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Spectral Leakage

Spectral Leakage. Pp289 jdsp. Freq of kth sample, No centering. Centering and the PSD. PSDs are centered if the DC component is at N/2, where N= number of samples. Before and after centering. How do I center the FFT?. UlawCodec ulc = new UlawCodec();

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Spectral Leakage

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  1. Spectral Leakage Pp289 jdsp

  2. Freq of kth sample, No centering

  3. Centering and the PSD • PSDs are centered if the DC component is at N/2, where N= number of samples. • Before and after centering

  4. How do I center the FFT? UlawCodec ulc = new UlawCodec(); • final double[] doubleArray = ulc.getDoubleArray(); • double d[] = Mat1.truncateToIntegralPowerOfTwo(doubleArray); • FFT1dDouble f = new FFT1dDouble(); • double in_im[] = new double[d.length]; • Mat1.centering(d); • f.computeForwardFFT(d, in_im); • f.normalize(); • final double[] psd = f.getPSDNotNormalized(); • Graph.graph(psd, "", "", "psd"); • f.computeBackwardFFT(d, in_im); • Mat1.centering(d); • UlawCodec ulc2 = new UlawCodec(f.getRealData()); • ulc2.play();

  5. How do I implement centering? • public static void centering(double r[]) { • int s = 1; • for (int i = 0; i < r.length; i++) { • s = -s; • r[i] *= s; • } • }

  6. How do I compute freqs?

  7. Computing Freqs • The center freq should be zero. • But what about 4001 and 3999?

  8. Thus every bin gives one freq • Bins give a unique freq. • K=n/2=0 hz • K=n = 4000 hz • K=0= -4000 hz

  9. Real signals • Real signals have zero complex components • Real signals have symmetrical PSDs • Real signals are positive on the right • Real signals are negative on the left • One bin maps to one frequency

  10. Spectral leakage • frequency analysis effect • Applies to: • finite-length signals • finite-length segments • energy has "leaked" out of the original signal spectrum into other buckets.

  11. Windowing • window function (also known as an apodization function or tapering function) • a function that is zero-valued outside of some chosen interval

  12. Windowing • windowing is the root cause of spectral leakage

  13. Does Windowing Really Help? • Non-rectangular window functions actually increase the total leakage • they can also redistribute it to places where it does the least harm, depending on the application • windowing predistorts the input samples so that the spectral leakage is evened out

  14. For example

  15. Frequency Bin • Often the kth array element in the frequency domain is referred to as a frequency bin

  16. The 400 Hz spectrum • we expect the maximum amplitude to occur at. k=102, and 103<- energy spreads around two bins

  17. Spectral Leakage • When a single freq leaks into more than one bucket in the fft. • This occurs because of rectangular windowing function use in a STFT • It smearing the PSD across multiple buckets.

  18. To limit spectral leakage • Use a non-rectangular windowing function • A function that is limited in time and zero everywhere else. • Finite support. • An a-periodic function

  19. Select a windowing function • Use either speed • Or accuracy • Or ease of implementation • Do a straight multiply of the window and your input signal. • You may need to do this several times, if you don’t want to loose data…

  20. But I need speed! • If you don’t overlap, you will loose data, but you will gain speed. • You need to know the nature of the input signal!

  21. There are many windows • Hanning • Blackman • Bartlet • Hamming • Etc.

  22. Windows

  23. Code for Hanning • public static double[] makeHanning(int n) { • double window[] = new double[n]; • double arg = 2.0 * Math.PI / (n - 1); • for (int i = 0; i < n; i++) • window[i] = 0.5 - 0.5 * Math.cos(arg * i); • return window; • }

  24. Lyon Window

  25. Lyon Window Code • public static double[] makeLyon(int n) { • double window[] = new double[n]; • double u = 0.0; • double du = 2.0 / n; • for (int i = 0; i < n / 2; i++) { • window[i] = y5(0, 1.0, u); • u += du; • } • u = 0; • for (int i = n / 2; i < n; i++) { • window[i] = y5(1.0, 0.0, u); • u += du; • } • return window; • }

  26. y5 • public static double y5(double y0, double y1, double u) { • double t2 = u * u; • double t3 = t2 * t2; • return (6 * y1 - 6 * y0) * t3 * u + • (-15 * y1 + 15 * y0) * t3 + • (10 * y1 - 10 * y0) * t2 * u + y0; • }

  27. Lyon vs Hanning Window

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