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# Slides1 - Discrete-time filters convolution Fourier...

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Unformatted text preview: Discrete-time filters: convolution; Fourier transform; lowpass and highpass filters Input → x[n] Filter Output → y[n] n denotes the time variable: {…, -2, -1, 0, 1, 2, …} x[n] denotes the sequence of input values: {…, x[-2], x[-1], x[0], x[1], x[2], …} y[n] denotes the sequence of output values: {…, y[-2], y[-1], y[0], y[1], y[2], …} Assume that a) the principle of superposition holds ⇔ system is linear, i.e. combining any two inputs in the form Ax1[n] + Bx2[n] results in an output of the form Ay1[n] + By2[n] 2 b) the behavior of the system does not change with time, i.e. a delayed version of any input xd[n] = x[n - d] produces an output with a corresponding delay yd[n] = y[n – d] Under these conditions, the system can be characterized by its response, h[n], to a unit impulse, δ[n], which is applied at time n = 0, 1 i.e. the particular input Unit x[n] = δ[n] L L Impulse -2 -1 0 1 2 n produces the output y[n] = h[n] h[0] h[1] h[-1] h[-2] h[2] L L -2 -1 0 1 2 n Impulse Response 3 The general input ∞ x[n] = ∑ x[k] δ[n – k] k = -∞ will thus produce the output ∞ y[n] = ∑ x[k]h[n-k] k = -∞ Convolution sum 4 Discrete time Fourier transform ∞ X(ω) = ∑ x[n] e-iωn n = -∞ Inverse x[n] = 1 2π π ⌠ X(ω) eiωn dω ⌡ -π 5 Frequency Response Suppose that we have the particular input x[n] = eiωn H(ω) eiωn eiωn What is the output? → → y[n] = ∑ h[k] x [n - k] k = eiωn ∑ h[k] e-iωk k 14243 14243 H(ω) Frequency Response 6 Convolution Theorem π A general input 1⌠ x[n] = 2π ⌡ X(ω) eiωn dω -π will thus produce the output π 1 y[n] = 2π ⌠ X(ω) H(ω) eiωn dω → Y(ω) = X(ω) H(ω) ⌡ - π 14243 Y(ω) Convolution Convolution of sequences x[n] and h[n] is denoted by h[n] ∗ x[n] = ∑ x[k] h[n - k] = y[n] (say) k 7 Matrix form: OOO Oh[0] h[-1] h[-2] Oh[1] h[0] h[-1] h[-2] h[2] h[1] h[0] h[-1] h[-2] h[2] h[1] h[0] h[-1]O h[2] h[1] h[0] O O O O 1444442444443 Toeplitz matrix M x[-2] x[-1] x [0] x [1] x[2] M = M y[-2] y[-1] y[0] y[1] y[2] M 8 Convolution is the result of multiplying polynomials: (… + h[-1]z + h[0] +h[1]z-1 + …) (… + x[-1]z + x[0] + x[1]z-1 + …) = (… + y[-1]z + y[0] + y[1]z-1 + …) Example: 3 12 62 6 14 -3 4 -10 -9 ↑↑↑ z-5 z-4 z-3 3 2 1 -5 2 2 4 -1 -1 5 -2 -20 8 0 4 00 -17 13 -2 ↑ z-2 01 -5 4 2 x[n] 1 0 -1 1 3 h[n] 2 2 13 14 6 ↑↑ z-1 z0 y[n] 0 12 -2 -17 4 3 -9 5 9 Discrete Time Filters (summary) Discrete Time: x[n] → h[n] y[n] → y[n] = ∑ x[k] h [n-k] k (Convolution) Discrete –time Fourier transform X(ω) = ∑ x[n] e-iωn n Frequency domain representation Y(ω) = H(ω) • X(ω) (Convolution theorem) 10 10 Toeplitz Matrix representation: M y[-2] y[-1] y[0] y[1] y[2] M = OOO Oh[0] h[-1] h[-2] Oh[1] h[0] h[-1] h[-2] h[2] h[1] h[0] h[-1] h[-2] h[2] h[1] h[0] h[-1]O h[2] h[1] h[0] O O O O M x[-2] x[-1] x [0] x [1] x[2] M Filter is causal if y[n] does not depend on future values of x[n]. Causal filters have h[n] = 0 for n < 0. 11 11 Filters a) Lowpass filter example: y[n] = ½ x[n] + ½ x [n-1] Filter representation: y[n] x[n] → h[n] → y[n] = ∑ x[k] h [n-k] k Impulse response is ½½ h[n] L L 01 n 12 12 Frequency Response is H(ω) = ∑ h[k]e-iωk k = ½ + ½ e-iω Rewrite as H(ω) = H(ω) eiφ(ω) 2 H(ω) = cos( ω /2) e-iω//2 ; -π ≤ ω ≤ π 13 13 b) Highpass Filter Example y[n] = ½ x[n] - ½ x[n-1] Impulse response is ½ h[n] L L 0 1 n -½ 14 14 Frequency response is H(ω) = ½ - ½ e-iω = i sin (ω/2) e-iω/2 14243 14243 2 sin (ω/2) e–i(π//2 = 2 sin (ω/2) ei(π//2 + ω/2) - ω/2) ; -π ≤ ω < 0 ; 0<ω≤ π 15 15 ...
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## This note was uploaded on 12/04/2011 for the course ESD 18.327 taught by Professor Gilbertstrang during the Spring '03 term at MIT.

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