Try these:
t = 0 * pi:0.119:4 * pi;
a = 57;
phase_angle = 0.26;
y = a*sin((67*pi*t) + phase_angle);
y_ofst = y-40;
zci = @(v) find(v(:).*circshift(v(:), [-1 0]) <= 0); % Returns Approximate Zero-Crossing Indices Of Argument Vector
y40 = zci(y_ofst);
for k = 1:numel(y40)
idxrng = y40(k)+[-1,1]
B = [t(idxrng); ones(size(t(idxrng)))].' \ y_ofst(idxrng).' % Linear Regression
xxct(k) = -B(2)/B(1)
end
figure
plot(t,y, xxct,40+[0 0],'pg')
alternatively:
for k = 1:numel(y40)
idxrng = y40(k)+[-1,1]
xxct(k) = interp1(y(idxrng), t(idxrng), 40) % Interpolation
end
figure
plot(t,y, xxct,40+[0 0],'pg')
Choose the most appropriate option for your application. Both give the same result.
