I want to calculate the angle between 2 vectors V = [Vx Vy Vz] and B = [Bx By Bz].
is this formula correct?
VdotB = (Vx*Bx + Vy*By + Vz*Bz)
Angle = acosd (VdotB / norm(V)*norm(B))
and is there any other way to calculate it?
My question is not for normalizing the vectors or make it easier. I am asking about how to get the angle between this two vectors
Based on this link, this seems to be the most stable solution:
atan2(norm(cross(a,b)), dot(a,b))
There are a lot of options:
a1 = atan2(norm(cross(v1,v2)), dot(v1,v2))
a2 = acos(dot(v1, v2) / (norm(v1) * norm(v2)))
a3 = acos(dot(v1 / norm(v1), v2 / norm(v2)))
a4 = subspace(v1,v2)
All formulas from this mathworks thread. It is said that a3 is the most stable, but I don't know why.
For multiple vectors stored on the columns of a matrix, one can calculate the angles using this code:
% Calculate the angle between V (d,N) and v1 (d,1)
% d = dimensions. N = number of vectors
% atan2(norm(cross(V,v2)), dot(V,v2))
c = bsxfun(#cross,V,v2);
d = sum(bsxfun(#times,V,v2),1);%dot
angles = atan2(sqrt(sum(c.^2,1)),d)*180/pi;
The traditional approach to obtaining an angle between two vectors (i.e. arccos(dot(u, v) / (norm(u) * norm(v))), as presented in some of the other answers) suffers from numerical instability in several corner cases. The following code works for n-dimensions and in all corner cases (it doesn't check for zero length vectors, but that's easy to add). See notes below.
% Get angle between two vectors
function a = angle_btw(v1, v2)
% Returns true if the value of the sign of x is negative, otherwise false.
signbit = #(x) x < 0;
u1 = v1 / norm(v1);
u2 = v2 / norm(v2);
y = u1 - u2;
x = u1 + u2;
a0 = 2 * atan(norm(y) / norm(x));
if not(signbit(a0) || signbit(pi - a0))
a = a0;
elseif signbit(a0)
a = 0.0;
else
a = pi;
end;
end
This code is adapted from a Julia implementation by Jeffrey Sarnoff (MIT license), in turn based on these notes by Prof. W. Kahan (page 15).
You can compute VdotB much faster and for vectors of arbitrary length using the dot operator, namely:
VdotB = sum(V(:).*B(:));
Additionally, as mentioned in the comments, matlab has the dot function to compute inner products directly.
Besides that, the formula is what it is so what you are doing is correct.
This function should return the angle in radians.
function [ alpharad ] = anglevec( veca, vecb )
% Calculate angle between two vectors
alpharad = acos(dot(veca, vecb) / sqrt( dot(veca, veca) * dot(vecb, vecb)));
end
anglevec([1 1 0],[0 1 0])/(2 * pi/360)
>> 45.00
The solution of Dennis Jaheruddin is excellent for 3D vectors, for higher dimensional vectors I would suggest to use:
acos(min(max(dot(a,b)/sqrt(dot(a,a)*dot(b,b)),-1),1))
This fixes numerical issues which could bring the argument of acos just above 1 or below -1. It is, however, still problematic when one of the vectors is a null-vector. This method also only requires 3*N+1 multiplications and 1 sqrt. It, however also requires 2 comparisons which the atan method does not need.
Related
Finding m and c for an equation y = mx + c, with the help of math and plots.
y is data_model_1, x is time.
Avoid other MATLAB functions like fitlm as it defeats the purpose.
I am having trouble finding the constants m and c. I am trying to find both m and c by limiting them to a range (based on smart guess) and I need to deduce the m and c values based on the mean error range. The point where mean error range is closest to 0 should be my m and c values.
load(file)
figure
plot(time,data_model_1,'bo')
hold on
for a = 0.11:0.01:0.13
c = -13:0.1:-10
data_a = a * time + c ;
plot(time,data_a,'r');
end
figure
hold on
for a = 0.11:0.01:0.13
c = -13:0.1:-10
data_a = a * time + c ;
mean_range = mean(abs(data_a - data_model_1));
plot(a,mean_range,'b.')
end
A quick & dirty approach
You can quickly get m and c using fminsearch(). In the first example below, the error function is the sum of squared error (SSE). The second example uses the sum of absolute error. The key here is ensuring the error function is convex.
Note that c = Beta(1) and m = Beta(2).
Reproducible example (MATLAB code[1]):
% Generate some example data
N = 50;
X = 2 + 13*random(makedist('Beta',.7,.8),N,1);
Y = 5 + 1.5.*X + randn(N,1);
% Example 1
SSEh =#(Beta) sum((Y - (Beta(1) + (Beta(2).*X))).^2);
Beta0 = [0.5 0.5]; % Initial Guess
[Beta SSE] = fminsearch(SSEh,Beta0)
% Example 2
SAEh =#(Beta) sum(abs(Y-(Beta(1) + Beta(2).*X)));
[Beta SumAbsErr] = fminsearch(SAEh,Beta0)
This is a quick & dirty approach that can work for many applications.
#Wolfie's comment directs you to the analytical approach to solve a system of linear equations with the \ operator or mldivide(). This is the more correct approach (though it will get a similar answer). One caveat is this approach gets the SSE answer.
[1] Tested with MATLAB R2018a
I have a 30x30 matrix as a base matrix (OD_b1), I also have two base vectors (bg and Ag). My aim is to optimize a matrix (X) who's dimensions are 30X30 such that:
1) the squared difference between vector (bg) and vector of sum of all the columns is minimized.
2)the squared difference between vector (Ag) and vector of sum of all rows is minimized.
3)the squared difference between the elements of matrix (X) and matrix (OD_b1) is minimized.
The mathematical form of the equation is as follows:
I have tried this:
fun=#(X)transpose(bg-sum(X,2))*(bg-sum(X,2))+ (Ag-sum(X,1))*transpose(Ag-sum(X,1))+sumsqr(X_b-X);
[val,X]=fmincon(fun,OD_b1,AA,BB,Aeq,beq,LB,UB)
I don't get errors but it seems like it's stuck.
Is it because I have too many variables or is there another reason?
Thanks in advance
This is a simple, unconstrained least squares problem and hence has a simple solution that can be expressed as the solution to a linear system.
I will show you (1) the precise and efficient way to solve this and (2) how to solve with fmincon.
The precise, efficient solution:
Problem setup
Just so we're on the same page, I initialize the variables as follows:
n = 30;
Ag = randn(n, 1); % observe the dimensions
X_b = randn(n, n);
bg = randn(n, 1);
The code:
A1 = kron(ones(1,n), eye(n));
A2 = kron(eye(n), ones(1,n));
A = (A1'*A1 + A2'*A2 + eye(n^2));
b = A1'*bg + A2'*Ag + X_b(:);
x = A \ b; % solves A*x = b
Xstar = reshape(x, n, n);
Why it works:
I first reformulated your problem so the objective is a vector x, not a matrix X. Observe that z = bg - sum(X,2) is equivalent to:
x = X(:) % vectorize X
A1 = kron(ones(1,n), eye(n)); % creates a special matrix that sums up
% stuff appropriately
z = A1*x;
Similarly, A2 is setup so that A2*x is equivalent to Ag'-sum(X,1). Your problem is then equivalent to:
minimize (over x) (bg - A1*x)'*(bg - A1*x) + (Ag - A2*x)'*(Ag - A2*x) + (y - x)'*(y-x) where y = Xb(:). That is, y is a vectorized version of Xb.
This problem is convex and the first order condition is a necessary and sufficient condition for the optimum. Take the derivative with respect to x and that equation will define your solution! Sample example math for almost equivalent (but slightly simpler problem is below):
minimize(over x) (b - A*x)'*(b - A*x) + (y - x)' * (y - x)
rewriting the objective:
b'b- b'Ax - x'A'b + x'A'Ax +y'y - 2y'x+x'x
Is equivalent to:
minimize(over x) (-2 b'A - 2y'*I) x + x' ( A'A + I) * x
the first order condition is:
(A'A+I+(A'A+I)')x -2A'b-2I'y = 0
(A'A+I) x = A'b+I'y
Your problem is essentially the same. It has the first order condition:
(A1'*A1 + A2'*A2 + I)*x = A1'*bg + A2'*Ag + y
How to solve with fmincon
You can do the following:
f = #(X) transpose(bg-sum(X,2))*(bg-sum(X,2)) + (Ag'-sum(X,1))*transpose(Ag'-sum(X,1))+sum(sum((X_b-X).^2));
o = optimoptions('fmincon');%MaxFunEvals',30000);
o.MaxFunEvals = 30000;
Xstar2 = fmincon(f,zeros(n,n),[],[],[],[],[],[],[],o);
You can then check the answers are about the same with:
normdif = norm(Xstar - Xstar2)
And you can see that gap is small, but that the linear algebra based solution is somewhat more precise:
gap = f(Xstar2) - f(Xstar)
If the fmincon approach hangs, try it with a smaller n just to gain confidence that my linear algebra based solution is more precise, way way faster etc... n = 30 is solving a 30^2 = 900 variable optimization problem: not easy. With the linear algebra approach, you can go up to n = 100 (i.e. 10000 variable problem) or even larger.
I would probably solve this as a QP using quadprog using the following reformulation (keeping the objective as simple as possible to make the problem "less nonlinear"):
min sum(i,v(i)^2)+sum(i,w(i)^2)+sum((i,j),z(i,j)^2)
v = bg - sum(c,x)
w = ag - sum(r,x)
Z = xbase-x
The QP solver is more precise (no gradients using finite differences). This approach also allows you to add additional bounds and linear equality and inequality constraints.
The other suggestion to form the first order conditions explicitly is also a good one: it also has no issue with imprecise gradients (the first order conditions are linear). I usually prefer a quadratic model because of its flexibility.
I asked this question in Math Stackexchange, but it seems it didn't get enough attention there so I am asking it here. https://math.stackexchange.com/questions/1729946/why-do-we-say-svd-can-handle-singular-matrx-when-doing-least-square-comparison?noredirect=1#comment3530971_1729946
I learned from some tutorials that SVD should be more stable than QR decomposition when solving Least Square problem, and it is able to handle singular matrix. But the following example I wrote in matlab seems to support the opposite conclusion. I don't have a deep understanding of SVD, so if you could look at my questions in the old post in Math StackExchange and explain it to me, I would appreciate a lot.
I use a matrix that have a large condition number(e+13). The result shows SVD get a much larger error(0.8) than QR(e-27)
% we do a linear regression between Y and X
data= [
47.667483331 -122.1070832;
47.667483331001 -122.1070832
];
X = data(:,1);
Y = data(:,2);
X_1 = [ones(length(X),1),X];
%%
%SVD method
[U,D,V] = svd(X_1,'econ');
beta_svd = V*diag(1./diag(D))*U'*Y;
%% QR method(here one can also use "\" operator, which will get the same result as I tested. I just wrote down backward substitution to educate myself)
[Q,R] = qr(X_1)
%now do backward substitution
[nr nc] = size(R)
beta_qr=[]
Y_1 = Q'*Y
for i = nc:-1:1
s = Y_1(i)
for j = m:-1:i+1
s = s - R(i,j)*beta_qr(j)
end
beta_qr(i) = s/R(i,i)
end
svd_error = 0;
qr_error = 0;
for i=1:length(X)
svd_error = svd_error + (Y(i) - beta_svd(1) - beta_svd(2) * X(i))^2;
qr_error = qr_error + (Y(i) - beta_qr(1) - beta_qr(2) * X(i))^2;
end
You SVD-based approach is basically the same as the pinv function in MATLAB (see Pseudo-inverse and SVD). What you are missing though (for numerical reasons) is using a tolerance value such that any singular values less than this tolerance are treated as zero.
If you refer to edit pinv.m, you can see something like the following (I won't post the exact code here because the file is copyrighted to MathWorks):
[U,S,V] = svd(A,'econ');
s = diag(S);
tol = max(size(A)) * eps(norm(s,inf));
% .. use above tolerance to truncate singular values
invS = diag(1./s);
out = V*invS*U';
In fact pinv has a second syntax where you can explicitly specify the tolerance value pinv(A,tol) if the default one is not suitable...
So when solving a least-squares problem of the form minimize norm(A*x-b), you should understand that the pinv and mldivide solutions have different properties:
x = pinv(A)*b is characterized by the fact that norm(x) is smaller than the norm of any other solution.
x = A\b has the fewest possible nonzero components (i.e sparse).
Using your example (note that rcond(A) is very small near machine epsilon):
data = [
47.667483331 -122.1070832;
47.667483331001 -122.1070832
];
A = [ones(size(data,1),1), data(:,1)];
b = data(:,2);
Let's compare the two solutions:
x1 = A\b;
x2 = pinv(A)*b;
First you can see how mldivide returns a solution x1 with one zero component (this is obviously a valid solution because you can solve both equations by multiplying by zero as in b + a*0 = b):
>> sol = [x1 x2]
sol =
-122.1071 -0.0537
0 -2.5605
Next you see how pinv returns a solution x2 with a smaller norm:
>> nrm = [norm(x1) norm(x2)]
nrm =
122.1071 2.5611
Here is the error of both solutions which is acceptably very small:
>> err = [norm(A*x1-b) norm(A*x2-b)]
err =
1.0e-11 *
0 0.1819
Note that use mldivide, linsolve, or qr will give pretty much same results:
>> x3 = linsolve(A,b)
Warning: Matrix is close to singular or badly scaled. Results may be inaccurate. RCOND = 2.159326e-16.
x3 =
-122.1071
0
>> [Q,R] = qr(A); x4 = R\(Q'*b)
x4 =
-122.1071
0
SVD can handle rank-deficiency. The diagonal matrix D has a near-zero element in your code and you need use pseudoinverse for SVD, i.e. set the 2nd element of 1./diag(D) to 0 other than the huge value (10^14). You should find SVD and QR have equally good accuracy in your example. For more information, see this document http://www.cs.princeton.edu/courses/archive/fall11/cos323/notes/cos323_f11_lecture09_svd.pdf
Try this SVD version called block SVD - you just set the iterations equal to the accuracy you want - usually 1 is enough. If you want all the factors (this has a default # selected for factor reduction) then edit the line k= to the size(matrix) if I recall my MATLAB correctly
A= randn(100,5000);
A=corr(A);
% A is your correlation matrix
tic
k = 1000; % number of factors to extract
bsize = k +50;
block = randn(size(A,2),bsize);
iter = 2; % could set via tolerance
[block,R] = qr(A*block,0);
for i=1:iter
[block,R] = qr(A*(A'*block),0);
end
M = block'*A;
% Economy size dense SVD.
[U,S] = svd(M,0);
U = block*U(:,1:k);
S = S(1:k,1:k);
% Note SVD of a symmetric matrix is:
% A = U*S*U' since V=U in this case, S=eigenvalues, U=eigenvectors
V=real(U*sqrt(S)); %scaling matrix for simulation
toc
% reduced randomized matrix for simulation
sims = 2000;
randnums = randn(k,sims);
corrrandnums = V*randnums;
est_corr_matrix = corr(corrrandnums');
total_corrmatrix_difference =sum(sum(est_corr_matrix-A))
I have a state space system with matrices A,B,C and D.
I can either create a state space system, sys1 = ss(A,B,C,D), of it or compute the transfer function matrix, sys2 = C*inv(z*I - A)*B + D
However when I draw the bode plot of both systems, they are different while they should be the same.
What is going wrong here? Does anyone have a clue? I know btw that the bodeplot generated by sys1 is correct.
The system can be downloaded here: https://dl.dropboxusercontent.com/u/20782274/system.mat
clear all;
close all;
clc;
Ts = 0.01;
z = tf('z',Ts);
% Discrete system
A = [0 1 0; 0 0 1; 0.41 -1.21 1.8];
B = [0; 0; 0.01];
C = [7 -73 170];
D = 1;
% Set as state space
sys1 = ss(A,B,C,D,Ts);
% Compute transfer function
sys2 = C*inv(z*eye(3) - A)*B + D;
% Compute the actual transfer function
[num,den] = ss2tf(A,B,C,D);
sys3 = tf(num,den,Ts);
% Show bode
bode(sys1,'b',sys2,'r--',sys3,'g--');
Edit: I made a small mistake, the transfer function matrix is sys2 = C*inv(z*I - A)*B + D, instead of sys2 = C*inv(z*I - A)*B - D which I did wrote done before. The problem still holds.
Edit 2: I have noticted that when I compute the denominator, it is correct.
syms z;
collect(det(z*eye(3) - A),z)
Your assumption that sys2 = C*inv(z*I- A)*B + D is incorrect. The correct equivalent to your state-space system (A,B,C,D) is sys2 = C*inv(s*I- A)*B + D. If you want to express it in terms of z, you'll need to invert the relationship z = exp(s*T). sys1 is the correct representation of your state-space system. What I would suggest for sys2 is to do as follows:
sys1 = ss(mjlsCE.A,mjlsCE.B,mjlsCE.C,mjlsCE.D,Ts);
sys1_c = d2c(sys1);
s = tf('s');
sys2_c = sys1_c.C*inv(s*eye(length(sys1_c.A)) - sys1_c.A)*sys1_c.B + sys1_c.D;
sys2_d = c2d(sys2_c,Ts);
That should give you the correct result.
Due to inacurracy of the inverse function extra unobservable poles and zeros are added to the system. For this reason you need to compute the minimal realization of your transfer function matrix.
Meaning
% Compute transfer function
sys2 = minreal(C*inv(z*eye(3) - A)*B + D);
What you are noticing is actually a numerical instability regarding pole-zero pair cancellations.
If you run the following code:
A = [0, 1, 0; 0, 0, 1; 0.41, -1.21, 1.8] ;
B = [0; 0; 0.01] ;
C = [7, -73, 170] ;
D = 1 ;
sys_ss = ss(A, B, C, D) ;
sys_tf_simp = tf(sys_ss) ;
s = tf('s') ;
sys_tf_full = tf(C*inv(s*eye(3) - A)*B + D) ;
zero(sys_tf_simp)
zero(sys_tf_full)
pole(sys_tf_simp)
pole(sys_tf_full)
you will see that the transfer function formulated by matrices directly has a lot more poles and zeros than the one formulated by MatLab's tf function. You will also notice that every single pair of these "extra" poles and zeros are equal- meaning that they cancel with each other if you were to simply the rational expression. MatLab's tf presents the simplified form, with equal pole-zero pairs cancelled out. This is algebraically equivalent to the unsimplified form, but not numerically.
When you call bode on the unsimplified transfer function, MatLab begins its numerical plotting routine with the pole-zero pairs not cancelled algebraically. If the computer was perfect, the result would be the same as in the simplified case. However, numerical error when evaluating the numerator and denominators effectively leaves some of the pole-zero pairs "uncancelled" and as many of these poles are in the far right side of the s plane, they drastically influence the output behavior.
Check out this link for info on this same problem but from the perspective of design: http://ctms.engin.umich.edu/CTMS/index.php?aux=Extras_PZ
In your original code, you can think of the output drawn in green as what the naive designer wanted to see when he cancelled all his unstable poles with zeros, but the output drawn in red is what he actually got because in practice, finite-precision and real-world tolerances prevent the poles and zeros from cancelling perfectly.
Why is an unobservable / uncontrollable pole? I think this issue comes only because the inverse of a transfer function matrix is inaccurate in Matlab.
Note:
A is 3x3 and the minimal realization has also order 3.
What you did is the inverse of a transfer function matrix, not a symbolic or numeric matrix.
# Discrete system
Ts = 0.01;
A = [0 1 0; 0 0 1; 0.41 -1.21 1.8];
B = [0; 0; 0.01];
C = [7 -73 170];
D = 1;
z = tf('z', Ts)) # z is a discrete tf
A1 = z*eye(3) - A # a tf matrix with a direct feedthrough matrix A
# inverse it, multiply with C and B from left and right, and plus D
G = D + C*inv(A1)*B
G is now a scalar (SISO) transfer function.
Without "minreal", G has order 9 (funny, I don't know how Matlab computes it, perhaps the "Adj(.)/det(.)" method). Matlab cannot cancel the common factors in the numerator and the denominator, because z is of class 'tf' rather than a symbolic variable.
Do you agree or do I have misunderstanding?
I have a square matrix that represents the frequency counts of co-occurrences in a data set. In other words, the rows represent all possible observations of feature 1, and the columns are the possible observations of feature 2. The number in cell (x, y) is the number of times feature 1 was observed to be x at the same time feature 2 was y.
I want to calculate the mutual information contained in this matrix. MATLAB has a built-in information function, but it takes 2 arguments, one for x and one for y. How would I manipulate this matrix to get the arguments it expects?
Alternatively, I wrote my own mutual information function that takes a matrix, but I'm unsure about its accuracy. Does it look right?
function [mutualinfo] = mutualInformation(counts)
total = sum(counts(:));
pX = sum(counts, 1) ./ total;
pY = sum(counts) ./ total;
pXY = counts ./ total;
[h, w] = size(counts);
mutualinfo = 0;
for row = 1:h
for col = 1:w
mutualinfo = mutualinfo + pXY(row, col) * log(pXY(row, col) / (pX(row)*pY(col)));
end;
end;
end
I don't know of any built-in mutual information functions in MATLAB. Perhaps you got a hold of one of the submissions from the MathWorks File Exchange or some other third-party developer code?
I think there may be something wrong with how you are computing pX and pY. Plus, you can vectorize your operations instead of using for loops. Here's another version of your function to try out:
function mutualInfo = mutualInformation(counts)
pXY = counts./sum(counts(:));
pX = sum(pXY,2);
pY = sum(pXY,1);
mutualInfo = pXY.*log(pXY./(pX*pY));
mutualInfo = sum(mutualInfo(:));
end