I have a rather large piece of code at the end of which I need to apply ifft to an array ZF. ZF is the fft of the columns of a matrix Z.
The problem is, the ifft turns out to be a complex double array (which it shouldn't if I did everything correctly) so I need a debugging tool to check at which point of the numerous manipulations I performed on the original DFT I lost the Hermitian Symmetry.
The array is a double 1024x1376
my attempt at this is:
for i=2:N/2
Herm=ZF(i,:)+ZF(N-i+2,:);
end
N=1024 is the number of rows and if Herm is purely real then I have the Symmetry.
Any thoughts? I'm still on the first month of MATLAB so if I'm way off feel free to tell me.
Related
I encountered some numerical questions when running simulation on MatLab. Here please find the questions:
I found that A*A' (a matrix times its transpose) is not guaranteed to be symmetric in MatLab. Can I know what is the reason? And because I will have A*C*A', where C is a symmetric matrix, and I would like to keep A*C*A' as symmetric. Is there any method to fix the numerical difference created by the transpose operation?
I implemented a for loop in Matlab to compute a set of matrices. Small numerical difference (around 10^(-10)) in each round accumulates to the next run, and it finally diverges after around 30 rounds. Is there any method to fix small error in each run and do not affect the result at the same time.
Thank you for reading my questions!
"I found that A*A' (a matrix times its transpose) is not guaranteed to be symmetric in MatLab."
I would dispute that statement as written. The MATLAB parser is smart enough to recognize that the operands of A*A' are the same and call a symmetric BLAS routine in the background to do the work, and then manually copy one triangle into the other resulting in an exactly symmetric result. Where one usually gets into trouble is by coding something that the parser cannot recognize. E.g.,
A = whatever;
B = whatever;
X = A + B;
(A+B) * (A+B)' <-- MATLAB parser will call generic BLAS routine
X * X' <-- MATLAB parser will call symmetric BLAS routine
In the first matrix multiply above, the MATLAB parser may not be not smart enough to recognize the symmetry so a generic matrix multiply BLAS routine (e.g., dgemm) could be called to do the work and the result is not guaranteed to be exactly symmetric. But in the second matrix multiply above the MATLAB parser does recognize the symmetry and calls a symmetric BLAS matrix multiply routine.
For the ACA' case, I don't know of any method to force MATLAB to generate an exact symmetric result. You could manually copy one resulting triangle into the other after the fact. I suppose you could also factor C into two parts X*X' and then regroup but that seems like too much work for what you are trying to do.
Solve Ax = b. Real double. A is overdetermined Mx2 with M >> 2. b is Mx1. I've run a ton of data against mldivide, and the results are excellent. I wrote a mex routine with MKL LAPACKE_dgels and it's nowhere near as good. The results have a ton of noise and the underlying signal is barely there. I checked the routine against the MKL example results first. I've searched through the mldivide doc (flowchart) and the SO questions. All I found is Matlab uses QR factorization for overdetermined rectangular.
What should I try next? Am I using the wrong LAPACK routine? Please help guide me in the right direction.
Update:
To within E-15 floating point difference on the solution vector, Intel MKL LAPACKE_dgels has the same result as Matlab mldivide for real double overdetermined (rectangular) problems. As far as I can tell, this is the QR method used.
Beware the residuals returned from this dgels. They do not equate to b - Ax. Many of them are close to this value, whereas some are far from it.
The problem was not the solution x, rather the returned residuals from DGELS. This routine's outputs are modify-in-place on the input array pointers. The MKL doc says the input array b is overwritten with the output vector x for the first N rows, then the residuals in N+1 to M. I confirmed this with my code.
The mistake was in aligning the b[N+1] residuals to original inputs b[1], and making further algorithmic decisions on that. The correct alignment of residual to original input is b[1] to b[1]. The first N residuals are not available; you have to compute those afterwards.
The doc doesn't say they are residuals per se, rather specifically
the residual sum of squares for the solution in each column is given by the sum of squares of modulus of elements n+1 to m in that column.
I have a problem.
I have a task to write an own fft2 without using for-loops in Matlab.
There is a formula for computing this task:
F(u,v) = sum (0 to M-1) {sum(o to N-1) {f(m,n)*e^(-i*2pi*(um/M + vn/N))}}
Or for better reading:
http://www.directupload.net/file/d/3808/qs3r9ogz_png.htm
It is easy to do it with two for-loops but I have no idea how to do this without these loops, absolutely no idea.
We get no help by the teaching personal. They don't even give a hint or a reference to a book, where we could read about it.
Now, I want to try to get help here.
Are you familiar with the matrix form of DFT? have a look here: http://en.wikipedia.org/wiki/DFT_matrix
You can do something similar in order to get a matrix form for 2D DFT.
You need to transformation matrices. The first is a N-by-N DFT matrix that operates on the columns of f, as explained in the link above. Next you need another M-byM DFT matrix the operates on the rows of f. Finally, you transformed signal is given by
F = Wm * f * Wn;
without any loops.
Note that the DFT matrix can be constructed also without loop by using something like
(1:M)*((1:M)')
Just a little correction in Thp's answer: (1:M)*((1:M)') is not the right way to create the matrix, but (1:M)'*(1:M) is the correct way.
I have a little code, that should implement cepstrum deconvolution for minimum phase FIR filter design, but being nonmatlab guy I'm struggling with understanding it. Can someone help?
wn = [ones(1,m)
2*ones((n+odd)/2-1,m)
ones(1-rem(n,2),m)
zeros((n+od d)/2-1,m)];
y = real(ifft(exp(fft(wn.*real(ifft(log(abs(fft(x)))))))));
Mainly I don't understand the first line, ".*" symbol in second line and also probably at some point there should be conversion from real to complex domain in the second line, but I have no idea where. Any ideas?
In the first line you are constructing the matrix wn row by row.
.* operator means element-wise multiplication. * alone would mean matrix multiplication.
In fact you should pay attention to the size of x and wn which must be the same for the element-wise multiplication to have sense.
actually there isn't any conversion from real to complex in the second line. There are the functions log, fft, ifft that may return complex values depending on the input.
You can access the Matlab help by the commands help or doc (for example doc ones should produce the documentation of the ones function - this produce a matrix filled with ones of the size specified by it's arguments).
To quickly summon the help when you're inspecting some code you can use Matlab's inline help by pressing the button F1 when the cursor is at the end of a function name (just before the parenthesis).
I have a matrix of size 10800x10800 in Matlab and I compute its inverse
directly with the function:
inv(A)
It takes 3 to 4 minutes just one such computation. And that is part of an
iterative algorithm which needs more than 20 iterations, so overall things would
be very slow. Is there a better way to do this? Maybe some mathematical formulas
or maybe a better Matlab function?
Edit: The matrix is diagonal. Each iteration the diagonal elements are updated
based on formulas for fitting a factor analyzer. But that is irrelevant, the
important thing is that it is a diagonal matrix and it changes each iteration.
THanks
If your matrix is indeed diagonal, you can obviously just do
Ainv = diag(1./diag(A));
which should be very fast.
The backslash operator \ is said to be faster and also could be more accurate. Without MATLAB really I cannot tell, but you could try to run A \ eye(10800) instead of inv(A), and see if it works out.