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一种基于SEC‑DED的星载MIMO检测器的抗SEU方法,其特征是:步骤1.1:进行编码操作;将进入MIMO检测器的数据拆分为等长的两部分:数据A和数据B,再将数据A和数据B分别截断形成数据a<sub>11</sub>、数据a<sub>21</sub>和数据b<sub>11</sub>、数据b<sub>21</sub>;将数据a<sub>11</sub>、数据a<sub>21</sub>形成矩阵<maths num="0001" id="cmaths0001"><math><![CDATA[<mrow><mfenced open='[' close=']'><mtable><mtr><mtd><msubsup><mi>a</mi><mn>11</mn><mo>*</mo></msubsup></mtd><mtd><msubsup><mi>a</mi><mn>12</mn><mo>*</mo></msubsup></mtd></mtr><mtr><mtd><msubsup><mi>a</mi><mn>21</mn><mo>*</mo></msubsup></mtd><mtd><msubsup><mi>a</mi><mn>22</mn><mo>*</mo></msubsup></mtd></mtr></mtable></mfenced><mo>,</mo></mrow>]]></math><img file="FDA0000536428740000011.GIF" wi="222" he="141" /></maths>其中,<maths num="0002" id="cmaths0002"><math><![CDATA[<mrow><msubsup><mi>a</mi><mn>11</mn><mo>*</mo></msubsup><mo>=</mo><msub><mi>a</mi><mn>11</mn></msub><mo>,</mo><msubsup><mi>a</mi><mn>12</mn><mo>*</mo></msubsup><mo>=</mo><msub><mi>a</mi><mn>21</mn></msub><mo>,</mo><msubsup><mi>a</mi><mn>21</mn><mo>*</mo></msubsup><mo>=</mo><msub><mi>a</mi><mn>21</mn></msub><mo>,</mo><msubsup><mi>a</mi><mn>22</mn><mo>*</mo></msubsup><mo>=</mo><msub><mi>a</mi><mn>11</mn></msub><mo>;</mo></mrow>]]></math><img file="FDA0000536428740000012.GIF" wi="785" he="70" /></maths>将数据b<sub>11</sub>、数据b<sub>21</sub>形成矩阵<maths num="0003" id="cmaths0003"><math><![CDATA[<mrow><mfenced open='[' close=']'><mtable><mtr><mtd><msub><mi>b</mi><mn>11</mn></msub></mtd></mtr><mtr><mtd><msub><mi>b</mi><mn>21</mn></msub></mtd></mtr></mtable></mfenced><mo>;</mo></mrow>]]></math><img file="FDA0000536428740000013.GIF" wi="126" he="137" /></maths>按照海明校验码设计原则,在矩阵<maths num="0004" id="cmaths0004"><math><![CDATA[<mfenced open='[' close=']'><mtable><mtr><mtd><msubsup><mi>a</mi><mn>11</mn><mo>*</mo></msubsup></mtd><mtd><msubsup><mi>a</mi><mn>12</mn><mo>*</mo></msubsup></mtd></mtr><mtr><mtd><msubsup><mi>a</mi><mn>21</mn><mo>*</mo></msubsup></mtd><mtd><msubsup><mi>a</mi><mn>22</mn><mo>*</mo></msubsup></mtd></mtr></mtable></mfenced>]]></math><img file="FDA0000536428740000014.GIF" wi="196" he="141" /></maths>中加入两行带有矩阵行列信息的校验码数据,形成矩阵<maths num="0005" id="cmaths0005"><math><![CDATA[<mrow><mfenced open='[' close=']'><mtable><mtr><mtd><msubsup><mi>a</mi><mn>11</mn><mo>*</mo></msubsup></mtd><mtd><msubsup><mi>a</mi><mn>12</mn><mo>*</mo></msubsup></mtd></mtr><mtr><mtd><msubsup><mi>a</mi><mn>21</mn><mo>*</mo></msubsup></mtd><mtd><msubsup><mi>a</mi><mn>22</mn><mo>*</mo></msubsup></mtd></mtr><mtr><mtd><msub><mi>v</mi><mn>11</mn></msub></mtd><mtd><msub><mi>v</mi><mn>12</mn></msub></mtd></mtr><mtr><mtd><msub><mi>w</mi><mn>11</mn></msub></mtd><mtd><msub><mi>w</mi><mn>12</mn></msub></mtd></mtr></mtable></mfenced><mo>,</mo></mrow>]]></math><img file="FDA0000536428740000015.GIF" wi="222" he="265" /></maths>其中,<maths num="0006" id="cmaths0006"><math><![CDATA[<mrow><msub><mi>v</mi><mn>11</mn></msub><mo>=</mo><msubsup><mi>a</mi><mn>11</mn><mo>*</mo></msubsup><mo>+</mo><msubsup><mi>a</mi><mn>21</mn><mo>*</mo></msubsup><mo>;</mo><msub><mi>v</mi><mn>12</mn></msub><mo>=</mo><msubsup><mi>a</mi><mn>12</mn><mo>*</mo></msubsup><mo>+</mo><msubsup><mi>a</mi><mn>22</mn><mo>*</mo></msubsup><mo>;</mo></mrow>]]></math><img file="FDA0000536428740000016.GIF" wi="550" he="85" /></maths><maths num="0007" id="cmaths0007"><math><![CDATA[<mrow><msub><mi>w</mi><mn>11</mn></msub><mo>=</mo><msubsup><mi>a</mi><mn>11</mn><mo>*</mo></msubsup><mo>-</mo><msubsup><mi>a</mi><mn>21</mn><mo>*</mo></msubsup><mo>;</mo><msub><mi>w</mi><mn>12</mn></msub><mo>=</mo><msubsup><mi>a</mi><mn>12</mn><mo>*</mo></msubsup><mo>-</mo><msubsup><mi>a</mi><mn>22</mn><mo>*</mo></msubsup><mo>;</mo></mrow>]]></math><img file="FDA0000536428740000017.GIF" wi="549" he="84" /></maths>然后将矩阵<maths num="0008" id="cmaths0008"><math><![CDATA[<mfenced open='[' close=']'><mtable><mtr><mtd><msubsup><mi>a</mi><mn>11</mn><mo>*</mo></msubsup></mtd><mtd><msubsup><mi>a</mi><mn>12</mn><mo>*</mo></msubsup></mtd></mtr><mtr><mtd><msubsup><mi>a</mi><mn>21</mn><mo>*</mo></msubsup></mtd><mtd><msubsup><mi>a</mi><mn>22</mn><mo>*</mo></msubsup></mtd></mtr><mtr><mtd><msub><mi>v</mi><mn>11</mn></msub></mtd><mtd><msub><mi>v</mi><mn>12</mn></msub></mtd></mtr><mtr><mtd><msub><mi>w</mi><mn>11</mn></msub></mtd><mtd><msub><mi>w</mi><mn>12</mn></msub></mtd></mtr></mtable></mfenced>]]></math><img file="FDA0000536428740000018.GIF" wi="201" he="264" /></maths>和矩阵<maths num="0009" id="cmaths0009"><math><![CDATA[<mfenced open='[' close=']'><mtable><mtr><mtd><msub><mi>b</mi><mn>11</mn></msub></mtd></mtr><mtr><mtd><msub><mi>b</mi><mn>21</mn></msub></mtd></mtr></mtable></mfenced>]]></math><img file="FDA0000536428740000019.GIF" wi="92" he="136" /></maths>相乘得输出矩阵<maths num="0010" id="cmaths0010"><math><![CDATA[<mrow><mfenced open='[' close=']'><mtable><mtr><mtd><msub><mi>c</mi><mn>11</mn></msub></mtd></mtr><mtr><mtd><msub><mi>c</mi><mn>21</mn></msub></mtd></mtr><mtr><mtd><msub><mi>v</mi><mn>1</mn></msub></mtd></mtr><mtr><mtd><msub><mi>w</mi><mn>1</mn></msub></mtd></mtr></mtable></mfenced><mo>,</mo></mrow>]]></math><img file="FDA00005364287400000110.GIF" wi="125" he="265" /></maths>表示如下:<maths num="0011" id="cmaths0011"><math><![CDATA[<mrow><mfenced open='[' close=']'><mtable><mtr><mtd><msub><mi>c</mi><mn>11</mn></msub></mtd></mtr><mtr><mtd><msub><mi>c</mi><mn>21</mn></msub></mtd></mtr><mtr><mtd><msub><mi>v</mi><mn>1</mn></msub></mtd></mtr><mtr><mtd><msub><mi>w</mi><mn>1</mn></msub></mtd></mtr></mtable></mfenced><mo>=</mo><mfenced open='[' close=']'><mtable><mtr><mtd><msubsup><mi>a</mi><mn>11</mn><mo>*</mo></msubsup></mtd><mtd><msubsup><mi>a</mi><mn>12</mn><mo>*</mo></msubsup></mtd></mtr><mtr><mtd><msubsup><mi>a</mi><mn>21</mn><mo>*</mo></msubsup></mtd><mtd><msubsup><mi>a</mi><mn>22</mn><mo>*</mo></msubsup></mtd></mtr><mtr><mtd><msub><mi>v</mi><mn>11</mn></msub></mtd><mtd><msub><mi>v</mi><mn>12</mn></msub></mtd></mtr><mtr><mtd><msub><mi>w</mi><mn>11</mn></msub></mtd><mtd><msub><mi>w</mi><mn>12</mn></msub></mtd></mtr></mtable></mfenced><mo>×</mo><mfenced open='[' close=']'><mtable><mtr><mtd><msub><mi>b</mi><mn>11</mn></msub></mtd></mtr><mtr><mtd><msub><mi>b</mi><mn>21</mn></msub></mtd></mtr></mtable></mfenced><mo>;</mo></mrow>]]></math><img file="FDA00005364287400000111.GIF" wi="470" he="264" /></maths>步骤1.2:进行检错操作;定义标志量m和标志量n,其中,m=c<sub>11</sub>+c<sub>21</sub>‑v<sub>1</sub>n=c<sub>11</sub>‑c<sub>21</sub>‑w<sub>1</sub>通过标志量m和标志量n检错的方法如下:情况A1:如果m=0,并且n=0,则没有发生错误;情况B1:如果m=0,并且n≠0,则w<sub>1</sub>发生错误;情况C1:如果m≠0,并且n=0,则v<sub>1</sub>发生错误;情况D1:如果m≠0,并且n≠0,则计算m+n;当m+n=0时,c<sub>21</sub>计算错误;当m+n≠0时,c<sub>11</sub>计算错误;步骤1.3:进行纠错操作;在情况D1中:如果发现c<sub>21</sub>计算错误,则用c<sub>21</sub>减去标志量m就可以获得正确的c<sub>21</sub>;如果发现c<sub>11</sub>计算错误,则用c<sub>11</sub>减去标志量m就可以获得正确的c<sub>21</sub>;步骤1.4:进行模块重新加载操作;如果w<sub>1</sub>或v<sub>1</sub>发生错误,对MIMO检测器的校验模块重新加载,如果c<sub>21</sub>或c<sub>11</sub>发生错误,对MIMO检测器的乘法器模块重新加载。 |