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Unformatted text preview: Speech Processing ShortTime Fourier Transform Analysis and Synthesis February 13, 2012 Veton Këpuska 2 ShortTime Fourier Transform Analysis and Synthesis MinimumPhase Synthesis Speech & Audio Signals are varying and can be considered stochastic signals that carry information. This necessitates shorttime analysis since a single Fourier transform (FT) can not characterize changes in spectral content over time (i.e., timevarying formants and harmonics) Discretetime shorttime Fourier transform (STFT) consists of separate FT of the signal in the neighborhood of that instant. FT in the STFT analysis is replaced by the discrete FT (DFT) ⇒ Resulting STFT is discrete in both time and frequency. ⇒ Discrete STFT vs. ⇒ Discretetime STFT which is continuous in frequency. In linear Prediction and Homomorphic Processing, underlying model of the source/filter is assumed. This leads to: Model based analysis/synthesis, also note that Analysis methods presented implicitly both used short time analysis methods (to be presented). In ShortTime Analysis systems no such restrictions apply. February 13, 2012 Veton Këpuska 3 ShortTime Analysis (STFT) Two approaches of STFT are explored: 1. Fouriertransform & 2. Filterbank February 13, 2012 Veton Këpuska 4 FourierTransform View Recall (from Chapter 3): w[n] is a finitelength, symmetrical sequence (i.e., window) of length N w . w[n] ≠ 0 for [0, N w1] w[n] – Analysis window or Analysis Filter ( 29 [ ] [ ] ∑ ∞∞ == m n j e m n w m x n X ϖ , February 13, 2012 Veton Këpuska 5 FourierTransform View x[n] – timedomain signal f n [m]=x[m]w[nm]  Denotes shorttime section of x[m] at point n. That is, signal at the frame n. X(n, ϖ )  Fourier transform of f n [m] of shorttime windowed signal data. Computing the DFT: ( 29 ( 29 k N n X k n X π ϖ 2  , , = = February 13, 2012 Veton Këpuska 6 FourierTransform View Thus X(n,k) is STFT for every ϖ =(2 π /N)k Frequency sampling interval = (2 π /N) Frequency sampling factor = N DFT: ( 29 [ ] [ ] ∑ ∞∞ == m km N j e m n w m x k n X π 2 , February 13, 2012 Veton Këpuska 7 FourierTransform View February 13, 2012 Veton Këpuska 8 Example 7.1 Let x[n] be a periodic impulse train sequence: Also let w[n] be a triangle of length P: ∑ ∞∞ == l lP n n x ] [ ] [ δ P 2P 3PP n … P/2+1P/2 Ppoints n February 13, 2012 Veton Këpuska 9 Example 7.1 lP j l m m j l m m j e lP n w e m n w lP m e m n w m x n X ) ( ] [ ) ( ] [ ] [ ) , ( ϖ δ∞∞ = ∞∞ =∞∞ = ∞∞ =∑ ∑ ∑ ∑= == Nonzero only for m=lP Window located at lP & Linear phase ϖ lP February 13, 2012...
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 Fall '10
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 Shorttime Fourier transform, Veton Këpuska

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