sec4solns - ECE 2200 Signals and Information Section 4(Week...

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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 = 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 < 0 1 for n 0 (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.
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ECE 2200 Signals and Information – Section 4 (Week 5 Feb 22-25) Page 2 1 4 0 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) < 0 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 0 2 4 6 8 10 0 0.2 0.4 0.6 0.8 1 Figure 2: Plot of y [ n ].
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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 ] = 0 for n < 0 n + 1 L for 0 n L - 1 1 for n L which defines y [ n ] as a piece-wise discrete function. 3. A linear time-invariant system with input x [ n ] and output y [ n ] is described by the difference equation y [ n ] = 3 . 4 x [ n ] - 5 . 2 x [ n - 1] + 3 . 4 x [ n - 2] (i) Determine the unit impulse response of this system.
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