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.Exercise Ch.66.5.The frequency response function of the bandpass filter is shown in following figure,(a) The figure above can be take as the result of the convolution of the frequency response function a lowpass filter with and. Their functions in time domain are and .Their relation in time domain is As the function of the bandpass filter in time domain is , the is . (b). The is a sinc function and the first zero is at. As increases, become more close to the orign. h(t) can be take as a modulated signal and the carrier signal is . So h(t) gets more concentrated about the orgin. The following figures are the h(t) for an , respectively.6.7.We have a lowpass filter with impulse respone h(t) and the frequency response function:If Since (a) The bandpass filter has the passband frequencies of 1000Hz and 3000Hz.(b) The bandpass filter has stopband frequencies of 800Hz and 3200Hz.Exercise Ch.77.4.(b)If , Then and So, the Nyquist rate is .(c) If , then , So, the Nyquist rate is 2.7.5. We know the frequency response function of from the Fig.7.3 on P.518 in text book, .As and , then. The frequency profile for k = 0 in is the same as that in .So the constraints on the frequency response of a filter that give as its output is shown as following, when is the input,where .7.6. In the system the x1(t) is band limited to w1, the x2(t) is band limited to w2, andSo, the Nyquist rate is 2w1 +2w2. The maximum sampling interval is
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