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#1 Help Me ! » Singular Value Decomposition of noisy data: null space » 2010-11-11 23:44:50

Replies: 0

Hi all

I am trying to understand the following problem a bit further, but I lack algebra background. Say that I have a noisy sequence of measurements (a vector) of length N (say it is a column vector, Nx1), that can be written as:

is an ideal vector of samples, and
is additive white Gaussian noise. So, the components
are iid, of zero mean and have variance

Now, we filter the noisy samples using:

is a matrix of size (PxN). The rows of the matrix are orthogonal, but not orthonormal, and so the filtered noise
becomes coloured Gaussian noise.

Moreover, suppose that now I form a Toeplitz matrix using the filtered measurements


The matrix can of any size ((P-M+1)x(M+1)), not necessarily square. And it is also not a condition that P>2M, even though the example I wrote is like that.

The idea is that, now, I have the following:

It is easy to prove that matrix

is diagonal, but not a multiple of the identity matrix. In fact, assume that the matrix
is such that the diagonal values of
are symmetric and decreasing from the ends to the central value.

Now, consider that doing SVD on

non-zero entries as singular values of
. This means that the null-space of
is spanned by the columns of
related to the (M-K) zero entries in

But, does anyone know how the SVD of

would relate to that of
? If the noise was not coloured, it is straightforward to see that the null-eigenspace is preserved. But I don't know what happens with coloured noise.

Thanks in advance

#2 Re: Help Me ! » Unusual equality :)) » 2010-03-23 01:50:07


I think there's only 1 solution, and I wrote it down, z = -W(-1)


#3 Re: Help Me ! » Unusual equality :)) » 2010-03-20 10:47:10

Hi there!

I actually got help from, and then checked what did the result mean. The idea is to use what is called the Lambert W function, which, by definition (Wikipedia) allows to express any complex number as:

Then, for your equation you have to do:

But any number can be written using the W function, so:

By equating both expressions, then:

which actually has |z| = 1.3745, so the condition is satisfied.

The other problem I cannot get it yet. I will keep trying smile

Ok, so I would say you have to do this numerically. First plot both functions, to have an idea. then use Newton-Rapson's method to obtain when they intersect. For instance, to calculate the area in the interval [0, 3pi/2], first lets find where they cross:

Say we want to find

Using N-R, from a guess xn, we obtain a better guess doing:

Choose as x0 = 4.5, and a few iterations give x = 4.4934

I have calculated only the area between [0, 4.4934], which is:

Hope this helps!

#4 Re: Help Me ! » Physics !! » 2009-12-22 03:42:24

Are you sure about the distances of the antinode to the sources?

I don't know exactly what does 2nd order antinode mean. If it would be the 1st antinode in the 2nd harmonic of the standing wave, the following should be true:

And it doesn't make sense.

If the 2nd order antinode means the 2nd antinode for a certain harmonic [n], then the following holds:


And neither of both makes sense, since [n] should be an integer.

#5 Re: Help Me ! » Physics » 2009-12-22 02:59:31

I am not sure why they give you the speed, because if you have the wavelength, isn't it straightforward?:

#6 Re: Help Me ! » Physics » 2009-12-22 02:51:48


Lets see if this works. We have the following relation between the wave speed [c] (m/s), the wavelength [lambda] (m) and the frequency [f] (Hz):

Then, for air you use c = 344 (m/s):

Now, for steel you can calculate:

which gives a speed in steel equal to:

We now have all the information to answer the time it takes to travel s = 84 (m):
(a) For air

(b) For steel

#7 Re: Help Me ! » Trigonometry » 2009-09-21 07:56:11


For the first question, be careful, because the cosine rule is not like that... (maybe dividing all the terms in the rhs rather than only the last one) Calling the triangle sides a = AB, b = BC and c = CA, and the angle between a and c being alpha:

For the second question, I attach a diagram for what I understand it is described. The cosine rule can again be applied to the diagram, using exactly the same notation as above. Then we get:

Then, the distance is 3.32 + 3.2 + 2.5 = 9.02


#8 Re: Help Me ! » move a point in 3d space » 2009-09-10 20:43:47


I am sure there will be an easier way to solve it, but I will try to say how would I do it.

I am going to assume you have an original point in the XY plane, heading north, which I will consider in the negative part of the X axis. Then, the original coordinates would be (-x, 0, 0), being the distance to the origin of coordinates equal to x.

I also consider that rotation from north and clockwise is keep the point in the XY plane, so rotate in Z an angle q1 = -33. And then doing the same for a "dip"angle, I assume it is rotate with respect to X, an angle q2.

First, you have to calculate the new coordinates of the point, for example from

1. Rotation in Z:

2. Likewise, rotation in X:

Yo can actually find the 2 matrices, multiply one with the other and obtain the total displacement.

Now have point (x'', y'', z''). Then, we have to move it in the direction normal to certain plane, a distance of 50. Ok, the equation characterizing a plane using a vector normal to it, (a, b, c) that passes through point (x0, y0, z0) is:

And the line passing through (x'', y'', z'') normal to the plane can be written using an independent variable t:

In total, you have to calculate what new point (x''', y''', z''') is at distance 50 from (x'', y'', z''):

So the point must satisfy the equation of the line and of the distance.

And that's what I can say with the info you provided smile


#9 Re: Help Me ! » 3D Shapes - Trigonometry » 2009-09-10 06:35:59


I guess the only angle left to calculate with respect to the base is that formed by the edges of the pyramid. This can be done knowing that its projection into the base has length b, that can be computed from the base length a:

Then, the new angle, the height of the pyramid and b can be related:

Hope this is it!

#10 Re: Help Me ! » Finding the domain of a changed pdf » 2009-09-01 21:32:37

Haha, you're welcome!

It is curious that you calculated the area underneath f(y) correctly, and not for f(x). That for f(y) is much more difficult! smile


#11 Re: Help Me ! » Finding the domain of a changed pdf » 2009-09-01 20:53:14


The area underneath the first curve is equal to 1:

And the domain is the mapping between x and y:


#12 Re: Help Me ! » Approximating Ln(2) » 2009-09-01 19:28:07


You can use Taylor series approximation around a = 1 for example, or any other a closer to 2:

Then, at x = 2

The result given is using up to 1/10. It is still not very precise, since ln(2) = 0.6931, but you can keep on adding and substracting fractions to get it more accurate.

I guess that if you approximate it closer to 2, for example a = 1.9, it will converge faster, but then you have to use powers in your calculator (or do many more multiplications smile )

Actually, I think that approximating around 1 is the only solution to only use (+, -, ×, ÷), because if doing it around a = 1.9, you need to know f(1.9) = ln(1.9). However, with a = 1, I guess ln(1) = 0 is well known smile


#13 Re: Help Me ! » changing variables in a pdf » 2009-09-01 19:15:32


I actually hadn't realized how you had used that formula, and not I see it, so go for both smile


#14 Re: Help Me ! » changing variables in a pdf » 2009-09-01 04:51:57


You have to differentiate in order to obtain the pdf out of the cdf!

I think then they results match!


#15 Re: Help Me ! » Meters Square Help » 2009-09-01 02:49:14


60mm = 0.06m
100mm = 0.1m

area = h x w = 0.06 x 0.1 = 0.006 m^2


#16 Re: Help Me ! » Changing the variables of a pdf » 2009-08-31 19:45:06


If you would like to understand HOW to calculate f(y), from Wikipedia []:

The probability contained in a differential area must be invariant under change of variables y = g(x)

This was for the simple case of a monotonic function g. There is a general expression in Wikipedia as well.

If you would like to know WHAT does it mean, I can only provide an example: Random Number Generation. Rather than explaining it myself, I think this other article will be useful []


#17 Re: Help Me ! » Changing the values of a, b and c » 2009-08-26 00:30:48


I would say that, if for example you have:

you can write

So, changing c and b effectively moves the origin of coordinates for the function

The same happens with

And if you check the derivative:

you see that the parameter a controls the rate of change of y.

Then, for the exponentiation function, you can check the derivative as well

In this case, for 0 < a < 1, increasing a makes the function to decrease slower (lna < 0, and increasing a makes the lna to be smaller in module. But be careful, because in this case the function itself is decreasing). And for 1 < a < inf, increasing a makes the function to increase faster (lna > 0, and increasing a makes lna to grow).

Hope it helps!

#18 Re: Help Me ! » integrate » 2009-08-24 11:13:24


I guess you need the steps, but I cannot figure them out now smile

Sometimes, the result can give some clue, even though it's not the case for me now. Anyway, Mathematica says:

That's a weird result, though! Hope you get luckier than me smile


#20 Re: Help Me ! » Precalculus » 2009-08-22 10:33:06


The first one I would say:

So every requirement holds.

The second:

Hope it helps!


#21 Re: Help Me ! » On the maximum of real-valued functions with complex image » 2009-08-22 06:12:00


If you want the maximum of the modulus, first of all you should calculate the absolute value of g:

After that, you can differentiate that rather ugly expression smile, solve it equal to 0 and find for what k is maximum


#22 Re: Help Me ! » Modulo Help » 2009-08-16 02:53:51


I think in this website they explain congruences very well

Suppose a, b and m are any integers with m not zero, then we say a is congruent to b modulo m if m divides a-b. We write this as

    a = b (mod m).

For your example, use the transitive property: If a = b (mod m) and b = c (mod m), then a = c (mod m).

Then, lambda = 0.25 (mod 23) and 0.25 = 6 (mod 23), then lambda = 6 (mod 23). Where the second is true because 23 / (0.25 - 6) = -4


#23 Re: Exercises » Trigonometry Proofs » 2009-08-15 19:57:16


I think I got the first one:



#24 Re: Help Me ! » Sum of n-th degrees » 2009-08-14 07:33:19


Apparently, according to


where Bk is the kth Berboulli number.

And in


#25 Re: Help Me ! » Algebra headache » 2009-08-13 19:41:01

Yes, thanks Macy!

I like the other approaches smile You can always learn many new things in this forum, even if you got the correct answer!


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