摘要：MATLAB?是一個強(qiáng)有力的工具，可用來快速分析從模數(shù)轉(zhuǎn)換器(ADC)輸出所捕獲的數(shù)據(jù)。本應(yīng)用筆記演示了如何使用MATLAB來突破邏輯分析儀存儲深度的局限。描述并比較了三種數(shù)碼拼接方式(基本，超前和反轉(zhuǎn))。并給出了三種方法所得的結(jié)果。

介紹

要評估一個高速模數(shù)轉(zhuǎn)換器(ADC)的性能，就需要捕獲其數(shù)字輸出碼，然后進(jìn)行分析。邏輯分析儀的存儲深度常常成為一個重要局限，妨礙系統(tǒng)捕獲足夠的數(shù)據(jù)點(diǎn)，以生成高分辨率FFT，或者精確的INL/DNL圖。解決該問題的一個簡單辦法是使用某種數(shù)學(xué)工具，例如MATLAB? (圖1)，將多組數(shù)據(jù)連接起來。連接數(shù)據(jù)的一個缺點(diǎn)是，通常會在兩組數(shù)據(jù)之間的連接點(diǎn)出現(xiàn)很大的不連續(xù)性。盡管不連續(xù)性對INL/DNL圖的影響極小，但對高分辨率FFT而言，幾乎是毀滅性的(圖2)。


圖1. 連接后的數(shù)據(jù)在兩組數(shù)據(jù)之間出現(xiàn)不連續(xù)。


圖2. a) 捕獲單組16384點(diǎn)數(shù)據(jù)并分析；b) 捕獲兩組8192點(diǎn)數(shù)據(jù)，連接，然后分析“拼接”技術(shù)。

有一種辦法可以消除不連續(xù)性，就是在各組數(shù)據(jù)中尋找相同的點(diǎn)簇(一般為3到4個點(diǎn))，然后在這些點(diǎn)將兩組數(shù)據(jù)“拼接”在一起(圖3)。最簡單的“拼接”方法是，記錄下第一組數(shù)據(jù)中的最后四個點(diǎn)，然后在第二組數(shù)據(jù)里尋找相同的點(diǎn)簇。相同點(diǎn)簇出現(xiàn)在第二組數(shù)據(jù)中的位置稱為“拼接點(diǎn)”。第二組數(shù)據(jù)中在拼接點(diǎn)之前的所有數(shù)據(jù)均被舍棄；第二組數(shù)據(jù)中的剩余部分與第一組數(shù)據(jù)合并。這種技術(shù)即所謂的基本數(shù)碼拼接，實現(xiàn)起來非常簡單，可以在MATLAB中非?？斓剡\(yùn)行。


圖3. 基本數(shù)碼拼接后得到的最終“拼接”數(shù)組。

采用基本拼接方法拼接數(shù)據(jù)時，有時必須丟掉第二組數(shù)據(jù)中的近一半，才能找到與第一組數(shù)據(jù)最后四個點(diǎn)相匹配的一簇點(diǎn)。作為另一種選擇，丟掉第一組數(shù)據(jù)尾部的幾個點(diǎn)，常常有助于找到更靠近第二組數(shù)據(jù)起點(diǎn)的拼接點(diǎn)(圖4)。然而，通過丟掉第一組數(shù)據(jù)尾部、第二組數(shù)據(jù)頭部的部分采樣點(diǎn)來尋找匹配點(diǎn)的方法實現(xiàn)起來比較困難。這種處理被稱為超前數(shù)碼拼接。理想拼接點(diǎn)應(yīng)該能夠保留盡可能多的數(shù)據(jù)點(diǎn)，尋找這樣的拼接點(diǎn)需要認(rèn)真的考慮和一定的編程技巧。正確地實現(xiàn)之后，超前拼接技術(shù)通常能夠得到兩組小數(shù)組所含數(shù)據(jù)點(diǎn)總數(shù)的至少90%。


圖4. 采用超前拼接技術(shù)尋找理想拼接點(diǎn)，最終得到“拼接”后的數(shù)組。

將第二組數(shù)據(jù)(數(shù)組B)拼接到第一組數(shù)據(jù)(數(shù)組A)之前被稱為反轉(zhuǎn)拼接，這種方式有可能得到更大的拼接數(shù)組(圖5)。但是，這種技術(shù)會使處理時間翻倍，因為必須在A領(lǐng)先于B和B領(lǐng)先于A兩種情況下尋找拼接點(diǎn)。此外，當(dāng)與其他拼接技術(shù)一起使用時，反轉(zhuǎn)拼接方式所帶來的好處通常很少。因此，對于較慢的PC，反轉(zhuǎn)拼接技術(shù)大幅度增加的處理時間開銷較之它所帶來的好處而言并不太值。表1詳細(xì)比較了這三種代碼拼接方法。


圖5. 反轉(zhuǎn)拼接的處理時間加倍，但常常收效甚微。

表1. 三種拼接技術(shù)對比*

Stitch Technique	Size of Final Data Set					Description
	Data Set Numbers			# of codes (averaged)	% of two data sets (averaged)
	1 + 2	3 + 4	1 + 4
Concatenate?	N/A			16384	100%	Will produce erroneous FFT; however, INL/DNL can be extracted from this data.
Basic	11060	8192?	14384	11212	68.4%	FFT is useable for calculating figures of merit.
Reverse	11060	8192?	14384	11212	68.4%
Advanced	13790	16046	16022	15286	93.3%
Advanced + Reverse	15427	16176	16022	15875	96.9%

*采用上述拼接技術(shù)對兩組8K (8192個)數(shù)據(jù)進(jìn)行拼接。為確保準(zhǔn)確性，采用四組8192點(diǎn)數(shù)據(jù)(分別編號為1至4)重復(fù)本測試。每組測試所合成的數(shù)據(jù)取平均后列于測試數(shù)據(jù)的右側(cè)。
? 直接連接總能得到100%的數(shù)據(jù)。
? 無法拼接數(shù)據(jù)。

MATLAB函數(shù)說明

本文后附的MATLAB代碼(附錄A和B中的StitchMatrices和FindStitchPoint)將上述論點(diǎn)結(jié)合到一個易于使用的函數(shù)中。這些函數(shù)可接受兩組數(shù)據(jù)(MATLAB中的單列矩陣)和幾個輸入變量(用來選擇超前/反轉(zhuǎn)拼接功能)。FindStitchPoint例程用來確定拼接點(diǎn)在數(shù)組A和B中的偏移量。StitchMatrices例程則根據(jù)FindStitchPoint例程給出的偏移量對兩組數(shù)據(jù)A和B進(jìn)行舍棄和組合。同時，最終數(shù)據(jù)的拼接點(diǎn)被記錄在PrevStitchBins數(shù)組中，以便于后續(xù)處理。當(dāng)拼接多組數(shù)據(jù)時，PrevStitchBins可保存老的拼接點(diǎn)。

結(jié)論

拼接兩組數(shù)據(jù)可以得到一組理想的結(jié)果。圖6給出了三組8192點(diǎn)數(shù)據(jù)使用上述拼接技術(shù)拼接起來(使用5個拼接點(diǎn))后的FFT圖。所得的FFT幾乎與前面圖2a所示，基于16384個連續(xù)點(diǎn)所得結(jié)果相同。


圖6. 數(shù)碼拼接后得到精確的FFT圖。

附錄A：StitchMatrices例程(StitchMatrices.m)

function [StitchedMatrix, StitchBins] = StitchMatrices(MatrixA, ...
   				MatrixB, StitchNumber, PrevStitchBins, ...
				AdvCodeStitchEnabled, ReverseStitchEnabled);
%Stitch Matrices Function
%Revision 1.0
%
%By Donald Schelle, May 2005
%Maxim Integrated Products
%120 San Gabriel Drive
%Sunnyvale, CA, 94086
%
%This function will take two matrices (MatrixA and MatrixB), find a 
%given number (StitchNumber) of identical points in each and 
%concatenate the two matrices into one.
%
%Inputs = MatrixA, MatrixB (Data Matrices)
%		    StitchNumber (Number of points to match)
%			 PrevStitchBins (Bins of Previous Stitches in MatrixA)
%			 AdvStitchEnabled (0 = NO, 1 = YES)
%			 ReverseStitchEnabled (0 = NO, 1 = YES)
%Output = StitchedMatrix (MatrixA + MatrixB)
%			 StitchBins (bins of StitchedMatrix where the two 
%						    matrices were joined.)
%
%If the matrices can not be joined the function will output a NaN
%for both the StitchedMatrix variable and the StitchBins variable
%--------------------------------------------------------------------------

%Check to see that there are at least TWO StitchNumber Points
if StitchNumber < 2,
   %Requested less than 2 stitch points
   StitchedMatrix = NaN;
   StitchBins = NaN;
   return;
end;

%Calculate Size of MatrixA and MatrixB
[SizeA, Junk] = size(MatrixA);
[SizeB, Junk] = size(MatrixB);

%Find the Stitch Points in MatrixB
[NormalA, NormalB] = FindStitchPoint(MatrixA, MatrixB, ...
   					StitchNumber, AdvCodeStitchEnabled);
%Calculate the size of the NormalStitched Matrix
NormalStitchedSize = NormalA + SizeB - NormalB + 1;
                            
%Check to see if the reverse function is enabled
if ReverseStitchEnabled == 1,
   %Find Stitch Points for Reverse Matrices
   [ReverseB, ReverseA] = FindStitchPoint(MatrixB, MatrixA, ...
      					StitchNumber, AdvCodeStitchEnabled);
   %Calculate the size of the Revered Stitched Matrix
   ReverseStitchedSize = ReverseB + SizeA - ReverseA + 1;
else
   %Set Values to defaults
   ReverseStitchedSize = NaN;		%MatrixB/A Stitch Size
   ReverseA = NaN;
   ReverseB = NaN;
end;

%Check to if it's possible to stitch two matrices
if isnan(NormalStitchedSize) & isnan(ReverseStitchedSize) == 1,
   %The two matrices could not be stitched
   StitchedMatrix = NaN;
   StitchBins = NaN;
   return;
end;

%--------------------- Normal Matrix Stitching Routine ---------------
if (NormalStitchedSize >= ReverseStitchedSize)| ...
   			isnan(ReverseStitchedSize) == 1,
   %Stitch MatrixB to the end of MatrixA
   StitchedMatrix = cat(1, MatrixA(1:NormalA), MatrixB(NormalB:SizeB));
   
   %Update Stitch Bins
   if isnan(PrevStitchBins) == 1,
      %There are no previous stitch bins
      StitchBins = [NormalA, NormalA + StitchNumber - 1];
   else
      %There are previous stitch bins
      %Check for Snipped Stitches
      [SizeStitchBins, Junk] = size(PrevStitchBins);
      
      while (PrevStitchBins(SizeStitchBins, 2) > (NormalA - 1)),
         %Second Bin is snipped from matrix. Check if first bin is snipped.
         if (PrevStitchBins(SizeStitchBins, 1) > (NormalA - 1)),
            %First Bin is snipped too. Delete Bin Pair
            PrevStitchBins = PrevStitchBins(1:(SizeStitchBins-1),:);
         else
            %First Bin is not snipped but second bin is snipped
            %Shrink Stitch Size
            PrevStitchBins(SizeStitchBins, 2) = NormalA - 1;
         end;
         
         %Calculate size of new PrevStitchBin Matrix
         [SizeStitchBins, Junk] = size(PrevStitchBins);
      end;
               
      %Insert New StitchBins      
      [SizeStitchBins, Junk] = size(PrevStitchBins);
      StitchBins = PrevStitchBins;
      StitchBins(SizeStitchBins + 1, :) = ...
      [NormalA, NormalA + StitchNumber - 1];
      
      %Check to see if the last two stitches need to be combined
      [SizeStitchBins, Junk] = size(StitchBins);
      if StitchBins(SizeStitchBins,1) == ...
            (StitchBins((SizeStitchBins - 1),2) + 1),
         %Combine Stitches
         StitchBins((SizeStitchBins - 1),2) = StitchBins((SizeStitchBins),2);
         %Shorten StitchBin Matrix
         StitchBins = StitchBins(1:(SizeStitchBins - 1),:);
      end;
      
   end;   
end;


%--------------------- Reverse Matrix Stitching Routine ---------------
if (ReverseStitchedSize >= NormalStitchedSize)| ...
   			isnan(NormalStitchedSize) == 1,
   %Stitch MatrixA to the end of MatrixB
   StitchedMatrix = cat(1,MatrixB(1:ReverseB), MatrixA(ReverseA:SizeA));
   
   %Update Stitch Bins
   if isnan(PrevStitchBins) == 1,
      %There are no previous stitch bins
      StitchBins = [ReverseB, ReverseB + StitchNumber - 1];
   else
      %There are previous stitch bins
      %Check for Snipped Stitches            
      while (PrevStitchBins(1,1) < (ReverseA + StitchNumber - 1)),
         %First Bin is snipped from matrix. Check if second is snipped
         if (PrevStitchBins(1,2) < (ReverseA + StitchNumber - 1)),
            %Second Bin is snipped too. Delete Bad Pair
            [SizeStitchBins, Junk] = size(PrevStitchBins);
      	     PrevStitchBins = PrevStitchBins(2:SizeStitchBins, :);
         else
            %Second Bin is not snipped, but first bin is snipped
            %Shrink Old Stitch Size
            PrevStitchBins(1,1) = ReverseA + StitchNumber - 1;
         end;
      end;
      
      %Offset Stitch Bins by inserted amount
      StitchBins = PrevStitchBins + ReverseB - ReverseA + 1; 
      %Make Room for new StitchBins
      [SizeStitchBins, Junk] = size(PrevStitchBins);
      StitchBins(2:SizeStitchBins+1, :) = StitchBins;
      %Insert New Stitch Bins
      StitchBins(1,:) = [ReverseB, ReverseB + StitchNumber - 1];
                  
      %Combine close stitches
      if StitchBins(1,2) == StitchBins(2,1) - 1,
         %Combine Stitches
         StitchBins(2,1) = StitchBins(1,1);
         %Shrink Stitch Bins Matrix
	 [SizeStitchBins, Junk] = size(StitchBins);
         StitchBins = StitchBins(2:SizeStitchBins,:);   
      end;
      
   end;
end;

附錄B：FindStitchPoint例程(FindStitchPoint.m)

function [OutputBinA, OutputBinB]=FindStitchPoint(MatrixA, MatrixB, ...
   				MatchNumber, AdvancedStitchFindEnabled)
%Find Stitch Points Function
%Revision 1.0
%
%By Donald Schelle, May 2005
%Maxim Integrated Products
%120 San Gabriel Drive
%Sunnyvale, CA, 94086
%
%This function will find the IDEAL stitch point in Matrix B given
%the number of data points to match
%
%Inputs = MatrixA
%			 MatrixB
%			 Number of Records to Match
%			 Advanced Stitch Find Enabled (0 = NO, 1 = YES)
%Output = (OutputBinA) End Bin of MatrixA to stitch data
%			 (OutputBinB) Start Bin of Matrix B to stitch data
%
%If no bins are found, the function will output a NaN
%--------------------------------------------------------------------------

%Do argument error checking to see if there is enough arguments
if nargin < 2,
   %The user has not supplied enough arguments
   disp('Function requires TWO Matrices');
   OutputBinA = NaN;
   OutputBinB = NaN;
   return;
elseif nargin < 3,
   disp('Select a number of points to match');
   OutputBinA = NaN;
   OutputBinB = NaN;
   return;
elseif nargin == 3,
   %Advanced code stitching is NOT enabled
   OutputBinA = NaN;
   AdvancedStitchFindEnabled=0;
end;

%Ensure that Matrix A and B are single ROW matrices
[row col] = size(MatrixA);
if row > col, MatrixA = MatrixA'; end;
[row col] = size(MatrixB);
if row > col, MatrixB = MatrixB'; end;

%Determine Size of Matrices
[Junk, SizeA] = size(MatrixA);
[Junk, SizeB] = size(MatrixB);

%Initialize OutputBinB to NaN (which means that NO stitch points are found)
OutputBinB = NaN;
%Set initial size of BinA
BinA = SizeA - MatchNumber + 1;
%Initialize BinStop Variable
BinStop = SizeA-100;

%Loop to search through Matrix B numerous times.  This loop is only
%excuted once if Advanced Stitch Find is disabled.  The loop will stop when
%the 'ideal' stitch point is found
while BinA > BinStop, 
   %Stuff the Match Numbers into a separate Matrix
	MatchMatrix = MatrixA(BinA:BinA+MatchNumber - 1);
   
   %Find all bins in MatrixB that match the first number of the Match Matrix
	MatchedBins = find(MatrixB == MatchMatrix(1));
   
   %Compare the 2nd through nth number of the Match Matrix with the
   %prospective series of numbers in MatrixB
   
   %Calculate the size of the Matched Bins Matrix
   [Junk, SizeMatchedBins] = size(MatchedBins);
   
   %The advanced stitch mode optimizes search time by eliminating
   %bad stitch points that would result in the final concatenated
   %matrix being smaller than the last set of stitch points
   if isnan(OutputBinB) == 0,
      %A Stitch Point exists from a previous run.  Elimiate bad stitch points
      %Calculate critical Stitch Point
      MatrixSize = OutputBinA + (SizeB-OutputBinB) + 1;
      CriticalBin = BinA + SizeB - MatrixSize - 1;
      %Find maximum number in the MatchMatrix
      BadBin = find(MatchedBins > CriticalBin);
      %Eliminate Bad Bins (if there are any)
      if isempty(BadBin) == 0,
         MatchedBins = MatchedBins(1:BadBin(1) - 1);
      end;
      %Calculate size of new Matched Bins Matrix
      [Junk, SizeMatchedBins] = size(MatchedBins);       
   end;
   
   %loop to cycle through initial matched bins
   for i=1:SizeMatchedBins,
      %Check to make sure that there isn't a MatrixB overrun
      if (MatchedBins(i) + MatchNumber - 1) > SizeB,
         break;
      end;
      
      %Assume that next few codes will match and set StitchBinGood = true
      StitchBinGood = 1;	   
      %Initialize MatchMatrixCounter
      Count = 1;
      %Cycle through MatrixB and compare Numbers with the MatchMatrix
      for j=MatchedBins(i):(MatchedBins(i) + MatchNumber - 1),
         if MatchMatrix(Count)==MatrixB(j),
            %Number is good, continue and check next number
	     Count = Count + 1;
	 else
            %Number is bad, break loop and try next sequence
            StitchBinGood = 0;
	     break;
	 end;
      end;
	   
      if StitchBinGood == 1,
         %The optimal (first) stitch has been found
         %Record the End bin of MatrixA
         %Record the Start bin of MatrixB
         OutputBinA = BinA;
         OutputBinB = MatchedBins(i) + 1;
         %Calculate the size of the joined Matrix and a new BinStop#
         BinStop = OutputBinA-OutputBinB+1;
         break;
      end;  
   end;
   
   if AdvancedStitchFindEnabled == 1,
      %Advanced Stitch Find is enabled and we should make a new match
      %matrix and search for these numbers
      BinA = BinA - 1;
   else
      %Advanced Stitch Find is disabled and we should end the loop
      break;
   end;
end;

%Check to see if NO Bins Matched
if isnan(OutputBinB) == 1,
   %NO Bins matched
   OutputBinA = NaN;
end;