{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

slides15

# slides15 - Course 18.327 and 1.130 Course Wavelets and...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Course 18.327 and 1.130 Course Wavelets and Filter Banks Signal and Image Processing: finite length signals; boundary filters and boundary wavelets; wavelet compression algorithms. Finite-Length Signals y-2 y-1 y0 y1 o yN-1 o = h0 h1 h2 h0 h1 y[n] h-1 h0 h-1 h1 h0 h-1 rr r x-2 x-1 x0 x1 o xN-1 o unknown LMN H(ω) x[n] length N (finite length) unknown 2 1) zero-padding x[n] m -2 -1 0 1 2 3 m n filtered by [1, 1] y[n] m n -2 -1 0 1 2 3 filtered by [1, -1] y[n] m -2 -1 0 1 2 3 n artificial edge resulting from zero-padding 3 2) Periodic Extension … x[N] = x[0] x[n] … … n - 1 0 1 2 … N-1 N wrap-around o y0 y1 o = yN-1 N-output h1 h0 h-1 h-2 m h2 h1 h1 h0 h-1 m h3 h2 h-1 h-2 m h2 h1 h0 xN-2 xN-1 x0 x1 x2 o xN-1 x0 o circulant matrix = H 4 What is the eigenvector for the circulant matrix? [ 1 eiω ei2ω m ei(N-1)ω ] T We need eiNω = 1 = ei0ω ∴ Nω = 2πk , ω= 2πk N discrete set of ω’s For the 0th row, N-1 H[k] = ∑ h[n] e-i 2πk Nn n=0 5 111m 1 w w2 [H] 1 w2 w4 oo o 1 wN-1 1 H[0] H[1] wN-1 r w2(N-1) = [F] o H[N-1] k=0k=1 k=N-1 LOOOOMOOOON 2π iN w=e F HF = FΛ Λ contains the Fourier coefficients 2πk Nn H[k] = ∑ n h[n]e-i ∑ ∑ h[n - x[e-i If x[ = ei n n 2πk0 N 2πk Nn = H[k]X[k] H[k]X[k] ⇒ X[k] = δ[k – k0] ⇒ H[k]X[k] = H[k0]X[k] 6 3) Symmetric Extension 1) Whole point symmetry – when filter is whole point symmetric. 2) Half point symmetry – when filter is half point symmetric. e.g. Whole point symmetry: filter and signal h1x2 + h0x1 + h1x0 h0 h1 h1x1 + h0x0 + h1x1 h1 h0 h1 = h1x0 + h0x1 + h1x2 h1 h0 h1 r rr x2 x1 x0 x1 x2 7 e.g. whole point symmetry – filter, half-point symmetry - signal h1x2 + h1x1 + h1x0 + h1x0 + h0x1 + h1x0 h0x0 + h1x0 h0x0 + h1x1 h0x1 + h1x2 Half point symmetry h1 h0 h1 h1 h0 h1 = h1 h0 h1 r rr x2 x1 x0 x0 x1 x2 Whole point symmetry 8 Downsampling a whole-point symmetric signal with even length N at the left boundary: x x -2 -1 0 1 ⇒ still whole-point symmetric after ↓2. 2 at the right boundary: x x x ⇒ half-point symmetric after ↓2. N-1 odd E.g. 9/7 filter: whole-point symmetric use the above extension for signal ⇒ N N/2 exactly N/2 9 Downsample a half-point symmetric signal x ⇒ nothing guaranteed x x -3 -2 -1 0 1 2 Linear-phase filters H(ω) = A(ω)e-iωα 1) half-point symmetric, α = fraction 2) whole-point symmetric, α = integer Symmetric extension of finite-length signal X(ω) = B(ω)e-iωβ 10 10 The output: Y(ω) = H(ω)X(ω) W W H H W H H W W H W H W = whole-point symmetry H = half-point symmetry The above extensions ensure the continuity of function values at boundaries, but not the continuity of derivatives at boundaries. 11 11 4) Polynomial Extrapolation (not useful in image processing) • Useful for PDE with boundary conditions. x0 x1 01 x2 x3 4 coefficients ⇒ fits up to 3rd order polynomials. 23 a + bn + cn2 + dn3 = x(n) 1000 1111 1 2 22 23 1 3 32 33 LOMON A a x0 b = x1 c x2 d x3 12 12 Then, x–1 = [1 -1 1 -1] PDE a b = [1 -1 1 -1] A–1 c d x0 x1 x2 x3 f(x) = ∑ ckφ(x – k) k Assume f(x) has polynomial behavior near boundaries p-1 ∑ αixi = f(x) = ∑ ckφ(x – k) k i=0 {φ (• - k)} orthonormal p-1 ⇒ ∑ αi ∫ φ(x – k)xidx = ck i=0 LOMON µik 13 13 µ0 µ1 m µ p-1 0 0 0 µ0 1 µ1 µ2 1 1 m o α0 c0 α1 c1 = o αp-1 o cp-1 Using the computed αi’s, we can extrapolate, α0 e.g. c–1 = [µ01 µ1 1 m µp-1] o – – –1 αp-1 DCT idea of symmetric extension cf. DFT X[k] = ∑ n complex-valued 2πk x[n]e-i N n Want real-valued results. Want 14 14 1m 0 N-1 N m 2N-1 2N DFT of this extended signal: N-1 2πk –i 2N n ∑ x[n]e 2N-1 +∑ LOOMOON n=N n=0 N-1 N-1 2πk π -i 2N (2N-1-m) ∑ x[m]e m=0 = ∑ x[n] n=0 2πk x[2N-1-n]e-i 2N n 2πk -i 2πk n {e 2N + 2πk e-i 2N (2N-1-n)} N-1 X(k) ck ∑ √2 x[n]cos πk (n+½) m DCT – II used in JPEG N N n=0 1 ck = √2 k=0 1 k = 1, 2, …, N - 1 15 15 ...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern