Find zero crossings in each row of a matrix (321 x 123 double)

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NotA_Programmer
NotA_Programmer 2022-5-9
评论： Star Strider 2022-5-9
How can I find the no. of times the signal crosses zero. Signal is in form of a matrix (321 x 123 double), where each row represents the data one signal.

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Star Strider
Star Strider 2022-5-9
Ran in:
t = linspace(0, 5, 123); % Create Data
A = sin(randi(9,5,1)*2*pi*t/5 + 2*pi*randi(9,5,1)) % Sine Curves With Varying Phases
A = 5×123
-0.0000 0.1028 0.2046 0.3041 0.4005 0.4925 0.5794 0.6602 0.7339 0.7998 0.8573 0.9057 0.9445 0.9733 0.9917 0.9997 0.9970 0.9838 0.9601 0.9263 0.8827 0.8297 0.7679 0.6979 0.6206 0.5367 0.4471 0.3528 0.2547 0.1539 -0.0000 0.3041 0.5794 0.7998 0.9445 0.9997 0.9601 0.8297 0.6206 0.3528 0.0515 -0.2547 -0.5367 -0.7679 -0.9263 -0.9970 -0.9733 -0.8573 -0.6602 -0.4005 -0.1028 0.2046 0.4925 0.7339 0.9057 0.9917 0.9838 0.8827 0.6979 0.4471 -0.0000 0.3528 0.6602 0.8827 0.9917 0.9733 0.8297 0.5794 0.2547 -0.1028 -0.4471 -0.7339 -0.9263 -0.9997 -0.9445 -0.7679 -0.4925 -0.1539 0.2046 0.5367 0.7998 0.9601 0.9970 0.9057 0.6979 0.4005 0.0515 -0.3041 -0.6206 -0.8573 -0.0000 0.3041 0.5794 0.7998 0.9445 0.9997 0.9601 0.8297 0.6206 0.3528 0.0515 -0.2547 -0.5367 -0.7679 -0.9263 -0.9970 -0.9733 -0.8573 -0.6602 -0.4005 -0.1028 0.2046 0.4925 0.7339 0.9057 0.9917 0.9838 0.8827 0.6979 0.4471 -0.0000 0.3041 0.5794 0.7998 0.9445 0.9997 0.9601 0.8297 0.6206 0.3528 0.0515 -0.2547 -0.5367 -0.7679 -0.9263 -0.9970 -0.9733 -0.8573 -0.6602 -0.4005 -0.1028 0.2046 0.4925 0.7339 0.9057 0.9917 0.9838 0.8827 0.6979 0.4471
for k = 1:size(A,1)
zc{k} = find(diff(sign(A(k,:)),[],2)); % Zero-Crossings For Row 'k'
end
figure
plot(t,A(1,:))
hold on
plot(t(zc{1}), zeros(size(zc{1})),'+r')
hold off
grid
figure
plot(t,A(5,:))
hold on
plot(t(zc{5}), zeros(size(zc{5})),'+r')
hold off
grid
Use interp1 in a loop for each zero crossing in each row to find the exact ‘x’ values for each zero-crossing.
.

更多回答（1 个）

Torsten
Torsten 2022-5-9
  1 个评论
NotA_Programmer
NotA_Programmer 2022-5-9
Torsten, thanks for your response, but I want a seperate matrix that contains locations or indices where signal crosses 0.
The sinal is of (321 x 123 double) matrix and I want a matrix of (321 x location_where_signal_crosses_zero) but while using the below code I am getting zm of different dimension
t = [0:0.048780:6];
y = Data; %Data (321 x 123 double)
zci = @(v) find(v(:).*circshift(v(:), [-1 0]) <= 0);
zx = zci(y);

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