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h. Lecture 5: Block Processing for FIR Filters. Block processing methods recorded data: x = {x 0 x 1 x 2 … x L-1 } length: L x = L system impulse response: h = {h 0 h 1 h 2 h 3 … h M } length: L h = M+1 output data: y = {y 0 y 1 y 2 y3 … y Ly-1 }
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h Lecture 5: Block Processing for FIR Filters • Block processing methods • recorded data: x = {x0 x1 x2 … xL-1} • length: Lx = L • system impulse response: h = {h0 h1 h2 h3 … hM} • length: Lh = M+1 • output data: y = {y0 y1 y2 y3 … yLy-1} • length: Ly = Lx + Lh - 1 • key question: how do we process h and x to compute y? x y EE421, Lecture 5
Block Processing • Convolution tabley(0) = h(0)x(0)y(1) = h(0)x(1) + h(1)x(0)y(2) = h(0)x(2) + h(1)x(1) + h(2)x(0)y(3) = h(0)x(3) + h(1)x(2) + h(2)x(1) + h(3)x(0)y(4) = h(0)x(4) + h(1)x(3) + h(2)x(2) + h(3)x(1) + h(4)x(0) y(0) y(1) y(2) y(3) EE421, Lecture 5 y(4)
Block Processing • Convolution table • example: x = {1 2 1 3}, h = {4 2 1} n=0 n=0 y(0) = 4 y(1) = 10 y(2) = 9 y(3) = 16 y(4) = 7 y(5) = 3 y = {4 10 9 16 7 3} n=0 EE421, Lecture 5
Block Processing • LTI Formy(n) = … + x(0)h(n) + x(1)h(n-1) + x(2)h(n-2) + x(3)h(n-3) + … y(0) y(1) y(2) y(3) y(4) EE421, Lecture 5
Block Processing • LTI form • example: x = {1 2 1 3}, h = {4 2 1} n=0 n=0 4 10 9 16 7 3 0 y = {4 10 9 16 7 3} n=0 EE421, Lecture 5
Block Processing • Matrix form: EE421, Lecture 5
Block Processing • Matrix form • example: x = {1 2 1 3}, h = {4 2 1} n=0 n=0 y = {4 10 9 16 7 3} n=0 EE421, Lecture 5
h(2) h(1) h(0) h(2) h(1) h(0) h(2) h(1) h(0) h(2) h(1) h(0) h(2) h(1) h(0) h(2) h(1) h(0) y(0) y(1) y(2) y(3) y(4) y(5) Block Processing • Flip-and-slide form:y(n) = h(0)x(n) + h(1)x(n-1) + h(2)x(n-2) + … + h(L)x(n-L) EE421, Lecture 5
1 2 4 1 2 4 1 2 4 1 2 4 1 2 4 1 2 4 4 10 9 16 7 3 Block Processing • Flip-and-slide form: • example: x = {1 2 1 3}, h = {4 2 1} n=0 n=0 y = {4 10 9 16 7 3} n=0 EE421, Lecture 5
Block Processing • Programming considerations • convolution table formfunction y = conv(h, x) Lh = length(h);Lx = length(x);Ly = Lh+Lx-1;y = zeros(1,Ly);for n = 0:Ly-1 y(n+1) = 0; for i = 0:Lh-1 for j = 0:Lx-1 if (i+j == n) y(n+1) = y(n+1) + h(i+1)*x(j+1); end; end endend All these for loops will be inefficient in Matlab! EE421, Lecture 5
Block Processing • Programming considerations • LTI formfunction y = conv(h, x)Lh = length(h);Lx = length(x);Ly = Lh+Lx-1; y = zeros(1,Ly); for j = 0:Lx-1 shifth = [zeros(1, j) h zeros(1,Ly-Lh-j)]; y = y + x(j+1)*shifth;end Only 1 for loop! EE421, Lecture 5