數字信號處理基礎答案第四章sol

1、Chapter 4 Solutions4.1(a)The pass band gain for this filter is unity. The gain drops to 0.707 of this value at 2400 Hz and 5200 Hz. Thus, the frequencies passed by the filter lie in the range 2400 to 5200 Hz.(b)The filter is a band pass filter.(c)The bandwidth is the range of frequencies for which t

2、he gain exceeds 0.707 of the maximum value, or 5200 2400 = 2800 Hz.4.2A low pass filter passes frequencies between DC and its cut-off frequency. The bandwidth is identical to the cut-off frequency. Thus, the cut-off frequency is 2 kHz.4.3The maximum pass band gain of the filter is 20 dB. The bandwid

3、th is defined as the range of frequencies for which the gain is no more than 3 dB below the pass band gain, or 17 dB. This gain occurs at the cut-off frequency of 700 Hz. For a high pass filter, the bandwidth is the range of frequencies between the cut-off frequency, 700 Hz, and the Nyquist frequenc

4、y (equal to half the sampling rate), 2 kHz. The bandwidth is 1300 Hz.4.4The low pass filter has a cut-off frequency of 150 Hz and bandwidth 150 Hz. The band pass filter has cut-off frequencies at 250 Hz and 350 Hz for a bandwidth of 100 Hz. The high pass filter has a cut-off frequency of 400 Hz and

5、a bandwidth of 100 Hz, which extends from its cut-off frequency to the Nyquist limit at half the sampling rate.4.5(a)The low pass filter output is on the left. The high pass filter output is on the right.nxn(b)An approximation to the original vowel signal can be found by adding the high and low pass

6、 waveforms together.4.6(a)linear(b)non-linear(c)non-linear(d)linear4.7Since the new input is shifted to the right by two positions from the original input, the new output is shifted to the right by two positions from the original output.ynn4.8(a)yn = 0.25yn1 + 0.75xn 0.25xn1(b)yn = yn1 xn 0.5xn14.9(

7、a)The system is non-recursive. b0 = b1 = b2 = 1/3(b)The system is recursive.a0 = 1, a1 = 0.2, b0 = 1(c)The system is recursive.a0 = 1, a1 = 0.5, b0 = 1, b1 = 0.44.10(a)n0123456789yn1.00.13.01.70.80.10.00.00.00.0ynn(b)n0123456789yn1.00.63.364.023.412.051.230.740.440.27ynn(c) n0123456789yn1.01.64.444.

8、103.293.663.292.962.672.40ynn(d) n0123456789yn0.50.51.52.51.50.50.00.00.00.0ynn4.11n0123456789xn1.001.0001.0001.0001.0000.0000.0000.0000.0000.000n0123456789yn1.000.7500.8130.7970.8010.200.0500.0130.0030.0014.12 n0123456789xn2.000.001.000.000.000.000.000.000.000.00n0123456789yn0.60.50.10.250.10.00.00

9、.00.00.0ynn4.13The overall input xn for any sampling instant is the sum of the inputs x1n and x2n. This overall input is applied to the difference equation in the normal way to obtain outputs.n0123456789xn0.000.8071.2001.0070.4000.5000.6000.7000.8000.900n0123456789yn0.000.8070.8370.4670.0530.320.375

10、0.4300.4850.544.14n0123456789xn0.000.3940.6320.7770.8650.9180.9500.9700.9820.989n0123456789yn0.000.3940.3170.5230.4460.5610.5020.5690.5270.5684.15xnyn+delay0.51.00.8delaydelayxn1xn34.16yn = 0.5yn2 + 1.2xn 0.6xn1 + 0.3xn24.17yn = 2.1xn1 1.5xn24.18wn = xn + 0.3wn1 0.1wn2yn = 0.8wn 0.4wn24.19The differ

11、ence equation for the first second-order section isy1n = 0.1xn + 0.2xn1 + 0.1xn2The difference equation for the second second-order section isyn = y1n + 0.3y1n2Substituting the first equation into the second givesyn = (0.1xn + 0.2xn1 + 0.1xn2) + 0.3(0.1xn2 + 0.2xn3 + 0.1xn4) = 0.1xn + 0.2xn1 + 0.07x

12、n2 + 0.06xn3 + 0.03xn44.20+0.3delaydelayxnyn0.50.24.21The direct form 2 equations are:wn = xn + 1.2wn1 0.5wn2yn = wn 0.2 wn14.22(a)yn = 0.14 yn1 0.38 yn2 + xnynxnwndelaydelay0.140.38+(b)wn = xn 0.14wn1 0.38wn2yn = wnNote that the difference equation diagram for this part is the same as that for part

13、 (a).ynxnwndelaydelay0.140.38+4.23The first ten samples of the impulse response are:n0123456789hn1.01.20.80.40.0 0.00.00.00.00.04.24From the figure, the filter has a finite impulse response. It may be described as a sum of impulse function as:hn = 0.5n + 0.4n1 + 0.3n2 + 0.2n2The difference equation

14、has the parallel form:yn = 0.5xn + 0.4xn1 + 0.3xn2 + 0.2xn34.25The impulse response is finite, with samples as listed in the table.n0123456hn1.0000.3000.0900.0270.0000.0000.000The impulse response samples for a FIR filter serve directly as bk coefficients, soyn = xn + 0.3xn1 + 0.09xn2 + 0.027xn3This

15、 result may also be seen by writing the impulse response in terms of impulse functions:hn = n + 0.3n1 + 0.09n2 + 0.027n34.26 n01234hn 1.00000.30000.24000.19200.1536n56789hn0.12290.09830.07860.06290.05034.27The impulse response may be found from the difference equation ashn = 0.5hn1 + n 0.8n1 The ste

16、p response may be found fromsn = 0.5sn1 + un 0.8un1 or by finding a cumulative sum of the impulse response samples.n0123456789hn1.001.3000.6500.3250.1630.0810.0410.0200.0100.005sn1.000.3000.3500.0250.1880.1060.1470.1270.1370.1324.28The difference equation for a five-term moving average filter isThe

17、impulse response,is plotted below.hnn4.29The impulse response belongs to a non-recursive filter because, after a finite number of samples, the output settles to zero permanently.4.30(a)The response to an impulse function is, by definition, the impulse response. Therefore, the answer to (a) is provid

18、ed in the question.(b)The signal xn consists of two impulse functions with different amplitudes and locations. The response to this input will be the same combination of impulse responses, that is,yn = 0.8hn + 0.5hn10.8hnn0.5hn1n0.8hn+0.5hn1nThe output samples are listed in the following table:n0123456789yn3.24.43.11.80.50.00.00.00.00.04.31The step response can be obtained from sn = un 0.5un1 0.7un2The first ten samples are:n0123456789sn1.000.500.200.200.200.200.200.200.200.204.32(a)The impulse respo