Info
This question is locked. Öffnen Sie sie erneut, um sie zu bearbeiten oder zu beantworten.
Please help me convert equation to matlab code.
8 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Deal all.
I need you help to convert this equation to matlab code
I spend a lot of time to write it but it doesn't work. Thank you.
1 Kommentar
Walter Roberson
am 1 Apr. 2018
Are you permitted to use the symbolic toolbox?
Is the question about providing some kind of symbolic proof, or is it about calculation of the formula using finite precision and a particular numeric input?
Antworten (8)
Roger Stafford
am 1 Apr. 2018
N = 100; % <-- Choose some large number
s = x;
for n = 2*N-1:-2:1
s = x - s*x^2/((n+2)*(n+1));
end
(I think you meant to take the limit as N approaches infinity, not x.)
0 Kommentare
kalai selvi
am 15 Sep. 2020
pls answer this question ...how to write the equation into code
2 Kommentare
Walter Roberson
am 15 Sep. 2020
π is written as pi in MATLAB.
exp of an expression is written as exp(expression) in MATLAB.
is written as sqrt(expression) in MATLAB.
kalai selvi
am 16 Sep. 2020
How to write a code on IOTA filter in fbmc system
2 Kommentare
Walter Roberson
am 16 Sep. 2020
Bearbeitet: Walter Roberson
am 17 Sep. 2020
Warning: pudn has questionable security. Take precautions when you access it.
Kunwar Pal Singh
am 26 Apr. 2021
please answer this....how to write this equation into MATLAB CODE
1 Kommentar
Walter Roberson
am 26 Apr. 2021
%these variables must be defined in a way appropriate for your situation
S_N = rand() * 10
theta = randn() * 2 * pi
l = randi([2 10])
b_1 = rand()
c_11 = rand()
t_year = randi([1950 2049])
d_11 = rand()
t_1 = rand()
t_x = t_1 + rand()
lambda_a = randi([500 579])
LOTF_a = rand()
P = rand()
K_l = rand()
k_0 = rand()
t_tau = randi(10)
overhaulcost_a = 1000 + rand()*100
%the work
syms t
part1 = int(S_N .* cos(theta) .* l .* b_1 .* t_year .* d_11, t, t_1, t_x);
part2 = int(lambda_a .* LOTF_a, t, t_1, t_x);
part3 = int(P*K_l .* t_year + P .* k_0 .* l .* l .* t_tau, t, t_1, t_x);
part4 = overhaulcost_a ;
result = part1 - part2 - part3 - part4;
Jakub Laznovsky
am 19 Mai 2021
Hi guys, can you please help me with conversion this piece of code to the mathematical equation?
It i a simple 3D mask proceeding the image, and searching for adjoining number one and number two. Thank you in advance.
Code:
m1=[0 0 0; 0 1 0; 0 0 0];
m2=[0 1 0; 1 1 1; 0 1 0];
mask=zeros(3,3,3);
mask(:,:,1)=m1;mask(:,:,2)=m2;mask(:,:,3)=m1;
for i=2:size(image,1)-1
for j=2:size(image,2)-1
for k=2:size(image,3)-1
help_var=image(i-1:i+1,j-1:j+1,k-1:k+1);
if sum(unique(help_var(mask==1)))==3
new_image(i,j,k)=3; %marks adjoining pixel with number 3
end
end
end
end
4 Kommentare
Adhin Abhi
am 4 Jan. 2022
(λlog vmax−log vmin) /(vmax−vmin )
1 Kommentar
Walter Roberson
am 4 Jan. 2022
(lambda .* log(vmax) - log(vmin)) ./ (vmax - vmin)
Lukasz Sarnacki
am 17 Aug. 2022
Please help
5 Kommentare
Lukasz Sarnacki
am 17 Aug. 2022
Thay all are arrays
A(x; y) is the average intensity of the fringe image
B(x; y) is the so-called intensity modulation.
ϕ is the corresponding wrapped phase
Walter Roberson
am 18 Aug. 2022
syms n N integer
syms A(x,y) B(x,y) phi(x,y)
Pi = sym(pi)
I(n,x,y) = A(x, y) + B(x,y) * cos(phi(x,y) - 2*Pi*n/N)
numerator = simplify(symsum(I(n, x, y) .* sin(2*Pi*n/N), n, 0, N-1))
denominator = simplify(symsum(I(n, x, y) .* cos(2*Pi*n/N), n, 0, N-1))
eqn = phi(x,y) == atan(numerator ./ denominator)
simplify(eqn)
This question is locked.
Siehe auch
Kategorien
Mehr zu Assumptions finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!