Question

# Problem 5 The L-point averaging filter is a low pass filter. Its passband width is controlled by L, being inversely proportional to L. It is also possible to create a filter whose passband is centered around some frequency other than zero. One simple way to do this is to define the impulse response of an L-point FIR as: h[n]=\beta \cos \left(\omega_{0} n\right) where L is the filter length, and wo is the center frequency that defines the frequency location of the passband. For example, we pick wo = 0.2 n if we want the peak of the filter's passband to be centered at 0.27. Also, it is possible to choose B so that the maximum value of the frequency response magnitude will be one. The bandwidth of the band pass filter is controlled by L; the larger the value of L, the narrower the bandwidth.Generate a band pass filter that will pass a frequency component at we = 0.2. Make the filter length (L) equal to 51. Figure out the value of 3 so that the maximum value of the frequency response magnitude will be one. Make a plot of the frequency response magnitude and phase. Hint: use MATLAB's freqz() function to calculate these values.The passband of the BPF filter is defined by the region of the frequency response where|H(ej)| is close to its maximum value of one. Typically, the passband width is defined as the length of the frequency region where |H(e")| is greater than 1//2 = 0.707. Note: you can use MATLAB's find function to locate those frequencies where the magnitude satisfies|H(ej«)| > 0.707.Use the plot of the frequency response for the length-51 band pass filter from first part to determine the passband width. (If the sampling rate is fs = 8000 HZ, determine the analog frequency components that will be passed by this band pass filter. Use the passband width and also the center frequency of the BPF to make this calculation.  Fig: 1  Fig: 2  Fig: 3  Fig: 4