How could I use if function to update the newest maximum value

Hi all,

I am going to use photodiode to monitor my laser and capture the max power during the machine is operating.

So, I write the code to generate similar signal and want to build a tool for me can always update the maximum value in the real time.

I have thought about using "if" to update the max value when the latest one is higher but not sure how to implemente it?

Here is my code below, I used rand() to produce random amplitude signal:

fs = 50000; % 50 kHz frequency
Ts = 1/fs*10^9; % sample rate in neno seconds
t = 1:Ts;
pulse = t<=5;
rand_amp = rand(10,1);
sig = pulse.*rand_amp;
sig = reshape(sig', [], 1);
t_total = 1:numel(sig);
plot(t_total, sig);
%find the maximum
maxPulse= max(sig);
xlabel('Time (ns)');
ylabel('Amplitude');

Answer 1

[Ameer Hamza]

Try this

fs = 50000; % 50 kHz frequency
Ts = 1/fs*10^9; % sample rate in neno seconds
t = 1:Ts;
pulse = t<=5;
rand_amp = rand(10,1);
sig = pulse.*rand_amp;
sig = reshape(sig', [], 1);

max_val = zeros(size(sig)); % save maximum value at each time step;
max_val(1) = sig(1); % first maximum value is the first sample of sig
for i=2:numel(max_val)
    if sig(i) > max_val(i-1)
        max_val(i) = sig(i);
    else
        max_val(i) = max_val(i-1);
    end
end
t_total = 1:numel(sig);

figure;
plot(t_total, sig);
xlabel('Time (ns)');
ylabel('Amplitude');

figure;
plot(t_total, max_val);
xlabel('Time (ns)');
ylabel('Amplitude');

[Chen Kevin]

Thanks again Ameer! It works as I expected!

[Ameer Hamza]

Glad to be of help.

[Chen Kevin]

Hi Ameer,

Sorry for annoying, but it is possible to add more lines in order obtain the average of pulse height and standard deviation?

In addition, average value of pulse width and standard deviation?

[Ameer Hamza]

How do you define the average of pulse height? The average of all the pulse height will just be a constant.

[Chen Kevin]

Sorry, maybe using "Pulse height" will let you misunderstand.

For example, get the average value of these 10 pulses and so on.

[Ameer Hamza]

Do you want something like this

fs = 50000; % 50 kHz frequency
Ts = 1/fs*10^9; % sample rate in neno seconds
t = 1:Ts;
pulse = t<=5;
rand_amp = rand(10,1);
sig = pulse.*rand_amp;
sig = reshape(sig', [], 1);
max_val = zeros(size(sig)); % save maximum value at each time step;
max_val(1) = sig(1); % first maximum value is the first sample of sig
for i=2:numel(max_val)
    if sig(i) > max_val(i-1)
        max_val(i) = sig(i);
    else
        max_val(i) = max_val(i-1);
    end
end
t_total = 1:numel(sig);
figure;
hold on;
plot(t_total, sig);
plot(t_total, mean(rand_amp)*ones(size(t_total)));
plot(t_total, var(rand_amp)*ones(size(t_total)));
legend({'signal', 'mean', 'variance'})
xlabel('Time (ns)');
ylabel('Amplitude');
figure;
plot(t_total, max_val);
xlabel('Time (ns)');
ylabel('Amplitude');

[Chen Kevin]

That is really good! Thanks again!

[Chen Kevin]

Hi Ameer,

Sorry for bother you again, I am wonder if I could also have the latest minimum pulse width, mean value and standard deviation?

[Ameer Hamza]

The pulse width of all the pulses is equal. How do you define minimum value in that case?

[Chen Kevin]

No problem, because in the realistic experiment, the laser pulse width definetly will have little varying during operating. Maybe is not suitable in this case but it is possible to implement this machnisim at first?

[Ameer Hamza]

In this case, pulse width is always 5 ns, so you can just plot a line at 5

plot(t_total, 5*ones(size(t_total)));

[Chen Kevin]

Hi Ameer, Thanks! It works!

I also want to calculate the ratio between each amplitude to max, so I try to divide them with max but it shown the error:

How could I fix that?

Matrix dimensions must agree.

Error in constantpulse (line 27)

fac= amp./max_val

fs = 50000; % 50 kHz frequency
Ts = 1/fs*10^9; % sample rate in neno seconds
t = 1:Ts;
pulse = t<=5;
amp = [0.1, 0.3, 0.5, 0.5, 0.6, 1, 0.2, 0.8, 0.1, 0.4].';
sig = pulse.*amp;
sig = reshape(sig', [], 1);
t_total = 1:numel(sig);
max_val = zeros(size(sig)); % save maximum value at each time step;
max_val(1) = sig(1); % first maximum value is the first sample of sig
for i=2:numel(max_val)
    if sig(i) > max_val(i-1)
        max_val(i) = sig(i);
    else
        max_val(i) = max_val(i-1);
    end
end
t_total = 1:numel(sig);
figure;
plot(t_total, sig);
xlabel('Time (ns)');
ylabel('Amplitude');
figure;
plot(t_total, max_val);
xlabel('Time (ns)');
ylabel('Amplitude');
fac= amp./max_val

[Ameer Hamza]

I think you should use sig instead of amp

fac = sig./max_val;

[Chen Kevin]

I tried it, but I think it divided to the wrong number...

[Ameer Hamza]

You can try this

mask = sig~=0;
fac = zeros(size(sig));
fac(mask) = sig(mask)./max_val(mask);

[Chen Kevin]

Still gave lots of 1...

fs = 50000; % 50 kHz frequency
Ts = 1/fs*10^9; % sample rate in neno seconds
t = 1:Ts;
pulse = t<=5;
amp = [0.1, 0.3, 0.5, 0.5, 0.6, 1, 0.2, 0.8, 0.1, 0.4].';
sig = pulse.*amp;
sig = reshape(sig', [], 1);
t_total = 1:numel(sig);
max_val = zeros(size(sig)); % save maximum value at each time step;
max_val(1) = sig(1); % first maximum value is the first sample of sig
for i=2:numel(max_val)
    if sig(i) > max_val(i-1)
        max_val(i) = sig(i);
    else
        max_val(i) = max_val(i-1);
    end
end
t_total = 1:numel(sig);
figure;
plot(t_total, sig);
xlabel('Time (ns)');
ylabel('Amplitude');
figure;
plot(t_total, max_val);
xlabel('Time (ns)');
ylabel('Amplitude');
mask = sig~=0;
fac = zeros(size(sig));
fac(mask) = sig(mask)./max_val(mask);
ra=fac(mask);

[Ameer Hamza]

This is because you used

amp = [0.1, 0.3, 0.5, 0.5, 0.6, 1, 0.2, 0.8, 0.1, 0.4].';

which is an increasing signal to 1. So upto that point, all the maximum values are equal to the signal itself.

[Chen Kevin]

I think I finally got what I want

amp1= [amp].';
fac= amp1./max_val;

It generate a 2000000x10 array, if I am only interest to the final value set, how could I extract them?

[Ameer Hamza]

What do you mean by "final value set"? I am not sure how to use this matrix.