How to use for or while loop for certain range?

조회 수: 22 (최근 30일)
Asrorkhuja Ortikov
Asrorkhuja Ortikov 2020년 6월 30일
답변: Asrorkhuja Ortikov 2020년 7월 1일
I want to add or substract a certain number to my array cell (result(i)) based on corresponding binary(0 or 1) pattern I have . Condition: if binary_pattern is 1 then it should increase with speed of 0.75 per minute, if zero it should decrease with the same speed(-0.75). But when binary_pattern is 1 and inside the range 15 and 20 it should increase by 1 not 0.75. Which means when result reaches 20, has to decrease by and increase again when hits 15 with the speed of 1. Any advices on that? I have tried with the code below, but due to I'm new to Matlab couldn't get my head around it. Thanks in advance!
binary_pattern = zeros(100,1);
binary_pattern(1:25) = 1;
binary_pattern(50:75) = 1;
binary_pattern(90:100) = 1; % to create binary pattern
minutes = [1:100]'; % scale of minutes
result = zeros(100,1);
result(1,1) = 5; % initial condition of a state
for i = 1:100
if binary_pattern(i) == 1
result(i+1) = result(i) + 0.75;
if result(i)>= 20
result(i+1) = result(i) - 1;
elseif result(i)<= 15
result(i+1) = result(i) + 1;
else
end
else
result(i+1) = result(i) - 0.75;
end
end
result(end,:) = [];
plot(minutes,result)

채택된 답변

Asrorkhuja Ortikov
Asrorkhuja Ortikov 2020년 7월 1일
binary_pattern = zeros(100,1);
binary_pattern(1:25) = 1;
binary_pattern(50:75) = 1;
binary_pattern(90:100) = 1; % to create binary pattern
minutes = [1:100]'; % scale of minutes
result = zeros(100,1);
result(1,1) = 15; % initial condition of a state
DeS = 0;
for i = 1:100
if binary_pattern(i) == 1
result(i+1) = result(i) + 1;
if DeS == -1
result(i+1) = result(i) - 1;
end
if result(i+1)>20
result(i+1)= result(i) - 1;
DeS = -1;
end
if result(i+1)<15
result(i+1) = result(i) + 1;
DeS = 0;
end
else % pattern is 0
result(i+1) = result(i) - 0.75;
end
end
result(end,:) = [];
plot(minutes,result)

추가 답변 (1개)

SC
SC 2020년 6월 30일
I've edited the code, according to your logic (from what I've understood)
(Try explaining your logic better)
see if this works, incase:
zerosmatrix = zeros(100,1);
zerosmatrix(1:25) = 1;
zerosmatrix(50:75) = 1;
zerosmatrix(90:100) = 1;
timeframe = [1:100]';
state = zeros(100,1);
for i = 1:100
if zerosmatrix(i) == 1
state(i+1) = state(i) + 0.75;
if state(i)>= 15 && state(i)<= 20
state(i+1) = state(i) + 1;
elseif state(i)>= 20
state(i+1) = state(i) - 1;
end
else
state(i+1) = state(i) - 0.75;
end
end
state(end,:) = [];
plot(timeframe,state)
  댓글 수: 5
SC
SC 2020년 7월 1일
Hey that is because of the binary pattern, please try another binary pattern and check.
Asrorkhuja Ortikov
Asrorkhuja Ortikov 2020년 7월 1일
@ SC I did, but it is still not working? Did you check it yourself first? Here I took different approach, have a look:
binary_pattern = zeros(100,1);
binary_pattern(1:25) = 1;
binary_pattern(50:75) = 1;
binary_pattern(90:100) = 1; % to create binary pattern
minutes = [1:100]'; % scale of minutes
result = zeros(100,1);
result(1,1) = 15; % initial condition of a state
DeS = 0;
for i = 1:100
if binary_pattern(i) == 1
result(i+1) = result(i) + 1;
if DeS == -1
result(i+1) = result(i) - 1;
end
if result(i+1)>20
result(i+1)= result(i) - 1;
DeS = -1;
end
if result(i+1)<15
result(i+1) = result(i) + 1;
DeS = 0;
end
else % pattern is 0
result(i+1) = result(i) - 0.75;
end
end
result(end,:) = [];
plot(minutes,result)

댓글을 달려면 로그인하십시오.

카테고리

Help CenterFile Exchange에서 Antenna and Array Analysis에 대해 자세히 알아보기

제품


릴리스

R2020a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by