使用滞后来减少噪声的信号阈值

Question

我正在使用简单的阈值来检测信号中的零边缘。由于噪声可能非常强烈，因此我正在使用滞后来改善结果。尽管这有很大帮助，但它也会减慢很多。有没有办法改善这一点，甚至是一种更好的方法来计算这一点？目前，我正在使用直线循环方法：

% generate a signal
t = linspace(0, 10, 1000);
y = sin(2 * pi * t); 
% constants
threshold = 0;
hyst = 0.1;
% thresholding
yth = zeros(size(y));
for i = 2:length(y)
    if yth(i - 1) > 0.5 
        yth(i) = y(i) > (threshold - hyst);
    else
        yth(i) = y(i) > (threshold + hyst);
    end 
end

与yth = y > threshold相比，这要慢得多。

Answer 1

可以通过预先计算yth(i)的两种可能性来获得改进(仅适用于大型输入；在R2017B上(：

function yth = idea1()
  yth = false(size(y)); % change #1 - `false` vs `zeros`
  c1 = y > (th - hyst); % change #2 - precompute c1, c2
  c2 = y > (th + hyst);
  for k = 2:numel(y)
      if yth(k - 1) 
          yth(k) = c1(k);
      else
          yth(k) = c2(k);
      end 
  end
end

通常，可以通过矢量化获得很大的改进。要介绍此问题，我们必须了解切换逻辑，因此让我们将其插入文字：

• 从开始模式2 。
• 模式1 ：从c1获取值。只要c1包含 true值 - 拿走。当遇到 false值时，切换到模式2 。
• 模式2 ：从c2获取值。只要c2包含false值 - 取之量即可。当遇到true值时，切换到模式1 。

因此，如果我们能找到过渡地点，我们实际上就完成了。

经过一定的反复试验，虽然我无法以提高性能的方式摆脱循环，但我确实得出了可能的结论(idea2(。此外，查看编码的正确yth，我想出了一些相当不错的近似值(idea3和idea4( - 尽管这些需要对其他输入进行调整。

也许有人可以使用它来找到一个更聪明的实现，并具有较少的冗余计算。我的完整代码在下面提供。编码算法的RLE是根据此答案进行了调整的，并从这里解码了。

function q48637952
% generate a signal
t = linspace(0, 10, 1000).';
y = sin(2 * pi * t); 
% constants
th = 0;
hyst = 0.1;
%% Comaprison:
% Correctness:
R = {originalIdea(), idea1(), idea2()};
assert(isequal(R{:}));
% Runtime:
T = [timeit(@originalIdea,1), timeit(@idea1,1), timeit(@idea2,1)];
disp(T);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function yth = originalIdea()
  yth = zeros(size(y));
  for i = 2:length(y)
      if yth(i - 1) > 0.5 
          yth(i) = y(i) > (th - hyst);
      else
          yth(i) = y(i) > (th + hyst);
      end 
  end
end
function yth = idea1()
  yth = false(size(y)); % change #1 - `false` vs `zeros`
  c1 = y > (th - hyst); % change #2 - precompute c1, c2
  c2 = y > (th + hyst);
  for k = 2:numel(y)
      if yth(k - 1) 
          yth(k) = c1(k);
      else
          yth(k) = c2(k);
      end 
  end
end
function yth = idea2()
  c = [y > (th - hyst), y > (th + hyst)];
  % Using Run-length encoding:
  [v1,l1] = rlEncode(c(:,1));
  [v2,l2] = rlEncode(c(:,2));
  rle = cat(3,[v1 l1 cumsum(l1)],[v2 l2 cumsum(l2)]);
  % rle(:,:,2) is similar to rle(:,:,1) with small changes:
  %  - col1 is circshifted by 1 and 
  %  - col2 has the 1st and last elements switched
  newenc = reshape([rle(1:2:end,3,2), rle(1:2:end,3,1)].', [], 1);
  yth = rlDecode(rle(:,1,2), [newenc(1); diff(newenc(1:end-1))]);
end
function yth = idea3()
  % Approximation of yth, should be almost indistinguishable with high resolution
  yth = [0 0 repmat(repelem([1,0],50), 1, ceil(numel(y)/(2*50)) )].';
  % The amount of zeros at the beginning as well as the value "50" are problem-specific 
  % and need to be computed using signal-specific considerations
  yth = yth(1:1000);
end
function yth = idea4()
  % Another approximation
  yth = circshift(y > th, 1);
  % The value "1" is problem-specific
end
end % q48637952
%% RLE (de)compression:
function [vals, lens] = rlEncode(vec)
  J = find(diff([vec(1)-1; vec(:)]));
  vals = vec(J);
  lens = diff([J; numel(vec)+1]);
end
function vec = rlDecode(vals, lens)
  l = cumsum([ 1; lens(:) ]);
    z = zeros(1, l(end)-1);
    z(l(1:end-1)) = 1;
    vec = vals(cumsum(z));
end