Digital Signal Processing - Finite impulse response of LTI systems
In this chapter, block processing and sample
processing methods applied to FIR filtering
and Convolution. Several computational
aspects of convolution equations are
considered:
* Direct form
* Convolution table
* LTI form
* Matrix form
* Flip-and-slide form
* Overlap-add block convolution form.
* Z-Transform (discussed in the Z-transform
Chapter)
By Assoc.Prof.Dr. Thuong Le-Tien 1 DIGITAL SIGNAL PROCESSING FINITE IMPULSE RESPONSE OF LTI SYSTEMS Lectured by: Assoc. Prof. Dr. Thuong Le-Tien National Distinguished Lecturer HCMC September, 2011 1 Practical DSP methods fall in two basis classes: 1. Block Processing Methods 2. Sample Processing Methods In block processing methods, data are collected and processed in blocks (DFT/FFT spectrum computation, speech analysis and synthesis, image processing. In sample processing methods, data are processed one at a time (real-time applications, audio effects processing , digital controls, 2By Assoc.Prof.Dr. Thuong Le-Tien In this chapter, block processing and sample processing methods applied to FIR filtering and Convolution. Several computational aspects of convolution equations are considered: * Direct form * Convolution table * LTI form * Matrix form * Flip-and-slide form * Overlap-add block convolution form. * Z-Transform (discussed in the Z-transform Chapter) 3By Assoc.Prof.Dr. Thuong Le-Tien 1. Block processing methods 1.1. Convolution Sampling time interval, T=1/fs. Number of time samples: L = TLfs x(n) = [x0, x1, , xL-1] where n = 0, 1, , L – 1: The direct and convolution forms Convolution table form mm mnhmxmnxmhny )()()()()( )()()()( . njijxihny ji 4By Assoc.Prof.Dr. Thuong Le-Tien 1.2 DIRECT FORM Consider a causal FIR filter of order M with impulse response h(n), h = [h0, h1, , hM] where n = 0, 1, , M the length of the filter or the number of filter coefficients LH = M + 1 The output of the filter: m mnxmhny )()()( 5By Assoc.Prof.Dr. Thuong Le-Tien with conditions 0 m M and 0 n – m L – 1 m n L – 1 + m The limit of output index n: 0 m n L – 1 + m L – 1 + M 0 n L – 1 + M y = [y0, y1, y2, , yL – 1 + M] The length of output: Ly = L + M Ly = Lx + Lh –1 6By Assoc.Prof.Dr. Thuong Le-Tien M must satisfy simultaneously the inequalities 0 m M n – L + 1 m n 7By Assoc.Prof.Dr. Thuong Le-Tien It follows: max(0, n – L + 1 ) m min(n,M) The direct form of convolution Example: an order 3 filter and a length 5-input signal: h = [h0, h1, h2, h3] x = [x0, x1, x2, x3, x4] y = h * x = [y0, y1, y2, y3, y4, y5, y6, y7] ),min( )1,0max( )()()( Mn Lnm mnxmhny 8By Assoc.Prof.Dr. Thuong Le-Tien Equation applied: for n= 0, 1, 2,.7 max (0, 0 – 4 ) m min(0, 3) => m = 0 max (0, 1 – 4 ) m min(1, 3) => m = 0, 1 max (0, 2 – 4 ) m min(2, 3) => m = 0,1 ,2 max (0, 3 – 4 ) m min(3, 3) => m = 0, 1, 2, 3 max (0, 4 – 4 ) m min(4, 3) => m = 0, 1, 2, 3 max (0, 5 – 4 ) m min(5, 3) => m = 1, 2, 3 max (0, 6 – 4 ) m min(6, 3) => m = 2, 3 max (0, 7 – 4 ) m min(7, 3) => m = 3 i.e. n = 5, y5 = h1x4 + h2x3 + h3x2 7...,,1,0)()()( )3,min( )4,0max( nmnxmhny n nm 9By Assoc.Prof.Dr. Thuong Le-Tien All the output samples: y0 = h0x0 y1 = h0x1 + h1x0 y2 = h0x2 + h1x1 + h2x0 y3 = h0x3 + h1x2 + h2x1 + h3x0 y4 = h0x4 + h1x3 + h2x2 + h3x1 y5 = h1x4 + h2x3 + h3x2 y6 = h2x4 + h3x3 y7 = h3x4 10By Assoc.Prof.Dr. Thuong Le-Tien 1.3. Convolution table: 11By Assoc.Prof.Dr. Thuong Le-Tien Example: Find the convolution of the following filter and input signal h = [1, 2, -1, 1] x = [1, 1, 2, 1, 2, 2, 1, 1] y = [1, 3, 3, 5, 3, 7, 4, 3, 3, 0, 1] Ly = L + M = 8 + 3 = 11 12By Assoc.Prof.Dr. Thuong Le-Tien 1.4. LTI form: h = [h0, h0, h2, h3] x = [x0, x1, x2, x3, x4] Input x can be rewritten as a linear combination of delayed impulses x = x0[1, 0, 0, 0, 0] + x1[0, 1, 0, 0, 0] + x2[0, 0, 1, 0, 0] + x3[0, 0, 0, 1, 0] + x4[0, 0, 0, 0, 1] x(n)=x0(n)+x1(n–1)+x2(n–2)+x3(n–3)+x4(n–4) Then: y(n)=x0h(n)+x1h(n–1)+x2h(n–2)+x3h(n–3)+x4h(n–4) 13By Assoc.Prof.Dr. Thuong Le-Tien We can present the input and output signals as blocks: 14By Assoc.Prof.Dr. Thuong Le-Tien LTI form of convolution 15By Assoc.Prof.Dr. Thuong Le-Tien Example: h = [1, 2, -1, 1] and x = [1, 1, 2, 1, 2, 2, 1, 1] 16By Assoc.Prof.Dr. Thuong Le-Tien The LTI form can also be written in a form similar by determine the proper limits of For n = 0, 1, , L + M – 1 17By Assoc.Prof.Dr. Thuong Le-Tien 1.5. Matrix form Linear matrix form: y = Hx The filter matrix H must be rectangular with dimensions: Ly * Lx = (L + M) * L H is also called the TOEPLITZ MATRIX in the sense of that it has the same entry a long each diagonal Hx x x x x x h hh hhh hhhh hhhh hhh hh h y y y y y y y y y 4 3 2 1 0 3 23 123 0123 0123 012 01 0 7 6 5 4 3 2 1 0 0000 000 00 0 0 00 000 0000 18By Assoc.Prof.Dr. Thuong Le-Tien Example: There is also a matrix Form written as follows 19By Assoc.Prof.Dr. Thuong Le-Tien 1.6. Flip and slide yn = h0xn + h1xn-1 + h2xn-2 + + hMxn-M the first M outputs correspond to the input-on transient behavior of the filter, and the last M outputs beyond the end of the input data are the input-off transients. The remains are the steady-state outputs. 20By Assoc.Prof.Dr. Thuong Le-Tien 1.7. Transient and Steady-State Behavior Length of input signal is L, filter order M, the output can be separated into 3 parts 0 n < M (input on transient) M n L – 1 (steady state) L – 1 < n L – 1 + M (input off transient) 21By Assoc.Prof.Dr. Thuong Le-Tien Example: An IIR filter has xung h(n) = (0.75)nu(n). Using convolution to find y(n) when the inputs are: a) x(n) = u(n) b) x(n) = (-1)nu(n) c) x(n) = u(n) – u(n – 25) Find the steady state response for each case. Solve: a) n m n m ny 00 )( m)-u(m)u(n(0.75)m)-h(m)x(n n n m n n 0 1 )75.0(34 75.01 )75.0(1n(0.75) 4 75.01 1 )( nylim 22By Assoc.Prof.Dr. Thuong Le-Tien b) Steady state response: n m 0 ( mn (0.75)-1) n m n m ny 00 )( m-nm(-1)(0.75) m)-h(m)x(n (0.75) 7 3 + 7 4 (-1) = 75.01 )75.0(1 (-1) nn 1 n n 7 4 )1( 75.01 1 1)(- y(n) n n 23By Assoc.Prof.Dr. Thuong Le-Tien n Lnm m n Lnm mnmn xhy )1,0max()1,0max( )75.0( c) Two cases: 24By Assoc.Prof.Dr. Thuong Le-Tien n nn m ny )75.0(34 75.01 )75.0(1 (0.75) 1 0 m n nm ny 24 25 24-nm 0.75-1 (0.75)-1 (0.75) = (0.75) 1.8. Overlap-Add Block Convolution Method Overlap-add convolution method 25By Assoc.Prof.Dr. Thuong Le-Tien Example: 26By Assoc.Prof.Dr. Thuong Le-Tien 2. Sample processing method Adder Multiplier Delay 27By Assoc.Prof.Dr. Thuong Le-Tien FIR Filtering in Direct Form Direct form realization of third-order filter. 28By Assoc.Prof.Dr. Thuong Le-Tien
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