I'm trying to simulate nonlinear vehicle braking system with ode45.
I take a short time to learning MATLAB. then, I don't know why error occur.
It would be very appreciated if you could point out errors, and tell me how to solve.
code 1. main script
code 2. function
code 3. errors
clear;
global m f Jw rw Fz Cd p A bw fw Nw g uw seta Te Tb T p0
x = [0 0.025 0.05 0.1 0.125 0.15 0.175 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1]
y = [0 0.225 0.45 0.65 0.685 0.705 0.69 0.68 0.65 0.635 0.63 0.6275 0.625 0.6225 0.62 0.6175 0.615 0.6125 0.610 0.6075 0.6050 0.6 0.5975 0.5950]
%plot(x,y)
p0=polyfit(x,y,6)
%y=polyval(p,x)
m = 1400; f = 0.01; Jw = 0.65; rw = 0.31; Fz = 3560.0; Cd = 0.5; p = 1.202; A = 1.95;
bw = 0.0; fw = 0.0; Nw = 4; g = 9.81; uw = 0.0; seta = 0.0; Te = 0.0; Tb = 1000.0; T = Te - Tb;
[t,i] = ode45(#dott,[0.0 1.0],[20 20]);
plot(t,i);
axis([0 1 0 20]);
legend('x1','x2');
function xdot = dott(t,x)
global m f Jw rw Fz Cd p A Nw Te Tb p0
X = [0 0.025 0.05 0.1 0.125 0.15 0.175 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1];
Y = [0 0.225 0.45 0.65 0.685 0.705 0.69 0.68 0.65 0.635 0.63 0.6275 0.625 0.6225 0.62 0.6175 0.615 0.6125 0.610 0.6075 0.6050 0.6 0.5975 0.5950];
%UNTITLED2 Summary of this function goes here
%Detailed explanation goes here
xdot = zeros(2,1);
Y=p0(1,1)*((x(2)-x(1))/x(1))^6+p0(1,2)*((x(2)-x(1))/x(1))^5+p0(1,3)*((x(2)-x(1))/x(1))^4+p0(1,4)*((x(2)-x(1))/x(1))^3+p0(1,5)*((x(2)-x(1))/x(1))^2+p0(1,6)*((x(2)-x(1))/x(1))^1+p0(1,1)*((x(2)-x(1))/x(1));
xdot(1)=(-0.5*p*Cd*A(x(2)/(1+Y)*rw)*(x(2)/(1+Y)*rw)-f*m+Nw*Y*Fz)/(rw*m);
xdot(2)=(Te-Tb-rw*Y*Fz)/Jw;
end
??? Subscript indices must either be real positive integers or logicals.
Error in ==> dott at 9
xdot(1)=(-0.5*p*Cd*A(x(2)/(1+Y)*rw)*(x(2)/(1+Y)*rw)-f*m+Nw*X*Fz)/(rw*m);
Error in ==> odearguments at 98
f0 = feval(ode,t0,y0,args{:}); % ODE15I sets args{1} to yp0.
Error in ==> ode45 at 172
[neq, tspan, ntspan, next, t0, tfinal, tdir, y0, f0, odeArgs, odeFcn, ...
Error in ==> a at 14
[t,i] = ode45(#dott,[0.0 1.0],[20 20]);
I'm pretty sure your error is in this statement:
A(x(2)/(1+Y)*rw)
The way you write it, you're trying to use x(2)/(1+Y)*rw as an index to the scalar A. I guess you want to multiply is this way:
... A * (x(2) / (1 + Y) * rw) ...
To make the code more readable:
Use spaces. A long compact line is really hard to read.
Split long lines into several using three dots ...
Something like this is easier to read in my opinion:
Y = p0(1,1) * ((x(2) - x(1)) / x(1))^6 + p0(1,2) * ...
((x(2) - x(1)) / x(1))^5 + p0(1,3) * ((x(2) - x(1)) / x(1))^4 ...
+ p0(1,4) * ((x(2) - x(1)) / x(1))^3 + p0(1,5) * ((x(2) - x(1)) / ...
x(1))^2 + p0(1,6) * ((x(2) - x(1)) / x(1))^1 + p0(1,1) * ...
((x(2) - x(1)) / x(1));
Related
I am trying to optimise the running time of my code by getting rid of some for loops. However, I have a variable that is incremented in each iteration in which sometimes the index is repeated. I provide here a minimal example:
a = [1 4 2 2 1 3 4 2 3 1]
b = [0.5 0.2 0.3 0.4 0.1 0.05 0.7 0.3 0.55 0.8]
c = [3 5 7 9]
for i = 1:10
c(a(i)) = c(a(i)) + b(i)
end
Ideally, I would like to compute it by writting:
c(a) = c(a) + b
but obviously it would not give me the same results since I have to recalculate the value for the same index several times so this way to vectorise it would not work.
Also, I am working in Matlab or Octave in case that this is important.
Thank you very much for any help, I am not sure that it is possible to be vectorise.
Edit: thank you very much for your answers so far. I have discovered accumarray, which I did not know before and also understood why changing the for loop between Matlab and Octave was giving me such different times. I also understood my problem better. I gave a too simple example which I thought I could extend, however, what if b was a matrix?
(Let's forget about c at the moment):
a = [1 4 2 2 1 3 4 2 3 1]
b =[0.69 -0.41 -0.13 -0.13 -0.42 -0.14 -0.23 -0.17 0.22 -0.24;
0.34 -0.39 -0.36 0.68 -0.66 -0.19 -0.58 0.78 -0.23 0.25;
-0.68 -0.54 0.76 -0.58 0.24 -0.23 -0.44 0.09 0.69 -0.41;
0.11 -0.14 0.32 0.65 0.26 0.82 0.32 0.29 -0.21 -0.13;
-0.94 -0.15 -0.41 -0.56 0.15 0.09 0.38 0.58 0.72 0.45;
0.22 -0.59 -0.11 -0.17 0.52 0.13 -0.51 0.28 0.15 0.19;
0.18 -0.15 0.38 -0.29 -0.87 0.14 -0.13 0.23 -0.92 -0.21;
0.79 -0.35 0.45 -0.28 -0.13 0.95 -0.45 0.35 -0.25 -0.61;
-0.42 0.76 0.15 0.99 -0.84 -0.03 0.27 0.09 0.57 0.64;
0.59 0.82 -0.39 0.13 -0.15 -0.71 -0.84 -0.43 0.93 -0.74]
I understood now that what I would be doing is rowSum per group, and given that I am using Octave I cannot use "splitapply". I tried to generalise your answers, but accumarray would not work for matrices and also I could not generalise #rahnema1 solution. The desired output would be:
[0.34 0.26 -0.93 -0.56 -0.42 -0.76 -0.69 -0.02 1.87 -0.53;
0.22 -1.03 1.53 -0.21 0.37 1.54 -0.57 0.73 0.23 -1.15;
-0.20 0.17 0.04 0.82 -0.32 0.10 -0.24 0.37 0.72 0.83;
0.52 -0.54 0.02 0.39 -1.53 -0.05 -0.71 1.01 -1.15 0.04]
that is "equivalent" to
[sum(b([1 5 10],:))
sum(b([3 4 8],:))
sum(b([6 9],:))
sum(b([2 7],:))]
Thank you very much, If you think I should include this in another question instead of adding the edit I will do so.
Original question
It can be done with accumarray:
a = [1 4 2 2 1 3 4 2 3 1];
b = [0.5 0.2 0.3 0.4 0.1 0.05 0.7 0.3 0.55 0.8];
c = [3 5 7 9];
c(:) = c(:) + accumarray(a(:), b(:));
This sums the values from b in groups defined by a, and adds that to the original c.
Edited question
If b is a matrix, you can use
full(sparse(repmat(a, 1, size(b,1)), repelem(1:size(b,2), size(b,1)), b))
or
accumarray([repmat(a, 1, size(b,1)).' repelem(1:size(b,2), size(b,1)).'], b(:))
Matrix multiplication and implicit expansion and can be used (Octave):
nc = numel(c);
c += b * (1:nc == a.');
For input of large size it may be more memory efficient to use sparse matrix:
nc = numel(c);
nb = numel(b);
c += b * sparse(1:nb, a, 1, nb, nc);
Edit: When b is a matrix you can extend this solution as:
nc = numel(c);
na = numel(a);
out = sparse(a, 1:na, 1, nc, na) * b;
Based on the following code:
clear vars;
close all;
x1 = [0 0 0.01 0.09 0.1 0.11 0.2 0.3 0.35 0.50 0.64 0.8 1]
y1 = [0.05 0.10 0.15 0.20 0.25 0.30 0.38 0.42 0.45 0.48 0.52 0.86 1]
x2 = [0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0.9 0.9 1]
y2 = [0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0.9 0.9 1]
plot(x1, y1); hold on;
plot(x2, y2);
I need to calculate the area (green area) between the two curves, for example:
How can I calculate it?
This area is the difference of the two curves integral in the specified domain between each intersection (as mentioned by MBo). Hence, you can find the intersections using InterX and then use trapz to do this:
P = InterX([x1;y1],[x2;y2]);
area = 0;
% for each segment
% each segment is between P(1,i) and P(1, i+1)
% So we can find xsegments with idx = find(x < P(1,i+1) && x > P(1,i)) and [P(1,i) x(idx) P(1,i+1)]
% ...
area = area + abs(trapz(xsegment1i,ysegment1i) - trapz(xsegment2i,ysegment2i));
Since one of the curves is a straight line you can rotate then add up the areas from the new x axis.
The line is at 45 degrees. So the rotation matrix is
cos 45 sin 45
-sin 45 cos 45
Multiply each point in the second curve by that matrix. That gives points with the line as the new x axis. Now use area of the triangle (0.5 * width * height) to add up the areas of the fragments.
I have astraightline with the following data
xinter=[1.13 1.36 1.62 1.81 2.00 2.30 2.61 2.83 3.05 3.39]
yinter=[0.10 0.25 0.40 0.50 0.60 0.75 0.90 1.00 1.10 1.25]
and I want to find the intersection with a result of an interpolated data
of a curve such as below
a50= [0.77 0.73 0.77 0.85 0.91 0.97 1.05 1.23 1.43 1.53 1.62 1.71 1.89 2.12 2.42];
a25= [0.51 0.60 0.70 0.80 0.85 0.90 0.96 1.09 1.23 1.30 1.36 1.41 1.53 1.67];
vel25=[0.43 0.35 0.30 0.27 0.25 0.24 0.22 0.21 0.22 0.24 0.25 0.27 0.30 0.35];
vel50=[0.68 0.57 0.49 0.43 0.40 0.38 0.36 0.34 0.36 0.38 0.40 0.43 0.49 0.57 0.68 ];
% back up original data, just for final plot
bkp_a50 = a50 ; bkp_vel50 = vel50 ;
% make second x vector monotonic
istart = find( diff(a50)>0 , 1 , 'first') ;
a50(1:istart-1) = [] ;
vel50(1:istart-1) = [] ;
% prepare a 3rd dimension vector (from 25 to 50)
T = [repmat(25,size(a25)) ; repmat(40,size(a50)) ] ;
% merge all observations together
A = [ a25 ; a50] ;
V = [vel25 ; vel50] ;
% find the minimum domain on which data can be interpolated
% (anything outside of that will return NaN)
Astart = max( [min(a25) min(a50)] ) ;
Astop = min( [max(a25) max(a50)] ) ;
% use the function 'griddata'
[TI,AI] = meshgrid( 25:40 , linspace(Astart,Astop,10) ) ;
VI = griddata(T,A,V,TI,AI) ;
% plot all the intermediate curves
%plot(AI,VI)
hold on
% the original curves
%plot(a25,vel25,'--k','linewidth',2)
%plot(bkp_a50,bkp_vel50,'--k','linewidth',2)
% Highlight the curve at T = 30 ;
c30 = find( TI(1,:) == 40 ) ;
plot(AI(:,c30),VI(:,c30),'--r','linewidth',2)
xinter=[1.13 1.36 1.62 1.81 2.00 2.30 2.61 2.83 3.05 3.39]
yinter=[0.10 0.25 0.40 0.50 0.60 0.75 0.90 1.00 1.10 1.25]
x1inter=(AI(:,c30))';
y1inter=(VI(:,c30))';
yy2 = interp1(xinter, yinter, x1inter,'spline')
plot(xinter,yinter, '--k','linewidth',2)
idx = find((y1inter - yy2) < eps, 1); %// Index of coordinate in array
px = x1inter(idx)
py = y1inter(idx)
plot(px, py, 'ro', 'MarkerSize', 18)
But there is an error in the result when I modify x1inter
You can use piecewise polynomial curvefitting and the fzero function to find the intersection point:
pp1 = pchip(xinter,yinter); % Curve 1
pp2 = pchip(AI(:,c30),VI(:,c30)); % Curve 2
fun = #(x) ppval(pp1,x) - ppval(pp2,x); % Curve to evaluate
xzero = fzero(fun,mean(xinter)) % intersection x value
yzero = ppval(pp1,xzero)
plot(xzero, yzero, 'bo', 'MarkerSize', 18)
I am new to MATLAB and I need help. I have 3 matrices (A, B, and C) and I want to create a new matrix average_ABC that contains average values.
A = [ 0.3 0.5 0.9
0.14 0.36 0.1
0.9 0.5 0.14]
B = [ 0.8 0.9 0.14
0.1 0.25 0.4
0.8 0.14 0.25]
C = [0.25 0.3 0.47
0.12 0.3 0.2
0.14 0.56 0.9]
The resulting matrix will be
average_matrix = [ 0.3 0.5 0.47
0.12 0.25 0.2
0.8 0.5 0.25]
Please, any suggestion, how can I do it?
You can first concatenate your matrices along the third dimension (using cat) and then compute whatever you want using the dim parameter that is available for most functions to specify that you want to perform that operation along the third dimension.
Also you've stated that you want the average (mean), but based on your example you actually want the median. Either way, we can compute them using this method.
data = cat(3, A, B, C);
% Compute the mean
mean(data, 3)
% 0.45 0.56667 0.50333
% 0.12 0.30333 0.23333
% 0.61333 0.4 0.43
% Compute the median (which seems to be what you actually want)
median(data, 3)
% 0.3 0.5 0.47
% 0.12 0.3 0.2
% 0.8 0.5 0.25
I hope this will work
average_matrix=(A+B+C)/3.;
I have a matrix proba (size :10 * 5).
proba=[0.5 0.3 0.8 0.9 0.8;
0.50 0.36 0.58 0.58 0.98;
0.1 0.25 0.6 0.8 0.9;
0.5 0.3 0.8 0.9 0.8;
0.2 0.9 0.58 0.58 0.69;
0.58 0.14 0.1 0.2 0.3;
0.25 0.9 0.8 0.7 0.5;
0.58 0.69 0.25 0.1 0.1;
0.1 0.25 0.36 0.2 0.3;
0.5 0.3 0.8 0.9 0.8 ];
I want to transform this matrix into a text file (proba.txt) with which the index column is written and the value of the column for each line as follows :
1 0.5 2 0.3 3 0.8 4 0.9 5 0.8
1 0.50 2 0.36 3 0.58 4 0.58 5 0.98
.
.
.
1 0.5 2 0.3 3 0.8 4 0.9 5 0.8
Please I need help, how can I do it?thanks in advance
You can use this function, it is useful for every matrix.
function data = addIndex(X)
[r, c] = size(X);
index = ones(r, 1);
data = zeros(r, 2 * c);
for i = 1:c
data(:, 2 * i - 1) = i .* index;
data(:, 2 * i) = X(:, i);
end
dlmwrite('proba.txt', data, '\t')
end
you can easily do this using dlmwrite, but first you want to add the column of indexes in front of your matrix:
function result = writematrix(proba)
rowind = 1:size(proba,2);
for t = 1:size(proba,1);
C(t,:,:) = [rowind',proba(t,:)']';
D(t,:) = C(t(:),:);
end
dlmwrite('filename.txt',D,'\t') %//I assume you want tab delimiter, if you want space, it is ' ' instead
%//dlmwrite('filename.txt',D,' ')
end
Note that this will write the text file into your local directory, and that it only works for numerical values, not strings, for strings, it is better to use csvwrite.
EDIT : Ops, didn't read the question fully, this should now work fine.