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1 Lecture Notes in Speech Production Speech Coding and Speech Recognition Mark Hasegawa-Johnson University of Illinois at Urbana-Champaign February

2 2 M. Hasegawa-Johnson. DRAFT COPY.

3 Chapter Basics of Digital Signal Processing. LTI Systems x(n)! h(n)! y(n) (.).. What is an LTI System? ffl Linearity If x (n)! y (n) andx 2 (n)! y 2 (n) then ax (n)+bx 2 (n)! ay (n)+by 2 (n) (.2) ffl Time-Invariance..2 Impulse Response If x(n)! y(n) then x(n m)! y(n m) (.3) An LTI system is completely characterized by the impulse response h(n). The output resulting from any input can be derived by convolution:..3 Eigenfunctions (CT) y(t) = h(t) Λ x(t) = (DT) y(n) = h(n) Λ x(n) = m= x(fi)h(t fi)dfi (.4) x(m)h(n m) (.5) Exponentials and sinusoids are the eigenfunctions of LTI systems meaning that the following rule holds: (.6) Input is Sinusoid at! ) Output is Sinusoid at! (different phase different amplitude) (.7) For example if H(s) =A(s)e jφ(s) then x(t) =e st! h(t)! y(t) =H(s )e st = A(s )e st+jφ(s) (.8) 3

4 4 M. Hasegawa-Johnson. DRAFT COPY..2 Transforms General Transform Continuous Time (s = ff + jω) Discrete Time (z = e st ) Laplace Transform (s) = x(t) = x(t)e st dt Z ff+j (s)e st ds 2ßj ff j Z Transform (z) = x(n) = x(n)z n n= Z (z)z n dz 2ßj C Evaluate at s = j2ßf (.9) z = e j! (.0) Signal is Periodic in Time F 0 = T 0 = N 0 T (.) Transform is Sampled in Frequency N length(x(n)) (.2) Fourier Transform (f) = x(t) = Fourier Series x(t)e j2ßft dt (f)e j2ßft df C k = T 0 ZT 0 x(t)e j2ßkf0t dt x(t) = k= C k e j2ßkf0t DTFT DFS DFT (e j! ) = x(n) = 2ß c k = N 0 N 0 x(n) = (k) = x(n)e j!n n= Z ß (e j! )e j!n d! ß N 0 k=0 N n=0 x(n) = N n=0 x(n)e j2ßkn=n0 c k e j2ßkn=n0 x(n)e j2ßkn=n N (k)e j2ßkn=n k=0.2. Fourier Series: Dirichlet's Conditions Fourier series is defined if: ffl x(t) has a finite number of discontinuities. ffl x(t) has a finite number of minima and maxima. ffl x(t) is absolutely integrable (x(n) is absolutely summable): Z T 0 jx(t)jdt < ; N Z Transform and DFT: Region of Convergence Zfa n u(n)g = az n= undefined; diverges jx(n)j < (.3) jzj > jaj jzj»jaj (.4) Zf a n undefined; diverges u( n )g = az jzj jaj jzj < jaj DFT defined iff Z transform converges on the unit circle (x(n) absolutely summable). (.5)

5 Lecture Notes in Speech... DRAFT COPY. 5.3 Transform Properties Real in A Conjugate Symmetric in B A=Time B=Frequency x(n) = real (k) = Λ ( k) A=Frequency B=Time x(n) = x Λ ( n) (k) = real Linearity N =max(n ;N 2 ) (.6) ax (n)+bx 2 (n) a (k)+b 2 (k) Convolve ina Multiply in B x (n) flλ x 2 (n) (k) 2 (k) x (n)x 2 (n) N (k) flλ 2 (k) Shift in A Modulation in B x((n m)) N e j2ßkm N (k) e j2ßin N x(n) ((k i)) N Sample in A Periodic in B P x(n) r ffi(n rm) (e j! ) flλ M Pr ffi! 2ßr M x(n) Λ P r ffi(n rn 0) P (e j! ) N 0 r ffi! 2ßr N 0 x (n) flλ x 2 (n) =x (n) Λ x 2 (n) iff N length(x (n)) + length(x 2 (n)) (.7).4 Sampling Theorem Sampling in Time = Periodic Repetition in Frequency x(t) = n x d (n)ffi(t nt ) (.8) (.9) (f) = (f + F s ) (.20) Sampling frequency F s ==T. Choose F s 2F max. 2F max is called the Nyquist frequency. x(t)! LPF at F max! x d (n) =x(nt )! x d (n) (.2) x d (n)! x(nt )=x d (n)! LPF at F max! x(t) (.22) Periodic Repetition in Time = Sampling in Frequency

6 6 M. Hasegawa-Johnson. DRAFT COPY. x(t) = x(t + T 0 ) (.23) (.24) (f) = C k ffi(f kf 0 ) (.25) k Fundamental frequency F 0 ==T 0..5 Downsampling We can reduce the amount of information stored on the computer with an algorithm like this:. Throw away M out of every M samples: x d (n) =x(n) r x(n) n = :::;M;2M;3M;::: ffi(n rm) = 0 else (.26) 2. Change the axis labels: y(n) =x d (nm) Y (e j! )= d (e j!=m ) (.27) The second step doesn't change the information in the signal but the first step does. sampling in time equals periodic repetition in frequency so So we get aliasing unless d (e j! )=(e j! ) Λ M k ffi! 2ßk M = M k Remember that j(! 2ßk (e M ) ) (.28) d (e j! )=0 for!> ß M (.29) This is just like resampling the signal at F s (new) = F s (old)=m. The solution is just like an A/D except that both input and output are digital: x(n)! LPF at ß=M! y(n) =x(nm)! y(n) (.30).6 Upsampling Suppose we want to increase the sampling rate: F s (new) = L F s (old). We start by relabeling the time and frequency axes like so: x(n=l) n = :::; L; 0;L;2L; : : : v u (n) = ; V 0 else u (e j! )=(e jl! ) (.3) V u (e j! ) is periodic with period 2ß=L. To get rid of the extra copies of the spectrum so it is periodic with period 2ß welpf: x(n)! v u (n) =x(n=l)! LPF at ß=L! v(n) (.32) But remember that the transform of an ideal LPF at frequency ß=L is h(n) = sin(ßn=l) ßn (.33)

7 Lecture Notes in Speech... DRAFT COPY. 7 The maximum value of this filter is h(0) = =L which means that if we use this filter we will wind up multiplying the entire signal by =L!! Therefore we must throw in a scaling factor of L: x(n)! v u (n) =x(n=l)! LPF at ß=L! Multiply by L! v(n) (.34) The inverse transform of an ideal LPF multiplied by L is Lh(n) = sin(ßn=l) ßn=L (.35) So if we use the algorithm given above we get an output signal v(n) which looks like x(n=l) v(n) = interpolated values n = :::; L; 0;L;2L;::: else (.36)

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