of your own insights into the equations.

2 Krishna Pillai November 26, 2008 at 6:14 am

@lvs: Thanks . This blog is my humble way of sharing what I know and hopefully learn more during the process.

I started off some time last year by writing posts on SER/BER computation of different modulation schemes in AWGN
http://www.dsplog.com/tag/awgn/

Later moved on to Rayleigh multipath channel model
http://www.dsplog.com/tag/Rayleigh/

Then started on receive diversity (EGC, MRC, selection combining) and transmit diversity (alamouti STBC)
http://www.dsplog.com/category/diversity/

And recently to MIMO
http://www.dsplog.com/tag/mimo/

Once in a while tries to discuss topics on negative frequency etc
http://www.dsplog.com/category/miscellaneous/

I sure do hope to make the contents more readable/friendly by putting some real world examples. Need to think of a way to fit in.

Btw, you have a really good blog out there. Nice long articles. I have bookmarked for future read.

Thanks.

3 Jarrod November 29, 2008 at 11:52 pm

Thanks a lot for this blog. I am curious how your model of the Zero Forcing equilizer would differ if the channel was ISI instead of the MIMO?

4 Krishna Pillai December 2, 2008 at 6:06 am

@Jarrod: As is the case with MIMO, if there is ISI, then Zero Forcing equaliser does not be the optimal way to equalize. Conceptually ISI can be thought of as interference in the time domain where as MIMO can be thought of as interference in the spatial domain.

5 mimo February 6, 2009 at 5:51 pm

Hello Krishna,

thanks for this blog, I am researching signal processing for MIMO. I have utilized vblast, qostbc, with mmse but I don’t know that algorithm can I utilized for wideband. Could you help me? Do you have some example code for ISI?

Thank you.

6 Krishna Pillai February 10, 2009 at 8:00 pm: @mimo: Sorry, I have not studied the case where there is ISI plus MIMO. Most of my MIMO analysis was assuming flat fading. Kindly do share your results when they are available. Good luck

Reply

7 karl December 13, 2008 at 2:24 pm

Krishna, why you don’t halved total transmit power for each transmit antenna like the case of alamouti matlab code and you still can compare with the 1×1 case?.

thanks.

8 Krishna Pillai December 13, 2008 at 6:41 pm: @karl: Well, recall that in the MIMO case we require only half the time to transmit the symbols.

Assuming P is the transmit power from a single antenna in a 1×1 case and we require time T to send N symbols. The energy
consumed in E_{1×1| = PT
For the 2×2 MIMO, the time taken is only T/2 and we assume that we transmit from both the antennas at power P each. The energy consumed is E_{2×2} = 2P*T/2 = PT
Since the total energy consumed is the same, I think it is fair to compare with the SISO 1×1 case.
Do you agree to my perspective? Kindly share your thoughts

Reply

9 Sky Stradlin December 14, 2008 at 5:52 am

Zero Forcing and MMSE equalizer, is it a receiver structure? Coz i know there is ZF/MMSE receiver, so I’m confusing. Thanks

10 Krishna Pillai December 14, 2008 at 7:12 am: @ Sky Stradlin: Yes, Zero Forcing and MMSE classifies two different types of receiver structure (with MMSE performing better than Zero Forcing in MIMO case). Can you please point me to the literature which discuss ZF/MMSE receiver.

Reply

11 Sky Stradlin December 15, 2008 at 12:10 am

Sorry, I mean the ZF equalizer you did in this post, is it should be ZF receiver?

12 Krishna Pillai December 16, 2008 at 6:05 am: @Sky Stradlin: I think ZF equalizer and ZF receiver are used interchangably.

Reply

13 me January 14, 2009 at 9:06 pm

thanx for your articles…
hope it can help me….;-)

14 Dwade January 19, 2009 at 9:24 pm

Dear Krishna,
In fully correlated channel, i.e. correlation coefficient nearing 1 (0.9999) at both Tx and Rx, do you think ZF will outperform MMSE in this case?

15 Krishna Pillai January 22, 2009 at 3:32 am: @Dwade: I do not have results to share. However, from a quick googling found the following article:
Advanced receiver design boosts performance
http://www.eetimes.com/news/design/showArticle.jhtml?articleID=206104566

In the article, author is saying that both MMSE and ZF suffers when the channel is correlated. He is suggesting that Maximum Likelihood Decoder (MLD) performs the best in that scenario.

Reply

16 Dwade February 2, 2009 at 8:16 pm

Hi Krishna
Thanks a lot for the helpful comment. Really appreciate it!

17 kamran February 23, 2009 at 2:55 pm: thanx dude

Reply

18 MIMO February 6, 2009 at 7:49 pm

Hello Krishna,

I am researching in MIMO wideband and ISI. Could you help me and tell me some algortihm for decode.

thanks you.

19 John Titor February 7, 2009 at 8:50 pm

How come the performance of ZF is the same as SISO? Since in the case of ZF, there is an interference between the transmitted symbols at each recive antenna.

Thanks alot

20 Krishna Pillai February 10, 2009 at 8:08 pm

@John: Note that we receive two copies of the transmit symbol at the receiver (one on each receive antenna). The presence of two copies helps to achieve the ZF performance eventhough there is interference.
Do you agree?

21 John Titor February 12, 2009 at 8:37 pm

Yep, but since the matrix W is also applied to the noise, so there should be some noise enhancement. I think that would decrease the performance of ZF?

22 Krishna Pillai February 19, 2009 at 6:09 am: @John: Yes, there is noise enhancement. But I would think that is case for 1×1 too. Hence 1×1 ZF can compare with 2×2 ZF.

As an additional thought:
If I see 1×1 with ZF equalization and MMSE equalization, I find that the equalized constellaion has a lower EVM/MER in the presence of MMSE. However, the raw BER is the same. I fail to understand that behaviour. Can you please share comments on that aspect?

Reply

23 communications_engineer May 15, 2009 at 9:00 pm

Hi. Can you tell me how you got this equation

W = (H^H*H)^-1 H^H

and what is means. Especially what is H^H

Its the only part that I can’t understand. How would I write if I only considered one path?

24 Krishna Pillai May 20, 2009 at 5:28 am

@communications_engineer: The term H^H is Hermitian operator – in simple terms its the conjugate transpose of a matrix.
If there is only one path, then the equation reduces to H*/|H|^2, where H* is the conjuagate of H. Agree?

25 communications_engineer May 22, 2009 at 8:36 pm

Umm… I’ll try to get some references to understand more….

Krishna, as I have written before I’m working on CDMA MUD, so I want to use ZF equalization (I now its not good, but only for the sake of it), so I’ll use ZF before the de-correlation (I’m using eq y1 = A1*r2 + A2*r2*sgma+noise, verdu’s book), so ZF before decision making.

r = conv(ytx,chan); % received signal
eq=ifft(fft(ytx) ./ fft(chan));
r_eq = conv(r, eq); % equalizer output

Would this be the right Matlab code to use? Where ytx is the spread signal of two users. If I use AWGN then I’ll just add wgn.

I’m kinda confused as you can probably see, so any suggestions?

26 Krishna Pillai May 31, 2009 at 8:02 pm: @communications_engineer: I did not quite understand your way of doing equalization. In equalization, we want to make the channel response close to an impulse, agree?

Reply

27 yazeed May 18, 2009 at 12:39 am

Hi
in the case of 2×2 MIMO with ZF equalization what the changes in the case of 8×8 or 12×12 MIMO,i.e how i can find inv(H^H*H)*H.

28 Krishna Pillai May 20, 2009 at 5:37 am: @yazeed: Well in the case of 8×8 or 12×12 MIMO, the equation still holds good. However, now you need to find the inverse of that 8×8 or 12×12 matrix.

Reply

29 amrali May 26, 2009 at 3:07 am

hi , please, could give me address of books or litrature ,which talk about mimo matrix.
I want to understand the matrix operation on MIMO.

Thanx

30 Krishna Pillai May 31, 2009 at 8:21 pm: @amrali: The book
Fundamentals of Wireless Communication, David Tse, Pramod Viswanath has a very good discussion on MIMO matrices. Hope this helps.

Reply

31 Rahul June 16, 2009 at 5:28 pm

Hi,
Can you tell me how to modfiy your code in cas I want to use it for a general Nt x Nr case and with 16 QAM Modulation..
Both for ZF and MMSE Equalizer

32 Saria July 6, 2009 at 8:26 pm

Dear Mr. Krishna, you are really helping students a lot. Please, I would like to know, why you used the operation sum(h(:,2,:).*conj(h(:,2,:)),1);to calculate each of the cofficents, because c*conj(c) will result in (abs(C)).^2 or in other worlds the magintude squared of each of the column elements. also, why taking summation?. Thank you again and best wishes.

33 Krishna Pillai July 15, 2009 at 4:36 am: @Saria: If you see the equation, the term (H^H*H) has |h_{1,2}|^2 + |h_{2,2}|^2. Hence the need for the sum.

Reply

34 Khattak July 20, 2009 at 4:13 pm

Thanks Krishna i got a lot of help from your blog and i am realy very immpressed of you. Can you tell me how to modfiy your code in case I want to use it for 4×4 and 4×3 MIMO system.

35 Krishna Pillai July 20, 2009 at 7:28 pm: @Khattak: Thanks. Well, I have not written posts on the 4×3 or 4×4 MIMO systems. Anyhow, I hope that you will be able to adapt the existing code to the above configurations. Good luck for you algorithm explorations.

Reply

36 khattak July 21, 2009 at 5:48 pm

ok thanks

37 amrali August 2, 2009 at 2:06 am

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

AWGN=10^(-Eb_N0_dB(ii)/20)*n;

WHY ?
why not , AWGN=sqrt(10^(-Eb_N0_dB(ii)/10))*n?
Thanx

38 Krishna Sankar August 5, 2009 at 5:49 am: @amrali: The noise signal n is a voltage signal. So when we convert the dB ratio to be applied to a voltage signal, the conversion happens by 10^(dB/20). Hope this helps.

Reply

39 hamidreza August 2, 2009 at 2:14 am

hello
my name is hamidreza
i am communication student
i work on mimo channel and i need mimo-ofdm matlab code
i would appreciate your help

40 Krishna Sankar August 5, 2009 at 5:50 am: @hamidreza: Glad to hear from you. Though I have not discussed posts on MIMO + OFDM, I have posts discussing OFDM and MIMO independently @
http://www.dsplog.com/tag/ofdm
http://www.dsplog.com/tag/mimo

Hope this helps

Reply

41 SARA' August 5, 2009 at 5:57 pm

Hi Krishna,
Thanks very much for the intersting topics. Could I ask what will be the case if we have AWGN channel. In this case h11, h12, h21 and h22 will be 1. and the inverse of H will Inf.
Could you explain in more details how this could be solved for AWGN channel
Regards

42 Krishna Sankar August 10, 2009 at 5:49 am: @SARA: If we assume that the AWGN channel is formed by connecting via cable, then we can assume that h11=1, h22=1, h12=0, h21=0.
i.e. formation of two parallel AWGN channels. Agree?

Reply

43 SARA' August 10, 2009 at 2:03 pm

Hi Krishna,

Thanks very much for the reply. The case that am considering is that
y1=x1+x2+n1
y2=x1+x2+n2
the channel is AWGN channel and I want to apply the ZF or MMSE equilizer to de-multiplex the channels. I this case, I assume that h11, h12, h21 and h22 are equal to 1. Therefore, the inverse of the channel matrix will be INF?
If you could help me to understand how to apply the MIMO detectors for AWGN channel?

for the two parallel AWGN channel
y1=x1+n1
y2+x2+n2

Thanks very much for the reply
waiting for your kind answer

44 Krishna Sankar August 11, 2009 at 5:09 am

@SARA: If you have
y1=x1+x2+n1
y2=x1+x2+n2
the you wont be able to solve for x1 and x2, given that you know y1 and y2. Think of your linear algebra classes. To solve for two unknowns (x1, x2), we need two equations (y1, y2). If the two equations carry ’same/similar’ information, then there is no way that we can use them for resolving y1, y2. In MIMO communication parlance, having such a channel is called rank deficient channel.

45 aydar October 26, 2009 at 6:19 pm

The problem you have is that you assume your both MIMO users (if it spatial MIMO case for example) have identical channel. That means both MIMO users are completely correlated. Your channel matrix is [1 1 ; 1 1] => you cant perform pseudoinverse inv(H^H*H). what you could do is set a phase difference between Rx antennas for one user, let’s say its ‘pi’, so your channel would be H=[1 1; 1 -1] => fully uncorrelated. It is a common problem for MMSE/ZF receivers when channels are correlated. Whereas, MLD could still exploit symbol constellation and outperform MMSE/ZF.

46 Krishna Sankar November 7, 2009 at 7:12 am: @aydar: In general, I agree with the intent of the comment. However, just to add that I have not assumed that the channel matrix is identical. I assume that the channel matrix takes tap from Rayleigh distributed channel.

Reply
47 aydar November 12, 2009 at 5:51 pm: oopps sorry, i think H=[1 1; 1 -1] IS still correlated, but anyway we can perform inverse (you can think of it as MMSE performs spatial separation or beamforming).

Reply

48 mohanad August 19, 2009 at 7:53 pm

Hi krishna…
Can you please tell me Why you are using
y = squeeze(sum(h.*sMod,2)) + 10^(-Eb_N0_dB(ii)/20)*n;
instead of :-
y = squeeze(sum(h.*sMod,1)) + 10^(-Eb_N0_dB(ii)/20)*n;
becauase the using of sum(s,2) will sum the raws of information bits
like this:-
ip =[0 0 1 1];
s=[-1 -1 1 1];
x=sum(s,2)
will result in x=0 and that will not make any sense to me?

49 Krishna Sankar August 22, 2009 at 5:09 am: @mohanad: The way the matrix multiplication is organized for each time slot is as follows:
h = [h11 h12 ; h21; h22] and s = [s1 s2;s1 s2]
We know that y = [h11*s1 + h12s2; h21*s1 + h22*s2]
So, how do we achieve that?
We do a dot product of h and s, and the sum in the column dimension. To the sum in the column dimension, we have the parameter sum(,2).

Hope this helps.

Reply

50 sam2 August 23, 2009 at 8:03 pm

Hi Mr krishna
I have an question about pseudo inverse matrix;
We know that we want to get an identity matrix from the equation WH=I.
So we can get the same result by using another equation which is
W=inv(H) , because [ inv(H)*H=I ]
(I consider this equation[W= inv(H)*H ] in my work and it gave me the same result.)
Can u pls explain why we can not use such equation??

51 Krishna Sankar August 24, 2009 at 5:04 am

@sam2: The inv(H) is not defined for a non-square matrix. Hence the pseudo inverse is a more general definition which works for non-square matrix, and scales down to inverse operation for a square matrix.
Agree?

52 aydar October 26, 2009 at 6:26 pm

Yep. You can use inv(H) only when the number of Tx and Rx antennas are equal.

53 Krishna Sankar November 7, 2009 at 7:13 am: @aydar: agree. we use pseudo inverse for non-square channels

Reply

54 mohanad August 24, 2009 at 10:59 pm

It is honor to me to know such a vast experience man like you ….
Thanks for all special efforts you have done ….

55 sam2 August 25, 2009 at 11:24 pm

I agree with you
your answer seems to be right
thank u Mr krishna

56 Street hawk August 31, 2009 at 6:32 pm

If I consider the SIMO case (1 Tx * 2 Rx), then what would be the Zero Forcing equalization matrix W?

And also for SIMO case with correlated fading which receiver structure would be better, i.e. Zero forcing/ MMSE or other kind?

57 Krishna Sankar September 7, 2009 at 5:31 am

@Street hawk: If you have 1 transmit and 2 receive antennas and the channel is flat fading, then the ZF equalization structure reduces to Maximal Ratio Combining receiver.
http://www.dsplog.com/2008/09/28/maximal-ratio-combining/

Note: If you have only one transmit antenna, then there is no interference to suppress.

58 aydar October 26, 2009 at 6:50 pm

Hi, Krishna.

That what was bothering me for some long time – difference between ZF and MRC. So, you mean there is no MRC for MIMO?

@Street hawk: well, if you have correlated fading then probably ML-based receivers would be a good choice. Or you could use correlation in order to perform beamforming to suppress inter-cell interference (like IRC receiver based on e.g. MMSE or MLD).

Cheers

59 Krishna Sankar November 8, 2009 at 7:13 am: @aydar: If there are multiple copies of the same information coming even in a MIMO link, we need to use MRC. For eg, a three receive, 2 transmit case.

Reply

60 chen October 16, 2009 at 7:29 pm

Tow questions about your code:
Q1, I think the channel need to be normalized which means h*conj(h’)=1, but it seems you did not do that, how do you think about the channel matrix normalization as I described.

Q2, To calculate the inverse matrix, in your code, why do not you directly use the matlab ‘inv()’ method but implement your self with a block of code?

61 Krishna Sankar October 17, 2009 at 4:18 am: @chen: My replies:
1/ The E{h*h’} over all channel realization (and not per realization) is unity.

2/ As I recall, the inv() operator works only a for two dimensional matrix, and I would have to put a for loop to perform the equalization for all symbols. To speeden things up, I used my own inv() and performed the whole story using matrix algebra.

Reply

62 minh October 24, 2009 at 5:17 pm

hi Krishna Sankar, you do all the simulations in flat-fading environment, how about frequency selective channels which are more realistic than the former? In fact, I use ITU channel models in my simulation but it’s difficult to me to apply your codes with more taps. Would you please help me with that problem. Thank you in advance

63 aydar October 26, 2009 at 7:19 pm

Well, i think there is no formula for uncoded BER at multi-tap Rayleigh fading channels
what will change is that if you perform equalization in time domain then you would have to do filtering (convolution) which is of course more complex than simple multiplication for one-tap fading, but if you use systems like LTE, WiMAX (OFDM, SC-FDMA based) where you have cyclic prefix and freq.domain pilots for channel estimation then there is no problem for simple frequency domain equalization.
Another thing, ITU channel were originally for SISO systems, they have no information about spatial correlation between antennas (antenna spacing, polarization, angular spread, AoA etc.). But this model can be extended to MIMO channel models with the definition of a per-tap spatial correlation.

64 Krishna Sankar November 8, 2009 at 7:14 am: @aydar: Ok

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65 Krishna Sankar October 27, 2009 at 5:38 am

@minh: Adding simulations with frequency selective channel is my next step

66 jhon November 1, 2009 at 5:32 pm

Hi krishna
Nice blog sir…i have only one doubt..

you are using
Inverse of a [2x2] matrix [a b; c d] = 1/(ad-bc)[d -b;-c a]

method to calculate inverse of (H^H*H)..

My question is since we are finding invers of 2X2 Matrix.. why cant we use

Inverse of a [2x2] matrix [a b; c d] = 1/(ad-bc)[d -b;-c a]

method directly to calculate inverse..

why are we going through

(H^H*H)^-1 *H^H route to calculate inverse..

i am sure there is some reason for that

Thanks

67 Krishna Sankar November 8, 2009 at 8:36 am: @jhon: 1/H is same as H^H/(H^H*H) i.e multiplying by H^H in numerator and denominator. This makes the matrix general for any non square matrix.
Just a quick matlab code snippet:
h = rand(2,2)+j*rand(2,2)
w1 = inv(h’*h)*h’
w2 = inv(h)

can see w1 and w2 are same.

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68 bopuhafs November 12, 2009 at 9:23 pm

pourqoui vous etuliser le signe moin (-) dans 10^(-Eb_N0_dB(ii)/20)

69 Krishna Sankar December 3, 2009 at 5:19 am: @bopuhafs: To make the noise power to be lower than the signal power

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70 boukhari November 15, 2009 at 3:34 pm

why you used the signe negatif (-) in 10^(-Eb_N0_dB(ii)/20)

71 Krishna Sankar December 3, 2009 at 5:41 am: @boukhari: To make the noise power to be lower than the signal power

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72 peet November 21, 2009 at 7:38 pm

Hi everybody;

my question is: If we use the same technique (e.g spatial multiplexing with the ZF receiver, or Alamouti scheme ) in the conventional MIMO performing over the flat fading channel and in MIMO-OFDM performing over the frequency selective channel, should be the results same?

I mean,… MIMO is designed for flat fading channels and OFDM transform frequency selective channel into several flat fading channel so the performance should be same, but I am not sure…

Can you please share comments on that aspect?

73 Krishna Sankar December 6, 2009 at 4:29 pm

@peet: Yes. The usage of OFDM need not change the underlying BER.
Please check the post on
a) BER for BPSK in AWGN (with OFDM)
http://www.dsplog.com/2008/06/10/ofdm-bpsk-bit-error/
b) BER for BPSK in multipath channel with OFDM (same as flat fading Rayleigh channel)
http://www.dsplog.com/2008/08/26/ofdm-rayleigh-channel-ber-bpsk/

74 peet January 26, 2010 at 2:12 pm: thanks!!!….

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75 greg November 24, 2009 at 8:27 pm

have you tried adapting this bpsk signal to a multi-user cdma signal

76 Krishna Sankar December 7, 2009 at 4:21 am: @greg: Sorry, no.

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77 greg November 24, 2009 at 11:10 pm

have you thought of CDMA instead of QAM for your MIMO work

78 Krishna Sankar December 7, 2009 at 4:23 am: @greg: Well, I do not understand how CDMA can be substituted for QAM. CDMA is a multiuser technique where users are separated by codes, whereas QAM is a amplitude+phase modulation technique.

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79 bouhafs December 13, 2009 at 7:31 pm

I need the matlab program calculates the”error rate”of binary systems (SISO, Simo, miso mimo) COMPARISON

80 Krishna Sankar December 22, 2009 at 5:33 am: @bouhafs: Please have a look at
http://www.dsplog.com/tag/diversity/
http://www.dsplog.com/tag/mimo

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81 bouhafs December 13, 2009 at 7:33 pm

help me please :my question is I need the matlab program calculates the’binary ‘error rate ”BER”of systems (SISO, Simo, miso mimo) COMPARISON

82 joey January 5, 2010 at 8:21 pm

have u considered the 4×4 case. how could thisbe extended to it

83 AceKris January 6, 2010 at 8:11 pm

How can I make sure that E[||H|| * ||H||] =1 ?
ie Expectation of square of Mod of the Channel matrix is equal to one.
This is the first condition that has to be met before I start to simulate.
Is the term [||H|| * ||H||] relating to the a single channel realization or an ensemble of them. In anycase how do I generate H satisfying the above condition and verify this the same. H is a Rayleigh fading channel initially and later on (not related to the first one) I will have to look into the cascaded rayleigh fading channel.

Thanks and Regards

84 Sadaf January 7, 2010 at 4:48 pm

hey karishna your codes and articls help alooooooooot
thank you sooo very much

85 Pushpa February 1, 2010 at 9:07 pm

Interesting article.
Can you help me in getting MATLAB code for MIMO frame synchronization?

MIMO with Zero Forcing equalizer

2×2 MIMO channel

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Zero forcing (ZF) equalizer for 2×2 MIMO channel

BER with ZF equalizer with 2×2 MIMO

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