how to multiply data to the aplitude of each pluse of multiple cycles of a signal??little emergency please reply fast

i have a pulse train of 500 cycles of half cycle sinus i need to multiply data(1)*amplitude of first pulse of p ,data[2] *amplitude of second pulse so on.....

please help me out im struggling to solve this from many days its little emergency please help me out ... thanks in advance

fs=44100;
ts=(1/fs);
fc=4000;
tp=2.2676e-3;
for n=1:10
   ak=ones(10);
t = 0:ts:n*tp;
p = abs(sin(pi*440*t));
subplot(3,1,2)
plot(t,p)
end
data = randi([0,1],1,10);

[Walter Roberson]

http://www.mathworks.com/matlabcentral/answers/29922-why-your-question-is-not-urgent-or-an-emergency

Answer 1

[Adam Danz]

Before we get into the solution, here's a few comments about the code.

• Your loop isn't doing anything useful. It's overwritten on each iteration and the final result only shows the last iteration when n=10. So, the code below just shows the results when n=10.
• The amplitude of each peak is clearly 1 +/- some tiny variation.
• Since your 'data' vector is only 0s and 1s and since you're multiplying that by the amplitudes which are all ~1, then the 'data' vector won't change unless you want to account for the tiny variation.

If you'd like to get the actual amplitude of your peaks which include the tiny variations around 1, you can use findpeaks().

% your code, simplified 
fs=44100;
ts=(1/fs);
fc=4000;
tp=2.2676e-3;
n = 10;
t = 0:ts:n*tp;
p = abs(sin(pi*440*t));
subplot(3,1,2)
plot(t,p)
data = randi([0,1],1,10);

% Find the amp of each peak
amps = findpeaks(p);

% multiply data(n) and peak(n)
out = data .* amps; 

[Shwetha Padmanabha]

no i didnt obtatain expected answer

[Adam Danz]

What did you expect?

How did you use the "out" variable to produce your plot?

"i need to multiply data(1)*amplitude of first pulse of p ,data[2] *amplitude of second pulse so on....."

That's what my answer does. If you need something different from that, you'll need to explain it.

[Shwetha Padmanabha]

[Shwetha Padmanabha]

need to solve using this equation tried all possible ways still couldnt get the answer it would very helpful if can solve

[Adam Danz]

Your question was:

"i need to multiply data(1)*amplitude of first pulse of p ,data[2] *amplitude of second pulse so on....."

That's what my answer does. I calculates the amp of each pulse and multiplies it by the same index value in 'data'. Nowhere in your question do you mention plotting or the need to flip the sign of the curve at particular locations.

"no i didnt obtatain expected answer"

That's because you changed your question. Judging from the picture, it seems that you need to apply the 'out' vector to determine whether the sin of each negative portion of the sin curve should be flipped to positive. I'm not sure how that relates to the equation you shared.

For every period of sin(pi*440*t), there will be a positive half and a negative half. If out(n) is >0 then you should flip that section of the curve.

[Adam Danz]

In response to your comment written as an answer, it's still very unclear what the goal is. The equation that was shared was not explained and doesn't exist anywhere in the code.

Judging from the image, my only guess is that you have a pulse train and a vector of 1s and 0s (or maybe 1s and -1s) that determines the sign of each pulse.

The code below does that. As you can see in the figure below, when data is 1, the pulse is positive. When data is 0 the pulse is negative. You can adapt this as needed.

% your code, simplified
fs=44100;
ts=(1/fs);
fc=4000;
tp=2.2676e-3;
n = 10;
t = 0:ts:n*tp;
s = sin(pi*440*t); 
p = abs(s);
% subplot(3,1,2)    %not needed
% plot(t,p)         %not needed
data = randi([0,1],1,10);
% multiply data(n) and peak(n)
% out = data .* amps; % Not used

% Find the amp of each peak
% amps = findpeaks(p);   % Not Used

% Find troughs
[~,troughLoc] = findpeaks(-p);
troughLoc = unique([1,troughLoc,numel(p)]); 

% For every 1 in 'data', flip the pulse sign to positive 
p2 = -abs(s); 
for i = 1:numel(data)
    if data(i)==1
        p2(troughLoc(i):troughLoc(i+1)) = abs(p2(troughLoc(i):troughLoc(i+1))); 
    end
end

clf()
plot(p2)
title(sprintf('Data = [%s]',num2str(data)))
hold on
plot(xlim(),[0,0],'k-') %line at y=0
plot([troughLoc;troughLoc],ylim(), 'k:')

[Adam Danz]

Shwetha Padmanabha's answer moved here as a comment.

Thank you!!!!

[Adam Danz]

Glad I could help.

Answer 2

[Shwetha Padmanabha]

this represent the pulse train

clc;
clear all;
close all;
fs=44100;
ts=(1/fs);
fc=4000;
tp=2.2676e-3;
tsingle = 0:ts:tp;
for n=1:500
t = 0:ts:n*tp;
p = abs(sin(pi*440*t));
subplot(3,1,2)
plot(t,p)
end

data = randi([0,1],1,20);
a=[];
for k=1:2:20
if data(k)==0 && data(k+1)==0
a=[a 1+i];
elseif data(k)==0 && data(k+1)==1
a=[a -1+i];
elseif data(k)==1 && data(k+1)==0
a=[a -1-i];
elseif data(k)==1 && data(k+1)==1
a=[a 1+i];
end
end
well let me be clear this the actual code each pulse  should carry one symboli.e. is 'a' tried all possible ways i couldnt solve even ohers couldnt deadline is near it would be great help if you cansolve 

[Adam Danz]

@ Shwetha Padmanabha, the problem is still very unlcear. I really have no idea what the goals is or the question.

I took a guess at what you're trying to do and added a comment in my answer above. I hope that's helpful. A mini lesson here is the importance of clearly defining a problem and fully understanding it before seeking a solution.

"If I had only one hour to save the world, I would spend fifty-five minutes defining the problem, and only five minutes finding the solution." -Albert Einstein