I am trying to calculate the Mean Squared Error in Vitis HLS. I am using hls::pow(...,2) and divide by n, but all I receive is a negative value for example -0.004. This does not make sense to me. Could anyone point the problem out or have a proper explanation for this??
Besides calculating the mean squared error using hls::pow does not give the same results as (a - b) * (a - b) and for information I am using ap_fixed<> types and not normal float or double precision
Thanks in advance!
It sounds like an overflow and/or underflow issue, meaning that the values reach the sign bit and are interpreted as negative while just be very large.
Have you tried tuning the representation precision or the different saturation/rounding options for the fixed point class? This tuning will depend on the data you're processing.
For example, if you handle data that you know will range between -128.5 and 1023.4, you might need very few fractional bits, say 3 or 4, leaving the rest for the integer part (which might roughly be log2((1023+128)^2)).
Alternatively, if n is very large, you can try a moving average and calculate the mean in small "chunks" of length m < n.
p.s. Getting the absolute value of a - b and store it into an ap_ufixed before the multiplication can already give you one extra bit, but adds an instruction/operation/logic to the algorithm (which might not be a problem if the design is pipelined, but require space if the size of ap_ufixed is very large).
Given a matrix's LDLT decomposition, I would like to modify the diagonal - for example floor all the values. Is there a way to do this with eigen?
To be clear, I can do:
auto ldlt_ = matrix.ldlt();
and I would like to follow up with:
ldlt_.vectorD().cwiseMax(Vector::Constant(n,epsilon))
before solving a problem:
ldlt_.solve(a)
I don't see any non const accessors to the vectorD member - what am I missing?
No, you cannot do that, and I don't think that's a good idea to increase small (or negative) diagonal entries this way. If there are too small entries, the usual approach is either to ignore them (default behavior of LDLT::solve), or to redo the factorization with matrix+eps*I. Anyway, if you really want to tweak D yourself, then you have to implement your own solve function.
These days I am trying to redo shock spectrum of single degree of freedom system using Sympy. The problem can reduce to find maximum value of a function. Following are two cases I cannot figure out how to do.
The first one is
tau,t,t_r,omega,p0=symbols('tau,t,t_r,omega,p0',positive=True)
h=expand(sin(omega*(t-tau)))
f=simplify(integrate(p0*tau/t_r*h,(tau,0,t_r))+integrate(p0*h,(tau,t_r,t)))
The final goal is to obtain maximum absolute value of f (The variable is t). The direct way is
df=diff(f,t)
sln=solve(simplify(df),t)
simplify(f.subs(t,sln[1]))
Here is the result, I tried many ways, but I can not simplify any further.
Therefore, I tried another way. Because I need the maximum absolute value and the location where abs(f) is maximum happens at the same location of square of f, we can calculate square of f first.
df=expand_trig(diff(expand(f)**2,t))
sln=solve(df,t)
simplify(f.subs(t,sln[2]))
It seems the answer is almost the same, just in another form.
The expected answer is a sinc function plus a constant as following:
Therefore, the question is how to get the final presentation.
The second one may be a little harder. The question can be reduced to find the maximum value of f=sin(pi*t/t_r)-T/2/t_r*sin(2*pi/T*t), in which t_r and T are two parameters. The maximum located at different peak when the ratio of t_r and T changes. And I do not find a way to solve it in Sympy. Any suggestion? The answer can be represented in following figure.
The problem is the log(exp(I*omega*t_r/2)) term. SymPy is not reducing this to I*omega*t_r/2. SymPy doesn't simplify this because in general, log(exp(x)) != x, but rather log(exp(x)) = x + 2*pi*I*n for some integer n. But in this case, if you replace log(exp(I*omega*t_r/2)) with omega*t_r/2 or omega*t_r/2 + 2*pi*I*n, it will be the same, because it will just add a 2*pi*I*n inside the sin.
I couldn't figure out any functions that force this simplification, but the easiest way is to just do a substitution:
In [18]: print(simplify(f.subs(t,sln[1]).subs(log(exp(I*omega*t_r/2)), I*omega*t_r/2)))
p0*(omega*t_r - 2*sin(omega*t_r/2))/(omega**2*t_r)
That looks like the answer you are looking for, except for the absolute value (I'm not sure where they should come from).
Typically, the re-estimation iterative procedure stops when lambda.bar - lambda is less than some epsilon value.
How exactly does one determine this epsilon value? I often only see is written as the general epsilon symbol in papers, and never the actual value used, which I assume would change depending on the data.
So, for instance, if the lambda value of my first iteration was 5*10^-22, second iteration was 1.3*10^-15, third was 8.45*10^-15, fourth was 1.65*10^-14, etc., how would I determine when the algorithm needed no more iteratons?
Moreover, what if I were to apply the same alogrithm to a different datset? would I need to change my epsilon definitions?
Sorry for the long question. Pretty puzzled by it... :)
"how would I determine when the algorithm needed no more iteratons?"
When you get a "good-enough" result within a reasonable amount of time. ;-)
"Moreover, what if I were to apply the same alogrithm to a different datset? would I need to
change my epsilon definitions?"
Yes, most probably.
If you can afford it, you can just let it iterate until the updated value <= the old value (it could be < due to floating point error). I would be inclined to go with this until I ran out of patience or cpu budget.
I apologize for being a bit verbose in advance: if you want to skip all the background mumbo jumbo you can see my question down below.
This is pretty much a follow up to a question I previously posted on how to compare two 1D (time dependent) signals. One of the answers I got was to use the cross-correlation function (xcorr in MATLAB), which I did.
Background information
Perhaps a little background information will be useful: I'm trying to implement an Independent Component Analysis algorithm. One of my informal tests is to (1) create the test case by (a) generate 2 random vectors (1x1000), (b) combine the vectors into a 2x1000 matrix (called "S"), and multiply this by a 2x2 mixing matrix (called "A"), to give me a new matrix (let's call it "T").
In summary: T = A * S
(2) I then run the ICA algorithm to generate the inverse of the mixing matrix (called "W"), (3) multiply "T" by "W" to (hopefully) give me a reconstruction of the original signal matrix (called "X")
In summary: X = W * T
(4) I now want to compare "S" and "X". Although "S" and "X" are 2x1000, I simply compare S(1,:) to X(1,:) and S(2,:) to X(2,:), each which is 1x1000, making them 1D signals. (I have another step which makes sure that these vectors are the proper vectors to compare to each other and I also normalize the signals).
So my current quandary is how to 'grade' how close S(1,:) matches to X(1,:), and likewise with S(2,:) to X(2,:).
So far I have used something like: r1 = max(abs(xcorr(S(1,:), X(1,:)))
My question
Assuming that using the cross correlation function is a valid way to go about comparing the similarity of two signals, what would be considered a good R value to grade the similarity of the signals? Wikipedia states that this is a very subjective area, and so I defer to the better judgment of those who might have experience in this field.
As you might realize, I'm not coming from a EE/DSP/statistical background at all (I'm a medical student) so I'm going through a sort of "baptism through fire" right now, and I appreciate all the help I can get. Thanks!
(edit: as far as directly answering your question about R values, see below)
One way to approach this would be to use cross-correlation. Bear in mind that you have to normalize amplitudes and correct for delays: if you have signal S1, and signal S2 is identical in shape, but half the amplitude and delayed by 3 samples, they're still perfectly correlated.
For example:
>> t = 0:0.001:1;
>> y = #(t) sin(10*t).*exp(-10*t).*(t > 0);
>> S1 = y(t);
>> S2 = 0.4*y(t-0.1);
>> plot(t,S1,t,S2);
These should have a perfect correlation coefficient. A way to compute this is to use maximum cross-correlation:
>> f = #(S1,S2) max(xcorr(S1,S2));
f =
#(S1,S2) max(xcorr(S1,S2))
>> disp(f(S1,S1)); disp(f(S2,S2)); disp(f(S1,S2));
12.5000
2.0000
5.0000
The maximum value of xcorr() takes care of the time-delay between signals. As far as correcting for amplitude goes, you can normalize the signals so that their self-cross-correlation is 1.0, or you can fold that equivalent step into the following:
ρ2 = f(S1,S2)2 / (f(S1,S1)*f(S2,S2);
In this case ρ2 = 5 * 5 / (12.5 * 2) = 1.0
You can solve for ρ itself, i.e. ρ = f(S1,S2)/sqrt(f(S1,S1)*f(S2,S2)), just bear in mind that both 1.0 and -1.0 are perfectly correlated (-1.0 has opposite sign)
Try it on your signals!
with respect to what threshold to use for acceptance/rejection, that really depends on what kind of signals you have. 0.9 and above is fairly good but can be misleading. I would consider looking at the residual signal you get after you subtract out the correlated version. You could do this by looking at the time index of the maximum value of xcorr():
>> t = 0:0.001:1;
>> y = #(a,t) sin(a*t).*exp(-a*t).*(t > 0);
>> S1=y(10,t);
>> S2=0.4*y(9,t-0.1);
>> f(S1,S2)/sqrt(f(S1,S1)*f(S2,S2))
ans =
0.9959
This looks pretty darn good for a correlation. But let's try fitting S2 with a scaled/shifted multiple of S1:
>> [A,i]=max(xcorr(S1,S2)); tshift = i-length(S1);
>> S2fit = zeros(size(S2)); S2fit(1-tshift:end) = A/f(S1,S1)*S1(1:end+tshift);
>> plot(t,[S2; S2fit]); % fit S2 using S1 as a basis
>> plot(t,[S2-S2fit]); % residual
Residual has some energy in it; to get a feel for how much, you can use this:
>> S2res=S2-S2fit;
>> dot(S2res,S2res)/dot(S2,S2)
ans =
0.0081
>> sqrt(dot(S2res,S2res)/dot(S2,S2))
ans =
0.0900
This says that the residual has about 0.81% of the energy (9% of the root-mean-square amplitude) of the original signal S2. (the dot product of a 1D signal with itself will always be equal to the maximum value of cross-correlation of that signal with itself.)
I don't think there's a silver bullet for answering how similar two signals are with each other, but hopefully I've given you some ideas that might be applicable to your circumstances.
A good starting point is to get a sense of what a perfect match will look like by calculating the auto-correlations for each signal (i.e. do the "cross-correlation" of each signal with itself).
THIS IS A COMPLETE GUESS - but I'm guessing max(abs(xcorr(S(1,:),X(1,:)))) > 0.8 implies success. Just out of curiosity, what kind of values do you get for max(abs(xcorr(S(1,:),X(2,:))))?
Another approach to validate your algorithm might be to compare A and W. If W is calculated correctly, it should be A^-1, so can you calculate a measure like |A*W - I|? Maybe you have to normalize by the trace of A*W.
Getting back to your original question, I come from a DSP background, so I get to deal with fairly noise-free signals. I understand that's not a luxury you get in biology :) so my 0.8 guess might be very optimistic. Perhaps looking at some literature in your field, even if they aren't using cross-correlation exactly, might be useful.
Usually in such cases people talk about "false acceptance rate" and "false rejection rate".
The first one describes how many times algorithm says "similar" for non-similar signals, the second one is the opposite.
Selecting a threshold thus becomes a trade-off between these criteria. To make FAR=0, threshold should be 1, to make FRR=0 threshold should be -1.
So probably, you will need to decide which trade-off between FAR and FRR is acceptable in your situation and this will give the right value for threshold.
Mathematically this can be expressed in different ways. Just a couple of examples:
1. fix some of rates at acceptable value and minimize other one
2. minimize max(FRR,FAR)
3. minimize aFRR+bFAR
Since they should be equal, the correlation coefficient should be high, between .99 and 1. I would take the max and abs functions out of your calculation, too.
EDIT:
I spoke too soon. I confused cross-correlation with correlation coefficient, which is completely different. My answer might not be worth much.
I would agree that the result would be subjective. Something that would involve the sum of the squares of the differences, element by element, would have some value. Two identical arrays would give a value of 0 in that form. You would have to decide what value then becomes "bad". Make up 2 different vectors that "aren't too bad" and find their cross-correlation coefficient to be used as a guide.
(parenthetically: if you were doing a correlation coefficient where 1 or -1 would be great and 0 would be awful, I've been told by bio-statisticians that a real-life value of 0.7 is extremely good. I understand that this is not exactly what you are doing but the comment on correlation coefficient came up earlier.)