I'm doing a research about map projections. I use this script:
koordinat = [0 102;...
0 103.5;...
-1 103.5;...
-1 102];
lat = koordinat(:,1);
lon = koordinat(:,2);
mstruct = defaultm('utm');
mstruct.geoid = referenceEllipsoid('wgs84','meters');
mstruct.zone = utmzone(koordinat);
mstruct = defaultm(mstruct);
[x,y] = mfwdtran (mstruct,lat,lon);
format long g
luas = polyarea(x,y)
By using the "defaultm" and "mfwdtran" function, I can calculate the projected coordinate. Buut, my fellow researcher is feeling doubtful about these 2 Matlab functions. Do you know how to see the script of these two functions?
I already opened the defaultm.m and mfwdtran.m, but I didn't see anything like an m-file from mathworks.com called "deg2utm.m".
Any suggestion will helps.
Thank you
Related
So on this problem it seems pretty straight forward we are given
mean of x = 10,281 and sigma of x = 4112.4
We are asked to determine P(X<15,000)
Now I thought the code for this in matlab should be super straightforward
mu = 10281
sigma = 4112.4
p = logncdf(15000,10281,4112.4)
However this gives
p = .0063
The given answer is .8790 and just looking at p you can tell it is wrong because we are at 15000 which is over the mean which means it should be above .5. What is the deal with this function?
I saw somewhere you might need to take the exp(15000) for x in the function that results in a probability of 1 which is too high.
Any pointers would be much appreciated
%If X is lognormally distributed with parameters:-
mu = 10281;
sigma = 4112.4;
%then log(X) is normally distributed with following parameters:
mew_actual = log((mu^2)/sqrt(sigma^2+mu^2));
sigma_actual = sqrt(log((sigma^2)/(mu^2) +1));
Now you can use either of the following to compute CDF:-
p = cdf('Normal',log(15000),mew_actual,sigma_actual)
or
p=logncdf(15000,mew_actual,sigma_actual)
which gives 0.8796
(which I believe is the correct answer)
The answer given to you is 0.8790 because if you solve the question by hand, you'll get something like: z = 1.172759 and when you look this value in the table, you can only find z = 1.17(without the rest of decimal places) and for which φ(z)=0.8790.
You can verify the exact answer using this calculator. The related screenshot is attached below:
I have some functions are depending on each ther , the functions are from this book page 136 http://www.cs.helsinki.fi/u/ahyvarin/papers/bookfinal_ICA.pdf .. I functions are presented below , How to write following functions in matlab ??
y(t) = W(t-1)*x(t)
h(t) = P(t-1)*y(t)
P(t)=(1/B)*Tri[P(t-1)-m(t)*h^T(t)]
m(t) = h(t)/(B+y^T(t))*h(t))
e(t) = x(t)-W^T(t-1)*y(t)
W(t) = W(t-1) + m(t)*e^T(t)
It is solving the weight matrix W(t) iteratively .. I tried to do like this in matlab but I did not work so may be you can advice to correct the code :
for i=1:10
e=randn(3,5000);
A=[1 0 0;-0.5 0.5 0;0.3 0.1 0.1];
x=A*e;
y(t) = W(t-1)*x(t)
h(t) = P(t-1)*y(t)
P(t)=(1/B)*Tri[P(t-1)-m(t)*h^T(t)]
m(t) = h(t)/(B+y^T(t))*h(t))
e(t) = x(t)-W^T(t-1)*y(t)
W(t) = W(t-1) + m(t)*e^T(t)
end
Thanks
Ok. I can't really understand what you want, but your code shows that you don't understand some moments. I will try to clarify some moments to you:
for i = 2:10
x = rand(3);
y = W(:,:,i-1)*x;
h = P(:,:,i-1)*y;
m=h/(1+y'*h);
P(:,:,i)=P(:,:,i-1)*m*h';
e=x-W(:,:,i-1)'*y;
W(:,:,i)=W(:,:,i-1)+m*e';
end
You must go something like this: 1. you calculate x and use it to calculate other functions.
2. all of them are matrices. So you need to define it first. For example y = ones(3) etc. 3.Thats not y^T or e^T. Its transposing. If you do not feel difference it's early for you to solve this task :)
And the last: Tri function will create a some kind of problems to you, but it's defined at 136 page.
P.S. i missed beta becouse of don't know what is it :)
I have a trajectory defined in Lat/Long coordinates in "n" points. I need to estimate its length over the surface of the earth:
Lat = [la1 la2 la3 la4 la5 la6];
Lon = [lo1 lo2 lo3 lo4 lo5 lo6];
How can I do this in Matlab? I've tried to use the command distance, but it seems it doesn't do this?
Thanks!
[arclen,az] = distance(lat1,lon1,lat2,lon2)
[arclen,az] = distance(lat1,lon1,lat2,lon2,ellipsoid)
Check the matlab documentation
Something similar to this shouldn't be very hard to write:
sum = 0;
for i=1:(size(Lat)-1)
sum = sum + distance(Lat(i),Lon(i),Lat(i+1),Lon(i+1));
//end (Sorry I haven't used matlab in over a year, and forgot the syntax)
dear stackoverflow members, i have a matlab homework. and im having trouble understanding how this piece of code:
m = [ones(1,t0/(3*ts)),-2*ones(1,t0/(3*ts)),zeros(1,t0/(3*ts)+1)];
was put and what does it do? .
It is supposed to plot this function: http://s22.postimg.org/8e3ieonoh/Untitled.gif
Here's the full code:
t0 = 0.15;
ts = 0.001;
fc = 250;
fs = 1/ts;
df = 0.3;
t = [0:ts:t0];
m = [ones(1,t0/(3*ts)),-2*ones(1,t0/(3*ts)),zeros(1,t0/(3*ts)+1)];
Thanks.
This creates the time sequence as a long array. ones(1,t0/(3*ts)) means when t<=t0/3, the function obtain 1; -2*ones(1,t0/(3*ts)) indicates the second value in the original function within the range of [t0/3,t0/3*2]. zeros(1,t0/(3*ts)+1) is the rest part of function. The time interval between two consecutive points in your curve is ts.
To avoid the warning, use
m = [ones(1,floor(t0/(3*ts))),-2*ones(1,floor(t0/(3*ts))),zeros(1,floor(t0/(3*ts))+1)];
I have to build a function from tabulated values (two columns) which are written in a text file. The process to make it is the following:
Use the command importdata to read the data file
Xp = importdata('Xp.dat','\t',1);
Store each column in a variable
x = Xp(1:18304,1);
y = Xp(1:18304,2);
Make a curve fitting with both variables
ft = fittype('linearinterp');
datos.f_Xp = fit(x,y,ft);
However, when I am profiling the code I have found out that my bottleneck are the built-in functions fittype.fittype, fittype.evaluate, cfit.feval, ppval and cfit.subsref
which are related to the curve fitting. So I ask myself how I should manage the tabulated values for improving my code.
you're trying to fit 18304 data points to a curve. Also, you're using linearinterp... which means a routine is being run in a piecewise fashion. if you want to make the code faster use less datapoints.
Or perhaps try:
ft = fittype('poly1');
Not sure is it will be the answer you need as I don't have access to the data
May be "Eval" function could work in your case,
some simple example :
A = '1+4'; eval(A)
ans =
5
P = 'pwd'; eval(P)
ans =
/home/myname
and a bit more advanced!
for n = 1:12
eval(['M',int2str(n),' = magic(n)'])
end
Also, it has a sister name "feval"
guess, what does it do !
[V,D] = feval('eig',A)
[V,D] = eig(A)
and here
function plotf(fun,x)
y = feval(fun,x);
plot(x,y)
You are right ! all are equivalent,
check out here and find more relevant function