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bee
Joined: 29 Nov 2003
Posts: 126
Helped: 2
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 31 Oct 2007 22:32 MMSE standard form ??? |
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do anyone has got the proof or have any idea of getting this proof that MMSE equation i.e. min E [|| WY-X ||^2] can be standerized to ( (H'H+σ²)^-1 )H'
where ' ----> hermettian transpose
^2 ---> rasie to the power 2
^-1 ----> raise to the power -1
σ ------> noise variance
E [] -----> mean
tthe system considered above is Y = HX+N , where X is transmiites signal, Y is recieved, H is the channel matrix and N is white gaussian noise. W is the MMSE filter coefficent.
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vkekk
Joined: 30 Mar 2007
Posts: 173
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| 02 Nov 2007 11:13 Re: MMSE standard form ??? |
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I got this principle expression for MMSE receiver is
G=H*(H*H + No/Po I )^-1 *x
Where H -- channel matrix
I --- identity matrix
X --- received vector
Iam not understand about the above expression That is what is No/Po and why multiplying with I , how can we achevied MMSE in the above expression etc.... Can you please help me ...
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eroica
Joined: 07 Sep 2007
Posts: 7
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| 03 Nov 2007 3:54 Re: MMSE standard form ??? |
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From MMSE sense, you can induce the following relationship.
E[|WY - X|^2] ==> E[|WY - X|Y'] = 0 ==> WE[YY'] - E[XY'] = 0 ==> W = E[XY']/E[YY']
Here, E[XY'] is the cross-correlation, while E[YY'] is the auto-correlation.
So, E[XY'] = XX'H + XN' and E[YY'] = XX'HH' + XHN' + X'H'N + NN'
Since X and N are independent and uncorrelated, XN' = X'N = 0, and NN' = sigma^2
Therefore, W = XX'H'/(XX'HH' + sigma^2).
If you send 2-PAM, then XX' = 1. Now you get W = H'/(HH' + sigma^2)
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vkekk
Joined: 30 Mar 2007
Posts: 173
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| 12 Nov 2007 7:38 Re: MMSE standard form ??? |
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G=H*(H*H + No/Po I )^-1 *x
Can you help me to understand the above expression...
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vkekk
Joined: 30 Mar 2007
Posts: 173
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| 04 Dec 2007 10:20 Re: MMSE standard form ??? |
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| Help pls...
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spweda
Joined: 28 Jun 2001
Posts: 195
Helped: 4
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| 09 Dec 2007 7:02 Re: MMSE standard form ??? |
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From the orthogonal principle, E{(WY-X)Y^H} = 0, Y=HX+n
we could get WE(YY^H)=E(XY^H), assuming E(XN^H)=0, i.e., signal and noise are independent, and mean of the noise is zero
E(YY^H)=HE(XX^H)H^H+E(nn^H), Assumped that E(XX^H)=PI, and E(NN^H)=NI, i.e., signal is an independent random vector and noise is also an independent random vector, W=H^H(HH^H+N/PI)^(-1). where P/I reprsents the signal-to-noise ratio. Here, ^H=^(*')
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Shwan
Joined: 26 Nov 2007
Posts: 13
Location: UK
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| 23 Feb 2008 18:13 MMSE standard form ??? |
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It is clear that the MMSE is one of the Equalization technique, and the equation is as follow:
Y(n)=X(n).[H*(n)/|H(n)|.^2+(Es/N0)^-1]
Where Y(n) is the received signal
X(n) is the transmitted signal
H(n) is the Faded signal (channel Impulse resp[onse)
Es is a symbol energy for QPSK
N0 is the noise.
I hope this will help
Sh
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dwade
Joined: 14 Feb 2008
Posts: 5
Location: UK
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| 03 Mar 2008 17:56 Re: MMSE standard form ??? |
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I have the same problem as you, but stumble across this book:
Multiantenna wireless communication systems by Sergio Barbarossa (Artech House)
refer pg 94-96 will help you to understand more
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Amine_z
Joined: 14 Mar 2008
Posts: 4
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| 08 Apr 2008 22:19 Re: MMSE standard form ??? |
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hi 
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