3
$\begingroup$

How would the LMS equalizer dimensions change for the MISO case?

LMS adaptive filters are typically described for equalizing a single input signal, $x(t)$. Can the LMS algorithm be modified in the MISO case to perform diversity combining when the transmitter has $N=1$ antennas, the receiver has $M>1$ antennas?

For the LMS algorithm, assume $n$ is the number of the current input sample and $p$ is the number of filter taps.

$$ \begin{aligned} \mathbf x(n) &= \begin{bmatrix}x(n)& x(n-1)& \cdots & x(n-p+1)\end{bmatrix}^T \\ e(n) &= d(n) - \hat h(n)^Hx(n) \\ \hat h(n+1) &= \hat h(n) + \mu e^*(n)x(n) \end{aligned} $$ where dimensions are:

$\mathbf x(n) $ = $p\times 1$ vector
$\hat h(n)^Hx(n) $ = $[1 \times p]$ * $[p \times 1]$ = $1\times1$
$e(n)$ = scalar
$\hat h(n)$ = $p\times 1$ vector

Now define the $M$ received vectors, $r(n)$, corresponding to the collected samples from each receiver, $r(n) = [r_1(n), r_2(n),...,r_M(n)]$

There are two ways I can think of to incorporate the additional received vectors.

The first way is to change the dimensions of $x(n)$ such that it is $p\times M$

$x(n) = [r_1(n), r_2(n),...,r_M(n)]^T$

This would result in the dimensions changing as follows:

$\mathbf x(n) $ = $p\times M$ matrix
$\hat h(n)^Hx(n) $ = $[M \times p]$ * $[p \times M]$ = $M \times M$ <-- should this be dot product or summed?
$e(n)$ = scalar
$\hat h(n)$ = $p\times M$ matrix

The second way is how one of the answers proposed, is to concatenate individual inputs into $x(n)$. This would result in a dimension of $Mp$.

$x(n) = [r_1(n) r_2(n),...,r_m(n),...r_m(n-p+1]^T$

In practice we almost always use a fractionally spaced equalizer so the technique would require to scale for that use case. The first way could be extended for the fractionally spaced equalizer case, but the second way doesn't seem like it could. Are either of these right, or is there another way, or is it not possible ?

Note: more details on LMS here.

$\endgroup$
4
  • $\begingroup$ I think you have some naming inconsistencies. Your title says MIMO, your text cites a receiver with multiple inputs and (presumably) one output, but you use "SIMO" to describe (presumably) your receiver. Do you mean to say "MISO" throughout? While that does suffer from its own ambiguity, I don't think I've ever seen cooking, Japanese or otherwise, discussed on this site. So "MISO" is probably safe. $\endgroup$
    – TimWescott
    Aug 6, 2022 at 19:23
  • $\begingroup$ Yes I meant MISO. One antenna at transmitter and multiple at receiver. I’ll correct it. $\endgroup$ Aug 6, 2022 at 23:06
  • $\begingroup$ Sorry I left this bit out before: please edit your question to reflect this. Stackexechange like things tidy... $\endgroup$
    – TimWescott
    Aug 6, 2022 at 23:43
  • $\begingroup$ Fixed. Also, this link describes SIMO the way i had it in my original question: evercomtech.com/understanding-siso-simo-miso-mimo $\endgroup$ Aug 7, 2022 at 2:05

2 Answers 2

0
$\begingroup$

I'm surprised I couldn't find a decent reference for this on the web.

Terminology gets difficult. Let $s(n)$ (a scalar) be the transmitted information. Let there be $M$ inputs. Let $\mathbf r(n)$ be the received vector at each instant, $$ \mathbf r(n) = \begin{bmatrix} r_1(n) \\ r_2(n) \\ \vdots \\ r_M(n) \end{bmatrix}, \tag 1$$ where each $r_m(n)$ is some linear combination of $\begin{matrix} s(n), & s(n-1), \cdots \end{matrix}$.

Then it ought to work if you let the $\mathbf x (n)$ above equal a concatenation of the individual inputs, i.e. $$\mathbf x (n) = \begin{bmatrix}r_1(n) & r_2(n) & \cdots & r_m(n) & r_1(n - 1) & \cdots & r_m(n-p+1)\end{bmatrix}^T. \tag 2$$

Then just do the rest "naturally", using the equations you cite above.

The length would become $Mp$, where $M$ is the number of inputs and $p$ is the number of points you take into the past.

$\endgroup$
7
  • $\begingroup$ I don’t see how this would result in the combining of the received signal vectors to output one waveform that is a linear combination of the inputs. $\endgroup$ Aug 7, 2022 at 1:02
  • $\begingroup$ Might be the typo in the definition for $\mathbf x(n)$ (now fixed). In a "regular" LMS filter, $\mathbf x(n)$ is a bunch of elements, each of which is a linear combination of $s(n), s(n-1), \cdots s(n - ??)$. In this filter, $\mathbf x(n)$ is a bunch of elements, each of which is a linear combination of $s(n), s(n-1), \cdots s(n - ??)$. So -- they should be equivalent. $\endgroup$
    – TimWescott
    Aug 7, 2022 at 2:15
  • $\begingroup$ how would this extend to a fractionally spaced LMS equalizer? Seems like it would fall part since you only same once per baud. $\endgroup$ Aug 7, 2022 at 2:19
  • $\begingroup$ That sounds like a separate question. $\endgroup$
    – TimWescott
    Aug 7, 2022 at 2:42
  • $\begingroup$ I can see why you want me to ask a separate question, but it's really an important detail to the original question, as LMS equalizers in communication systems are typically implemented at an oversampled rate. Might it be more appropriate if I edit this question to include that detail? $\endgroup$ Aug 7, 2022 at 12:49
0
$\begingroup$

My communication knowledge is rusty. Bit you are using some kind of known training sequence in a single transmitter antenna, to identify the linear convolutive system from Tx antenna to each Rx antenna so you can combine them using eg matched filtering?

If so, I think that each system can be identified using LMS simultaneously but independently?

$\endgroup$
2
  • $\begingroup$ Yes, there would be a known sequence like a training sequence or pilots to perform synchronization $\endgroup$ Aug 7, 2022 at 11:58
  • $\begingroup$ Then, as I said, I fail to see the problem. Identify the N paths from the 1 pilot using N LMS filters. Do whatever to combine the N Rx antennas with known convolutive distortion. $\endgroup$
    – Knut Inge
    Aug 8, 2022 at 7:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.