hi,
i want to write the following equation in a for loop to store values for 1000 rows..Can anyone help please?
z(k)=lamda*x(k)+(1-lamda)*z(k-1)

 채택된 답변

Jan
Jan 2017년 7월 8일

0 개 추천

z = zeros(1000, 22);
x = rand(1000, 22);
lamda = rand();
for k = 2 : 1000
z(k, :) = lamda * x(k, :) + (1-lamda) * z(k-1, :);
end

댓글 수: 4

Md. Tanjin Amin
Md. Tanjin Amin 2017년 7월 8일
Hi, the code runs ..still the result is not satisfactory....I am attaching the data file and expected result for lamda=0.03. I have done the calculation using excell. Can you look again and help please? Thank you.
Walter Roberson
Walter Roberson 2017년 7월 8일
Your expected values are wrong, not even self-consistent.
See attached code and plots.
Md. Tanjin Amin
Md. Tanjin Amin 2017년 7월 10일
Hi, I am sorry. You are right the expected values were wrong. I have attached the corrected expected values. This is right this time. Can you give some time and help please?
Walter Roberson
Walter Roberson 2017년 7월 10일
Your first two lines of X from X.mat are the same. That cannot be explained with any lamda other than 0 .
Your other lines of expected output cannot be explained with lamda = 0.03 : they all require lamda = 1.0003
Even then your expected outputs are frequently wrong in the 4th digit.
See attached, which is configured for 1.0003

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추가 답변 (1개)

Walter Roberson
Walter Roberson 2017년 7월 8일

0 개 추천

z(1) = rand();
x = randn(1,1000);
lamda = rand();
for k = 2 : 1000
z(k)=lamda*x(k)+(1-lamda)*z(k-1);
end

댓글 수: 1

Md. Tanjin Amin
Md. Tanjin Amin 2017년 7월 8일
Hi, Thank you. Atually, I should have been more specific. X is 1000*22 matrix and initial value of Z is zero (a 1*22 zero matrix)..Can you help please?

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카테고리

도움말 센터File Exchange에서 Loops and Conditional Statements에 대해 자세히 알아보기

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2017년 7월 8일

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2017년 7월 10일

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