EE 464 Short-Time Fourier Transform Fall and Spectrogram. Many signals of importance have spectral content that

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1 EE 464 Short-Time Fourier Transform Fall 2018 Read Text, Chapter 4.9. and Spectrogram Many signals of importance have spectral content that changes with time. Let xx(nn), nn = 0, 1,, NN 1 1 be a discrete-time signal, with sampling rate FF ss = 1 TT samples/sec, with NN 1 very large. The NN 1 -point DFT is defined as NN 1 1 XX(kk) = xx(nn)ee jj2ππππππ NN 1 nn=0, kk = 0,1,, NN 1 1 and with frequency resolution ff = 1 = FF ss Hz, and with NN 1 TT NN 1 XX ee jj2ππππππ = nn= xx(nn)ee jj2ππππππππ the discrete-time Fourier transform (DTFT) of xx(nn). The DFT is simply samples of the DTFT at ff = kk, kk = 0,1,, NN NN 1 TT 1 1 (provided xx(nn) is duration-limited to range [0, NN 1 TT] sec. Parseval s relation says that, in general, EE xx = 1 2TT xx(nn) 2 = TT XX ee jj2ππππππ 2 dddd nn= 1 2TT 1

2 and, specialized the DFT relationship, NN 1 1 xx(nn) 2 = 1 NN 1 nn=0 NN 1 1 XX(kk) 2. nn=0 The power in the signal is defined as NN 1 1 PP = 1 xx(nn) 2. NN 1 nn=0 Define the power density spectrum as SS XX (kk) = XX(kk) 2. NN 1 Then Parseval s relation says NN 1 1 PP = 1 NN 1 xx(nn) 2 nn=0 NN 1 1 = 1 NN 1 SS XX (kk). nn=0 Up to a scale constant, each value of XX(kk) 2 then represents the average signal power over the duration NN 1 TT sec in the frequency band of width ff around the frequency kk Hz. If NN 1 TT the duration NN 1 TT is large, then this is the average over a large time duration. 2

3 For example, consider a musical score. The score is divided into measures, each measure a collection of notes. A long score has many measures. We could count the number of occurrences of each note over a measure, multiple measures, or the entire score, and determine a note density spectrum. The frequencies are discrete. For one or a few measures such a note density spectrum reflects the relative frequency of occurrence of the different notes (sinusoidal frequencies) over the length of time corresponding to those measures. Taken over the entire musical score, the note density reflects the occurrence of notes on average over the entire piece of music. 3

4 The short-time Fourier transform (STFT) is used to determine the sinusoidal frequency content over local segments of time. This is done by partitioning the signal into segments (using a window) and then computing the DFT of each (windowed) segment. The segmentation can be thought of as multiplying the signal by a window function. xx mm (nn) = xx(nn)ww mm (nn) where ww mm (nn) = ww(nn (mm 1)NN), with ww(nn) a window function used for each segment. The windowing can be non-overlapping Or, more commonly, overlapping Let LL denote the length of a segment (including overlap, if present). Then the LL-point DFT is computed for each segment, resulting in XX mm (kk), kk = 0,1, LL 1, for mm = 1,2,. Typically, XX mm (kk), kk = 0,1, is studied, as m varies. 4

5 The STFT is typically presented as a 2-d or 3-d plot. The segment time duration is = NNNN sec, and the frequency resolution of the DFT is ff = 1 Hz, where and LL-point DFT is computed, and LL NN is the total number of samples overlapped. LLLL Spectrogram this is a 2-d plot of XX mm (kk) where kk varies along one dimension and mm along the other (e.g., in the figure above, kk varies vertically). The value of XX mm (kk) is usually denoted as a color or grey level. A 3-d plot is also possible, where the viewing perspective can be adjusted to better visualize the signal information. 5

6 Matlab has a function spectrogram(xx, NNNN, Overlap, ), where xx = signal to be analyzed NN ww is either a positive integer (analysis segment length) or the explicit window to be used (a vector); Overlap is the fraction of the length NN ww of window overlap at each end of the analysis segment (e.g., Overlap = 0.25 means 25% overlap at each end). In an integer is used for the window length, then the default is to use the Hamming window. Use the command >> help spectrogram to get further documentation. 6

7 Example 1. Consider a short audio signal consisting of the musical notes C, D, E, C, E, C, E. The notes have defined frequencies CC 4 = Hz, DD 4 = Hz, and EE 4 = Hz. The notes correspond to the beginning of the familiar song doe-ray-me. The audio file has sampling frequency FF ss = 16,000 samples/sec and the file length is NN TT = 96,008 samples. The figure below shows the first 2,000 samples of the audio file. This is not very informative, since the signal is simply a sinusoid of a fixed frequency over significant fractions of a second. 7

8 The DFT can be computed for the entire signal (96,008 samples) and the DFT magnitude is plotted below. The frequency resolution is ff = 1 = 16,000 = Hz. NN TT TT 96,008 Note that the three frequencies are clustered below 500 Hz, and some magnification of the frequency axis is necessary to clearly identify the individual frequencies. This is shown in the next figure. 8

9 Now the individual spectral peaks can be identified, and they provide a close match to the ideal musical scale frequencies listed above. Note: The signal generated used equal amplitude for each note. Why do the peak DFT magnitudes differ? The DFT magnitude plots above give information about the spectral content of the entire audio signal (of about 6 seconds duration). The frequency resolution is very fine (about Hz). However, there is not readily discernable information about the time-varying spectral nature of the signal. 9

10 A spectrogram is computed using an analysis segment size of 20 ms (corresponding to 320 samples), 25% window overlap, and the Hamming window, the figure below shows the 1 spectrogram. There are thus = 50 analysis segments per 20 ms second. The spectral resolution is 1 (20 ms) Hz. Modifying only the analysis window size to 0.5 sec, the spectrogram becomes, with resolution Hz. 10

11 Now, repeat, but using a rectangular window with an analysis segment size of 20 ms, 25% window overlap, yielding the figure below. There are thus 50 analysis segments per second and the spectral resolution is 1 (20 ms) Hz. Modifying only the analysis window size to 0.5 sec, the resolution is Hz, the spectrogram becomes 11

12 Recall three windows used in spectral analysis: rectangular, Hamming, and Blackman. Time window, ww(nn). (Note: DFT even.) Window Frequency Response Magnitude (Normalized T=1) 12

13 Example 2. Consider the narrow-band speech signal (FF ss = 8,000 samples/s, band-limited to about 3.5 khz) of length 18,000 samples shown below, together with its DFT magnitude. 13

14 Using an analysis segment size of 20 ms (corresponding to 160 samples) and 25% window overlap, the spectrograms using rectangular and Hamming windows are shown below. Note the decrease in spectral leakage, but broader spectral peaks, for the Hamming window case. 14

15 Using the Blackman window (same analysis segment size and overlap) results in the spectrogram below. Note the even broader spectral peaks. 15

16 As an alternative to the time-frequency spectrogram image presentation, the STFT information can also be presented as a 3-d plot, as shown below. For comparison, the corresponding spectrogram is shown below. 16

17 Example 3. Higher Frequency Signal Swainson s Thrush (FF ss = 22,050 samples/sec, 302,504 samples) The DFT is shown below. The frequency content of the signal extends to about 8 khz. 17

18 Using an analysis segment size of 20 ms (corresponding to 441 samples) and 25% window overlap, the spectrograms using rectangular, Hamming, and Blackman windows are shown below. Note the spectral leakage due to the rectangular window. 18

19 19

20 The 3-d STFT Magnitude plot is shown below. 20

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