数字信号处理答案10.doc

Chapter 10 Solutions 10.1 (a) The impulse response is given by h[n] = –0.8h[n–1] + 0.1h[n–2] + ?[n]. The first ten samples are listed in the table. n 0 1 2 3 4 h[n] 1.000 –0.800 0.740 –0.672 0.612 n 5 6 7 8 9 h[n] –0.557 0.506 –0.461 0.419 –0.381 (b) The impulse response contains an infinite number of non-zero terms. 10.2 ? 0 ? 0 6365 13255 21382 32000 47891 77255 160874 ? ? ? 10.3 (a)(i) Without pre-warping, the transfer function for the analog filter is The bilinear transformation gives (ii) The analog frequency 2.5 kHz is converted to a digital frequency ?p1 = =1.9635 rads. This frequency is pre-warped to the analog frequency = 23946 rad/sec, which makes the transfer function for the analog filter After the bilinear transformation, the digital transfer function is obtained: (b) The magnitude responses for both filters are shown below. The –3 dB frequency for the pre-warped filter is equal to the specified 2.5 kHz. |H(f)|Filter with pre-warping |H(f)| Filter with pre-warping Filter without pre-warping Filter without pre-warping f f 10.4 (a) The cut-off frequency for the analog filter is 1500/(2?) = 238.73 Hz. The digital frequency that corresponds to this analog frequency is ?p1 = = 0.1875 rads. The pre-warped analog cut-off frequency is = 1504.4 rad/sec, to give the transfer function . (b) Use the bilinear transformation to get the digital transfer function (c) The difference equation is y[n] = 0.8281y[n–1] + 0.0859x[n] + 0.0859x[n–1]. (d) The frequency response is . The magnitude response may be found by taking the magnitudes of this expression for several values of ?. It is plotted below against frequency in Hz. |H(f)| |H(f)| f f (e) The magnitude for the analog transfer function is The shape of the digital filter may be found by substituting: This function may be plotted for various values of ?. The magnitude response is plotted below against digital frequency in rads. When digital frequencies are converted to frequencies in Hz, the plot becomes i