Hi Community,
I have modelled a grid using a programmable three phase voltage source and there are three loads attached to it. Model & Parameter file is attached. Each load is defined by the rated active and reactive powers (P& Q) as follows –
Load 1: Rated P = 25 kW; Rated Q = 5 kVA
Load 2: Rated P = 30 kW; rated Q = 10 kVA
Load 3: Rated P = 125 kW; rated Q = 10 kVA
The loads 2 and 3 are switched off at T = 2 secs and switched back on at T = 4 secs. I am also measuring the powers at each of the individual loads as well as the total power supplied by grid.
When I use the voltage source rating = 2300V, the P & Q values are nearly the same (but not exact match) as the above specified parameters.

When I use voltage source = 400V (with everything else remaining the same) there is a huge drop in the measured P & Q values and I am unable to follow why is this happening?

Q1) Why are the measured P&Q so low for V = 400V vs V = 2300V? Is there some limitation on the power this grid can deliver? If yes, then how is this being limited and how can I determine what is the maximum power rating of the load that can be connected across this grid source?
Q2) Additionally in the first scenario with 2300V why are the measured P&Q not the exact match of rated power of loads slightly lesser. I assume some power is lost in "losses" but I canot figure out where are these losses happening?Hi Community,
I have modelled a grid using a programmable three phase voltage source and there are three loads attached to it. Model & Parameter file is attached. Each load is defined by the rated active and reactive powers (P& Q) as follows –
Load 1: Rated P = 25 kW; Rated Q = 5 kVA
Load 2: Rated P = 30 kW; rated Q = 10 kVA
Load 3: Rated P = 125 kW; rated Q = 10 kVA
The loads 2 and 3 are switched off at T = 2 secs and switched back on at T = 4 secs. I am also measuring the powers at each of the individual loads as well as the total power supplied by grid.
When I use the voltage source rating = 2300V, the P & Q values are nearly the same (but not exact match) as the above specified parameters.

When I use voltage source = 400V (with everything else remaining the same) there is a huge drop in the measured P & Q values and I am unable to follow why is this happening?

Q1) Why are the measured P&Q so low for V = 400V vs V = 2300V? Is there some limitation on the power this grid can deliver? If yes, then how is this being limited and how can I determine what is the maximum power rating of the load that can be connected across this grid source?
Q2) Additionally in the first scenario with 2300V why are the measured P&Q not the exact match of rated power of loads slightly lesser. I assume some power is lost in "losses" but I canot figure out where are these losses happening? Hi Community,
I have modelled a grid using a programmable three phase voltage source and there are three loads attached to it. Model & Parameter file is attached. Each load is defined by the rated active and reactive powers (P& Q) as follows –
Load 1: Rated P = 25 kW; Rated Q = 5 kVA
Load 2: Rated P = 30 kW; rated Q = 10 kVA
Load 3: Rated P = 125 kW; rated Q = 10 kVA
The loads 2 and 3 are switched off at T = 2 secs and switched back on at T = 4 secs. I am also measuring the powers at each of the individual loads as well as the total power supplied by grid.
When I use the voltage source rating = 2300V, the P & Q values are nearly the same (but not exact match) as the above specified parameters.

When I use voltage source = 400V (with everything else remaining the same) there is a huge drop in the measured P & Q values and I am unable to follow why is this happening?

Q1) Why are the measured P&Q so low for V = 400V vs V = 2300V? Is there some limitation on the power this grid can deliver? If yes, then how is this being limited and how can I determine what is the maximum power rating of the load that can be connected across this grid source?
Q2) Additionally in the first scenario with 2300V why are the measured P&Q not the exact match of rated power of loads slightly lesser. I assume some power is lost in "losses" but I canot figure out where are these losses happening? grid, programmable voltage source, active power, reactive power, impedance, power measurement MATLAB Answers — New Questions

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Hello all! I am doing a program about hair removal. And I found a code online but part of this code I couln’t understand. The final part(the loop part) was really confused me.
img_b is a binary image and after dilation we could get the img_b binary image, which the black part is skin and the white part is hair. Thus, the ‘for’ part in the final loop is to decide if this pixel is skin or hair, if the pixel is in the skin part then it keeps the same pixel value as the original image; if the pixel is in the hair part, then do ‘else’ and calculate it for removing hair.
So here comes the question, I do not understand the calculation part in ‘else’ and the meaning of parameter r.When I change the value of r into a bigger one, the running speed gets slower. So I wonder could anyone give me a guide about this part please?when I change the value of r into a bigger one, the running speed gets slower. Thank you!
Images are listed below: the first image is which we need to deal with and the second image is the processed image.

Updated on Mar. 7th 2019: Sorry guys forgot to put the otsu function. Already put it in case anyone need it.

if true
clear;
clc;
r=7;
se1_para = 3;
se2_para = 2;
img_old = imread(‘D:MATLAB1binpicsam3-2.jpg’);
figure(1),imshow(img_old);
[x,y,z] = size(img_old);
img = rgb2gray(img_old);
se1 = strel(‘disk’,se1_para);
img_c = imclose(img,se1)
figure(2), imshow(img_c,[]);
img_fur = double(img_c) – double(img);
figure(3),imshow(img_fur,[]);
[X Y]=meshgrid(1:x);
tt=(X-280).^2+(Y-280).^2<280^2;
thresh = otsu(img_fur(tt),sum(tt(:)));
img_b = (img_fur>thresh);
figure(4),imshow(img_b);
se2 = strel(‘disk’,se2_para);
img_b =imdilate(img_b,se2);
figure(5),imshow(img_b);
img_new = uint8(zeros(x,y,z));
for i = 1:x
for j = 1:y
if img_b(i,j) == false
img_new(i,j,:) = img_old(i,j,:);
else
ttt = img_old(max(1,i-r):min(i+r,x),max(j-r,1):min(j+r,y),:);
no_efficient_pix = cat(3,img_b(max(1,i-r):min(i+r,x),max(j-r,1):min(j+r,y)),not(tt(max(1,i-r):min(i+r,x),max(j-r,1):min(j+r,y))));
no_efficient_pix = any(no_efficient_pix,3);
ttt = ttt.*repmat(uint8(not(no_efficient_pix)),[1,1,3]);
efficient_pix_num = (2*r+1)^2-sum(no_efficient_pix(:));
img_new(i,j,:) = uint8(sum(sum(ttt))./efficient_pix_num);
end
end
end
figure(6),imshow(img_new,[]);
end

function thresh = otsu(data,pix_num)
img_var = zeros(256,1);
for i=1:256
     w0 = sum(sum(data<=i-1))./pix_num;
     w1 = 1-w0;
     u0 = sum(sum(data.*double(data<=i-1)))./(w0*pix_num);
     u1 = sum(sum(data.*double(data>i-1)))./(w1*pix_num);
     img_var(i) = w0.*w1.*((u0-u1).^2);
end
[~,I] = max(img_var);
thresh = I-1;
endHello all! I am doing a program about hair removal. And I found a code online but part of this code I couln’t understand. The final part(the loop part) was really confused me.
img_b is a binary image and after dilation we could get the img_b binary image, which the black part is skin and the white part is hair. Thus, the ‘for’ part in the final loop is to decide if this pixel is skin or hair, if the pixel is in the skin part then it keeps the same pixel value as the original image; if the pixel is in the hair part, then do ‘else’ and calculate it for removing hair.
So here comes the question, I do not understand the calculation part in ‘else’ and the meaning of parameter r.When I change the value of r into a bigger one, the running speed gets slower. So I wonder could anyone give me a guide about this part please?when I change the value of r into a bigger one, the running speed gets slower. Thank you!
Images are listed below: the first image is which we need to deal with and the second image is the processed image.

Updated on Mar. 7th 2019: Sorry guys forgot to put the otsu function. Already put it in case anyone need it.

if true
clear;
clc;
r=7;
se1_para = 3;
se2_para = 2;
img_old = imread(‘D:MATLAB1binpicsam3-2.jpg’);
figure(1),imshow(img_old);
[x,y,z] = size(img_old);
img = rgb2gray(img_old);
se1 = strel(‘disk’,se1_para);
img_c = imclose(img,se1)
figure(2), imshow(img_c,[]);
img_fur = double(img_c) – double(img);
figure(3),imshow(img_fur,[]);
[X Y]=meshgrid(1:x);
tt=(X-280).^2+(Y-280).^2<280^2;
thresh = otsu(img_fur(tt),sum(tt(:)));
img_b = (img_fur>thresh);
figure(4),imshow(img_b);
se2 = strel(‘disk’,se2_para);
img_b =imdilate(img_b,se2);
figure(5),imshow(img_b);
img_new = uint8(zeros(x,y,z));
for i = 1:x
for j = 1:y
if img_b(i,j) == false
img_new(i,j,:) = img_old(i,j,:);
else
ttt = img_old(max(1,i-r):min(i+r,x),max(j-r,1):min(j+r,y),:);
no_efficient_pix = cat(3,img_b(max(1,i-r):min(i+r,x),max(j-r,1):min(j+r,y)),not(tt(max(1,i-r):min(i+r,x),max(j-r,1):min(j+r,y))));
no_efficient_pix = any(no_efficient_pix,3);
ttt = ttt.*repmat(uint8(not(no_efficient_pix)),[1,1,3]);
efficient_pix_num = (2*r+1)^2-sum(no_efficient_pix(:));
img_new(i,j,:) = uint8(sum(sum(ttt))./efficient_pix_num);
end
end
end
figure(6),imshow(img_new,[]);
end

function thresh = otsu(data,pix_num)
img_var = zeros(256,1);
for i=1:256
     w0 = sum(sum(data<=i-1))./pix_num;
     w1 = 1-w0;
     u0 = sum(sum(data.*double(data<=i-1)))./(w0*pix_num);
     u1 = sum(sum(data.*double(data>i-1)))./(w1*pix_num);
     img_var(i) = w0.*w1.*((u0-u1).^2);
end
[~,I] = max(img_var);
thresh = I-1;
end Hello all! I am doing a program about hair removal. And I found a code online but part of this code I couln’t understand. The final part(the loop part) was really confused me.
img_b is a binary image and after dilation we could get the img_b binary image, which the black part is skin and the white part is hair. Thus, the ‘for’ part in the final loop is to decide if this pixel is skin or hair, if the pixel is in the skin part then it keeps the same pixel value as the original image; if the pixel is in the hair part, then do ‘else’ and calculate it for removing hair.
So here comes the question, I do not understand the calculation part in ‘else’ and the meaning of parameter r.When I change the value of r into a bigger one, the running speed gets slower. So I wonder could anyone give me a guide about this part please?when I change the value of r into a bigger one, the running speed gets slower. Thank you!
Images are listed below: the first image is which we need to deal with and the second image is the processed image.

Updated on Mar. 7th 2019: Sorry guys forgot to put the otsu function. Already put it in case anyone need it.

if true
clear;
clc;
r=7;
se1_para = 3;
se2_para = 2;
img_old = imread(‘D:MATLAB1binpicsam3-2.jpg’);
figure(1),imshow(img_old);
[x,y,z] = size(img_old);
img = rgb2gray(img_old);
se1 = strel(‘disk’,se1_para);
img_c = imclose(img,se1)
figure(2), imshow(img_c,[]);
img_fur = double(img_c) – double(img);
figure(3),imshow(img_fur,[]);
[X Y]=meshgrid(1:x);
tt=(X-280).^2+(Y-280).^2<280^2;
thresh = otsu(img_fur(tt),sum(tt(:)));
img_b = (img_fur>thresh);
figure(4),imshow(img_b);
se2 = strel(‘disk’,se2_para);
img_b =imdilate(img_b,se2);
figure(5),imshow(img_b);
img_new = uint8(zeros(x,y,z));
for i = 1:x
for j = 1:y
if img_b(i,j) == false
img_new(i,j,:) = img_old(i,j,:);
else
ttt = img_old(max(1,i-r):min(i+r,x),max(j-r,1):min(j+r,y),:);
no_efficient_pix = cat(3,img_b(max(1,i-r):min(i+r,x),max(j-r,1):min(j+r,y)),not(tt(max(1,i-r):min(i+r,x),max(j-r,1):min(j+r,y))));
no_efficient_pix = any(no_efficient_pix,3);
ttt = ttt.*repmat(uint8(not(no_efficient_pix)),[1,1,3]);
efficient_pix_num = (2*r+1)^2-sum(no_efficient_pix(:));
img_new(i,j,:) = uint8(sum(sum(ttt))./efficient_pix_num);
end
end
end
figure(6),imshow(img_new,[]);
end

function thresh = otsu(data,pix_num)
img_var = zeros(256,1);
for i=1:256
     w0 = sum(sum(data<=i-1))./pix_num;
     w1 = 1-w0;
     u0 = sum(sum(data.*double(data<=i-1)))./(w0*pix_num);
     u1 = sum(sum(data.*double(data>i-1)))./(w1*pix_num);
     img_var(i) = w0.*w1.*((u0-u1).^2);
end
[~,I] = max(img_var);
thresh = I-1;
end hair removal MATLAB Answers — New Questions

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I am working onI am working on I am working on notag MATLAB Answers — New Questions

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I get these errors when running this snippet of code that references another script (script attached as a matlab file)
[filt_in, data_in_segs_corrected, segment_params]=segment_filt_v2(data_to_segment,[],[],params.genome_build);
params.segment_params=segment_params;

Warning: The data contains more columns (number of channels) than rows (number of samples). Check that the data is not transposed inadvertently.
> In ctrlMsgUtils.warning (line 25)
In segment (line 76)
In segment_filt_v2>seg_fun (line 145)
In segment_filt_v2 (line 101)
Unrecognized function or variable ‘jmax’.

Error in segment (line 236)
hh=hist(:,jmax);

Error in segment_filt_v2>seg_fun (line 145)
thm=segment([data_in(idx+1:end),ones(length(data_in)-idx,1)],[0 1 1],R2);

Error in segment_filt_v2 (line 101)
data_in_segs_corrected{Chr}(in,c-4)=seg_fun(Data{Chr}(in,c), R2);I get these errors when running this snippet of code that references another script (script attached as a matlab file)
[filt_in, data_in_segs_corrected, segment_params]=segment_filt_v2(data_to_segment,[],[],params.genome_build);
params.segment_params=segment_params;

Warning: The data contains more columns (number of channels) than rows (number of samples). Check that the data is not transposed inadvertently.
> In ctrlMsgUtils.warning (line 25)
In segment (line 76)
In segment_filt_v2>seg_fun (line 145)
In segment_filt_v2 (line 101)
Unrecognized function or variable ‘jmax’.

Error in segment (line 236)
hh=hist(:,jmax);

Error in segment_filt_v2>seg_fun (line 145)
thm=segment([data_in(idx+1:end),ones(length(data_in)-idx,1)],[0 1 1],R2);

Error in segment_filt_v2 (line 101)
data_in_segs_corrected{Chr}(in,c-4)=seg_fun(Data{Chr}(in,c), R2); I get these errors when running this snippet of code that references another script (script attached as a matlab file)
[filt_in, data_in_segs_corrected, segment_params]=segment_filt_v2(data_to_segment,[],[],params.genome_build);
params.segment_params=segment_params;

Warning: The data contains more columns (number of channels) than rows (number of samples). Check that the data is not transposed inadvertently.
> In ctrlMsgUtils.warning (line 25)
In segment (line 76)
In segment_filt_v2>seg_fun (line 145)
In segment_filt_v2 (line 101)
Unrecognized function or variable ‘jmax’.

Error in segment (line 236)
hh=hist(:,jmax);

Error in segment_filt_v2>seg_fun (line 145)
thm=segment([data_in(idx+1:end),ones(length(data_in)-idx,1)],[0 1 1],R2);

Error in segment_filt_v2 (line 101)
data_in_segs_corrected{Chr}(in,c-4)=seg_fun(Data{Chr}(in,c), R2); segment MATLAB Answers — New Questions

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Followed the example of https://www.mathworks.com/help/matlab/ref/tiledlayout.html with ‘TileSpacing’, ‘none’option. The output is as image shown with huge gap in rows.
row = 8;
col = 5;
figure
tiledlayout(row,col,’TileSpacing’,’none’);
for ns = 1:nums
nexttile
imagesc(images(:,:,ns));colormap gray;axis off; axis square
endFollowed the example of https://www.mathworks.com/help/matlab/ref/tiledlayout.html with ‘TileSpacing’, ‘none’option. The output is as image shown with huge gap in rows.
row = 8;
col = 5;
figure
tiledlayout(row,col,’TileSpacing’,’none’);
for ns = 1:nums
nexttile
imagesc(images(:,:,ns));colormap gray;axis off; axis square
end Followed the example of https://www.mathworks.com/help/matlab/ref/tiledlayout.html with ‘TileSpacing’, ‘none’option. The output is as image shown with huge gap in rows.
row = 8;
col = 5;
figure
tiledlayout(row,col,’TileSpacing’,’none’);
for ns = 1:nums
nexttile
imagesc(images(:,:,ns));colormap gray;axis off; axis square
end tiledlayout, matlab MATLAB Answers — New Questions

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Using matlab on arch linux leads to very low quality graphics, I tried following "matlab support high dpi screends on linux", however it did not fix my issue

I tried changing the fonts to make it more readible, but that doesn’t completely work.
Even stranger, the editor seems perfectly fine, it’s just everything else that is low quality

Simulink looks much nicer, and it is readable, might have to zoom in to compare the two pictures
Could the issue be that I am using wayland instead of xorg, which is why the other tutorial wasn’t working?Using matlab on arch linux leads to very low quality graphics, I tried following "matlab support high dpi screends on linux", however it did not fix my issue

I tried changing the fonts to make it more readible, but that doesn’t completely work.
Even stranger, the editor seems perfectly fine, it’s just everything else that is low quality

Simulink looks much nicer, and it is readable, might have to zoom in to compare the two pictures
Could the issue be that I am using wayland instead of xorg, which is why the other tutorial wasn’t working? Using matlab on arch linux leads to very low quality graphics, I tried following "matlab support high dpi screends on linux", however it did not fix my issue

I tried changing the fonts to make it more readible, but that doesn’t completely work.
Even stranger, the editor seems perfectly fine, it’s just everything else that is low quality

Simulink looks much nicer, and it is readable, might have to zoom in to compare the two pictures
Could the issue be that I am using wayland instead of xorg, which is why the other tutorial wasn’t working? linux, display MATLAB Answers — New Questions

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I am trying to cut frequencies above the cutoff frequency using the lowpass function, but it is not working.
I am trying to cut the frequency by applying a lowpass filter to the tunnel current z multiplied by the reference signal sin(wt).
The cutoff frequency is set at 500 Hz, so when I graph the FFT, the peaks at frequencies higher than 500 Hz should disappear, but they do not.

%Parameter Setting
pixel_image = 256; %Input the number of pixels in the image obtained by raster scanning (input 2^n)
dr = 1/(2*sqrt(3)); %Enter dither circle radius [grid].
a_fast_grid = 10; %fast axis scanning range [grid]
a_slow_grid = 10; %Slow axis scanning range [grid]
fm=5000; %Dither circle modulation frequency [Hz]
fs= fm*240 ; %Sampling frequency [Hz]
f_fast = 10.2; %Input scanning frequency [Hz] (1 line scanning count in 1[s])
start_point_x = 0; %Input x-coordinate of scanning start point (input 1 if you want to move by 1[grid])
start_point_y = 0; %Input y-coordinate of scanning start point (input 1 if you want to move by 1[grid])

%Parameter setting for fast-axis triangular wave
amplitude_fast = a_fast_grid/2; %fast axis amplitude

%Parameter setting for slow-axis triangular wave
amplitude_slow = a_slow_grid/2; %slow axis amplitude
f_slow = (f_fast)/(2*pixel_image); %Slow axis triangular wave frequency

% Generation of time vectors
total_time=256/f_fast; %Total Scan Time
t = linspace(0, total_time, fs * total_time);

x_raster = start_point_x + amplitude_fast*(2/pi)*acos(cos(2*pi*f_fast*t));
y_raster = start_point_y + amplitude_slow*(2/pi)*acos(cos(2*pi*f_slow*t));

x_dither = dr*cos(2*pi*fm*t);
y_dither = dr*sin(2*pi*fm*t);

x = x_raster + x_dither;
y = y_raster + y_dither;

z1 = cos(2*pi*((x-y)/(sqrt(3))));
z2 = cos(2*pi*(2*y/(sqrt(3))));
z3 = cos(2*pi*((x+y)/(sqrt(3))));
z = (z1 + z2 + z3);

%Reference Signal Creation
w = 2*pi*fm ;
phi = 0 ;
reference_signal = sin(w*t+phi) ;

%Mixing signal creation
mixising_signal = z.* reference_signal ;

%Applying a low-pass filter
f_cutoff = 500 ; %Cutoff frequency [Hz] setting
lowpass_signal = lowpass(mixising_signal,f_cutoff,fs) ;

%The result of FFT.——————————————-
n = length(t); % Number of samples
f = (0:n-1)*(fs/n); % Frequency axis [Hz]
half_n = floor(n/2); % Half of data

fft_z = abs(fft(z) / n);
fft_mixising_signal = abs(fft(mixising_signal) / n);
fft_lowpass_signal = abs(fft(lowpass_signal) / n);

figure;
tiledlayout(3,1)
nexttile
plot(f(1:half_n), fft_z(1:half_n));
title(‘FFT of tunnel current’); xlabel(‘Frequency [Hz]’); ylabel(‘Amplitude’);
xlim([0 20000]),ylim([0 0.3]);

nexttile
plot(f(1:half_n), fft_mixising_signal(1:half_n));
title(‘FFT of mixing signal’); xlabel(‘Frequency [Hz]’); ylabel(‘Amplitude’);
xlim([0 20000]),ylim([0 0.3]);

nexttile
plot(f(1:half_n), fft_lowpass_signal(1:half_n));
title(‘FFT after applying low-pass’); xlabel(‘Frequency [Hz]’); ylabel(‘Amplitude’);
xlim([0 20000]),ylim([0 0.3]);I am trying to cut frequencies above the cutoff frequency using the lowpass function, but it is not working.
I am trying to cut the frequency by applying a lowpass filter to the tunnel current z multiplied by the reference signal sin(wt).
The cutoff frequency is set at 500 Hz, so when I graph the FFT, the peaks at frequencies higher than 500 Hz should disappear, but they do not.

%Parameter Setting
pixel_image = 256; %Input the number of pixels in the image obtained by raster scanning (input 2^n)
dr = 1/(2*sqrt(3)); %Enter dither circle radius [grid].
a_fast_grid = 10; %fast axis scanning range [grid]
a_slow_grid = 10; %Slow axis scanning range [grid]
fm=5000; %Dither circle modulation frequency [Hz]
fs= fm*240 ; %Sampling frequency [Hz]
f_fast = 10.2; %Input scanning frequency [Hz] (1 line scanning count in 1[s])
start_point_x = 0; %Input x-coordinate of scanning start point (input 1 if you want to move by 1[grid])
start_point_y = 0; %Input y-coordinate of scanning start point (input 1 if you want to move by 1[grid])

%Parameter setting for fast-axis triangular wave
amplitude_fast = a_fast_grid/2; %fast axis amplitude

%Parameter setting for slow-axis triangular wave
amplitude_slow = a_slow_grid/2; %slow axis amplitude
f_slow = (f_fast)/(2*pixel_image); %Slow axis triangular wave frequency

% Generation of time vectors
total_time=256/f_fast; %Total Scan Time
t = linspace(0, total_time, fs * total_time);

x_raster = start_point_x + amplitude_fast*(2/pi)*acos(cos(2*pi*f_fast*t));
y_raster = start_point_y + amplitude_slow*(2/pi)*acos(cos(2*pi*f_slow*t));

x_dither = dr*cos(2*pi*fm*t);
y_dither = dr*sin(2*pi*fm*t);

x = x_raster + x_dither;
y = y_raster + y_dither;

z1 = cos(2*pi*((x-y)/(sqrt(3))));
z2 = cos(2*pi*(2*y/(sqrt(3))));
z3 = cos(2*pi*((x+y)/(sqrt(3))));
z = (z1 + z2 + z3);

%Reference Signal Creation
w = 2*pi*fm ;
phi = 0 ;
reference_signal = sin(w*t+phi) ;

%Mixing signal creation
mixising_signal = z.* reference_signal ;

%Applying a low-pass filter
f_cutoff = 500 ; %Cutoff frequency [Hz] setting
lowpass_signal = lowpass(mixising_signal,f_cutoff,fs) ;

%The result of FFT.——————————————-
n = length(t); % Number of samples
f = (0:n-1)*(fs/n); % Frequency axis [Hz]
half_n = floor(n/2); % Half of data

fft_z = abs(fft(z) / n);
fft_mixising_signal = abs(fft(mixising_signal) / n);
fft_lowpass_signal = abs(fft(lowpass_signal) / n);

figure;
tiledlayout(3,1)
nexttile
plot(f(1:half_n), fft_z(1:half_n));
title(‘FFT of tunnel current’); xlabel(‘Frequency [Hz]’); ylabel(‘Amplitude’);
xlim([0 20000]),ylim([0 0.3]);

nexttile
plot(f(1:half_n), fft_mixising_signal(1:half_n));
title(‘FFT of mixing signal’); xlabel(‘Frequency [Hz]’); ylabel(‘Amplitude’);
xlim([0 20000]),ylim([0 0.3]);

nexttile
plot(f(1:half_n), fft_lowpass_signal(1:half_n));
title(‘FFT after applying low-pass’); xlabel(‘Frequency [Hz]’); ylabel(‘Amplitude’);
xlim([0 20000]),ylim([0 0.3]); I am trying to cut frequencies above the cutoff frequency using the lowpass function, but it is not working.
I am trying to cut the frequency by applying a lowpass filter to the tunnel current z multiplied by the reference signal sin(wt).
The cutoff frequency is set at 500 Hz, so when I graph the FFT, the peaks at frequencies higher than 500 Hz should disappear, but they do not.

%Parameter Setting
pixel_image = 256; %Input the number of pixels in the image obtained by raster scanning (input 2^n)
dr = 1/(2*sqrt(3)); %Enter dither circle radius [grid].
a_fast_grid = 10; %fast axis scanning range [grid]
a_slow_grid = 10; %Slow axis scanning range [grid]
fm=5000; %Dither circle modulation frequency [Hz]
fs= fm*240 ; %Sampling frequency [Hz]
f_fast = 10.2; %Input scanning frequency [Hz] (1 line scanning count in 1[s])
start_point_x = 0; %Input x-coordinate of scanning start point (input 1 if you want to move by 1[grid])
start_point_y = 0; %Input y-coordinate of scanning start point (input 1 if you want to move by 1[grid])

%Parameter setting for fast-axis triangular wave
amplitude_fast = a_fast_grid/2; %fast axis amplitude

%Parameter setting for slow-axis triangular wave
amplitude_slow = a_slow_grid/2; %slow axis amplitude
f_slow = (f_fast)/(2*pixel_image); %Slow axis triangular wave frequency

% Generation of time vectors
total_time=256/f_fast; %Total Scan Time
t = linspace(0, total_time, fs * total_time);

x_raster = start_point_x + amplitude_fast*(2/pi)*acos(cos(2*pi*f_fast*t));
y_raster = start_point_y + amplitude_slow*(2/pi)*acos(cos(2*pi*f_slow*t));

x_dither = dr*cos(2*pi*fm*t);
y_dither = dr*sin(2*pi*fm*t);

x = x_raster + x_dither;
y = y_raster + y_dither;

z1 = cos(2*pi*((x-y)/(sqrt(3))));
z2 = cos(2*pi*(2*y/(sqrt(3))));
z3 = cos(2*pi*((x+y)/(sqrt(3))));
z = (z1 + z2 + z3);

%Reference Signal Creation
w = 2*pi*fm ;
phi = 0 ;
reference_signal = sin(w*t+phi) ;

%Mixing signal creation
mixising_signal = z.* reference_signal ;

%Applying a low-pass filter
f_cutoff = 500 ; %Cutoff frequency [Hz] setting
lowpass_signal = lowpass(mixising_signal,f_cutoff,fs) ;

%The result of FFT.——————————————-
n = length(t); % Number of samples
f = (0:n-1)*(fs/n); % Frequency axis [Hz]
half_n = floor(n/2); % Half of data

fft_z = abs(fft(z) / n);
fft_mixising_signal = abs(fft(mixising_signal) / n);
fft_lowpass_signal = abs(fft(lowpass_signal) / n);

figure;
tiledlayout(3,1)
nexttile
plot(f(1:half_n), fft_z(1:half_n));
title(‘FFT of tunnel current’); xlabel(‘Frequency [Hz]’); ylabel(‘Amplitude’);
xlim([0 20000]),ylim([0 0.3]);

nexttile
plot(f(1:half_n), fft_mixising_signal(1:half_n));
title(‘FFT of mixing signal’); xlabel(‘Frequency [Hz]’); ylabel(‘Amplitude’);
xlim([0 20000]),ylim([0 0.3]);

nexttile
plot(f(1:half_n), fft_lowpass_signal(1:half_n));
title(‘FFT after applying low-pass’); xlabel(‘Frequency [Hz]’); ylabel(‘Amplitude’);
xlim([0 20000]),ylim([0 0.3]); low pass filter, matlab, signal processing MATLAB Answers — New Questions

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Dear colleagues,
I have problem with optimisation function fminimax, which I use as a variant for regression analyses made by Curve fitter.
In the attached file are the vectors for x, y and z = f(x,y).
My code is, as follow. Sorry, I don’t know how to paste it in appropriate format:
"x=x_vremetraene;
y=y_period_pretovarvane;
z=z_111;
f = @(p) abs(p(1)*(p(2)+y).^p(3)./(x+p(4)).^p(5) – z);
F = @(p) reshape(f(p),[],1);

p0 = [1 1 1 1 1];

F(p0)

ans =

1.0e+02 *

2.314576069270182
2.796122919180851
3.081916551260510
3.284514792242277
3.441464927295814
3.926826269823946
1.650257401771352
………………………..

%by this point everything is good

format long
[p,fval] = fminimax(F,p0)

Local minimum possible. Constraints satisfied.

fminimax stopped because the size of the current search direction is less than
twice the value of the step size tolerance and constraints are
satisfied to within the value of the constraint tolerance.

<stopping criteria details>

p =

1.0e+03 *

-0.320589519017169 -2.281076321933144 -0.703130918618337 -0.366998883941303 1.591728772382496

fval =

1.0e+02 *

2.317909402603515
2.801122919180851
3.088583217927177
………"
%The expected values for fval should be about 20. But they are as large as input z data.

Local minimum possible. Constraints satisfied.

I have tried p = lsqnonlin(F,p0) for preestimation of the parameters p
but it also gives me an error message.
Error using lsqncommon (line 38)
Objective function is returning Inf or NaN values at initial point. lsqnonlin cannot continue.

Error in lsqnonlin (line 264)
lsqncommon(funfcn,xCurrent,lb,ub,options,defaultopt,caller,…
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^Dear colleagues,
I have problem with optimisation function fminimax, which I use as a variant for regression analyses made by Curve fitter.
In the attached file are the vectors for x, y and z = f(x,y).
My code is, as follow. Sorry, I don’t know how to paste it in appropriate format:
"x=x_vremetraene;
y=y_period_pretovarvane;
z=z_111;
f = @(p) abs(p(1)*(p(2)+y).^p(3)./(x+p(4)).^p(5) – z);
F = @(p) reshape(f(p),[],1);

p0 = [1 1 1 1 1];

F(p0)

ans =

1.0e+02 *

2.314576069270182
2.796122919180851
3.081916551260510
3.284514792242277
3.441464927295814
3.926826269823946
1.650257401771352
………………………..

%by this point everything is good

format long
[p,fval] = fminimax(F,p0)

Local minimum possible. Constraints satisfied.

fminimax stopped because the size of the current search direction is less than
twice the value of the step size tolerance and constraints are
satisfied to within the value of the constraint tolerance.

<stopping criteria details>

p =

1.0e+03 *

-0.320589519017169 -2.281076321933144 -0.703130918618337 -0.366998883941303 1.591728772382496

fval =

1.0e+02 *

2.317909402603515
2.801122919180851
3.088583217927177
………"
%The expected values for fval should be about 20. But they are as large as input z data.

Local minimum possible. Constraints satisfied.

I have tried p = lsqnonlin(F,p0) for preestimation of the parameters p
but it also gives me an error message.
Error using lsqncommon (line 38)
Objective function is returning Inf or NaN values at initial point. lsqnonlin cannot continue.

Error in lsqnonlin (line 264)
lsqncommon(funfcn,xCurrent,lb,ub,options,defaultopt,caller,…
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Dear colleagues,
I have problem with optimisation function fminimax, which I use as a variant for regression analyses made by Curve fitter.
In the attached file are the vectors for x, y and z = f(x,y).
My code is, as follow. Sorry, I don’t know how to paste it in appropriate format:
"x=x_vremetraene;
y=y_period_pretovarvane;
z=z_111;
f = @(p) abs(p(1)*(p(2)+y).^p(3)./(x+p(4)).^p(5) – z);
F = @(p) reshape(f(p),[],1);

p0 = [1 1 1 1 1];

F(p0)

ans =

1.0e+02 *

2.314576069270182
2.796122919180851
3.081916551260510
3.284514792242277
3.441464927295814
3.926826269823946
1.650257401771352
………………………..

%by this point everything is good

format long
[p,fval] = fminimax(F,p0)

Local minimum possible. Constraints satisfied.

fminimax stopped because the size of the current search direction is less than
twice the value of the step size tolerance and constraints are
satisfied to within the value of the constraint tolerance.

<stopping criteria details>

p =

1.0e+03 *

-0.320589519017169 -2.281076321933144 -0.703130918618337 -0.366998883941303 1.591728772382496

fval =

1.0e+02 *

2.317909402603515
2.801122919180851
3.088583217927177
………"
%The expected values for fval should be about 20. But they are as large as input z data.

Local minimum possible. Constraints satisfied.

I have tried p = lsqnonlin(F,p0) for preestimation of the parameters p
but it also gives me an error message.
Error using lsqncommon (line 38)
Objective function is returning Inf or NaN values at initial point. lsqnonlin cannot continue.

Error in lsqnonlin (line 264)
lsqncommon(funfcn,xCurrent,lb,ub,options,defaultopt,caller,…
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ fminimax MATLAB Answers — New Questions

​

How can I find the bandwidth of the signal in the attached picture?How can I find the bandwidth of the signal in the attached picture? How can I find the bandwidth of the signal in the attached picture? digital signal processing MATLAB Answers — New Questions

​

Hi,

I have one set of RGB values

RGB=[85 270 100];

How to represent/visualise the colour in a say, 80 by 80 pixel colour patch?Hi,

I have one set of RGB values

RGB=[85 270 100];

How to represent/visualise the colour in a say, 80 by 80 pixel colour patch? Hi,

I have one set of RGB values

RGB=[85 270 100];

How to represent/visualise the colour in a say, 80 by 80 pixel colour patch? single set rgb, rgb visualisation MATLAB Answers — New Questions

​