how to generate a sequence x of length 200 samples,where the first 30 values equal 1 and the remainder have a value of 0.
MATLAB: Generating sequence
mat
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The 11- chip sequence that corresponds to a Barker code of length 11 is:
[1,0,1,1,0,1,1,1,0,0,0]
You can generate this sequence by using any 11 degree polynomial with the eleven states set up as the PN code values. In other words, the initial states parameter should be set to:
[1,0,1,1,0,1,1,1,0,0,0]
and the Generator polynomial can be any 11th order polynomial, such as [0 -11], for example.
If you want repeated sets of this same sequence, you will not be able to generate that using the PN block. You may however, send those values into the workspace and then use a From Workspace block with the 'Form output after final data value by' option set to 'CyclicRepetition'.
You can also use 'Barker Code Generator' block from 'Communications Blockset->Comm Sources->Sequence Generators' Library to achieve the functionality.
Here's the only way I think you can do your particular choice set - don't bother storing the choices in memory, just print them to the screen:
function solutionsFound = perms_bounded(choices, minBounds, maxBounds)% Since we're only printing solutions, let's use strings
if ~iscellstr(choices) choices = arrayfun(@num2str, choices, 'UniformOutput',false);end% Which choice indices can be placed in each location?
validInds = arrayfun(@(x)find(minBounds<=x & maxBounds>=x),1:length(choices),'UniformOutput',false);setLens = cellfun(@numel,validInds);n = length(setLens);iSet = zeros(1, n); % Which indices of each indice set are we up to?
candSet = zeros(1,n); % What's the combination we have so far?
solutionsFound = 0;rLevel = 1;while 1 iSet(rLevel) = iSet(rLevel)+1; if iSet(rLevel)>setLens(rLevel) % We've run out of indices for this location. Exit if we're
% finished, otherwise go back one rLevel and start again.
if all(iSet>=setLens), break, end iSet(rLevel) = 0; rLevel = rLevel-1; continue; end candidate = validInds{rLevel}(iSet(rLevel)); % Has this candidate already been used?
if any(candSet(1:rLevel-1)==candidate) continue; % Already in use, try the next candidate.
else % Place this candidate and move to the next level
candSet(rLevel) = candidate; if rLevel<n rLevel = rLevel+1; else solutionsFound = solutionsFound+1; fprintf('#%d: %s\n',solutionsFound, sprintf('%s ',choices{candSet})) end endendUsing the small set we get:
choices = {'A','B','C','D'}; minBounds = [1 1 2 3]; maxBounds = [4 3 4 4]; >> numSolutions = perms_bounded(choices, minBounds, maxBounds) #1: A B C D #2: A B D C #3: A C B D #4: B A C D #5: B A D C #6: B C A D #7: B C D A numSolutions = 7Using the large set we get:
>> perms_bounded(choices, minBounds, maxBounds)#1: a b c d e f g h i j k l m n o p q r s t u v w x y z aa ab ac ad ae af ag ah ai aj ak al am an ao ap aq ar as at au av aw ax ay #2: a b c d e f g h i j k l m n o p q r s t u v w x y z aa ab ac ad ae af ag ah ai aj ak al am an ao ap aq as ar at au av aw ax ay #3: a b c d e f g h i j k l m n o p q r s t u v w x y z aa ab ac ad ae af ag ah ai aj ak al am an ao ap ar aq as at au av aw ax ay #4: a b c d e f g h i j k l m n o p q r s t u v w x y z aa ab ac ad ae af ag ah ai aj ak al am an ao ap ar as aq at au av aw ax ay ...#49998: a b c d e f g h i j k l m n o p q r s t u v w x y aa z ab ac ag ad ae af ah ai al aj ak am an aq ao ap as ar at au av aw ax ay #49999: a b c d e f g h i j k l m n o p q r s t u v w x y aa z ab ac ag ad ae af ah ai al aj ak am aq an ao ap ar as at au av aw ax ay #50000: a b c d e f g h i j k l m n o p q r s t u v w x y aa z ab ac ag ad ae af ah ai al aj ak am aq an ao ap as ar at au av aw ax ay #50001: a b c d e f g h i j k l m n o p q r s t u v w x y aa z ab ac ag ad ae af ah ai al aj ak aq am an ao ap ar as at au av aw ax ay ...I've truncated the output after 50000 valid solutions. Note that choices a-through-y are still on their first valid combination. If you want to see all solutions, you may need a few weeks for it to run.
Thanks, Sven.
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