数字信号处理答案9.docx

1、数字信号处理答案9Chapter 9 Solutions9.1 Substituting hn for yn and n for xn: hn = 0.1n + 0.5n1 + 0.9n2 + 0.5n3 + 0.1n4n0123456789hn0.10.50.90.50.10.00.00.00.00.0The impulse response has a finite number of non-zero terms and is therefore FIR. The length of the impulse response is 5, which is one greater than

2、 the maximum delay in the filter. This is a general relationship for all impulse responses that start at n = 0: the length of the impulse response is one sample larger than the maximum delay in the filter.9.2 (a) (b) (c) (d) 9.3 (a) From question 2,0/8/43/8/25/83/47/820log|H()|0.02.916.914.516.920.5

3、16.931.416.9() ()06813512902345680|H()|()(b) In the pass band, which lies within the first “bump” of the magnitude spectrum, the phase spectrum follows a straight line. This means the phase is linear in this region, which in turn means that no distortion will occur.9.4 (a) The difference equation fo

4、r a three-term moving average filter isThe filter finds the average of each group of three input samples. The first sixteen outputs are:n01234567yn3.333.674.001.001.001.001.001.00n89101112131415yn1.001.001.004.004.004.001.001.00ynnThe filter has the effect of smoothing out the large spikes in the in

5、put. Boundary effects are evident in the first few samples of the output.(b) The frequency response for the three-term moving average filter is 0/4/23/4|H()|1.0000 0.80470.33330.13810.3333The magnitude response for the filter shows a cut-off frequency at approximately 1 radian. This point defines th

6、e pass band edge for this simple filter.|H()|(c) To examine the phase response in the pass band, phases between 0 and 1 radian are calculated in small steps.00.10.20.30.40.50.60.70.80.91.0()0.00.10.20.30.40.50.60.70.80.91.0The values in the table certainly appear to describe a straight line. The tru

7、e phase response, plotted below, confirms that phase is linear in the pass band.()9.5 For this pass band edge frequency the impulse response for an ideal low pass filter isThe sample values are given in the table. Note that at n = 0, it is easiest to evaluate h1n if it is re-expressed asAt n = 0, th

8、e sinc function has the value 1, so h10 = 0.5/.nh1n80.030170.016060.007550.038140.072430.105820.133910.152600.159210.152620.133930.105840.072450.038160.007570.016080.03019.6 (a) The impulse response for the ideal low pass filter is given byThe samples of the impulse response between n = 3 and n = 3

9、aren3210123h1n0.07500.15920.22510.25000.22510.15920.0750After truncation, the impulse response for the non-deal low pass filter consists of the seven non-zero samples listed in the table and zero samples for other values of n. The figure below shows the truncated impulse response.h1nn(b) The impulse

10、 response must be shifted three steps to the right to make it causal. The causal impulse response is shown in the figure. Its equation ishn = 0.0750n + 0.1592n1 + 0.2251n2 + 0.2500n3 + 0.2251n4 + 0.1592n5 + 0.0750n6hnn(c) The frequency response for the impulse response is:0/4/23/4|H()|1.16850.46220.

11、06830.03780.0319|H()|An ideal low pass filter with a cut-off frequency of /4 radians and a pass band gain identical to the non-ideal filter is superimposed on the plot.9.7 (a) The pass band ripple is the maximum deviation from unity gain in the pass band. The minimum gain the pass band is 0.962; the

12、 maximum is 1.078. Thus, the pass band ripple is p = 0.078.(b) The gain at the top of the firsts bump in the stop band is 0.1005. This is the stop band ripple s.(c) Since the pass band ripple is 0.078, the gain at the edge of the pass band is 1 0.078 = 0.922, or 0.705 dB.(d) The gain at the edge of

13、the stop band is 0.1005, or 19.96 dB.(e) The stop band attenuation is the difference between the gain at the edge of the pass band and the gain at the edge of the stop band, in dB. The stop band attenuation is 0.705 (19.96) = 19.26 dB.(f) The pass band ripple defines the edge of the pass band. It oc

14、curs at the frequency 0.672 rads. The edge of the stop band is the first frequency at which the filter gain drops below the stop band ripple. This frequency is 0.892 radians. The transition width is the difference between the stop band edge and the pass band edge, or 0.22 radians.(The magnitude resp

15、onse shown uses a 27-term rectangular window to create a low pass filter with its pass band edge at /4 radians.)9.8 (a) The pass band ripple is practically zero. The exact value of the gain at the edge of the pass band is 20log(1p) = 0.019 dB, for a pass band ripple of p = 0.002.(b) The stop band ri

16、pple is determined by the gain at the edge of the stop band, set by the highest side lobe. From the graph it appears to be about 52 dB. The exact value is 51.5 dB. Therefore 20logs = 51.5, which gives s = 0.0027.(c) The bandwidth is set by the 3 dB frequencies. Their exact values are 1.608 and 2.478

17、 radians, though they cannot be identified with this much precision from the graph. Using fS = 22000, these convert to analog frequencies 5630 and 8677 Hz, for a bandwidth of 3047 Hz.(d) The stop band attenuation is the negative of the gain at the edge of the stop band, or 51.5 dB.(e) The center fre

18、quency is 2.04 radians, or 7143 Hz.(f) The upper edge of the pass band, where the gain is 20log(1p) = 0.019 dB, occurs at 2.365 radians. The frequency at the edge of the stop band occurs at 2.661 radians. Thus, the upper pass band and stop band edges lie at 8281 and 9317 Hz, for a transition width o

19、f 1036 Hz.(The magnitude response shown uses a 71-term hamming window to create a band pass filter with pass band edges at 0.5 and 0.8 radians.)9.9 (a)|H(f)|f(b) Gain at the edge of pass band = 0.72 dB means 1 p = 0.92, or p = 0.08, which is also the required pass band ripple. Stop band attenuation

20、of 25 dB is the same as stop band gain of 25 dB, which equals 0.0562|H(f)|f9.10wn (a) n(b) wnn(c) wnn(d)wnn9.11 (a) 0/4/23/4|W()|9.01.01.01.01.0(b) 0/4/23/4|W()|4.02.00.00.00.0|W()|(c) 0/4/23/4|W()|4.41.760.080.080.08|W()|(d) 0/4/23/4|W()|3.362.00.320.00.0|W()|9.12 (a) A rectangular window gives the

21、 required stop band attenuation. Since N = 0.91(12000)/1000 = 10.92, N = 11 is the best choice.(b) A Hamming window is the best choice. Since N = 3.44(5000)/2000 = 8.6, N = 9 is the best choice.(c) As in (b), a Hamming window is the best choice. Since N = 3.44(5000)/500 = 34.4, N = 35 terms are need

22、ed.(d) The stop band attenuation is the difference between the pass band and stop band gains, or 40 dB. A Hanning window gives the required attenuation. The transition width is the distance in frequency between the stop band and pass band edges, or 1.5 kHz. From the table, N = 3.32(22000)/1500 = 48.

23、7. Since the number of terms must be odd, N = 49 is the best choice.9.13 (a)(i) Some pass band ripple is added to the sketch to show it exists.|H(f)|f(ii) A Hanning window can be used to meet the specifications. The transition width is 2.5 1 = 1.5 kHz. N = 3.32 (12000)/1500 = 26.6, or 27 terms must

24、be included.(iii) The pass band edge frequency used in the design should be (desired pass band edge) + (transition width/2) = 1000 + (1500/2) = 1750 Hz.(b)(i) Some pass band ripple is added to the sketch to show it exists.|H(f)|f(ii) A rectangular window can meet the specifications. N = 0.91(2000)/3

25、30 = 5.52. To be sure the specifications are met, choose N = 7.(iii) To obtain a pass band edge at 500 Hz, the design must proceed with the pass band edge frequency 500 + 330/2 = 665 Hz.9.14 (a) 50 zeros, 50 poles, 51 coefficients(b) 100 zeros, 100 poles, 101 coefficients9.15 The transition width is

26、 500 Hz. To get the pass band edge in the right place, the pass band edge frequency for the design should be pass band edge + transition width/2 = 3000 + 500/2 = 3250 Hz. The matching digital frequency is 1 = 2f/fS = 2(3250)/12000 = 0.5417 rads. Thus, the equation for the ideal low pass filter with

27、the correct pass band edge is h1n = sin(n1)/(n) = sin(0.5417n)/(n). This function is straightforward to evaluate, except at n = 0, where h10 = (1/)sinc(n1) = 0.5417.The required stop band attenuation of 20log(0.01) = 40 dB points to the Hanning window. This windows stop band attenuation leads to a s

28、top band ripple of 10(43/20) = 0.0071 which satisfies the stop band ripple requirements. The pass band ripple for this window is 0.06 dB or 0.007, which also satisfies the specifications. The transition width for the filter is 500 Hz. The number of terms required is 3.32(12000)/500 = 79.7. Using 79 terms will produce a filter slightly poorer than the one specified. Using 81 terms will produce a filter that exceeds the specifications. For this case, select 79 terms. The window function is wn = 0.5 + 0.5cos(2n/N1) = 0.5 + 0.5cos(2n/78), defined for n = 39 to n = 39.