Let the torus T^2 be the domain [0,1)^2 in R^2 with periodic boundary conditions. In general, the negative Sobolev norm H^(-1) of a function f defined on the torus can be computed as:
a dual norm from the H^1 space, i.e. H^(-1) is defined to be de dual of H^1 (note: we say g is in H^1 if grad g is in L^2).
in terms of the Fourier transform Ff like the sum over all frequencies k in Z^2 without k=(0,0) of |Ff(k)|^2/|k|^2.
I am running some numerical experiments in Matlab and I would like to compute the H^(-1) norm of some functions that I have obtained numerically and hence I have defined in a finite set of discrete points. The most straightforward way to compute it might be by means of the command fft2, since in the end my 2D function is just described by a matrix. My problem is that I do not know how the frequencies are labelled by this command, would it be enough to do
Ff = fft2(f);
sob_norm = 0;
for i=1:L
for j=1:L
if i==1 && j==1
sob_norm = sob_norm;
else
sob_norm = sob_norm + norm([i-1,j-1])^(-2)*norm(Ff(i,j))^2;
end
end
end
or should I scale the 1/|k|^2 in a different way? I guess there is something I am not understanding correctly because the scale of the norm is much bigger than it should.
I apologize whether this is not the most appropriate forum to ask about this, any help about a more convenient place it is very much appreciated too. Thanks a lot.
Related
I am working on a MR-physic simulation written in Matlab which simulates bloch's equations on an defined object. The magnetisation in the object is updated every time-step with the following functions.
function Mt = evolveMtrans(gamma, delta_B, G, T2, Mt0, delta_t)
% this function calculates precession and relaxation of the
% transversal component, Mt, of M
delta_phi = gamma*(delta_B + G)*delta_t;
Mt = Mt0 .* exp(-delta_t*1./T2 - 1i*delta_phi);
end
This function is a very small part of the entire code but is called upon up to 250.000 times and thus slows down the code and the performance of the entire simulation. I have thought about how I can speed up the calculation but haven't come up with a good solution. There is one line that is VERY time consuming and stands for approximately 50% - 60% of the overall simulation time. This is the line,
Mt = Mt0 .* exp(-delta_t*1./T2 - 1i*delta_phi);
where
Mt0 = 512x512 matrix
delta_t = a scalar
T2 = 512x512 matrix
delta_phi = 512x512 matrix
I would be very grateful for any suggestion to speed up this calculation.
More info below,
The function evovleMtrans is called every timestep during the simulation.
The parameters that are used for calling the function are,
gamma = a constant. (gyramagnetic constant)
delta_B = the magnetic field value
G = gradientstrength
T2 = a 512x512 matrix with T2-values for the object
Mstart.r = a 512x512 matrix with the values M.r had the last timestep
delta_t = a scalar with the difference in time since the last calculated M.r
The only parameters of these that changed during the simulation are,
G, Mstart.r and delta_t. The rest do not change their values during the simulation.
The part below is the part in the main code that calls the function.
% update phase and relaxation to calcTime
delta_t = calcTime - Mstart_t;
delta_B = (d-d0)*B0;
G = Sq.Gx*Sq.xGxref + Sq.Gz*Sq.zGzref;
% Precession around B0 (z-axis) and B1 (+-x-axis or +-y-axis)
% is defined clock-wise in a right hand system x, y, z and
% x', y', z (see the Bloch equation, Bloch 1946 and Levitt
% 1997). The x-axis has angle zero and the y-axis has angle 90.
% For flipping/precession around B1 in the xy-plane, z-axis has
% angle zero.
% For testing of precession direction:
% delta_phi = gamma*((ones(size(d)))*1e-6*B0)*delta_t;
M.r = evolveMtrans(gamma, delta_B, G, T2, Mstart.r, delta_t);
M.l = evolveMlong(T1, M0.l, Mstart.l, delta_t);
This is not a surprise.
That "single line" is a matrix equation. It's really 1,024 simultaneous equations.
Per Jannick, that first term means element-wise division, so "delta_t/T[i,j]". Multiplying a matrix by a scalar is O(N^2). Matrix addition is O(N^2). Evaluating exponential of a matrix will be O(N^2).
I'm not sure if I saw a complex argument in there as well. Does that mean complex matricies with real and imaginary entries? Does your equation simplify to real and imaginary parts? That means twice the number of computations.
Your best hope is to exploit symmetry as much as possible. If all your matricies are symmetric, you cut your calculations roughly in half.
Use parallelization if you can.
Algorithm choice can make a big difference, too. If you're using explicit Euler integration, you may have time step limitations due to stability concerns. Is that why you have 250,000 steps? Maybe a larger time step is possible with a more stable integration schema. Think about a higher order adaptive scheme with error correction, like 5th order Runge Kutta.
There are several possibilities to improve the speed of the code but all that I see come with a caveat.
Numerical ode integration
The first possibility would be to change your analytical solution by numerical differential equation solver. This has several advantages
The analytical solution includes the complex exponential function, which is costly to calculate, while the differential equation contains only multiplication and addition. (d/dt u = -a u => u=exp(-at))
There are plenty of built-in solvers for matlab available and they are typically pretty fast (e.g. ode45). The built-ins however all use a variable step size. This improves speed and accuracy but would be a problem if you really need a fixed equally spaced grid of time points. Here are unofficial fixed step solvers.
As a start you could also try to use just an euler step by replacing
M.r = evolveMtrans(gamma, delta_B, G, T2, Mstart.r, delta_t);
by
delta_phi = gamma*(delta_B + G)*t_step;
M.r += M.r .* (1-t_step*1./T2 - 1i*delta_phi);
You can then further improve that by precalculating all constant values, e.g. one_over_T1=1/T1, moving delta_phi out of the loop.
Caveat:
You are bound to a minimum step size or the accuracy suffers. Therefore this is only a good idea if you time-spacing is quite fine.
Less points in time
You should carfully analyze whether you really need so many points in time. It seems somewhat puzzling to me that you need so many points. As you know the full analytical solution you can freely choose how to sample the time and maybe use this to your advantage.
Going fortran
This might seem like a grand step but in my experience basic (simple loops, matrix operations etc.) matlab code can be relatively easily translated to fortran line-by-line. This would be especially helpful in addition to my first point. If you still want to use the full analytical solution probably there is not much to gain here because exp is already pretty fast in matlab.
I am trying to solve two equations with complex coefficients using ode45.
But iam getting an error message as "Inputs must be floats, namely single or
double."
X = sym(['[',sprintf('X(%d) ',1:2),']']);
Eqns=[-(X(1)*23788605396486326904946699391889*1i)/38685626227668133590597632 + (X(2)*23788605396486326904946699391889*1i)/38685626227668133590597632; (X(2)*23788605396486326904946699391889*1i)/38685626227668133590597632 + X(1)*(- 2500000 + (5223289665997855453060886952725538686654593059791*1i)/324518553658426726783156020576256)] ;
f=#(t,X)[Eqns];
[t,Xabc]=ode45(f,[0 300*10^-6],[0 1])
How can i fix this ? Can somebody can help me ?
Per the MathWorks Support Team, the "ODE solvers in MATLAB 5 (R12) and later releases properly handle complex valued systems." So the complex numbers are the not the issue.
The error "Inputs must be floats, namely single or double." stems from your definition of f using Symbolic Variables that are, unlike complex numbers, not floats. The easiest way to get around this is to not use the Symbolic Toolbox at all; just makes Eqns an anonymous function:
Eqns= #(t,X) [-(X(1)*23788605396486326904946699391889*1i)/38685626227668133590597632 + (X(2)*23788605396486326904946699391889*1i)/38685626227668133590597632; (X(2)*23788605396486326904946699391889*1i)/38685626227668133590597632 + X(1)*(- 2500000 + (5223289665997855453060886952725538686654593059791*1i)/324518553658426726783156020576256)] ;
[t,Xabc]=ode45(Eqns,[0 300*10^-6],[0 1]);
That being said, I'd like to point out that numerically time integrating this system over 300 microseconds (I assume without units given) will take a long time since your coefficient matrix has imaginary eigenvalues on the order of 10E+10. The extremely short wavelength of those oscillations will more than likely be resolved by Matlab's adaptive methods, and that will take a while to solve for a time span just a few orders greater than the wavelength.
I'd, therefore, suggest an analytical approach to this problem; unless it is a stepping stone another problem that is non-analytically solvable.
Systems of ordinary differential equations of the form
,
which is a linear, homogenous system with a constant coefficient matrix, has the general solution
,
where the m-subscripted exponential function is the matrix exponential.
Therefore, the analytical solution to the system can be calculated exactly assuming the matrix exponential can be calculated.
In Matlab, the matrix exponential is calculate via the expm function.
The following code computes the analytical solution and compares it to the numerical one for a short time span:
% Set-up
A = [-23788605396486326904946699391889i/38685626227668133590597632,23788605396486326904946699391889i/38685626227668133590597632;...
-2500000+5223289665997855453060886952725538686654593059791i/324518553658426726783156020576256,23788605396486326904946699391889i/38685626227668133590597632];
Eqns = #(t,X) A*X;
X0 = [0;1];
% Numerical
options = odeset('RelTol',1E-8,'AbsTol',1E-8);
[t,Xabc]=ode45(Eqns,[0 1E-9],X0,options);
% Analytical
Xana = cell2mat(arrayfun(#(tk) expm(A*tk)*X0,t,'UniformOutput',false)')';
k = 1;
% Plots
figure(1);
subplot(3,1,1)
plot(t,abs(Xana(:,k)),t,abs(Xabc(:,k)),'--');
title('Magnitude');
subplot(3,1,2)
plot(t,real(Xana(:,k)),t,real(Xabc(:,k)),'--');
title('Real');
ylabel('Values');
subplot(3,1,3)
plot(t,imag(Xana(:,k)),t,imag(Xabc(:,k)),'--');
title('Imaginary');
xlabel('Time');
The comparison plot is:
The output of ode45 matches the magnitude and real parts of the solution very well, but the imaginary portion is out-of-phase by exactly π.
However, since ode45's error estimator only looks at norms, the phase difference is not noticed which may lead to problems depending on the application.
It will be noted that while the matrix exponential solution is far more costly than ode45 for the same number of time vector elements, the analytical solution will produce the exact solution for any time vector of any density given to it. So for long time solutions, the matrix exponential can be viewed as an improvement in some sense.
I have the following differential equation which I'm not able to solve.
We know the following about the equation:
D(r) is a third grade polynom
D'(1)=D'(2)=0
D(2)=2D(1)
u(1)=450
u'(2)=-K * (u(2)-Te)
Where K and Te are constants.
I want to approximate the problem using a matrix and I managed to solve
the similiar equation: with the same limit conditions for u(1) and u'(2).
On this equation I approximated u' and u'' with central differences and used a finite difference method between r=1 to r=2. I then placed the results in a matrix A in matlab and the limit conditions in the vector Y in matlab and ran u=A\Y to get how the u value changes. Heres my matlab code for the equation I managed to solve:
clear
a=1;
b=2;
N=100;
h = (b-a)/N;
K=3.20;
Ti=450;
Te=20;
A = zeros(N+2);
A(1,1)=1;
A(end,end)=1/(2*h*K);
A(end,end-1)=1;
A(end,end-2)=-1/(2*h*K);
r=a+h:h:b;
%y(i)
for i=1:1:length(r)
yi(i)=-r(i)*(2/(h^2));
end
A(2:end-1,2:end-1)=A(2:end-1,2:end-1)+diag(yi);
%y(i-1)
for i=1:1:length(r)-1
ymin(i)=r(i+1)*(1/(h^2))-1/(2*h);
end
A(3:end-1,2:end-2) = A(3:end-1,2:end-2)+diag(ymin);
%y(i+1)
for i=1:1:length(r)
ymax(i)=r(i)*(1/(h^2))+1/(2*h);
end
A(2:end-1,3:end)=A(2:end-1,3:end)+diag(ymax);
Y=zeros(N+2,1);
Y(1) =Ti;
Y(2)=-(Ti*(r(1)/(h^2)-(1/(2*h))));
Y(end) = Te;
r=[1,r];
u=A\Y;
plot(r,u(1:end-1));
My question is, how do I solve the first differential equation?
As TroyHaskin pointed out in comments, one can determine D up to a constant factor, and that constant factor cancels out in D'/D anyway. Put another way: we can assume that D(1)=1 (a convenient number), since D can be multiplied by any constant. Now it's easy to find the coefficients (done with Wolfram Alpha), and the polynomial turns out to be
D(r) = -2r^3+9r^2-12r+6
with derivative D'(r) = -6r^2+18r-12. (There is also a smarter way to find the polynomial by starting with D', which is quadratic with known roots.)
I would probably use this information right away, computing the coefficient k of the first derivative:
r = a+h:h:b;
k = 1+r.*(-6*r.^2+18*r-12)./(-2*r.^3+9*r.^2-12*r+6);
It seems that k is always positive on the interval [1,2], so if you want to minimize the changes to existing code, just replace r(i) by r(i)/k(i) in it.
By the way, instead of loops like
for i=1:1:length(r)
yi(i)=-r(i)*(2/(h^2));
end
one usually does simply
yi=-r*(2/(h^2));
This vectorization makes the code more compact and can benefit the performance too (not so much in your example, where solving the linear system is the bottleneck). Another benefit is that yi is properly initialized, while with your loop construction, if yi happened to already exist and have length greater than length(r), the resulting array would have extraneous entries. (This is a potential source of hard-to-track bugs.)
By hypothesis my measured probability density functions (PDF) result from n convolutions of an elementary distribution (E).
I have two distributions the first (F) of which is supposed to have undergone more convolutions (m_1) than the second (G) (m_2 convolutions).
In fourier space:
F' = E'^m_1
G' = E'^m_2
As the two PDFs are constituted from the same elementary distribution, I should be able to be able to calculate the PDF of G from F
G' = F'^{m_1/m_2}
Taking the IFFT i should have a distribution that overlaps well with G.
A naive way of doing this would to be simply to calculate the FT of F and raise it to the power 1/integer and testing a range of integers.
My question are there any tricks for raising the Fourier transformed PDF to a fractional power. I have done so but the IFFT gives a distribution far from that which is expected. And strange aliasing errors.
I've included a padded vector as one might do if they were to do a convolution of two PDFS.
My normalization is based on the fact that the k=0 [ProbF(1,1)] wave vector gives the integral of the PDF which should be equal to one.
Of course, the hypothesis could be wrong, but it has all the reasons in the world to be valid.
My code
Inc = INC1 ; % BINS
null = zeros(1,length(Inc)); % PADDED PROB
Inc = [ Inc.*-1 (Inc) ]; % PADDED INC VECTOR
Prob = [ null heightProb1 ] ; % PADDED PROB VECTOR
ProbF = (fft(Prob)) ;
ProbFnorm = ProbF./ProbF(1,1) ; % NORMALIZED BY K=0 COMPONENT (integral of PDF =1)
m=.79 % POWER TO RAISE
ProbFtrans = ((ProbFnorm).^(m)); % 'DECONVOLUTION' IN FOURIER SPACE
ProbIF = (ifft((ProbFtrans)).*(ProbF(1,1))); % RETURN TO PROBABILITY SPACE
figure(2);
plot(Inc,ProbIF,'rs')
Thank you in advance for your help
Fourier coefficients are typically complex numbers (unless your function is symmetric).
You should be very careful when you raise complex numbers to fractional powers.
For example, consider
z=1.2 + i*0.65;
then raise z to power 4
>> z^4
ans =
-1.398293750000001e+00 + 3.174599999999999e+00i
and to power 8.
>> z^8
ans =
-8.122859748710933e+00 - 8.878046677500002e+00i
Then try obtain z^4 as (z^8)^(1/2)
>> (z^8)^(1/2)
ans =
1.398293750000001e+00 - 3.174600000000000e+00i
Surprise! You don't get z^4! (wrong sign)
If you avoid taking the fractional power and "rewind" z^8 by diving back by z you get back z^4 correctly:
>> (z^8)/z/z/z/z
ans =
-1.398293750000000e+00 + 3.174599999999998e+00i
The reason is in the definition of fractional powers in the complex plane. Fractional powers are multi-valued functions, which are made single-valued by introducing a branch-cut in the complex plane. The nth-root z^(1/n) has n possible values, and
matlab singles out one of these by interpreting the complex function z^(1/n) as its so-called principal branch. The main implication is that in the world of complex numbers ^1/n does not always invert ^n.
If this doesn't make any sense to you, you should probably review some basic complex analysis, but the bottom line is that fractional powers of complex numbers are tricky animals. Wherever possible you should try to work around fractional powers by using division (as show above).
I am not sure this will fix all of your problems, but from your description it looks like this is certainly one problem you have.
I am trying to fit a line to some data without using polyfit and polyval. I got some good help already on how to implement this and I have gotten it to work with a simple sin function. However, when applied to the function I am trying to fit, it does not work. Here is my code:
clear all
clc
lb=0.001; %lowerbound of data
ub=10; %upperbound of data
step=.1; %step-size through data
a=.03;
la=1482/120000; %1482 is speed of sound in water and 120kHz
ep1=.02;
ep2=.1;
x=lb:step:ub;
r_sq_des=0.90; %desired value of r^2 for the fit of data without noise present
i=1;
for x=lb:step:ub
G(i,1)= abs(sin((a/la)*pi*x*(sqrt(1+(1/x)^2)-1)));
N(i,1)=2*rand()-1;
Ghat(i,1)=(1+ep1*N(i,1))*G(i,1)+ep2*N(i,1);
r(i,1)=x;
i=i+1;
end
x=r;
y=G;
V=[x.^0];
Vfit=[x.^0];
for i=1:1:1000
V = [x.^i V];
c = V \ y;
Vfit = [x.^i Vfit];
yFit=Vfit*c;
plot(x,y,'o',x,yFit,'--')
drawnow
pause
end
The first two sections are just defining variables and the function. The second for loop is where I am making the fit. As you can see, I have it pause after every nth order in order to see the fit.
I changed your fit formula a bit, I got the same answers but quickly got
a warning that the matrix was singular. No sense in continuing past
the point that the inversion is singular.
Depending on what you are doing you can usually change out variables or change domains.
This doesn't do a lot better, but it seemed to help a little bit.
I increased the number of samples by a factor of 10 since the initial part of the curve
didn't look sampled highly enough.
I added a weighting variable but it is set to equal weight in the code below. Attempts
to deweight the tail didn't help as much as I hoped.
Probably not really a solution, but perhaps will help with a few more knobs/variables.
...
step=.01; %step-size through data
...
x=r;
y=G;
t=x.*sqrt(1+x.^(-2));
t=log(t);
V=[ t.^0];
w=ones(size(t));
for i=1:1:1000
% Trying to solve for value of c
% c that
% yhat = V*c approximates y
% or y = V*c
% V'*y = V'*V * c
% c = (V'*V) \ V'*y
V = [t.^i V];
c = (V'*diag(w.^2)*V ) \ (V'*diag(w.^2)*y) ;
yFit=V*c;
subplot(211)
plot(t,y,'o',t,yFit,'--')
subplot(212)
plot(x,y,'o',x,yFit,'--')
drawnow
pause
end
It looks like more of a frequency estimation problem, and trying to fit a unknown frequency
with polynomial tends to be touch and go. Replacing the polynomial basis with a quick
sin/cos basis didn't seem to do to bad.
V = [sin(t*i) cos(t*i) V];
Unless you specifically need a polynomial basis, you can apply your knowledge of the problem domain to find other potential basis functions for your fit, or to attempt to make the domain in which you are performing the fit more linear.
As dennis mentioned, a different set of basis functions might do better. However you can improve the polynomial fit with QR factorisation, rather than just \ to solve the matrix equation. It is a badly conditioned problem no matter what you do however, and using smooth basis functions wont allow you to accurately reproduce the sharp corners in the actual function.
clear all
close all
clc
lb=0.001; %lowerbound of data
ub=10; %upperbound of data
step=.1; %step-size through data
a=.03;
la=1482/120000; %1482 is speed of sound in water and 120kHz
ep1=.02;
ep2=.1;
x=logspace(log10(lb),log10(ub),100)';
r_sq_des=0.90; %desired value of r^2 for the fit of data without noise present
y=abs(sin(a/la*pi*x.*(sqrt(1+(1./x).^2)-1)));
N=2*rand(size(x))-1;
Ghat=(1+ep1*N).*y+ep2*N;
V=[x.^0];
xs=(lb:.01:ub)';
Vfit=[xs.^0];
for i=1:1:20%length(x)-1
V = [x.^i V];
Vfit = [xs.^i Vfit];
[Q,R]=qr(V,0);
c = R\(Q'*y);
yFit=Vfit*c;
plot(x,y,'o',xs,yFit)
axis([0 10 0 1])
drawnow
pause
end