Can anyone please help me to understand the below mentioned code?
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function [pwav]=p_wav(x,a_pwav,d_pwav,t_pwav,li)
l=li;
a=a_pwav;
x=x+t_pwav;
b=(2*l)/d_pwav;
n=100;
p1=1/l;
p2=0;
for i = 1:n
harm1=(((sin((pi/(2*b))*(b-(2*i))))/(b-(2*i))+(sin((pi/(2*b))*(b+(2*i))))/(b+(2*i)))*(2/pi))*cos((i*pi*x)/l);
function [qwav]=q_wav(x,a_qwav,d_qwav,t_qwav,li)
l=li;
x=x+t_qwav;
a=a_qwav;
b=(2*l)/d_qwav;
n=100;
q1=(a/(2*b))*(2-b);
q2=0;
for i = 1:n
harm5=(((2*b*a)/(i*i*pi*pi))*(1-cos((i*pi)/b)))*cos((i*pi*x)/l);
q2=q2+harm5;
end
qwav=-1*(q1+q2);
function [qrswav]=qrs_wav(x,a_qrswav,d_qrswav,li)
l=li;
a=a_qrswav;
b=(2*l)/d_qrswav;
n=100;
qrs1=(a/(2*b))*(2-b);
qrs2=0;
for i = 1:n
harm=(((2*b*a)/(i*i*pi*pi))*(1-cos((i*pi)/b)))*cos((i*pi*x)/l);
qrs2=qrs2+harm;
end
function [swav]=s_wav(x,a_swav,d_swav,t_swav,li)
l=li;
x=x-t_swav;
a=a_swav;
b=(2*l)/d_swav;
n=100;
s1=(a/(2*b))*(2-b);
s2=0;
for i = 1:n
harm3=(((2*b*a)/(i*i*pi*pi))*(1-cos((i*pi)/b)))*cos((i*pi*x)/l);
s2=s2+harm3;
end
swav=-1*(s1+s2);
function [twav]=t_wav(x,a_twav,d_twav,t_twav,li)
l=li;
a=a_twav;
x=x-t_twav-0.045;
b=(2*l)/d_twav;
n=100;
t1=1/l;
t2=0;
for i = 1:n
harm2=(((sin((pi/(2*b))*(b-(2*i))))/(b-(2*i))+(sin((pi/(2*b))*(b+(2*i))))/(b+(2*i)))*(2/pi))*cos((i*pi*x)/l);
t2=t2+harm2;
end
twav1=t1+t2;
twav=a*twav1;
qrswav=qrs1+qrs2;
function [uwav]=u_wav(x,a_uwav,d_uwav,t_uwav,li)
l=li;
a=a_uwav
x=x-t_uwav;
b=(2*l)/d_uwav;
n=100;
u1=1/l
u2=0
for i = 1:n
harm4=(((sin((pi/(2*b))*(b-(2*i))))/(b-(2*i))+(sin((pi/(2*b))*(b+(2*i))))/(b+(2*i)))*(2/pi))*cos((i*pi*x)/l);
u2=u2+harm4;
end
uwav1=u1+u2;
uwav=a*uwav1;
12 个评论
回答(1 个)
Walter Roberson
2016-9-18
You need to read the documentation that was provided as part of that File Exchange Submission as unfortunately there is no useful comments in the source.
The contribution includes files ECG.doc and ECG.pdf which describe the mathematics involved.
0 个评论
另请参阅
类别
在 Help Center 和 File Exchange 中查找有关 ECG / EKG 的更多信息
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