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sec4solns - ECE 2200 Signals and Information – Section...

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Unformatted text preview: ECE 2200 Signals and Information – Section 4 (Week 5 Feb 22-25) Page 1 1. The impulse response h [ n ] of an FIR filter is h [ n ] = δ [ n- 1]- 3 δ [ n- 3] + 0 . 1 δ [ n- 4] + 0 . 04 δ [ n- 7] Write the difference equation for the FIR filter. ————————————— ————————————— Solution (read “5-3 The General FIR Filter” and see “exam5 1.pdf” on the included disc): y [ n ] = M X k =0 b k x [ n- k ] = x [ n- 1]- 3 x [ n- 3] + 0 . 1 x [ n- 4] + 0 . 04 x [ n- 7] . with b = 0, b 1 = 1, b 2 = 0, b 3 =- 3, b 4 = 0 . 1, b 5 = 0, b 6 = 0 and b 7 = 0 . 04. 2. Consider the unit step signal u [ n ] = 0 for n < 1 for n ≥ (i) With u [ n ] as the input, compute and sketch the output y [ n ] for n =- 10 to n = 10 of the four-term running average filter with difference equation y [ n ] = 1 4 3 X k =0 x [ n- k ] (ii) Confirm your answer to part (i) using MATLAB. (iii) Derive a general formula for y [ n ] as the output of an L-term averager with difference equation y [ n ] = 1 L L- 1 X k =0 x [ n- k ] with a unit step signal u [ n ] as the input. ————————————— ————————————— Solution: (i) Substituting x [ n ] = u [ n ], the running average filter is then y [ n ] = 1 4 3 X k =0 u [ n- k ] = 1 4 u [ n ] + 1 4 u [ n- 1] + 1 4 u [ n- 2] + 1 4 u [ n- 3] which is a summation of individual step signals. ECE 2200 Signals and Information – Section 4 (Week 5 Feb 22-25) Page 2 1 4 1 2 3 4 1 1 2 3 10- 10 n y [ n ] Figure 1: Sketch of y [ n ]. (ii) MATLAB code and plot: 1 n =- 10:10; 2 % unit step signal: 3 %----------------- 4 for i = 1:length(n) 5 if n(i) < 6 u(i) = 0; 7 else 8 u(i) = 1; 9 end 10 end 11 12 % FIR filter: 13 %----------- 14 for k = 1:length(n) 15 try 16 y(k) = 1/4 * u(k) + 1/4 * u(k- 1) + 1/4 * u(k- 2) + 1/4 * u(k- 3); 17 catch 18 y(k) = 0; 19 end 20 end 21 stem(n,y, 'fill' , 'color' , 'k' ) n y [ n ] − 10 − 8 − 6 − 4 − 2 2 4 6 8 10 0.2 0.4 0.6 0.8 1 Figure 2: Plot of y [ n ]. ECE 2200 Signals and Information – Section 4 (Week 5 Feb 22-25) Page 3 (iii) By inspection of y [ n ] = 1 L L- 1 X k =0 u [ n- k ] for various L , a general formula for y [ n ] is y [ n ] = for n < n + 1 L for 0 ≤ n ≤ L- 1 1 for n ≥ L which defines y [ n ] as a piece-wise discrete function....
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This note was uploaded on 03/11/2010 for the course ECE 2200 taught by Professor Johnson during the Spring '05 term at Cornell.

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sec4solns - ECE 2200 Signals and Information – Section...

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