Question

Consider a filter characterized by the following impulse response: h [1, 2, -1, 1] Which of the following statements are true about the filter? Assume that the sampling frequency in this application is 8192 Hz. (You may use MATLAB to help you analyze this filter). o 1. The filter characterized by h = 1 2 1 1 Is a frequency selective FIR filter. In terms of the frequency response the ter is best characterized as a band-stop filter with a bandstop center frequency around 1500 Hz. O 2. The filter characterized by h 11 2-1 1] is a frequency selective FIR filter. In terms of the frequency response the filter is best characterized as a band-stop filter with a bandstop center frequency around 750 Hz. O 3. The filiter characterized by h -112-1 1] is a frequency selective IIR fiter because FIR filters are characterized by b filter coefficients. In terms of the frequency response the filter is best characterized as a band-stop filter with a bandstop center frequency around 1500 Hz. o 4. The filter characterized by h = 1 2 1 1 is a frequency selective FIR filter. In terms of the frequency response the filter is best characterized as a band-pass filter with a center frequency around 1500 Hz O 5 The filter characterized by h = 1 2 1 1 is a frequency selective FIR filter. In terms of the frequency response the filter is best characterized as a band pass filter with a center frequency around 750 Hz

Answer 1

Use MATLAB to find the FFT of the given impulse response. Pad the impulse response with zeroes to get better resolution in FFT zeros (1,60) to pad the impulse response Creating array of 0 zeros Z z]i response with zeros [1 2-1 1 Padding the impulse h fftshift (fft (h) ); Taking impulse response and center shifting the result plot ((-length (h) +1) /2: (length (h)) /2), abs (H)) %Plotting the Magnitude response xlim length (h) +1) /2, (length (h) ) /21) Setting X limits set (gca, 'XTick', length (h))/2:16: (length (h))/2) the fft of the H = wwww ww wwww0w wwwwe % Setting x axis divisions set (gca, 'XTickLabel',{'-pi',' pi/2','0','pi/2','pi'}) xlabel ('Digital Labelling x-axis ylabel ('|H (Omega') www wwwwwA -->'); Frequency Omega w w clo

Plot the magnitude response of the above code. 3.2 3 2.8 2.6 2,4 2.2 2 1.8 1.6 pi/2 pi/2 Di Digital Frequency Omega - We see that the magnitude response shows a dip at 2 8 We know that the digital frequency T corresponds to 2 Thus 3f 3x8192 3T 1536 Hz 2 x T 8 16 16 Thus, the filter is roughly a band stop filter with center frequency around 1500 Hz Hence, option (2) is correct. e6ouo)H