fft and discrete functions

Hi,

I have a vector that contains the time (epoch) of an occurrence (many occurrences per second, for 300 seconds), and I want to determine the frequency of these occurrences, that is, plot its spectrum.

What I have done so far is counted the number of occurrences per second, but its FFT does not give me what I want.

I think that what I really need is a "signal" that is 1 at the times specified in the vector and 0 otherwise (such as stem). The FFT of that signal will give me what I want. But how can I create such a signal? I know the stem function only plots it. How can I have the FFT of that "stem" signal?

Answer 1

[Star Strider]

You can probably do what you want with a combination of thresholding and logical indexing. The Signal Processing Toolbox findpeaks function could be helpful, and would be what I would use first.

I don’t know what your data are or what counts as an ‘occurrence’, so I cannot be more specific.

[Arquelau]

Hi Star Strider,

I dont have the Signal Processing Toolbox, so unfortunately that is not an option for me.

The 'occurrence' is a packet being transmitted or received in my network. I have a file that specifies the time in epoch that a packet is received or transmitted by my network interface. This file has 2 million lines, but a few lines of it are:

1482505376.158728000
1482505376.158767000
1482505376.159893000
1482505376.160085000
1482505376.160114000
1482505377.297492000
1482505377.297553000
1482505377.297561000
1482505377.556846000
1482505378.056147000

I want to determine the frequency on which packets are being transmitted or received by me. Can you help me with the standard MATLAB library?

Many thanks!!

[Star Strider]

My pleasure!

See if this does what you want:

PktTimes = [1482505376.158728000
            1482505376.158767000
            1482505376.159893000
            1482505376.160085000
            1482505376.160114000
            1482505377.297492000
            1482505377.297553000
            1482505377.297561000
            1482505377.556846000
            1482505378.056147000];

DPktTimes = diff(PktTimes);                                 % Differences In Packet Times
MinDPktTimes = min(DPktTimes);                              % Minimum Difference (Becomes Sampling Interval)
RngPktTimes = PktTimes(end)-PktTimes(1);                    % Range Of Packet Times

StartFromZero = PktTimes - PktTimes(1);                     % Set Initial Packet Time To Sero
Intvls = StartFromZero / MinDPktTimes;                      % Calculate Packet Times In Terms Of Intervals
OccVct([1; round(Intvls(2:end))]) = 1;                      % Create Vector Of Occurrences
T = (0:length(OccVct)-1)*MinDPktTimes;                      % Create Time Vector
L = length(T);                                              % Length Of Vectors

figure(1)
stem(T, OccVct)
grid
title('Packet Occurrences & Times')
xlabel('Time')

figure(2)
hist(StartFromZero, 100)
grid
xlabel('Time')
ylabel('Number of Occurences Per Interval Range')

Fs = 1/MinDPktTimes;                                        % Sampling Frequency
Fn = Fs/2;                                                  % Nyquist Frequency
FT_OV = fft(OccVct)/L;                                      % Fourier Transform Of ‘Occurrence Vector’
Fv = linspace(0, 1, fix(L/2)+1)*Fn;                         % Frequency Vector
Iv = 1:length(Fv);                                          % Index Vector

figure(3)
plot(Fv, abs(FT_OV(Iv))*2)
grid
title('Fourier Transform Of Occurrence Vector')
xlabel('Frequency')
ylabel('Magnitude')

[Arquelau]

That is absolutely perfect! I appreciate very much your help! Just one last thing: When running on the actual file, I get an error at the line

OccVct([1; round(Intvls(2:end))]) = 1;

The error says

Subscript indices must either be real positive integers or logicals.

Turns out that when running in the actual file,

round(Intvls(2:end))

happens to be

Inf

I tried to understand why, but I couldnt. I did noticed that its value increases really fast as more packet times are added to the vector. Do you know how to solve it?

EDIT: Oh well, I realized now that

MinDPktTimes

is 0 (because there are 2 packets transmitted at nearly the same time?), which causes the division to result in infinit. Adding 0.0001 to its value solved the problem.

Again, I really appreciate your effort and willingness.

[Star Strider]

My pleasure!

I am not certain where in your vector the Inf value is being calculated. It represents an overflow condition, so it occurs where the packet time is large enough that when divided by the sampling interval, it produces an infinite result. If it occurs at or near the end of the vector, I would replace it with the length of the vector by adding this line just after the ‘Intvls’ calculation assignment:

Intvls(isinf(Intvls)) = max(Intvls(isfinite(Intvls)));

If it occurs earlier in the vector, this will not produce correct data. In that instance, we need to find an alternative solution.

EDIT — If two packets are so close as to cause a nearly infinite result, either this:

MinDPktTimes = min(DPktTimes(DPktTimes > 0));               % Minimum Difference (Becomes Sampling Interval)

or this:

MinDPktTimes = min(DPktTimes(DPktTimes > 1E-8));            % Minimum Difference (Becomes Sampling Interval)

should solve the problem. You may have to tweak the threshold (here 1E-8) to get the result you want.

Another alternative is to sort the ‘DPktTimes’ sampling intervals and choose the first non-zero value as the correct one. With a large vector, this will slow your code.

Answer 2

[Image Analyst]

Why is the mean frequency simply not the count divided by the observation time? If you have 300 thousand in 300 seconds, they occurred at a rate of one thousand per second. What other information do you need?

[Arquelau]

I am trying to determine and identify some patterns in our network, and a simple frequency wouldnt provide us much information for analysis. A frequency spectrum (if applied in some specific packets) will help us identify if and when our netowrk is under attack.

But I appreciate your attention and collaboration on the issue!