As far as I know the IFFT of spectrum which amplitude part is even symmetric and phase part is odd symmetric should be real.
Let's consider this example:
signal_spectrum = [1 2+i 3+2*i 4+8*i 5 4-8*i 3-2*i 2-i 1];
It's clear that this spectrum meets two conditions that I listed above. When I perform IFFT using Matlab I obtain:
signal= ifft(signal_spectrum) =
2.7778
0.8003 - 0.2913i
-1.2861 + 1.0792i
0.5218 - 0.9038i
-0.0812 + 0.4604i
0.0976 + 0.5536i
-0.6329 - 1.0962i
1.3343 + 1.1196i
-2.5316 - 0.9214i
Obtained signal is complex-valued. Why? What's the problem?
Just remove the last 1 from the vector :)
The symmetry should be relative to the first element.
You can think of it like circular buffer or one period of a periodical signal.
Related
I am trying to check convolution theorem in MATLAB. I have a signal called sine_big_T. Then I have a filter called W. W and sine_big_T have the same length.
The convolution theorem says that fft(sine_big_T.*W) should be same as the convolution of fft(sine_big_T) with fft(W).
I am quite confused about this theorem. fft(sine_big_T.*W) will give me a array with length length(sine_big_T). However, conv(fft(sine_big_T), fft(W)) gives me a array with length length(sine_big_T) + length(W) - 2. I have tried the commend conv(fft(sine_big_T), fft(W), 'same'), but the result is still much different from fft(sine_big_T.*W).
T = 128;
big_T = 8*T;
small_T = T/8;
sine_big_T = zeros(1,129);
sine_small_T = zeros(1,129);
W = zeros(1,129);
for i = 0:T
sine_big_T(1, i+1) = sin(2*pi/big_T*i);
W(1, i + 1) = 1 - cos(2*pi/T * i);
end
figure
plot(1:129,fft(sine_big_T.*W));
I_fft = fft(sine_big_T);
W_fft = fft(W);
test = conv(I_fft, W_fft,'same');
figure
plot(1:length(I_fft), test)
From the theorem, the two plots should look same. But the result is not even close. I think the way I use conv is not correct. What is the right way to verify the theorem?
Using conv with 'same' is correct. You are seeing two things:
fft defines the origin in the first array element, not the middle of the domain. This does not play well with conv. Use this instead:
test = ifftshift( conv( fftshift(I_fft), fftshift(W_fft), 'same' ) );
The fftshift function shifts the origin to the middle of the array, where it is good for conv with 'same', and ifftshift shifts the origin back to the first element.
Normalization. The FFT is typically normalized so that multiplication in the frequency domain is convolution in the spatial domain. Since you are computing convolution in the frequency domain, the normalization is off. To correct the normalization, plot
plot(1:129,fft(sine_big_T.*W)*length(W));
This is my first using xcorr and I don't seem to understand it. I have a signal signalSubset within which I'm searching for another signal signalWhole. The search signal (signalSubset) is known to be within the signal being search through (signalWhole). Here's the code:
% signalSubset really is a contained within signalWhole
areSame = isequal(signalSubset, signalWhole(1 : length(signalSubset))); % true
% same length signals - finds best delay (which is zero)
[acor,lag] = xcorr(signalSubset, signalWhole(1 : length(signalSubset)));
[~,I] = max(abs(acor));
lagDiff = lag(I);
% different length signals - totally different delay of -1518
[acor,lag] = xcorr(signalSubset, signalWhole);
[~,I] = max(abs(acor));
lagDiff = lag(I);
Why is it when the signals have different lengths during xcorr that the index I is -1518 instead 0 as expected?
xcorr is quite a messy function - whenever I want to calculate correlation's, I prefer to use either SPSS or R.
Nevertheless, in order to use xcorr correctly, you need to pre-align the size of your signals beforehand. The function does not do it for you.
This alignment is just seing which is the shortest signal and add 0's to the start and end of the signal.
Then yes, you can calculate xcorr correctly.
Code:
if length(dist_RF)<length(dist_LF)
diferenca_tamanho=length(dist_LF)-length(dist_RF);
extra_zeros=floor(diferenca_tamanho/2);
quociente=mod(diferenca_tamanho,2);
% Adicionar zeros inicio
temp=dist_RF;
dist_RF=zeros((length(dist_RF)+2*extra_zeros+quociente),1);
dist_RF((extra_zeros+quociente+1):(length(temp)+(extra_zeros+quociente+1))-1,:)=temp;
elseif length(dist_RF)>length(dist_LF)
diferenca_tamanho=length(dist_RF)-length(dist_LF);
extra_zeros=floor(diferenca_tamanho/2);
quociente=mod(diferenca_tamanho,2);
% Adicionar zeros inicio e fim
temp=dist_LF;
dist_LF=zeros((length(dist_LF)+2*extra_zeros+quociente),1);
dist_LF((extra_zeros+quociente+1):(length(temp)+(extra_zeros+quociente+1))-1,:)=temp;
end
[correl,lags]=xcorr(dist_RF, dist_LF, 'coeff');
if (max(correl)>abs(min(correl)))
correlation_coef=max(correl);
else
correlation_coef=min(correl);
end
delay_signals=find(correlation_coef==correl);
delay_signals=lags(delay_signals);
cross_correlation_value=correlation_coef;
I have a dataset comprising of 30 independent variables and I tried performing linear regression in MATLAB R2010b using the regress function.
I get a warning stating that my matrix X is rank deficient to within machine precision.
Now, the coefficients I get after executing this function don't match with the experimental one.
Can anyone please suggest me how to perform the regression analysis for this equation which is comprising of 30 variables?
Going with our discussion, the reason why you are getting that warning is because you have what is known as an underdetermined system. Basically, you have a set of constraints where you have more variables that you want to solve for than the data that is available. One example of an underdetermined system is something like:
x + y + z = 1
x + y + 2z = 3
There are an infinite number of combinations of (x,y,z) that can solve the above system. For example, (x, y, z) = (1, −2, 2), (2, −3, 2), and (3, −4, 2). What rank deficient means in your case is that there is more than one set of regression coefficients that would satisfy the governing equation that would describe the relationship between your input variables and output observations. This is probably why the output of regress isn't matching up with your ground truth regression coefficients. Though it isn't the same answer, do know that the output is one possible answer. By running through regress with your data, this is what I get if I define your observation matrix to be X and your output vector to be Y:
>> format long g;
>> B = regress(Y, X);
>> B
B =
0
0
28321.7264417536
0
35241.9719076362
899.386999172398
-95491.6154990829
-2879.96318251964
-31375.7038251919
5993.52959752106
0
18312.6649115112
0
0
8031.4391233753
27923.2569044728
7716.51932560781
-13621.1638587172
36721.8387047613
80622.0849069525
-114048.707780113
-70838.6034825939
-22843.7931997405
5345.06937207617
0
106542.307940305
-14178.0346010715
-20506.8096166108
-2498.51437396558
6783.3107243113
You can see that there are seven regression coefficients that are equal to 0, which corresponds to 30 - 23 = 7. We have 30 variables and 23 constraints to work with. Be advised that this is not the only possible solution. regress essentially computes the least squared error solution such that sum of residuals of Y - X*B has the least amount of error. This essentially simplifies to:
B = X^(*)*Y
X^(*) is what is known as the pseudo-inverse of the matrix. MATLAB has this available, and it is called pinv. Therefore, if we did:
B = pinv(X)*Y
We get:
B =
44741.6923363563
32972.479220139
-31055.2846404536
-22897.9685877566
28888.7558524005
1146.70695371731
-4002.86163441217
9161.6908044046
-22704.9986509788
5526.10730457192
9161.69080479427
2607.08283489226
2591.21062004404
-31631.9969765197
-5357.85253691504
6025.47661106009
5519.89341411127
-7356.00479046122
-15411.5144034056
49827.6984426955
-26352.0537850382
-11144.2988973666
-14835.9087945295
-121.889618144655
-32355.2405829636
53712.1245333841
-1941.40823106236
-10929.3953469692
-3817.40117809984
2732.64066796307
You see that there are no zero coefficients because pinv finds the solution using the L2-norm, which promotes the "spreading" out of the errors (for a lack of a better term). You can verify that these are correct regression coefficients by doing:
>> Y2 = X*B
Y2 =
16.1491563400241
16.1264219600856
16.525331600049
17.3170318001845
16.7481541301999
17.3266932502295
16.5465094100486
16.5184456100487
16.8428701100165
17.0749421099829
16.7393450000517
17.2993993099419
17.3925811702017
17.3347117202356
17.3362798302375
17.3184486799219
17.1169638102517
17.2813552099096
16.8792925100727
17.2557945601102
17.501873690151
17.6490477001922
17.7733493802508
Similarly, if we used the regression coefficients from regress, so B = regress(Y,X); then doing Y2 = X*B, we get:
Y2 =
16.1491563399927
16.1264219599996
16.5253315999987
17.3170317999969
16.7481541299967
17.3266932499992
16.5465094099978
16.5184456099983
16.8428701099975
17.0749421099985
16.7393449999981
17.2993993099983
17.3925811699993
17.3347117199991
17.3362798299967
17.3184486799987
17.1169638100025
17.281355209999
16.8792925099983
17.2557945599979
17.5018736899983
17.6490476999977
17.7733493799981
There are some slight computational differences, which is to be expected. Similarly, we can also find the answer by using mldivide:
B = X \ Y
B =
0
0
28321.726441712
0
35241.9719075889
899.386999170666
-95491.6154989513
-2879.96318251572
-31375.7038251485
5993.52959751295
0
18312.6649114859
0
0
8031.43912336425
27923.2569044349
7716.51932559712
-13621.1638586983
36721.8387047123
80622.0849068411
-114048.707779954
-70838.6034824987
-22843.7931997086
5345.06937206919
0
106542.307940158
-14178.0346010521
-20506.8096165825
-2498.51437396236
6783.31072430201
You can see that this curiously matches up with what regress gives you. That's because \ is a more smarter operator. Depending on how your matrix is structured, it finds the solution to the system by a different method. I'd like to defer you to the post by Amro that talks about what algorithms mldivide uses when examining the properties of the input matrix being operated on:
How to implement Matlab's mldivide (a.k.a. the backslash operator "\")
What you should take away from this answer is that you can certainly go ahead and use those regression coefficients and they will more or less give you the expected output for each value of Y with each set of inputs for X. However, be warned that those coefficients are not unique. This is apparent as you said that you have ground truth coefficients that don't match up with the output of regress. It isn't matching up because it generated another answer that satisfies the constraints you have provided.
There is more than one answer that can describe that relationship if you have an underdetermined system, as you have seen by my experiments shown above.
I am having a hard time understanding what should be a simple concept. I have constructed a vector in MATLAB that is real and symmetric. When I take the FFT in MATLAB, the result has a significant imaginary component, even though the symmetry rules of the Fourier transform say that the FT of a real symmetric function should also be real and symmetric. My example code:
N = 1 + 2^8;
k = linspace(-1,1,N);
V = exp(-abs(k));
Vf1 = fft(fftshift(V));
Vf2 = fft(ifftshift(V));
Vf3 = ifft(fftshift(V));
Vf4 = ifft(ifftshift(V));
Vf5 = fft(V);
Vf6 = ifft(V);
disp([isreal(Vf1) isreal(Vf2) isreal(Vf3) isreal(Vf4) isreal(Vf5) isreal(Vf6)])
Result:
0 0 0 0 0 0
No combinations of (i)fft or (i)fftshift result in a real symmetric vector. I've tried with both even and odd N (N = 2^8 vs. N = 1+2^8).
I did try looking at k+flip(k) and there are some residuals on the order of eps(1), but the residuals are also symmetric and the imaginary part of the FFT is not coming out as fuzz on the order of eps(1), but rather with magnitude comparable to the real part.
What blindingly obvious thing am I missing?
Blindingly obvious thing I was missing:
The FFT is not an integral over all space, so it assumes a periodic signal. Above, I am duplicating the last point in the period when I choose an even N, and so there is no way to shift it around to put the zero frequency at the beginning without fractional indexing, which does not exist.
A word about my choice of k. It is not arbitrary. The actual problem I am trying to solve is to generate a model FTIR interferogram which I will then FFT to get a spectrum. k is the distance that the interferometer travels which gets transformed to frequency in wavenumbers. In the real problem there will be various scaling factors so that the generating function V will yield physically meaningful numbers.
It's
Vf = fftshift(fft(ifftshift(V)));
That is, you need ifftshift in time-domain so that samples are interpreted as those of a symmetric function, and then fftshift in frequency-domain to again make symmetry apparent.
This only works for N odd. For N even, the concept of a symmetric function does not make sense: there is no way to shift the signal so that it is symmetric with respect to the origin (the origin would need to be "between two samples", which is impossible).
For your example V, the above code gives Vf real and symmetric. The following figure has been generated with semilogy(Vf), so that small as well as large values can be seen. (Of course, you could modify the horizontal axis so that the graph is centered at 0 frequency as it should; but anyway the graph is seen to be symmetric.)
#Yvon is absolutely right with his comment about symmetry. Your input signal looks symmetrical, but it isn't because symmetry is related to origin 0.
Using linspace in Matlab for constructing signals is mostly a bad choice.
Trying to repair the results with fftshift is a bad idea too.
Use instead:
k = 2*(0:N-1)/N - 1;
and you will get the result you expect.
However the imaginary part of the transformed values will not be perfectly zero.
There is some numerical noise.
>> max(abs(imag(Vf5)))
ans =
2.5535e-15
Answer to Yvon's question:
Why? >> N = 1+2^4 N = 17 >> x=linspace(-1,1,N) x = -1.0000 -0.8750 -0.7500 -0.6250 -0.5000 -0.3750 -0.2500 -0.1250 0 0.1250 0.2500 0.3750 0.5000 0.6250 0.7500 0.8750 1.0000 >> y=2*(0:N-1)/N-1 y = -1.0000 -0.8824 -0.7647 -0.6471 -0.5294 -0.4118 -0.2941 -0.1765 -0.0588 0.0588 0.1765 0.2941 0.4118 0.5294 0.6471 0.7647 0.8824 – Yvon 1
Your example is not a symmetric (even) function, but an antisymmetric (odd) function. However, this makes no difference.
For a antisymmetric function of length N the following statement is true:
f[i] == -f[-i] == -f[N-i]
The index i runs from 0 to N-1.
Let us see was happens with i=2. Remember, count starts with 0 and ends with 16.
x[2] = -0.75
-x[N-2] == -x[17-2] == -x[15] = (-1) 0.875 = -0.875
x[2] != -x[N-2]
y[2] = -0.7647
-y[N-2] == -y[15] = (-1) 0.7647
y[2] == y[N-2]
The problem is, that the origin of Matlab vectors start at 1.
Modulo (periodic) vectors start with origin 0.
This difference leads to many misunderstandings.
Another way of explanation why linspace(-1,+1,N) is not correct:
Imagine you have a vector which holds a single period of a periodic function,
for instance a Cosinus function. This single period is one of a infinite number of periods.
The first value of your Cosinus vector must not be same as the last value of your vector.
However,that is exactly what linspace(-1,+1,N) does.
Doing so, results in a sequence where the last value of period 1 is the same value as the first sample of the following period 2. That is not what you want.
To avoid this mistake use t = 2*(0:N-1)/N - 1. The distance t[i+1]-t[i] is 2/N and the last value has to be t[N-1] = 1 - 2/N and not 1.
Answer to Yvon's second comment
Whatever you put in an input vector of a DFT/FFT, by theory it is interpreted as a periodic function.
But that is not the point.
DFT performs an integration.
fft(m) = Sum_(k=0)^(N-1) (x(k) exp(-i 2 pi m k/N )
The first value x(k=0) describes the amplitude of the first integration interval of length 1/N. The second value x(k=1) describes the amplitude of the second integration interval of length 1/N. And so on.
The very last integration interval of the symmetric function ends with same value as the first sample. This means, the starting point of the last integration interval is k=N-1 = 1-1/N. Your input vector holds the starting points of the integration intervals.
Therefore, the last point of symmetry k=N is a point of the function, but it is not a starting point of an integration interval and so it is not a member of the input vector.
You have a problem when implementing the concept "symmetry". A purely real, even (or "symmetric") function has a Fourier transform function that is also real and even. "Even" is the symmetry with respect to the y-axis, or the t=0 line.
When implementing a signal in Matlab, however, you always start from t=0. That is, there is no way to "define" the signal starting from before the origin of time.
Searching the Internet for a while lead me to this -
Correct use of fftshift and ifftshift at input to fft and ifft.
As Luis has pointed out, you need to perform ifftshift before feeding the signal into fft. The reason has never been documented in Matlab, but only in that thread. For historical reasons, outputs AND inputs of fft and ifft are swapped. That is, instead of ordered from -N/2 to N/2-1 (the natural order), the signal in time or frequency domain is ordered from 0 to N/2-1 and then -N/2 to -1. That means, the correct way to code is fft( ifftshift(V) ), but most people ignore this at most times. Why it's got silently ignored rather than raising huge problems is that most concerns have been put on the amplitude of signal, not phase. Since circular shifting does not affect amplitude spectrum, this is not a problem (even for the Matlab guys who have written the documentations).
To check the amplitude equality -
Vf2 = fft(ifftshift(V));
Vf5 = fft(V);
Va2 = abs(fftshift(Vf2));
Va5 = abs(fftshift(Vf5));
>> min(abs(Va2-Va5)<1e-10)
ans =
1
To see how badly wrong in phase -
Vp2 = angle(fftshift(Vf2));
Vp5 = angle(fftshift(Vf5));
Anyway, as I wrote in the comment, after copy&pasting your code into a fresh and clean Matlab, it gives 0 1 0 1 0 0.
To your question about N=even and N=odd, my opinion is when N=even, the signal is not symmetric, since there are unequal number of points on either side of the time origin.
Just add the following line after "k = linspace(-1,1,N);"
k(end)=[];
it will remove the last element of the array. This is defined to be symmetric array.
also consider that isreal(complex(1,0)) is false!!!
The isreal function just checks for the memory storage format. so 1+0i is not real in the above example.
You have define your function in order to check for real numbers (like this)
myisreal=#(x) all((abs(imag(x))<1e-6*abs(real(x)))|(abs(x)<1e-8));
Finally your source code should become something like this:
N = 1 + 2^8;
k = linspace(-1,1,N);
k(end)=[];
V = exp(-abs(k));
Vf1 = fft(fftshift(V));
Vf2 = fft(ifftshift(V));
Vf3 = ifft(fftshift(V));
Vf4 = ifft(ifftshift(V));
Vf5 = fft(V);
Vf6 = ifft(V);
myisreal=#(x) all((abs(imag(x))<1e-6*abs(real(x)))|(abs(x)<1e-8));
disp([myisreal(Vf1) myisreal(Vf2) myisreal(Vf3) myisreal(Vf4) myisreal(Vf5) myisreal(Vf6)]);
My objective is to classify non-speech signal for which I am using mfcc and dtw in java. However I am stuck in middle. I would appreciate any help.
I have evaluated 13 mfcc values for each frame however some values are negative, I am confused whether the process I am following is right or wrong. Currently I am using the code provided by JAudio. I have also tried other code, they give me negative values as well.
Secondly, I get 13 coefficients for each frame, considering 157 frames for a certain length of sample, I get 157 sets of 13 mfccs. I am having hard time how to use all the coefficients in DTW because dtw only gives closest distance between two time signals. I do have code of DTW to compare two time signals. I am not sure how to use all the mfccs values of the signal as features.
Is there some crucial step of classification I am missing? Please help me.
The use of DTW suppose to verify 2 audio sequences in your case. Thus, for the sequence to be verify you will have a matrix M1xN and for the query M2xN. This implies that your cost matrix will have M1xM2.
To construct the cost matrix you have to apply a distance/cost measure between the sequences, as cost(i,j) = your_chosen_multidimension_metric(M1[i,:],M2[j,:])
The resulted cost matrix will be 2D, and you could apply easily DTW.
I made a similar code for DTW based on MFCC. Below is the Python implementation which returs DTW score; x and y are the MFCC matrix of voice sequences, with M1xN and M2xN dimensions:
def my_dtw (x, y):
cost_matrix = cdist(x, y,metric='seuclidean')
m,n = np.shape(cost_matrix)
for i in range(m):
for j in range(n):
if ((i==0) & (j==0)):
cost_matrix[i,j] = cost_matrix[i,j]
elif (i==0):
cost_matrix[i,j] = cost_matrix[i,j] + cost_matrix[i,j-1]
elif (j==0):
cost_matrix[i,j] = cost_matrix[i,j] + cost_matrix[i-1,j]
else:
min_local_dist = cost_matrix[i-1,j]
if min_local_dist > cost_matrix[i,j-1]:
min_local_dist = cost_matrix[i,j-1]
if min_local_dist > cost_matrix[i-1,j-1]:
min_local_dist = cost_matrix[i-1,j-1]
cost_matrix[i,j] = cost_matrix[i,j] + min_local_dist
return cost_matrix[m-1,n-1]
Say you have N1 sets of 13 MFCCs each for the first signal and N2 sets of MFCCs for the second.
You should compute the distance between each set in from the first signal and each set from the second (you can use the Euclidian Distance for the distance between two 13-sized arrays)
This would leave you with an N1xN2 bidimensional array on which you should now apply the DTW.
Check out: http://code.google.com/p/aquila/
Specifically: http://code.google.com/p/aquila/source/browse/trunk/examples/dtw_distance/main.cpp which has an example codeof dtw distace calculation.