Short time fourier transform (STFT) frequency axis

Could someone please explain the frequency axis output of the short-time fourier transform spectrogram? The help menu indicates that the number of output columns will be equal to (# of input samples)/2. I would expect that the number of output frequency bins would be equal to half the size of the fft and that the bin width would be determined by the sample rate.

STFT is implemented using the windowed Fourier transform. Before applying the Fourier transform, the windowed signal can be zero-padded to get the same length as the input signal. So in this case, the FFT length is equal to the signal length. Such zero padding can get a smooth spectrogram for the detailed visual investigation of frequency contents.
NISH DSP
Medical ultrasound and signal processing

Similar Messages

  • Discrete time fourier transform

    Hi all,
    something I need your guidance once again. This time round I'm not sure whether it's wrong about my mathematical calculation towards DTFT or my program is wrong (that has been the case for most of the time).
    I'm supposed to perform a DTFT for a square wave and I had set the parameters to be '1' for the frequency and #s to be '100'.
    After performing the DTFT, the amplitude in the frequency domain is over 60.
    Can someone check for me where had gone wrong?
    Please find attached files for reference.
    Attachments:
    fourier front panel.JPG ‏57 KB

    Hi Gareth1983
    You are on the right track. But I think you need some better understanding of the DFT concept. This and http://zone.ni.com/devzone/cda/tut/p/id/4278, and this http://zone.ni.com/devzone/cda/tut/p/id/4541 will help you out. Remember that in newer labview versions the DFT output is a complex array. In older versions the output was split in a IM and RE part. But you will find the functions you need in the complex palette
    good luck with your school assignment
    Besides which, my opinion is that Express VIs Carthage must be destroyed deleted
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  • Short Time Energy Function

    Has anyone ever come across Short Time Processing methods? I would like to look into Short Time Energy Function, Short Time Average Zero Crossing Rate and Median Smoothing. I don't know a lot about them, but don't want to get too deep in before I find out it's going to be a nightmare to work on! I'm not sure if the STFT Spectogram.vi will help me, or whether that's something different. Thanks.

    Hello,
    The Short-Time Fourier Transform (STFT) allows you to analyze a signal in the time and frequency domain simultaneously, which allows you to better understand and process a signal. The STFT Spectogram.vi is available within the Signal Processing Toolset. For a better understanding of STFT or Joint Time Frequency Analysis (JTFA) take a look at the following knowledge bases:
    http://zone.ni.com/devzone/conceptd.nsf/webmain/0A3D35A85FC7D2EA862568D2004A966F?opendocument
    http://zone.ni.com/devzone/conceptd.nsf/webmain/53EB764F16B52B1C862568F8004EEA84?opendocument
    The Signal Processing Toolset User Manual (http://digital.ni.com/manuals.nsf/websearch/C355A4DB4B3DFAE086256A6A006A9170?OpenDocument&node=132100_US) is also very helpful.
    Regards,
    Jy
    oti F
    National Instruments

  • TFA STFT Spectrogram: log scale on frequency axis

    Hi,
    I'm trying to get my spectrogram to display a logarithmic scale on the frequency axis, but whenever I go to the graph properties->y-axis, turn off autoscaling, check the log scale box, hit okay, and then run the vi, I still have a linear scale.  I tried creating a property node for the spectrogram's y-axis and setting the scale to logarithmic, but this doesn't work either.  Does anyone have any suggestions? 
    Thanks a lot,
    Chris K.
    Attachments:
    Front Panel.png ‏267 KB

    log (0), log (1), log (2), log(3), ... ?
    The Intensity graph uses the array indices (0,1,2,3,...) for the axis, not the values of Y.  If your frequencies are 1, 2, 5, 10, 20, 50, 100, 200, 500, 1000, ..., you would need an array with >=1001 elements to get the axis values where you wanted them and it would still plot the axis linearly.  You can certainly put the Z-values for frequency 10 in the 4th column of the array.  To get the labels to display the Y-axis frequencies might require overlaying an X-Y graph with logarithmic mapping of the Y-axis on the Intensity graph.  By making the appropriate parts transparent, you should be able to get the Y-axis display you want.  Tedious, but probably not too difficult.
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  • Why not use Fourier transforms for filtering?

    Are there any resources for constructing filters using discrete Fourier transforms (DFTs)? Or is it as easy as it seems. For example, for lowpass filtering I obtain the DFT of my signal, zero out components above the selected cutoff frequency and then transform back to get my signal minus the high frequency (noise) components. It seems to behave correctly for the signal I've applied it to, but am I missing something. Upsides would appear to be (1) no time delay between the input and output and (2) the filter is sharp with no transition band. Downside is that this method is slower than using impulse filters, but are there other upsides, downsides, or considerations I'm missing?

    The technique of filtering using frequency domain truncation and IFFT is used in specific areas where computation time is essential (to avoid the time domain convolution with a long impulse response), like image processing. However the technique also has some downsides you should be aware of.
    The frequency domain filtering performed on a data record is totally equivalent to a circular convolution made on your time signal with the impulse response of your ideal low-pass filter (typically a sin(x)/x function). The fact what we are doing a circular convolution will result in unwanted artifacts and you'll see some "parts" of the beginning of your signal at the end of the filtered signal and vice-versa.
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    se) and extend it with zeros. After filtering you will see a sort of "pre-shoot" of your original signal at the very end of your signal. If you then rotate your time signal you'll discover that the signal is actually continuous (in a circular way).
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  • After fourier transform

    Hi all, A simple question do ask. I have this input signal which is being generated by Signal Generator by Duration.vi. I got another output signal generated after applying the Fourier transform to the signal above.
    I believe the signal generated by the first vi is to be a signal in the time domain, and the signal after applying the fourier transform, I get a waveform in the frequency domain. Is there any way where I can view the waveform under the same waveform graph?
    Thanks a lot in advance,
    Gareth

    Hello Gareth,
    Have you taken a look at this thread yet? It is about showing two waveforms on the same waveform graph. Briefly, you will need to use the Build Array function to combine the two data signals together (e.g. if you are using ExpressVIs, combine the two Dynamic Data Type outputs) using the Build Array function, and feed it into the graph. Some properties of the waveform graph needs to be modified to show separate axes for each signal.
    While this is technically possible, it might not be a very feasible idea as the x-axis will be common to both waveforms. Thus, you might not be able to view both waveforms clearly at one go. I have attached an example in this post. "Bothsignals.jpg" show both the time domain and FFT signals together in the same graph. However, if we zoom into the green area, we can see the actual time domain signal ("timedomain.jpg"). The time duration is about 100ms. But for the FFT, the x-axis should represent frequency (Hz), and is centered at 50Hz, which is much bigger than 100milli.
    Thus, I think it might be a better idea to separate the two. But this is a personal view.
    Best regards,
    Victor
    AE, NI ASEAN
    Attachments:
    bothsignals.jpg ‏148 KB
    timedomain.jpg ‏163 KB

  • Surface graph of a Fourier Transform

    Hi there -
    I want to make a surface graph of my Fourier Transform. My z-axis should be the amplitude coefficients, My y-axis should be Time and my x-axis should be Frequency. However, when I use the FFT vi I only get a z matrix of one column. Yet the surface graph VI requires a 2D array for z. How can I account for this?
    Thanks
    Cameron

    Duplicate post
    See http://forums.ni.com/ni/board/message?board.id=170&message.id=193793&jump=true

  • How to create fourier transform properties using labview

    Hi everyone
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    PLS HELP!!!
    Thanks

    i am refering to the fourier transform properties listed below:
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    x(t-a) ------- >  exp (-jwa) X(w) 
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  • Continuous Wavelet Transforma​tion frequency output in columns and rows?

    Hi
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    Best Regards
    Jesper

    Your current code can be replaced by an autoindexing FOR loop, you only need once instance of the "add array elements" operation. Easiest would be to acquire it as a 1D array with a constant number of samples then use "reshape array" to create a 2D array of the desired dimensions to be used for autoindexing as suggested.
    LabVIEW Champion . Do more with less code and in less time .

  • HOW TO SCALE THE CORRECT FREQUENCY AXIS?

    hello guys
    im trying to calculate the real cepstrum of a short speech signal for time/pitch scaling. cepstrum=ifft[log(abs ((fft))].
    my problem is 1) how can i correctly show the frequency axis (x-axis) of the fft result in Hz and not in time (samples) and then  2) the result of the cepstrum in time?
    i cant seem to be able to do that as you can see from my program below.
    can anybody help me please?
    kind regards
    alex
    p.s. the zip file contains my program with a sample sound file that can be processed with
    Attachments:
    TD-PSOLA ALGO.vi.zip ‏494 KB

    Yes I used the 512 samples.
    You can see the DF if you set the graph to a plot-style, then zoom in to a 0-500 Hz scale then you will see that at every 83 Hz a point.
     The reason your maximum is not half dF is that you have 512 samples, so 256 frequency bins times 86 =22016 BUT you start at zero so the max is a little lower
    255*86.13 is 21963 Hz
    About CSepstrum I don't know, I hardly ever used IFFT.
    Ton
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  • 2D Fourier Transform of an image to remove grating

    I'm trying to remove a grating that has been added to my image:
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    Unfortunately, I am failing at the first step: transforming it to the Fourier domain. This is my code:
    When I try to run it I get this error.
    Anyone have any ideas to fix this?
    Attachments:
    Fouriertransform.vi ‏12 KB

    Hi Choisai,
    I found a Knowledge Base that possibly will help you, check the following link: http://digital.ni.com/public.nsf/allkb/12039EFD213​AD08086257B6E0030FDC6?OpenDocument
    One more thing.. I do not see anywhere assigned path to get the image, which in future will throw an error for sure.
    Best,
    Bozhidar

  • Fourier Transform and the DAQ Assistant

    Hi,
    I'm trying to take the fourier transform of a analog voltage measurement in Labview 8.6 and I'm having a problem with the data.
    The measurement is an anaolog input voltage signal taken using the USB-6211 box.  I am acquiring the signal on ports 15 and 16 (differential) with the DAQ Assistant at a rate of 5KHz, 200 samples.  The setup works perfect thankfully, and a very nice waveform plot is generated on the VI.  I would now like to add a fourier transform of this plot to do some frequency analysis.
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    Hi
    Your thinking is not bad. You have only some scaling problems. You may toy with the vi I am sending. Change the log/linear control and see the result.
    You will find all information you need here http://zone.ni.com/devzone/cda/tut/p/id/4541 and http://zone.ni.com/devzone/cda/tut/p/id/4541. The dt value is the reciprocal of the sample frequency(1/x) The unit is second, and it will give you the time interval between two samples.
    Also note how I use the "Dynamic To Waveform Array.vi" to convert the dynamic data. This VI is not documented by NI, but you find it here ...vi.lib\express\express shared\transition.llb\Dynamic To Waveform Array.vi
    Good luck
    Besides which, my opinion is that Express VIs Carthage must be destroyed deleted
    (Sorry no Labview "brag list" so far)
    Attachments:
    Spectrum Analyzer.vi ‏32 KB

  • Passing data from Microphone to FFT (Fast Fourier Transform) Class

    I am quite new to Java and trying to find a way to use microphone and pass the data from microphone to a FFT Class and get the result back as one number (which will be the frequency). Because I am quite new, I am in need of some advice and description to solve this problem.
    I will be glad if anybody experienced on this issue can help me.
    Regards.

    You need to read a good bit about signal analysis before you tackle this.
    The FFT is an algorithm for the implementation of the Discret Fourier Transform (DFT). If you just perform an FFT of your data how are you then going to extract the frequency. For example, if you take the DFT of 1024 samples you will generate 1024 complex values with the to 512 pairs being complex conjugates of the bottom 512 pairs.
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    OK, so you have applied a window, performed the DFTand found the sample with the largest magnitude then how do you relate this frequency index to the real frequency?
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  • Real Fourier Transformation: confusing phase spectrum

    When setting sample rate and block size both to 1024, the Real Fourier Transformation of a sin-wave with Frequency=1 and PhaseAngel=0 calculates the phase spectrum as displayed in the left part of the image.
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    *) Why does the phase spectrum contain more than one value, when the input signal just contains one single frequency?
    *) Why do the phase spectra (is this the plural of "spectrum"?) look different for different sample rates / block sizes?
    Maybe someone is able to answer these questions
    Thanks!
    Roman

    This is not my area of expertise, so, I went to one of the experts, and he gave me this. I hope that it helps.
    *) Why is the phase of frequency 1 equals -90, when I set the phase angel
    of the input signal to 0?
    My explanation is that the phases are calculated relative to a cos-signal.
    Is that true?
    The important information of the
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    In a typical signal analysis you
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    *) Why do the phase spectra (is this the plural of "spectrum"?)
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    - cj
    Measurement Computing (MCC) has free technical support. Visit www.mccdaq.com and click on the "Support" tab for all support options, including DASYLab.

  • Fourier Transform of an Exponentia​l Function

    ok. this is something which i had tried earlier. I'm supposed to do a fourier transform of this signal.... exp(-alpha*t) where alpha is greater or less than 0.. the upper limit is inf and lower limit is 0. the waveform i got is a straight line, and there's this error -20003 i.e number of samples > 0.
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    Attachments:
    FT for real decaying exponential (eg 3-24) test v02.vi ‏25 KB

    I cannot find your function exp(-alpha*t) anywhere in your VI. -- Where should I define this?
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    What's up with the blue diagram constants. Make them orange to avoid coercions. - May I know how I can do this?
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