{"id":4336,"date":"2022-10-07T12:34:25","date_gmt":"2022-10-07T12:34:25","guid":{"rendered":"https:\/\/www.raymaps.com\/?p=4336"},"modified":"2024-12-09T06:34:58","modified_gmt":"2024-12-09T06:34:58","slug":"why-is-mimo-fading-capacity-higher-than-awgn-capacity","status":"publish","type":"post","link":"https:\/\/www.raymaps.com\/index.php\/why-is-mimo-fading-capacity-higher-than-awgn-capacity\/","title":{"rendered":"Why is MIMO Fading Capacity Higher than AWGN Capacity"},"content":{"rendered":"\n

In a previous post<\/a> we have seen that MIMO fading capacity is much higher than AWGN capacity with multiple antennas. How is this possible? How can randomness added by a fading channel help us? In this post we try to find the reason for this. Let\u2019s assume the following signal model for a Multi Input Multi Output antenna system.<\/p>\n\n\n\n

x=Hs+w<\/p>\n\n\n\n

Here s is the NT<\/sub> by 1 signal vector, w is the NR<\/sub> by 1 noise vector and H is the NR<\/sub> by NT<\/sub> channel matrix. The received signal vector is represented by x which has dimensions of NR<\/sub> by 1. In expanded form this can be written as (assuming NT<\/sub> =4 and NR<\/sub> =4):<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Let\u2019s assume that there is no fading and all entries of the channel matrix are equal to unity i.e. the channel provides no gain or phase rotation. Let\u2019s further assume that the four transmitted symbols are A, B, C and D (real modulation like PAM is assumed). Then in an AWGN channel we have:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Let\u2019s assume that A=1, B=3, C=5 and D=-1. Let\u2019s further assume that there is no noise i.e. vector w is all zeros. The received signal vector x would then look like: <\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

From linear algebra we know that to find four unknowns we need four independent equations. There is no way we can find the values of  A, B, C and D from the above equations. To simplify the above equations we have removed AWGN but even in presence of AWGN we will have the same predicament. <\/p>\n\n\n\n

This shows that in the absence of fading\nthere is no multiplexing gain however high the Signal to Noise Ratio is (in the\nabove example SNR is infinite). Now let\u2019s assume that we have a 4×4 MIMO fading\nchannel, which might now look something like this (a totally random matrix that\nthis author could think of):<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

 Substituting in the above linear model we have:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

We have assumed that we are operating in\nthe high SNR region where noise can be ignored. Now we have four nicely\narranged linear independent equations that can be used to find four independent\nvariables . So the fading channel\ndoes this magic to convert four dependent equations into four independent\nequations that can be easily solved (independence of equations can be checked\nby looking at the rank of channel matrix, which is 4 in this case). <\/p>\n\n\n\n

There are a number of methods to solve the above system of equations but my preference is using matrix inversion using MATLAB function pinv (pseudo inverse). In case AWGN noise is added, the received vector x would deviate from the above and we might get some errors in estimation of the signal vector. Now looking back at the no fading case, we ask the question: is their a way to arrange the equations such that four variables A, B, C and D can be found?  <\/p>\n","protected":false},"excerpt":{"rendered":"

From linear algebra we know that to find four unknowns we need four independent equations. There is no way we can find the values of \u00a0A, B, C and D from the above equations. To simplify the above equations we have removed AWGN but even in presence of AWGN we will have the same predicament. This shows that in the absence of fading there is no multiplexing gain however high the Signal to Noise Ratio is (in the above example SNR is infinite). <\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[12,22,9,81],"tags":[77,115,58,42,28],"class_list":["post-4336","post","type-post","status-publish","format-standard","hentry","category-ant","category-capacity","category-chancap","category-fundamentals","tag-awgn","tag-capacity","tag-fading","tag-mimo","tag-shannon-capacity"],"_links":{"self":[{"href":"https:\/\/www.raymaps.com\/index.php\/wp-json\/wp\/v2\/posts\/4336","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.raymaps.com\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.raymaps.com\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.raymaps.com\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.raymaps.com\/index.php\/wp-json\/wp\/v2\/comments?post=4336"}],"version-history":[{"count":15,"href":"https:\/\/www.raymaps.com\/index.php\/wp-json\/wp\/v2\/posts\/4336\/revisions"}],"predecessor-version":[{"id":4530,"href":"https:\/\/www.raymaps.com\/index.php\/wp-json\/wp\/v2\/posts\/4336\/revisions\/4530"}],"wp:attachment":[{"href":"https:\/\/www.raymaps.com\/index.php\/wp-json\/wp\/v2\/media?parent=4336"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.raymaps.com\/index.php\/wp-json\/wp\/v2\/categories?post=4336"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.raymaps.com\/index.php\/wp-json\/wp\/v2\/tags?post=4336"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}