As part of a course in signal processing at university, we have been asked to write an algorithm in Matlab to calculate the single sided spectrum of our signal using DFT, without using the fft() function built in to matlab. this isn't an assessed part of the course, I'm just interested in getting this "right" for myself. I am currently using the 2018b version of Matlab, should anyone find this useful.
I have built a signal of a 1 KHz and 2KHz sinusoid, phase shifted by 135 degrees (2*pi/3 rad).
then using the equations in 9.1 of Discrete time signal processing (Allan V. Oppenheim) and Euler's formula to simplify the exponent, I produce this code:
%%DFT(currently buggy)
n=0;m=0;
for m=1:DFT_N-1 %DFT_Fmin;DFT_Fmax; %scrolls through DFT m values (K in text.)
for n=1:DFT_N-1;%;(DFT_N-1);%<<redundant code? from Oppenheim eqn. 9.1 % eulers identity, K=m and n=n
X(m)=x(n)*(cos((2*pi*n*m)/DFT_N)-j*sin((2*pi*n*m)/DFT_N));
n=n+1;
end
%m=m+1; %redundant code?
end
This takes x as the input, in this case the signal mentioned earlier, as well as the resolution of the transform, as represented by the DFT_N, which has been initialized to 100. The output of this function, X, should be something in the frequency domain, but plotting X yields a circular plot slightly larger than the unit circle, and with a gap on the left hand edge.
I am struggling to see how I am supposed to convert this to the stem() plots as given by the in-built DFT algorithm.
Many thanks, J.
This is your bug:
replace X(m)=x(n)*(cos.. with X(m)=X(m)+x(n)*(cos..
For a given m, it does not integrate over the variable n, but overwrites X(m) only the last calculation for n = DFT_N-1.
Notice that integrating over n=1:DFT_N-1 omits one harmonic, i.e., the first one, exp(-j*2*pi). Replace
n=1:DFT_N-1 with n=1:DFT_N to include that. I would also replace m=1:DFT_N-1 with m=1:DFT_N for plotting concerns.
Also replace any 2*pi*n*m with 2*pi*(n-1)*(m-1) to get the phase right, since the first entry of X should correspond to zero-frequency, yielding sum_n x(n) * (cos(0) + j sin(0)) = sum_n x(n). If your signal x is real-valued then the zero-frequency component X(1) should be real-valued, angle(X(1))=0.
Last remark, don't forget to shift zero-frequency component to the center of the spectrum for better visibility, X = circshift(X,floor(size(X)/2));
If you are interested in the single-sided spectrum only, than you can just calculate X(m) for m=1:DFT_N/2 since X it is conjugate symmetric around m=DFT_N/2, i.e., X(DFT_N/2+m) = X(DFT_N/2-m)', due to exp(-j*(pi*n+2*pi/DFT_N*m)) = exp(-j*(pi*n-2*pi/DFT_N*m))'.
As a side note, for a given m this program calculates an inner product between the array x and another array of complex exponentials, i.e., exp(-j*2*pi/DFT_N*m*n), for n = 0,1,...,N-1. MATLAB syntax is very convenient for such calculations, and you can avoid this inner loop by the following command
exp(-j*2*pi/DFT_N*m*(0:DFT_N-1)) * x
where x is a column vector. Similarly, you can avoid the first loop too by expanding your complex exponential vector row-wise for every m, i.e., build the matrix exp(-j*2*pi/DFT_N*(0:DFT_N-1)'*(0:DFT_N-1)). Then your DFT is simply
X = exp(-j*2*pi/DFT_N*(0:DFT_N-1)'*(0:DFT_N-1)) * x
For single-sided spectrum, instead use
X = exp(-j*2*pi/DFT_N*(0:floor((DFT_N-1)/2))'*(0:DFT_N-1)) * x
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.
Using MATLAB I want to implement some kind of a spectral method. The idea is as following (described for a example which is working).
Dirichlet (and Neumann, and periodic) boundaries leads to eigenvalues in the fourier space of k=n*pi/L
Projecting all the linear operators in the fourier space to the discretized k-values:
e.g. L = -D*(k.*k) (for diffusion only)
Defining the propagator in time as P = exp( dt * L )
Calculating iteratively the evolution in time by uh_{n+1} = uh_n * P
return the calculated value to the real space every time I want to save the value by ifft( uh )
My question concerns another boundary conditions.
In my case I have Robin boundary conditions. So, the eigenvalues are defined through some weird equation of the form tan( x ) = x or the like. The problem of computing them is solved.
As I have the values, the step no. 2 and 3 is simple too, but:
For applying P on the fourier-transformed vector uh I have to ensure that my uh = fft(u) uses the same eigenvalues, which is not the case by default.
By default MATLAB uses equidistant modes for the fft.
Is there any simple trick for this?
Or, maybe, do I have any mistake in my thoughts?
I want to know the best way to fit a sine-wave with a distorted time base, in Matlab.
The distortion in time is given by a n-th order polynomial (n~10), of the form t_distort = P(t).
For example, consider the distortion t_distort = 8 + 12t + 6t^2 + t^3 (which is just the power series expansion of (t-2)^3).
This will distort a sine-wave as follows:
I want to be able to find the distortion given this distorted sine-wave. (i.e. I want to find the function t = G(t_distort), but t_distort = P(t) is unknown.)
If your resolution is high enough, then this is basically an angle-demodulation problem. The standard way to demodulate an angle-modulated signal is to take the derivative, followed by an envelope detector, followed by an integrator.
Since I don't know the exact numbers you're using, I'll show an example with my own numbers. Suppose my original timebase has 10 million points from 0 to 100:
t = 0:0.00001:100;
I then get the distorted timebase and calculate the distorted sine wave:
td = 0.02*(t+2).^3;
yd = sin(td);
Now I can demodulate it. Take the "derivative" using approximate differences divided by the step size from before:
ydot = diff(yd)/0.00001;
The envelope can be easily detected:
envelope = abs(hilbert(ydot));
This gives an approximation for the derivative of P(t). The last step is an integrator, which I can approximate using a cumulative sum (we have to scale it again by the step size):
tdguess = cumsum(envelope)*0.00001;
This gives a curve that's almost identical to the original distorted timebase (so, it gives a good approximation of P(t)):
You won't be able to get the constant term of the polynomial since we made our approximation from its derivative, which of course eliminates the constant term. You wouldn't even be able to find a unique constant term mathematically from just yd, since infinitely many values will yield the same distorted sine wave. You can get the other three coefficients of P(t) using polyfit if you know the degree of P(t) (ignore the last number, it's the constant term):
>> polyfit(t(1:10000000), tdguess, 3)
ans =
0.0200 0.1201 0.2358 0.4915
This is pretty close to the original, aside from the constant term: 0.02*(t+2)^3 = 0.02t^3 + 0.12t^2 + 0.24t + 0.16.
You wanted the inverse mapping Q(t). Can you do that knowing a close approximation for P(t) as found so far?
Here's an analytical driven route that takes asin of the signal with proper unwrapping of the angle. Then you can fit a polynomial using polyfit on the angle or using other fit methods (search for fit and see). Last, take a sin of the fitted function and compare the signal to the fitted one... see this pedagogical example:
% generate data
t=linspace(0,10,1e2);
x=0.02*(t+2).^3;
y=sin(x);
% take asin^2 to obtain points of "discontinuity" where then asin hits +-1
da=(asin(y).^2);
[val locs]=findpeaks(da); % this can be done in many other ways too...
% construct the asin according to the proper phase unwrapping
an=NaN(size(y));
an(1:locs(1)-1)=asin(y(1:locs(1)-1));
for n=2:numel(locs)
an(locs(n-1)+1:locs(n)-1)=(n-1)*pi+(-1)^(n-1)*asin(y(locs(n-1)+1:locs(n)-1));
end
an(locs(n)+1:end)=n*pi+(-1)^(n)*asin(y(locs(n)+1:end));
r=~isnan(an);
p=polyfit(t(r),an(r),3);
figure;
subplot(2,1,1); plot(t,y,'.',t,sin(polyval(p,t)),'r-');
subplot(2,1,2); plot(t,x,'.',t,(polyval(p,t)),'r-');
title(['mean error ' num2str(mean(abs(x-polyval(p,t))))]);
p =
0.0200 0.1200 0.2400 0.1600
The reason I preallocate with NaNand avoid taking the asin at points of discontinuity (locs) is to reduce the error of the fit later. As you can see, for a 100 points between 0,10 the average error is of the order of floating point accuracy, and the polynomial coefficients are as exact as you can have them.
The fact that you are not taking a derivative (as in the very elegant Hilbert transform) allows to be numerically exact. For the same conditions the Hilbert transform solution will have a much bigger average error (order of unity vs 1e-15).
The only limitation of this method is that you need enough points in the regime where the asin flips direction and that function inside the sin is well behaved. If there's a sampling issue you can truncate the data and only maintain a smaller range closer to zero, such that it'll be enough to characterize the function inside the sin. After all, you don't need millions op points to fit to a 3 parameter function.
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 doing a project where i find an approximation of the Sine function, using the Least Squares method. Also i can use 12 values of my own choice.Since i couldn't figure out how to solve it i thought of using Taylor's series for Sine and then solving it as a polynomial of order 5. Here is my code :
%% Find the sine of the 12 known values
x=[0,pi/8,pi/4,7*pi/2,3*pi/4,pi,4*pi/11,3*pi/2,2*pi,5*pi/4,3*pi/8,12*pi/20];
y=zeros(12,1);
for i=1:12
y=sin(x);
end
n=12;
j=5;
%% Find the sums to populate the matrix A and matrix B
s1=sum(x);s2=sum(x.^2);
s3=sum(x.^3);s4=sum(x.^4);
s5=sum(x.^5);s6=sum(x.^6);
s7=sum(x.^7);s8=sum(x.^8);
s9=sum(x.^9);s10=sum(x.^10);
sy=sum(y);
sxy=sum(x.*y);
sxy2=sum( (x.^2).*y);
sxy3=sum( (x.^3).*y);
sxy4=sum( (x.^4).*y);
sxy5=sum( (x.^5).*y);
A=[n,s1,s2,s3,s4,s5;s1,s2,s3,s4,s5,s6;s2,s3,s4,s5,s6,s7;
s3,s4,s5,s6,s7,s8;s4,s5,s6,s7,s8,s9;s5,s6,s7,s8,s9,s10];
B=[sy;sxy;sxy2;sxy3;sxy4;sxy5];
Then at matlab i get this result
>> a=A^-1*B
a =
-0.0248
1.2203
-0.2351
-0.1408
0.0364
-0.0021
However when i try to replace the values of a in the taylor series and solve f.e t=pi/2 i get wrong results
>> t=pi/2;
fun=t-t^3*a(4)+a(6)*t^5
fun =
2.0967
I am doing something wrong when i replace the values of a matrix in the Taylor series or is my initial thought flawed ?
Note: i can't use any built-in function
If you need a least-squares approximation, simply decide on a fixed interval that you want to approximate on and generate some x abscissae on that interval (possibly equally spaced abscissae using linspace - or non-uniformly spaced as you have in your example). Then evaluate your sine function at each point such that you have
y = sin(x)
Then simply use the polyfit function (documented here) to obtain least squares parameters
b = polyfit(x,y,n)
where n is the degree of the polynomial you want to approximate. You can then use polyval (documented here) to obtain the values of your approximation at other values of x.
EDIT: As you can't use polyfit you can generate the Vandermonde matrix for the least-squares approximation directly (the below assumes x is a row vector).
A = ones(length(x),1);
x = x';
for i=1:n
A = [A x.^i];
end
then simply obtain the least squares parameters using
b = A\y;
You can clearly optimise the clumsy Vandermonde generation loop above I have just written to illustrate the concept. For better numerical stability you would also be better to use a nice orthogonal polynomial system like Chebyshev polynomials of the first kind. If you are not even allowed to use the matrix divide \ function then you will need to code up your own implementation of a QR factorisation and solve the system that way (or some other numerically stable method).