I need help understanding the Pan Tompkins algorythm, and I\'d also like to know how to detect
QRS peaks of the ECG (with circles, for example). I\'m using the script in this website as a
guidance, but I have trouble understanding what it\'s doing. Perhaps there\'s a simpler way to do
it? http://cnx.org/contents/YR1BUs9_@1/QRS-Detection-Using-Pan-Tompki
Our teacher has otherwise explained to us FIR filters, and wants us to use them instead of the
low pass, high pass that are mentioned in the website and though I do understand how they work,
I don\'t really know which window is the best, or how many coefficients the filter needs. I\'ve
applying the cut-off frequencies 5-15 because I\'ve seen that\'s where the QRS peaks are
concentrated. Is there any way to improve this as well?
TB=1;
TBn=TB/Fs;
N2=ceil(3.3/TBn);
Fc1 =5;
Fc2 =15 ;
wc1 = 2*pi*Fc1;
wc2 = 2*pi*Fc2;
wcn1 = 2*pi*Fc1/Fs;
wcn2 = 2*pi*Fc2/Fs;
Bz=fir1(N2,[wcn1 wcn2]);
y3=filter(Bz,1,x1);
figure; plot(x1); figure; plot(y3);
Solution
Save the function and use it as given below ::
pan_tompkin(ECG1,200,1) 1st is ECG raw data , 2nd is sampling frequency, 3rd is
for plots -always set it to 1
Matlab Code ------------------------------------ function .
function [qrs_amp_raw,qrs_i_raw,delay]=pan_tompkin(ecg,fs,gr)
%% function [qrs_amp_raw,qrs_i_raw,delay]=pan_tompkin(ecg,fs)
% Complete implementation of Pan-Tompkins algorithm
%% Inputs
% ecg : raw ecg vector signal 1d signal
% fs : sampling frequency e.g. 200Hz, 400Hz and etc
% gr : flag to plot or not plot (set it 1 to have a plot or set it zero not
% to see any plots
%% Outputs
% qrs_amp_raw : amplitude of R waves amplitudes
% qrs_i_raw : index of R waves
% delay : number of samples which the signal is delayed due to the
% filtering
if ~isvector(ecg)
error(\'ecg must be a row or column vector\');
end
if nargin < 3
gr = 1; % on default the function always plots
end
ecg = ecg(:); % vectorize
%% Initialize
qrs_c =[]; %amplitude of R
qrs_i =[]; %index
SIG_LEV = 0;
nois_c =[];
nois_i =[];
delay = 0;
skip = 0; % becomes one when a T wave is detected
not_nois = 0; % it is not noise when not_nois = 1
selected_RR =[]; % Selected RR intervals
m_selected_RR = 0;
mean_RR = 0;
qrs_i_raw =[];
qrs_amp_raw=[];
ser_back = 0;
test_m = 0;
SIGL_buf = [];
NOISL_buf = [];
THRS_buf = [];
SIGL_buf1 = [];
NOISL_buf1 = [];
THRS_buf1 = [];
%% Plot differently based on filtering settings
if gr
if fs == 200
figure, ax(1)=subplot(321);plot(ecg);axis tight;title(\'Raw ECG Signal\');
else
figure, ax(1)=subplot(3,2,[1 2]);plot(ecg);axis tight;title(\'Raw ECG Signal\');
end
end
%% Noise cancelation(Filtering) % Filters (Filter in between 5-15 Hz)
if fs == 200
%% Low Pass Filter H(z) = ((1 - z^(-6))^2)/(1 - z^(-1))^2
b = [1 0 0 0 0 0 -2 0 0 0 0 0 1];
a = [1 -2 1];
h_l = filter(b,a,[1 zeros(1,12)]);
ecg_l = conv (ecg ,h_l);
ecg_l = ecg_l/ max( abs(ecg_l));
delay = 6; %based on the paper
if gr
ax(2)=subplot(322);plot(ecg_l);axis tight;title(\'Low pass filtered\');
end
%% High Pas.
Incorporate the SOR method in the multigridTest-m and apply the multig.pdfaartechindia
Incorporate the SOR method in the multigridTest.m and apply the multigridTest.m, to
determine the effects of changes in
Pre-smoothing iterations
Post-smoothing iterations
Multigrid cycle type
Iterative method
Number of grids (or grid hierarchy)
% SOR (Successive Over-Relaxation)
n = input('Enter number of equations, n: ');
A = zeros(n,n+1);
x1 = zeros(1,n);
A=[5 -2 3 -1;-3 9 1 2;2 -1 -7 3];
x1 = [0 0 0];
tol = input('Enter tolerance, tol: ');
m = input('Enter maximum number of iterations, m: ');
w = input('Enter the parameter w (omega): ');
k = 1;
while k <= m
err = 0;
for i = 1 : n
s = 0;
for j = 1 : n
s = s-A(i,j)*x1(j);
end
s = w*(s+A(i,n+1))/A(i,i);
if abs(s) > err
err = abs(s);
end
x1(i) = x1(i)+s;
end
if err <= tol
break;
else
k = k+1;
end
end
fprintf('The solution vector after %d iterations is :\n', k);
for i = 1 : n
fprintf(' %11.8f \n', x1(i));
end
%% Multigrid test case with visulization
% Author: Ben Beaudry
clc; clear; close all
load A_b.mat;
A = full(A); % Unsparce matrix to show full power of Multigrid
pre = 2; % Number of presmoothing iterations
post = 2; % Number of postsmoothing iterations
% cycle = 1; % Type of multigrid cycle (1=V-cycle, 2=W-cycle, 3=F-cycle)
smooth = 1; % Smoother type (1=Jacobi, 2=Gauss-Seidel)
grids = 5; % Max grids in cycle
maxit = 10; % Max iterations of solver
tol = 1e-02; % Tolerance of solver
%% Solvers
% solve directly
t=cputime;
x_D = A\b;
fprintf('Direct Solve Time: %g Seconds\n',cputime-t)
% V-Cycle
t=cputime;
[x_V,res_V,it_V] = multigrid(A,b,pre,post,1,smooth,grids,maxit,tol);
fprintf('V-Cycle Solve Time: %g Seconds\n',cputime-t)
% W-Cycle
t=cputime;
[x_W,res_W,it_W] = multigrid(A,b,pre,post,2,smooth,grids,maxit,tol);
fprintf('W-Cycle Solve Time: %g Seconds\n',cputime-t)
% F-Cycle
t=cputime;
[x_F,res_F,it_F] = multigrid(A,b,pre,post,3,smooth,grids,maxit,tol);
fprintf('F-Cycle Solve Time: %g Seconds\n',cputime-t)
% max it for iterative methods
it = 100;
% Gauss-Siedel
t=cputime;
L = tril(A);
U = triu(A,1);
x_G = zeros(length(b),1);
res_G= zeros(it,1);
for g = 1:it
x_G = L\(b-U*x_G);
r = b - A * x_G;
res_G(g) = norm(r);
if res_G(g) < tol
break;
end
end
fprintf('Gauss-Seidel Solve Time: %g Seconds\n',cputime-t)
% Jacobi
t=cputime;
d = diag(A);
dinv = 1./d;
LU = tril(A,-1)+triu(A,1);
U = triu(A,1);
x_J = zeros(length(b),1);
res_J= zeros(it,1);
for j = 1:it
x_J = dinv.*(b-(LU*x_J));
r = b - A * x_J;
res_J(j) = norm(r);
if res_J(j) < tol
break;
end
end
fprintf('Jacobi Solve Time: %g Seconds\n',cputime-t)
%% Plotting
figure;
hold on
plot(1:g,res_G(1:g),'o-','DisplayName','Guass-Seidel')
plot(1:j,res_J(1:j),'o-','DisplayName','Jacobi')
plot(1:it_V,res_V(1:it_V),'o-','DisplayName','V-Cycle Multigrid')
plot(1:it_F,res_F(1:it_F),'o-','DisplayName','F-Cycle Multigrid')
plot(1:it_W,res_W(1:it_W),'o-','DisplayName','W-Cycle Multigrid')
set(gca,'XLim',[0 100]);
grid on
legend('location','best')
ylabel('Relative Convergance')
xlabel('Iterations')
title('Convergance of Numerical System')
hold off
MULTIGRID FUNC.
Contents of the presentation:
- ABOUT ME
- Bisection Method using C#
- False Position Method using C#
- Gauss Seidel Method using MATLAB
- Secant Mod Method using MATLAB
- Report on Numerical Errors
- Optimization using Golden-Section Algorithm with Application on MATLAB
Incorporate the SOR method in the multigridTest-m and apply the multig.pdfaartechindia
Incorporate the SOR method in the multigridTest.m and apply the multigridTest.m, to
determine the effects of changes in
Pre-smoothing iterations
Post-smoothing iterations
Multigrid cycle type
Iterative method
Number of grids (or grid hierarchy)
% SOR (Successive Over-Relaxation)
n = input('Enter number of equations, n: ');
A = zeros(n,n+1);
x1 = zeros(1,n);
A=[5 -2 3 -1;-3 9 1 2;2 -1 -7 3];
x1 = [0 0 0];
tol = input('Enter tolerance, tol: ');
m = input('Enter maximum number of iterations, m: ');
w = input('Enter the parameter w (omega): ');
k = 1;
while k <= m
err = 0;
for i = 1 : n
s = 0;
for j = 1 : n
s = s-A(i,j)*x1(j);
end
s = w*(s+A(i,n+1))/A(i,i);
if abs(s) > err
err = abs(s);
end
x1(i) = x1(i)+s;
end
if err <= tol
break;
else
k = k+1;
end
end
fprintf('The solution vector after %d iterations is :\n', k);
for i = 1 : n
fprintf(' %11.8f \n', x1(i));
end
%% Multigrid test case with visulization
% Author: Ben Beaudry
clc; clear; close all
load A_b.mat;
A = full(A); % Unsparce matrix to show full power of Multigrid
pre = 2; % Number of presmoothing iterations
post = 2; % Number of postsmoothing iterations
% cycle = 1; % Type of multigrid cycle (1=V-cycle, 2=W-cycle, 3=F-cycle)
smooth = 1; % Smoother type (1=Jacobi, 2=Gauss-Seidel)
grids = 5; % Max grids in cycle
maxit = 10; % Max iterations of solver
tol = 1e-02; % Tolerance of solver
%% Solvers
% solve directly
t=cputime;
x_D = A\b;
fprintf('Direct Solve Time: %g Seconds\n',cputime-t)
% V-Cycle
t=cputime;
[x_V,res_V,it_V] = multigrid(A,b,pre,post,1,smooth,grids,maxit,tol);
fprintf('V-Cycle Solve Time: %g Seconds\n',cputime-t)
% W-Cycle
t=cputime;
[x_W,res_W,it_W] = multigrid(A,b,pre,post,2,smooth,grids,maxit,tol);
fprintf('W-Cycle Solve Time: %g Seconds\n',cputime-t)
% F-Cycle
t=cputime;
[x_F,res_F,it_F] = multigrid(A,b,pre,post,3,smooth,grids,maxit,tol);
fprintf('F-Cycle Solve Time: %g Seconds\n',cputime-t)
% max it for iterative methods
it = 100;
% Gauss-Siedel
t=cputime;
L = tril(A);
U = triu(A,1);
x_G = zeros(length(b),1);
res_G= zeros(it,1);
for g = 1:it
x_G = L\(b-U*x_G);
r = b - A * x_G;
res_G(g) = norm(r);
if res_G(g) < tol
break;
end
end
fprintf('Gauss-Seidel Solve Time: %g Seconds\n',cputime-t)
% Jacobi
t=cputime;
d = diag(A);
dinv = 1./d;
LU = tril(A,-1)+triu(A,1);
U = triu(A,1);
x_J = zeros(length(b),1);
res_J= zeros(it,1);
for j = 1:it
x_J = dinv.*(b-(LU*x_J));
r = b - A * x_J;
res_J(j) = norm(r);
if res_J(j) < tol
break;
end
end
fprintf('Jacobi Solve Time: %g Seconds\n',cputime-t)
%% Plotting
figure;
hold on
plot(1:g,res_G(1:g),'o-','DisplayName','Guass-Seidel')
plot(1:j,res_J(1:j),'o-','DisplayName','Jacobi')
plot(1:it_V,res_V(1:it_V),'o-','DisplayName','V-Cycle Multigrid')
plot(1:it_F,res_F(1:it_F),'o-','DisplayName','F-Cycle Multigrid')
plot(1:it_W,res_W(1:it_W),'o-','DisplayName','W-Cycle Multigrid')
set(gca,'XLim',[0 100]);
grid on
legend('location','best')
ylabel('Relative Convergance')
xlabel('Iterations')
title('Convergance of Numerical System')
hold off
MULTIGRID FUNC.
Contents of the presentation:
- ABOUT ME
- Bisection Method using C#
- False Position Method using C#
- Gauss Seidel Method using MATLAB
- Secant Mod Method using MATLAB
- Report on Numerical Errors
- Optimization using Golden-Section Algorithm with Application on MATLAB
Consider a 4-Link robot manipulator shown below. Use the forward kine.pdfmeerobertsonheyde608
Consider a 4-Link robot manipulator shown below. Use the forward kinematic D-H table and
write an m file that plots the manipulator. The instructions are given in the module 6. Submit
your solutions by the due date, in a single MATLAB m file.
Solution
Please give the kinetic D-H table else it would be difficult to code as we need to know the
rotation spin axis and other momentum of manipulator
Stating a general example code for manipulator with data
function X = fwd_kin(q,x)
% given a position in the configuration space, calculate the position of
% the end effector in the workspace for a two-link manipulator.
% q: vector of joint positions
% x: design vector (link lengths)
% X: end effector position in cartesian coordinates
% configuration space coordinates:
q1 = q(1); % theta 1
q2 = q(2); % theta 2
% manipulator parameters:
l1 = x(1); % link 1 length
l2 = x(2); % link 2 length
% calculate end effector position:
X = [l1*cos(q1) + l2*cos(q1+q2)
l1*sin(q1) + l2*sin(q1+q2)];
% SimulateTwolink.m uses inverse dynamics to simulate the torque
% trajectories required for a two-link planar robotic manipulator to follow
% a prescribed trajectory. It also computes total energy consumption. This
% code is provided as supplementary material for the paper:
%
% \'Engineering System Co-Design with Limited Plant Redesign\'
% Presented at the 8th AIAA Multidisciplinary Design Optimization
% Specialist Conference, April 2012.
%
% The paper is available from:
%
% http://systemdesign.illinois.edu/publications/All12a.pdf
%
% Here both the physical system design and control system design are
% considered simultaneously. Manipulator link length and trajectory
% specification can be specified, and torque trajectory and energy
% consumption are computed based on this input. It was found that maximum
% torque and total energy consumption calculated using inverse dynamics
% agreed closely with results calculated using feedback linearization, so
% to simplify optimization problem solution an inverse dynamics approach
% was used, which reduces the control design vector to just the trajectory
% design.
%
% In the conference paper several cases are considered, each with its own
% manipulator task, manipulator design, and trajectory design. The
% specifications for each of these five cases are provided here, and can be
% explored by changing the case number variable (cn).
%
% This code was incorporated into a larger optimization project. The code
% presented here includes only the analysis portion of the code, no
% optimization.
%
% A video illustrating the motion of each of these five cases is available
% on YouTube:
%
% http://www.youtube.com/watch?v=OR7Y9-n5SjA
%
% Author: James T. Allison, Assistant Professor, University of Illinois at
% Urbana-Champaign
% Date: 4/10/12
clear;clc
% simulation parameters:
p.dt = 0.0005; % simulation step size
tf = 2; p.tf = tf; % final time
p.ploton = 0; % turn off additional plotting capabilities
p.ploton2 = 0;
p.Tallow = 210; % maximum .
Glowworm swarm optimization (GSO) is a swarm intelligence based algorithm, introduced by K.N. Krishnanand and D. Ghose in 2005, for simultaneous computation of multiple optima of multimodal functions
In 9. 3-1 BALANCE SHEET The assets of Dallas & Associates consist e.pdfeyewatchsystems
In 9. 3-1 BALANCE SHEET The assets of Dallas & Associates consist entirely of current assets
and net plant and equipment. The firm has total assets of $2.5 million and net plant and
equipment equals $2 million. It has notes payable of $150,000, long-term debt of $750,000, and
total common equity of S1.5 million. The firm does have accounts payable and accruals on its
balance sheet. The finm only finances with debt and common equity, so it has no preferred stock
on its balance sheet. a. What is the company\'s total debt? b. What is the amount of total
liabilities and equity that appears on the firm\'s balance sheet? c. What is the balance of current
assets on the firm\'s balance sheet? d. What is the balance of current liabilities on the firm\'s
balance sheet? e. What is the amount of accounts payable and accruals on its balance sheet?
(Hint: Consider this as a single line item on the firm\'s balance sheet.) f. What is the firm\'s net
working capital? g. What is the firm\'s net operating working capital? h. What is the explanation
for the difference in your answers to parts f and g?
Solution
(a) Total Assets = $ 2.5 million, The firm\'s total assets are entirely composed of current assets
and net plant and equipment.
Net Plant and Equipment = $ 2 million and Current Assets = Total Assets - Net Plant and
Equipment = 2.5 - 2 = $ 0.5 million
Notes Payable = $ 150000 and Long-Term Debt = $ 750000
Total Common Equity = $ 1.5 million
Owner\'s Equity + Liabilities = Notes Payable + Long-Term Debt + Accruals + Accounts
Payable + Total Common Equity = Total Assets = $ 2.5 million
0.15 + 0.75 + Accounts Payable + Accruals + 1.5 = 2.5
Accounts Payable + Accruals = $ 0.1 million
Total Debt includes long-term capital financing debt and short-term interest bearing debt only.
Accounts Payable and Accruals do not form a part of a firm\'s total debt.
Hence, total debt = Notes Payable + Long-Term Debt = 150000 + 750000 = $ 900000
(b) Total Liabilities = Accounts Payable + Accruals + Long-Term Debt + Notes Payable = 0.1 +
0.75 + 0.15 = $ 1 million
Owner\'s Equity = Total Assets - Total Liabilities = 2.5 - 1 = $ 1.5 million
(c) Current Assets = Total Assets - Net Plant and Equipment = 2.5 - 2 = $ 0.5 million
(d) Current Laibility = Accruals + Accounts Payable = $ 0.1 million
NOTE: The current portion of notes payable(amount due within one year or accounting period)
is treated as current liability and the remaining portion is treated as non-current liability. As the
notes payable portion does not mention the amount due within a year, the entire value of notes
payable is treated as non-current laibility.
NOTE: Please raise separate queries for solutions to the remaining sub-parts..
I believe my chosen answer is correct, but Im not entirely positiv.pdfeyewatchsystems
I believe my chosen answer is correct, but I\'m not entirely positive. Any help is appreciated. The
enzyme glucose-6-phosphate dehydrogenase deficiency (G6PD) is inherited as a recessive gene
on the X chromosome in humans. Below is a pedigree of a family affected by G6PD. X-
represents a mutant X chromosome. What is I-1\'s genotype? XX. XX- X-X- could be A or B
Solution
G6PD is recessive traits disease and so in I-2 the Male would be having diseased X chromosome
recessive traits now for being disease in II a male must have a single disease X trait for that if
female is homozygous for non disease trait of heterzygous for disease trait in both condition in I
is possible.
So condition A and B both is correct.
More Related Content
Similar to I need help understanding the Pan Tompkins algorythm, and Id also .pdf
Consider a 4-Link robot manipulator shown below. Use the forward kine.pdfmeerobertsonheyde608
Consider a 4-Link robot manipulator shown below. Use the forward kinematic D-H table and
write an m file that plots the manipulator. The instructions are given in the module 6. Submit
your solutions by the due date, in a single MATLAB m file.
Solution
Please give the kinetic D-H table else it would be difficult to code as we need to know the
rotation spin axis and other momentum of manipulator
Stating a general example code for manipulator with data
function X = fwd_kin(q,x)
% given a position in the configuration space, calculate the position of
% the end effector in the workspace for a two-link manipulator.
% q: vector of joint positions
% x: design vector (link lengths)
% X: end effector position in cartesian coordinates
% configuration space coordinates:
q1 = q(1); % theta 1
q2 = q(2); % theta 2
% manipulator parameters:
l1 = x(1); % link 1 length
l2 = x(2); % link 2 length
% calculate end effector position:
X = [l1*cos(q1) + l2*cos(q1+q2)
l1*sin(q1) + l2*sin(q1+q2)];
% SimulateTwolink.m uses inverse dynamics to simulate the torque
% trajectories required for a two-link planar robotic manipulator to follow
% a prescribed trajectory. It also computes total energy consumption. This
% code is provided as supplementary material for the paper:
%
% \'Engineering System Co-Design with Limited Plant Redesign\'
% Presented at the 8th AIAA Multidisciplinary Design Optimization
% Specialist Conference, April 2012.
%
% The paper is available from:
%
% http://systemdesign.illinois.edu/publications/All12a.pdf
%
% Here both the physical system design and control system design are
% considered simultaneously. Manipulator link length and trajectory
% specification can be specified, and torque trajectory and energy
% consumption are computed based on this input. It was found that maximum
% torque and total energy consumption calculated using inverse dynamics
% agreed closely with results calculated using feedback linearization, so
% to simplify optimization problem solution an inverse dynamics approach
% was used, which reduces the control design vector to just the trajectory
% design.
%
% In the conference paper several cases are considered, each with its own
% manipulator task, manipulator design, and trajectory design. The
% specifications for each of these five cases are provided here, and can be
% explored by changing the case number variable (cn).
%
% This code was incorporated into a larger optimization project. The code
% presented here includes only the analysis portion of the code, no
% optimization.
%
% A video illustrating the motion of each of these five cases is available
% on YouTube:
%
% http://www.youtube.com/watch?v=OR7Y9-n5SjA
%
% Author: James T. Allison, Assistant Professor, University of Illinois at
% Urbana-Champaign
% Date: 4/10/12
clear;clc
% simulation parameters:
p.dt = 0.0005; % simulation step size
tf = 2; p.tf = tf; % final time
p.ploton = 0; % turn off additional plotting capabilities
p.ploton2 = 0;
p.Tallow = 210; % maximum .
Glowworm swarm optimization (GSO) is a swarm intelligence based algorithm, introduced by K.N. Krishnanand and D. Ghose in 2005, for simultaneous computation of multiple optima of multimodal functions
In 9. 3-1 BALANCE SHEET The assets of Dallas & Associates consist e.pdfeyewatchsystems
In 9. 3-1 BALANCE SHEET The assets of Dallas & Associates consist entirely of current assets
and net plant and equipment. The firm has total assets of $2.5 million and net plant and
equipment equals $2 million. It has notes payable of $150,000, long-term debt of $750,000, and
total common equity of S1.5 million. The firm does have accounts payable and accruals on its
balance sheet. The finm only finances with debt and common equity, so it has no preferred stock
on its balance sheet. a. What is the company\'s total debt? b. What is the amount of total
liabilities and equity that appears on the firm\'s balance sheet? c. What is the balance of current
assets on the firm\'s balance sheet? d. What is the balance of current liabilities on the firm\'s
balance sheet? e. What is the amount of accounts payable and accruals on its balance sheet?
(Hint: Consider this as a single line item on the firm\'s balance sheet.) f. What is the firm\'s net
working capital? g. What is the firm\'s net operating working capital? h. What is the explanation
for the difference in your answers to parts f and g?
Solution
(a) Total Assets = $ 2.5 million, The firm\'s total assets are entirely composed of current assets
and net plant and equipment.
Net Plant and Equipment = $ 2 million and Current Assets = Total Assets - Net Plant and
Equipment = 2.5 - 2 = $ 0.5 million
Notes Payable = $ 150000 and Long-Term Debt = $ 750000
Total Common Equity = $ 1.5 million
Owner\'s Equity + Liabilities = Notes Payable + Long-Term Debt + Accruals + Accounts
Payable + Total Common Equity = Total Assets = $ 2.5 million
0.15 + 0.75 + Accounts Payable + Accruals + 1.5 = 2.5
Accounts Payable + Accruals = $ 0.1 million
Total Debt includes long-term capital financing debt and short-term interest bearing debt only.
Accounts Payable and Accruals do not form a part of a firm\'s total debt.
Hence, total debt = Notes Payable + Long-Term Debt = 150000 + 750000 = $ 900000
(b) Total Liabilities = Accounts Payable + Accruals + Long-Term Debt + Notes Payable = 0.1 +
0.75 + 0.15 = $ 1 million
Owner\'s Equity = Total Assets - Total Liabilities = 2.5 - 1 = $ 1.5 million
(c) Current Assets = Total Assets - Net Plant and Equipment = 2.5 - 2 = $ 0.5 million
(d) Current Laibility = Accruals + Accounts Payable = $ 0.1 million
NOTE: The current portion of notes payable(amount due within one year or accounting period)
is treated as current liability and the remaining portion is treated as non-current liability. As the
notes payable portion does not mention the amount due within a year, the entire value of notes
payable is treated as non-current laibility.
NOTE: Please raise separate queries for solutions to the remaining sub-parts..
I believe my chosen answer is correct, but Im not entirely positiv.pdfeyewatchsystems
I believe my chosen answer is correct, but I\'m not entirely positive. Any help is appreciated. The
enzyme glucose-6-phosphate dehydrogenase deficiency (G6PD) is inherited as a recessive gene
on the X chromosome in humans. Below is a pedigree of a family affected by G6PD. X-
represents a mutant X chromosome. What is I-1\'s genotype? XX. XX- X-X- could be A or B
Solution
G6PD is recessive traits disease and so in I-2 the Male would be having diseased X chromosome
recessive traits now for being disease in II a male must have a single disease X trait for that if
female is homozygous for non disease trait of heterzygous for disease trait in both condition in I
is possible.
So condition A and B both is correct.
how does genies transcript illustrate the statement we learn to .pdfeyewatchsystems
how does genie\'s transcript illustrate the statement \"we learn to be human.\" aren\'t we human?
Solution
The transcript of genie illustrates that a wild child living in captivity, when given proper care and
attention can learn to be human. initially, she does not respond to the methods that were applied
for the betterment of her and to let her learn to talk. But slowly she was left to learn at her own
pace and understanding. The study on her suggested that we the humans have the innate ability
to learn words and use them correctly. So when genie was provided the situation she steadily
learned the ability to talk..
How many different alleles can a gene have Explain.SolutionEa.pdfeyewatchsystems
How many different alleles can a gene have? Explain.
Solution
Each gene can have many alleles. But a diploid organism typically contains two alleles in each
gene. This because in sexual reproduction one allele is received from each parent. If a gene
contains more than two alleles, it causes chromosomal abnormalities and thus diseases..
Comment on the behavior of the different types of flexures (crab-leg.pdfeyewatchsystems
Comment on the behavior of the different types of flexures (crab-leg, folded flexure, serpentine,
cartwheel, straight, notched). Which would have the highest cross-axis sensitivity? The lowest?
Which would you use for a device that experiences large displacements?
Solution
Serpentine have highest cross axis sensitivity.straight have lowest sensitivity. And crab leg is
gives promising displacements in three directions... So its best suited for device..
human color-blindness mutations result from A.reciprocal translocat.pdfeyewatchsystems
human color-blindness mutations result from: A.reciprocal translocation B. dosge compensation
C. Unequal crossing over D. Chromosome nondisjunction E. Fragile X syndrome I think answer
is B. dosage compensation???
Solution
Human Color blindness results from unequal crossing over . Gene\'s for red and green Color
vision lie adjacent on x chromosome . Evolving from a common ancestor they are similar enough
to mispair occasionally which results in the production of gametes lacking some or all of the one
of the two genes resulting in red green color blindness..
Generation 1 2 3 4 6 5 6 7 SolutionThe chances of their first c.pdfeyewatchsystems
Generation 1 2 3 4 6 5 6 7
Solution
The chances of their first child showing the trait is 50%
As this trait is being carried by the females in generation II, and in generation III due to the
congenial marriage among the family members the homozygosity of trait increases.
Due to which this trait again shows up again generation IV. If the individual IV-4 and IV-7 get
married, which is again congenial marriage among the family member. The risk factor for the
children having this trait increases..
Explain how cell division occurs in the germ (meiosis) cells of anim.pdfeyewatchsystems
Explain how cell division occurs in the germ (meiosis) cells of animals. Be sure to state and
describe each stage and phase of the cell cycle.
Solution
Germ cell division occurs by Meiosis. The division is explained with oogenesis as example.
Diploid primordial germ cells that form the germinal epithelium of fetal ovary divide repeatedly
by mitosis to produce diploid oogonia.
Diploid oogonia produced in groups just under the surface of the ovary. In contrast to
spermatogenesis, only one oogonium from each group normally grows into a diploid primary
oocyte. Each primary oocyte then enters prophase of Meiosis I and remains in this arrested state
until puberty.
As puberty begins, a primary oocyte completes Meiosis I each month to produce a secondary
oocyte and a polar body.
The secondary oocyte undergoes Meiosis II at fertilization to produce an ovum and a second
polar body..
Ecologists are sampling mice from different parks in New York City. M.pdfeyewatchsystems
Ecologists are sampling mice from different parks in New York City. Most of the mice these
researchers collect demonstrate straight\" tails; however, few mice exhibit \"curly\" tails. In this
scenario, a \"straight tail would be considered the henotype, and a \"curly\" tail would be
considered a phenotype. O heterogametic; homogametic O polymorphism; wild-type O mutant;
wild-type O wild-type; mutant None of the above.
Solution
The answer to your first question is wrong. Most of the researchers find mice with straight tails,
while a few find mice with curly tails. However, this finding does not prove that the curly-tailed
mice are mutants. A mutant phenotype is much rarer than the wild type. The curly phenotype
could simply be a recessive phenotype with a phenotypic frequency lower than that of the
straight phenotype, which might be the dominant trait. Thus, the answer should be \'None of the
above\'.
The answer to the second question is correct. Since the parents\' and the child\'s DNA sequences
differ in only a single nucleotide, it is a point mutation.
The answer to the third question is wrong. Since the flower colour is determined by a gene with
2 alleles, 1 of which is dominant, to get a red colour, the orchid just needs a single dominant
allele. Hence the answer is 1.
Your answer to the fourth question is wrong. The parent and daughter cells may not be
completely identical, and we must take into account the de-novo mutations. Therefore, the
answer should be \'The DNA sequences of the parent and daughter cells were largely similar,
with the exception of any de-novo mutations\'..
Describe the electronic configuration of a given element. Why are we.pdfeyewatchsystems
Describe the electronic configuration of a given element. Why are we interested in electrons (two
primary answers). (With regard to Human Anatomy and Physiology, what is the importance of
electrons?)
Solution
Electronic configuration is the arrangement of electron around the nucleus of an atom. Number
of electrons orbiting the nucleus determines the energy level of an atom. This electron makes the
bonds like, ionic bond, covalent bond and htdrogen bond. These bonds help in electrostatic
interactions between the molecules and helps molecules like haemoglobin, myoglobin to be
folded and form a three dimensional structure and function properly. these electrons helps in
phosphorylation, hydrolysisetc,..
Assets are listed on the balance sheet in order ofincreasing size .pdfeyewatchsystems
Assets are listed on the balance sheet in order of
increasing size and relative life.
decreasing size.
relative life.
decreasing liquidity.a.
increasing size and relative life.b.
decreasing size.c.
relative life.d.
decreasing liquidity.
Solution
Liquidity of the the assets means how quick an asset can be converted to cash. More quicker it
can be higher the liquidity.
Standard format of balance states assets to be reported on the basis of their liquidity. Assets with
highest liquidity should be reported on Top
Therefore in Assets we report first current assets and then PPE. Even in Current assets we report
Cash and Marketable securities (most liquid aset) on top. There fore
Assets are listed on the balance sheet in order of
decreasing liquidity.d.
decreasing liquidity..
1) Even the simplest types of infectious agents must have DNA, RNA, .pdfeyewatchsystems
1) Even the simplest types of infectious agents must have DNA, RNA, and protein.
a) Not all infectious agents have DNA, RNA, and protein. Some viruses have RNA and protein
only, viroids have only protein, and prions are composed only of RNA.
b) Not all infectious agents have DNA, RNA, and protein. Some viruses have DNA and protein
only, viroids have only protein, and prions are composed only of DNA.
c) Not all infectious agents have DNA, RNA, and protein. Some viruses have RNA and protein
only, viroids have only DNA, and prions are composed only of protein.
d) Not all infectious agents have DNA, RNA, and protein. Some viruses have DNA and protein
only, viroids have only RNA, and prions are composed only of protein.
2) A large amount of the DNA in eukaryotic cells has no function and is called \"junk DNA\".
a) Although some DNA in cells has been called \"junk DNA\" and have no function, their
concentraition maintains constant to allow other DNA act normally.
b) Although some DNA in cells has been called \"junk DNA\", it may simply be that we do not
yet understand its function.
c) Although some DNA in cells has been called \"junk DNA\", they have well known function to
store energy in the cells.
d) Although some DNA in cells has been called \"junk DNA\", but this is due to their mutation
and they could mutate again to restore their functions.
Solution
1. Even the simplest types of infectious agents must have DNA, RNA, and protein.
ANSWER: The above statement is false. This must be corrected as:
Even the simplest types of infectious agents must have either DNA and protein coat , RNA only
or Protein Coat only.
So the answer is:
D. Not all infectious agents have DNA, RNA, and protein. Some viruses have DNA and protein
only, viroids have only RNA, and prions are composed only of protein.
2.
Junk DNA/Non Coding DNA
The DNA sequences that does not code for a protein, that usually occurs in repetitive sequences
of nucleotides, and does not seem to serve any useful purpose are called Junk DNA.
So the answer is:
B. Although some DNA in cells has been called \"junk DNA\", it may simply be that we do not
yet understand its function..
1-A) What types of data document patterns and processes of evoluti.pdfeyewatchsystems
1-
A) What types of data document patterns and processes of evolution?
B) What does homology mean and how is it important with respect to evolution? How types of
homology are for providing evidence of evolution?
C) does the fossil record provide support for the of evolution? Why do sugar gliders and flying
squirrels look so much alike?
D) What do we call the similarity between flying squirrels and sugar gliders?
E) Provide a complete explanation for the question, what is evolution Vocabulary
Solution
Answer A) There are mainly four types of data which document patterns and process of
evolution; Homology, fossil record, direct observations and biogeography.
Answer B) Homology means sharing a similar structure due to relatedness. this is important with
respect to evolution because homology tells us that two different animals which shares the
homology structures had a common ancestor during evolution. there are two types of homology
in evolution; paralogs, which tells us that two copy of one gene located at two different positions
on genome inside a single organism has evolved as a duplication event and such genes are called
paralogous. while orthologs tells us that if two genes are similar in two different organism\'s
genomes then they were originated in their ancestor.
Answer C) Yes, fossil record is one of the data documentation for the evolutionary studies and
they help to find the missing links between various points of evolution. sugar gliders and flying
squirrels look alike because they share similarity in structures such as white belly, big eyes, same
size, and a thin layer of skin under the legs for gliding in the air.
Answer D) We call this kind of similarity in same functional structures as analogy..
Which of the following does not take place in a signaltransduction pa.pdfeyewatchsystems
Which of the following does not take place in a signaltransduction pathway? O An external
signaling molecule binds to a specific receptor. O An activated membrane-bound receptor
undergoes a conformational change. O Proteins in the pathway add or remove phosphate groups
from other proteins. O Activated proteins move into the nucleus to activate transcription factors.
O Proteins are activated when histones are deacetylated.
Solution
1) Protiens are activated when histones are deacetylated.
Histones are not involved in signal transduction.
2) The mutation could allow telomerase to be expressed when it should not.
Cancerous cells continously express telomerase as a result of which these cells have their
telomeres in tact and thus they constantly divide.
3)Retinoblsstoma results when an idividual inherits mutant gene from both the parents both of
which are necessary for cancer to develop.
4). DNA repair genes..
Which term indicates the evolution of many closely related species fr.pdfeyewatchsystems
Which term indicates the evolution of many closely related species from one or a few ancestral
species in a relatively short period of time? microevolution preadaptations adaptive radiation
allometric growth allopatric speciation A significant reduction in the amount of which gas
would allow more ultraviolet radiation to reach Earth\'s surface? oxygen methane ozone
cyanide carbon dioxide The most common outcome of the process in the accompanying figure
is no viable gametes. This happens because of hybrid inviability defective meiosis. gametic
isolation species incompatibility.
Solution
1. microevolution
2. defects in ozone gas layer allows more UV rays to come to the earth
3. because of defects in meiosis no viable gametes are formed.
you inoculate an LB tube with an aerobic bacterium and incubate it f.pdfeyewatchsystems
you inoculate an LB tube with an aerobic bacterium and incubate it for 24 hours. what kind of
growth do you expect to see and why?
Solution
If a liquid media in a tube is inoculated with aerobic bacteria and incubated for 24hrs, the growth
can be seen as a surface pellicle. Bacteria grows on the surface of the liquid media. This is
because it is the area where oxygen is available for the bacteria to utilize. Whereas anaerobic
bacteria grow at the bottom of the tube where air is absent..
What is meant by the relationship or the two strands of DNA The rig.pdfeyewatchsystems
What is meant by the relationship or the two strands of DNA? The right handed twisted
configuration of the strands. The arrangement branching pattern of the strands of DNA. The
arrangement where A is matched with T and G is matched with C. The staircase arrangement of
paired purine and pyrimidine bases linked by ionic bonds. The alignment or strands so one
strand starts with a 3. carbon, the other with a 5 carbon. integrative you have discovered a new
species bacteria. To begin your investigation of this content of ONA from the bacterial cells 10%
what is the adenine content? 10% 20% 40% 60% it is impossible to calculate this number for
eukaryotes. What happens during the process of transcription? Proteins are synthesized from
RNA molecules RNA is synthesized from protein DNA is synthesized from protein RNA is
synthesized from DNA Proteins are synthesized from DNA molecules What is a TATA box? a
nucleotide sequence for a codon on mRNA a sequence in the promoter of DNA a nucleotide
sequence for an anticodon on IRNA molecule a nucleotide sequence an indicating a stop codon
for translation a special protein in the enhancer region that is required for the initiation of
transcription What is an exon? A protein that is edited out of RNA after translation. A part of
RNA that is removed during the processing of an RNA molecule in the nucleus. A transfer RNA
that binds to the codon and carries an amino acid. A series of amino acids at the 3\' end of the
mRNA molecule that protect transcript. A part of an intact. Mature mRNA that leaves the
nucleus. What account for the uniform diameter of the DNA molecule? The two sides of the
DNA molecule are held together by hydrogen bonds. A purine always bonds with a pyrimidine.
One side of the DNA molecule has an unconnected 5\' phosphate group and the opposite end has
an unconnected 3\' hydroxyl group. The 3\' carbon of one deoxyribose and the 5\' carbon of
another deoxyribose bond together. The alternating sugar and phosphate backbone colls around
the outside of the hellx. If you isolate a single nucleotide from a nucleic acid chain and
determine that the nitrogenous ring structure is guanine, what can you say with certainty about
the type of molecule that the nucleotide could have come from? The molecule could be DNA
but not RNA. The molecule could be RNA but not DNA. The molecule is neither DNA nor
RNA. The molecule could be either DNA or RNA. The molecule could be double-stranded
DNA but not single-stranded DNA.
Solution
34. The correct answer is C.
35. The correct answer is B. ATP, RNA, and DNA
36. The correct answer is B. Purine always bonds with Pyrimidine
37. The correct answer is D. The molecule could either be DNA or RNA..
Which of the following categories of control can include the logical.pdfeyewatchsystems
Which of the following categories of control can include the logical mechanisms used to control
access and authenticate users?
A. Administrative
B. Clerical
C. Technical
D. Physical
Solution
Answer: C. Technical
Technical controls can be a software or hardware. They are some logical mechanisms that can be
used to control the access and authenticate users and identified unusual activities and restricted
unauthorized access.
Clerical is nonexistent category towards technical.
Physical controls include locks, guards, alarms and gates.
Admisirative controls are procedurals..
What is it, and how does Chromatic Dispersion Compensation work.pdfeyewatchsystems
What is it, and how does Chromatic Dispersion Compensation work?
Solution
Chromatic dispersion is a phenomenon that is an important factor in fiber optic communications.
It is the result of the different colors, or wavelengths, in a light beam arriving at their destination
at slightly different times. The result is a spreading, or dispersion, of the on-off light pulses that
convey digital information.
Chromatic dispersion is common place, as it is actually what causes rainbows - sunlight is
dispersed by droplets of water in the air. Sir Isaac Newton observed this phenomenon when he
passed sunlight through a prism and saw it diverge into a spectrum of different colors. This
dispersion occurs because different colors, or light frequencies, act slightly differently as they
pass through a medium such as glass. In fiber-based systems, an optical fiber, comprised of a
core and cladding with differing refractive index materials, inevitably causes some wavelengths
of light to travel slower or faster than others.
Chromatic dispersion is due to inherent property of silica fiber. The speed of a light wave is
dependent on the refractive index, , of the medium within which it is propagating. In silica fiber,
as well as many other materials, varies with wavelength. Therefore, different wavelength
channels will travel at slightly different speeds within the fiber. Because, even a pulse of light
from a laser has some spectral width, each of its wavelength components will travel at slightly
different speeds within the fiber. This results in a spreading of the transmission pulse as it travels
through the fiber..
True or FalseDue to developed molecular tools, functional genes c.pdfeyewatchsystems
True or False:
Due to developed molecular tools, functional genes can be identified directly in the clinical and
environmental samples.
Solution
True because the functional genes can be directly identified in samples by the metagenomic
arrays, using special biomarkers characterized by a unique order for the specific gene or gene
sequence, SNPs etc and their molecular structures. This biomarker array is sometimes followed
by the amplification tool and sequencing method for the entire DNA or RNA extracted from the
sample..
Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
Synthetic Fiber Construction in lab .pptxPavel ( NSTU)
Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
Introduction to AI for Nonprofits with Tapp NetworkTechSoup
Dive into the world of AI! Experts Jon Hill and Tareq Monaur will guide you through AI's role in enhancing nonprofit websites and basic marketing strategies, making it easy to understand and apply.
June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...Levi Shapiro
Letter from the Congress of the United States regarding Anti-Semitism sent June 3rd to MIT President Sally Kornbluth, MIT Corp Chair, Mark Gorenberg
Dear Dr. Kornbluth and Mr. Gorenberg,
The US House of Representatives is deeply concerned by ongoing and pervasive acts of antisemitic
harassment and intimidation at the Massachusetts Institute of Technology (MIT). Failing to act decisively to ensure a safe learning environment for all students would be a grave dereliction of your responsibilities as President of MIT and Chair of the MIT Corporation.
This Congress will not stand idly by and allow an environment hostile to Jewish students to persist. The House believes that your institution is in violation of Title VI of the Civil Rights Act, and the inability or
unwillingness to rectify this violation through action requires accountability.
Postsecondary education is a unique opportunity for students to learn and have their ideas and beliefs challenged. However, universities receiving hundreds of millions of federal funds annually have denied
students that opportunity and have been hijacked to become venues for the promotion of terrorism, antisemitic harassment and intimidation, unlawful encampments, and in some cases, assaults and riots.
The House of Representatives will not countenance the use of federal funds to indoctrinate students into hateful, antisemitic, anti-American supporters of terrorism. Investigations into campus antisemitism by the Committee on Education and the Workforce and the Committee on Ways and Means have been expanded into a Congress-wide probe across all relevant jurisdictions to address this national crisis. The undersigned Committees will conduct oversight into the use of federal funds at MIT and its learning environment under authorities granted to each Committee.
• The Committee on Education and the Workforce has been investigating your institution since December 7, 2023. The Committee has broad jurisdiction over postsecondary education, including its compliance with Title VI of the Civil Rights Act, campus safety concerns over disruptions to the learning environment, and the awarding of federal student aid under the Higher Education Act.
• The Committee on Oversight and Accountability is investigating the sources of funding and other support flowing to groups espousing pro-Hamas propaganda and engaged in antisemitic harassment and intimidation of students. The Committee on Oversight and Accountability is the principal oversight committee of the US House of Representatives and has broad authority to investigate “any matter” at “any time” under House Rule X.
• The Committee on Ways and Means has been investigating several universities since November 15, 2023, when the Committee held a hearing entitled From Ivory Towers to Dark Corners: Investigating the Nexus Between Antisemitism, Tax-Exempt Universities, and Terror Financing. The Committee followed the hearing with letters to those institutions on January 10, 202
Instructions for Submissions thorugh G- Classroom.pptxJheel Barad
This presentation provides a briefing on how to upload submissions and documents in Google Classroom. It was prepared as part of an orientation for new Sainik School in-service teacher trainees. As a training officer, my goal is to ensure that you are comfortable and proficient with this essential tool for managing assignments and fostering student engagement.
Biological screening of herbal drugs: Introduction and Need for
Phyto-Pharmacological Screening, New Strategies for evaluating
Natural Products, In vitro evaluation techniques for Antioxidants, Antimicrobial and Anticancer drugs. In vivo evaluation techniques
for Anti-inflammatory, Antiulcer, Anticancer, Wound healing, Antidiabetic, Hepatoprotective, Cardio protective, Diuretics and
Antifertility, Toxicity studies as per OECD guidelines
I need help understanding the Pan Tompkins algorythm, and Id also .pdf
1. I need help understanding the Pan Tompkins algorythm, and I'd also like to know how to detect
QRS peaks of the ECG (with circles, for example). I'm using the script in this website as a
guidance, but I have trouble understanding what it's doing. Perhaps there's a simpler way to do
it? http://cnx.org/contents/YR1BUs9_@1/QRS-Detection-Using-Pan-Tompki
Our teacher has otherwise explained to us FIR filters, and wants us to use them instead of the
low pass, high pass that are mentioned in the website and though I do understand how they work,
I don't really know which window is the best, or how many coefficients the filter needs. I've
applying the cut-off frequencies 5-15 because I've seen that's where the QRS peaks are
concentrated. Is there any way to improve this as well?
TB=1;
TBn=TB/Fs;
N2=ceil(3.3/TBn);
Fc1 =5;
Fc2 =15 ;
wc1 = 2*pi*Fc1;
wc2 = 2*pi*Fc2;
wcn1 = 2*pi*Fc1/Fs;
wcn2 = 2*pi*Fc2/Fs;
Bz=fir1(N2,[wcn1 wcn2]);
y3=filter(Bz,1,x1);
figure; plot(x1); figure; plot(y3);
Solution
Save the function and use it as given below ::
pan_tompkin(ECG1,200,1) 1st is ECG raw data , 2nd is sampling frequency, 3rd is
for plots -always set it to 1
Matlab Code ------------------------------------ function .
function [qrs_amp_raw,qrs_i_raw,delay]=pan_tompkin(ecg,fs,gr)
%% function [qrs_amp_raw,qrs_i_raw,delay]=pan_tompkin(ecg,fs)
% Complete implementation of Pan-Tompkins algorithm
%% Inputs
% ecg : raw ecg vector signal 1d signal
% fs : sampling frequency e.g. 200Hz, 400Hz and etc
% gr : flag to plot or not plot (set it 1 to have a plot or set it zero not
2. % to see any plots
%% Outputs
% qrs_amp_raw : amplitude of R waves amplitudes
% qrs_i_raw : index of R waves
% delay : number of samples which the signal is delayed due to the
% filtering
if ~isvector(ecg)
error('ecg must be a row or column vector');
end
if nargin < 3
gr = 1; % on default the function always plots
end
ecg = ecg(:); % vectorize
%% Initialize
qrs_c =[]; %amplitude of R
qrs_i =[]; %index
SIG_LEV = 0;
nois_c =[];
nois_i =[];
delay = 0;
skip = 0; % becomes one when a T wave is detected
not_nois = 0; % it is not noise when not_nois = 1
selected_RR =[]; % Selected RR intervals
m_selected_RR = 0;
mean_RR = 0;
qrs_i_raw =[];
qrs_amp_raw=[];
ser_back = 0;
test_m = 0;
SIGL_buf = [];
NOISL_buf = [];
THRS_buf = [];
SIGL_buf1 = [];
NOISL_buf1 = [];
THRS_buf1 = [];
3. %% Plot differently based on filtering settings
if gr
if fs == 200
figure, ax(1)=subplot(321);plot(ecg);axis tight;title('Raw ECG Signal');
else
figure, ax(1)=subplot(3,2,[1 2]);plot(ecg);axis tight;title('Raw ECG Signal');
end
end
%% Noise cancelation(Filtering) % Filters (Filter in between 5-15 Hz)
if fs == 200
%% Low Pass Filter H(z) = ((1 - z^(-6))^2)/(1 - z^(-1))^2
b = [1 0 0 0 0 0 -2 0 0 0 0 0 1];
a = [1 -2 1];
h_l = filter(b,a,[1 zeros(1,12)]);
ecg_l = conv (ecg ,h_l);
ecg_l = ecg_l/ max( abs(ecg_l));
delay = 6; %based on the paper
if gr
ax(2)=subplot(322);plot(ecg_l);axis tight;title('Low pass filtered');
end
%% High Pass filter H(z) = (-1+32z^(-16)+z^(-32))/(1+z^(-1))
b = [-1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 32 -32 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1];
a = [1 -1];
h_h = filter(b,a,[1 zeros(1,32)]);
ecg_h = conv (ecg_l ,h_h);
ecg_h = ecg_h/ max( abs(ecg_h));
delay = delay + 16; % 16 samples for highpass filtering
if gr
ax(3)=subplot(323);plot(ecg_h);axis tight;title('High Pass Filtered');
end
else
%% bandpass filter for Noise cancelation of other sampling frequencies(Filtering)
f1=5; %cuttoff low frequency to get rid of baseline wander
f2=15; %cuttoff frequency to discard high frequency noise
Wn=[f1 f2]*2/fs; % cutt off based on fs
4. N = 3; % order of 3 less processing
[a,b] = butter(N,Wn); %bandpass filtering
ecg_h = filtfilt(a,b,ecg);
ecg_h = ecg_h/ max( abs(ecg_h));
if gr
ax(3)=subplot(323);plot(ecg_h);axis tight;title('Band Pass Filtered');
end
end
%% derivative filter H(z) = (1/8T)(-z^(-2) - 2z^(-1) + 2z + z^(2))
h_d = [-1 -2 0 2 1]*(1/8);%1/8*fs
ecg_d = conv (ecg_h ,h_d);
ecg_d = ecg_d/max(ecg_d);
delay = delay + 2; % delay of derivative filter 2 samples
if gr
ax(4)=subplot(324);plot(ecg_d);axis tight;title('Filtered with the derivative filter');
end
%% Squaring nonlinearly enhance the dominant peaks
ecg_s = ecg_d.^2;
if gr
ax(5)=subplot(325);plot(ecg_s);axis tight;title('Squared');
end
%% Moving average Y(nt) = (1/N)[x(nT-(N - 1)T)+ x(nT - (N - 2)T)+...+x(nT)]
ecg_m = conv(ecg_s ,ones(1 ,round(0.150*fs))/round(0.150*fs));
delay = delay + 15;
if gr
ax(6)=subplot(326);plot(ecg_m);axis tight;title('Averaged with 30 samples length,Black
noise,Green Adaptive Threshold,RED Sig Level,Red circles QRS adaptive threshold');
axis tight;
end
%% Fiducial Mark
% Note : a minimum distance of 40 samples is considered between each R wave
% since in physiological point of view no RR wave can occur in less than
% 200 msec distance
[pks,locs] = findpeaks(ecg_m,'MINPEAKDISTANCE',round(0.2*fs));
%% initialize the training phase (2 seconds of the signal) to determine the THR_SIG and
5. THR_NOISE
THR_SIG = max(ecg_m(1:2*fs))*1/3; % 0.25 of the max amplitude
THR_NOISE = mean(ecg_m(1:2*fs))*1/2; % 0.5 of the mean signal is considered to be noise
SIG_LEV= THR_SIG;
NOISE_LEV = THR_NOISE;
%% Initialize bandpath filter threshold(2 seconds of the bandpass signal)
THR_SIG1 = max(ecg_h(1:2*fs))*1/3; % 0.25 of the max amplitude
THR_NOISE1 = mean(ecg_h(1:2*fs))*1/2; %
SIG_LEV1 = THR_SIG1; % Signal level in Bandpassed filter
NOISE_LEV1 = THR_NOISE1; % Noise level in Bandpassed filter
%% Thresholding and online desicion rule
for i = 1 : length(pks)
%% locate the corresponding peak in the filtered signal
if locs(i)-round(0.150*fs)>= 1 && locs(i)<= length(ecg_h)
[y_i x_i] = max(ecg_h(locs(i)-round(0.150*fs):locs(i)));
else
if i == 1
[y_i x_i] = max(ecg_h(1:locs(i)));
ser_back = 1;
elseif locs(i)>= length(ecg_h)
[y_i x_i] = max(ecg_h(locs(i)-round(0.150*fs):end));
end
end
%% update the heart_rate (Two heart rate means one the moste recent and the other selected)
if length(qrs_c) >= 9
diffRR = diff(qrs_i(end-8:end)); %calculate RR interval
mean_RR = mean(diffRR); % calculate the mean of 8 previous R waves interval
comp =qrs_i(end)-qrs_i(end-1); %latest RR
if comp <= 0.92*mean_RR || comp >= 1.16*mean_RR
% lower down thresholds to detect better in MVI
6. THR_SIG = 0.5*(THR_SIG);
%THR_NOISE = 0.5*(THR_SIG);
% lower down thresholds to detect better in Bandpass filtered
THR_SIG1 = 0.5*(THR_SIG1);
%THR_NOISE1 = 0.5*(THR_SIG1);
else
m_selected_RR = mean_RR; %the latest regular beats mean
end
end
%% calculate the mean of the last 8 R waves to make sure that QRS is not
% missing(If no R detected , trigger a search back) 1.66*mean
if m_selected_RR
test_m = m_selected_RR; %if the regular RR availabe use it
elseif mean_RR && m_selected_RR == 0
test_m = mean_RR;
else
test_m = 0;
end
if test_m
if (locs(i) - qrs_i(end)) >= round(1.66*test_m)% it shows a QRS is missed
[pks_temp,locs_temp] = max(ecg_m(qrs_i(end)+ round(0.200*fs):locs(i)-
round(0.200*fs))); % search back and locate the max in this interval
locs_temp = qrs_i(end)+ round(0.200*fs) + locs_temp -1; %location
if pks_temp > THR_NOISE
qrs_c = [qrs_c pks_temp];
qrs_i = [qrs_i locs_temp];
% find the location in filtered sig
if locs_temp <= length(ecg_h)
[y_i_t x_i_t] = max(ecg_h(locs_temp-round(0.150*fs):locs_temp));
7. else
[y_i_t x_i_t] = max(ecg_h(locs_temp-round(0.150*fs):end));
end
% take care of bandpass signal threshold
if y_i_t > THR_NOISE1
qrs_i_raw = [qrs_i_raw locs_temp-round(0.150*fs)+ (x_i_t - 1)];% save index of
bandpass
qrs_amp_raw =[qrs_amp_raw y_i_t]; %save amplitude of bandpass
SIG_LEV1 = 0.25*y_i_t + 0.75*SIG_LEV1; %when found with the second thres
end
not_nois = 1;
SIG_LEV = 0.25*pks_temp + 0.75*SIG_LEV ; %when found with the second
threshold
end
else
not_nois = 0;
end
end
%% find noise and QRS peaks
if pks(i) >= THR_SIG
% if a QRS candidate occurs within 360ms of the previous QRS
% ,the algorithm determines if its T wave or QRS
if length(qrs_c) >= 3
if (locs(i)-qrs_i(end)) <= round(0.3600*fs)
Slope1 = mean(diff(ecg_m(locs(i)-round(0.075*fs):locs(i)))); %mean slope of
the waveform at that position
Slope2 = mean(diff(ecg_m(qrs_i(end)-round(0.075*fs):qrs_i(end)))); %mean
8. slope of previous R wave
if abs(Slope1) <= abs(0.5*(Slope2)) % slope less then 0.5 of previous R
nois_c = [nois_c pks(i)];
nois_i = [nois_i locs(i)];
skip = 1; % T wave identification
% adjust noise level in both filtered and
% MVI
NOISE_LEV1 = 0.125*y_i + 0.875*NOISE_LEV1;
NOISE_LEV = 0.125*pks(i) + 0.875*NOISE_LEV;
else
skip = 0;
end
end
end
if skip == 0 % skip is 1 when a T wave is detected
qrs_c = [qrs_c pks(i)];
qrs_i = [qrs_i locs(i)];
% bandpass filter check threshold
if y_i >= THR_SIG1
if ser_back
qrs_i_raw = [qrs_i_raw x_i]; % save index of bandpass
else
qrs_i_raw = [qrs_i_raw locs(i)-round(0.150*fs)+ (x_i - 1)];% save index of
bandpass
end
qrs_amp_raw =[qrs_amp_raw y_i];% save amplitude of bandpass
SIG_LEV1 = 0.125*y_i + 0.875*SIG_LEV1;% adjust threshold for bandpass filtered sig
end
% adjust Signal level
SIG_LEV = 0.125*pks(i) + 0.875*SIG_LEV ;
end
9. elseif THR_NOISE <= pks(i) && pks(i) THR_SIG, that location is identified as a “QRS
complex
candidate” and the signal level (SIG_LEV) is updated:
SIG _ LEV = 0.125 ×CURRENTPEAK + 0.875× SIG _ LEV
If THR_NOISE < CURRENTPEAK < THR_SIG, then that location is identified as a
“noise peak” and the noise level (NOISE_LEV) is updated:
NOISE _ LEV = 0.125×CURRENTPEAK + 0.875× NOISE _ LEV
Based on new estimates of the signal and noise levels (SIG_LEV and NOISE_LEV,
respectively) at that point in the ECG, the thresholds are adjusted as follows:
THR _ SIG = NOISE _ LEV + 0.25 × (SIG _ LEV ? NOISE _ LEV )
THR _ NOISE = 0.5× (THR _ SIG)
These adjustments lower the threshold gradually in signal segments that are deemed to
be of poorer quality.
c) SEARCHBACK FOR MISSED QRS COMPLEXES - In the thresholding step above, if
CURRENTPEAK < THR_SIG, the peak is deemed not to have resulted from a QRS
complex. If however, an unreasonably long period has expired without an abovethreshold
peak, the algorithm will assume a QRS has been missed and perform a
searchback. This limits the number of false negatives. The minimum time used to trigger
a searchback is 1.66 times the current R peak to R peak time period (called the RR
interval). This value has a physiological origin - the time value between adjacent
heartbeats cannot change more quickly than this. The missed QRS complex is assumed
to occur at the location of the highest peak in the interval that lies between THR_SIG and
THR_NOISE. In this algorithm, two average RR intervals are stored,the first RR interval is
calculated as an average of the last eight QRS locations in order to adapt to changing heart
rate and the second RR interval mean is the mean
of the most regular RR intervals . The threshold is lowered if the heart rate is not regular
to improve detection.
d) ELIMINATION OF MULTIPLE DETECTIONS WITHIN REFRACTORY PERIOD - It is
impossible for a legitimate QRS complex to occur if it lies within 200ms after a previously
detected one. This constraint is a physiological one – due to the refractory period during
which ventricular depolarization cannot occur despite a stimulus[1]. As QRS complex
candidates are generated, the algorithm eliminates such physically impossible events,
thereby reducing false positives.
e) T WAVE DISCRIMINATION - Finally, if a QRS candidate occurs after the 200ms
refractory period but within 360ms of the previous QRS, the algorithm determines
10. whether this is a genuine QRS complex of the next heartbeat or an abnormally prominent
T wave. This decision is based on the mean slope of the waveform at that position. A slope of
less than one half that of the previous QRS complex is consistent with the slower
changing behaviour of a T wave – otherwise, it becomes a QRS detection.
Extra concept : beside the points mentioned in the paper, this code also
checks if the occured peak which is less than 360 msec latency has also a
latency less than 0,5*mean_RR if yes this is counted as noise
f) In the final stage , the output of R waves detected in smoothed signal is analyzed and double
checked with the help of the output of the bandpass signal to improve
detection and find the original index of the real R waves on the raw ecg
signal