数字信号处理基础答案第五章sol

1、Chapter 5 Solutions5.1(a)xn = n + 0.6065n1 + 0.3679n2(b)The only non-zero components in the sum are k = 0, k = 1 and k = 2.5.2The systems response to n is hn, while the response to n1 is hn1. Therefore, when the input xn = n + n1 is applied to the system, the output is yn = hn + hn1. The two signals

2、 hn and hn1 can be summed sample by sample.hnnhn1nhn + hn1n5.3(a)hn2hn2n(b)hn+1hn+1n(c)hnhnn(d)h3nh3nn5.4Starting with n = 0, the input xn has the sample values 0, 1 and 2. Starting with n = 0, the impulse response hn has the sample values 1, 2, 3 and 4. These samples are inserted into a convolution

3、 table. Beyond n = 5, all outputs are zero. This is a direct result of the finite lengths of the input and impulse response signals.xk012hk4321y0 = 0h1k4321y1 = 1h2k4321y2 = 4h3k4321y3 = 7h4k4321y4 = 10h5k4321y5 = 8h6k4321y6 = 05.5(a)xk1234567hk1000y0 = 0h1k1000y1 = 0h2k1000y2 = 0h3k1000y3 = 1h4k100

4、0y4 = 2h5k1000y5 = 3h6k1000y6 = 4(b)The impulse response hn = nk delays the signal by k steps.5.6(a)All output samples before n = 0 and after n = 8 are zero.xk1111111hk0.20.41?y0 = 1.0h1k0.20.41?y1 = 0.6h2k0.20.41y2 = 0.4h3k0.20.41y3 = 0.4h4k0.20.41y4 = 0.4h5k0.20.41y5 = 0.4h6k0.20.41y6 = 0.4h7k0.20

5、.41?y7 = 0.6h8k0.20.41?y8 = 0.2(b)The outputs influenced by boundary effects are indicated by question marks.5.7The convolution of the impulse response with the step function produces the step response. The samples in the step response are, of course, cumulative sums of the impulse response samples.

6、n0123456789hn1.0000.8500.7230.6140.5220.4440.3770.3210.2730.232sn1.0001.8502.5733.1873.7094.1524.5304.8505.1235.354n10111213141516171819hn0.1970.1670.1420.1210.1030.0870.0740.0630.0540.046sn5.5515.7185.8615.9816.0846.1726.2466.3096.3636.408After 17 samples, the step response samples change by less t

7、han 1%. This point may be considered the beginning of an approximate steady state.5.8 (a)xk010101010hk1y0 = 0.0h1k0.61y1 = 1.0h2k0.360.61y2 = 0.6h3k0.2160.360.61y3 = 0.64h4k0.1300.2160.360.61y4 = 0.384h5k0.0780.1300.2160.360.61y5 = 0.770h6k0.0470.0780.1300.2160.360.61y6 = 0.462h7k0.0280.0470.0780.13

8、00.2160.360.61y7 = 0.723h8k0.0170.0280.0470.0780.1300.2160.360.61y8 = 0.434The output values for n = 10 to n = 19 are added below, and all twenty output samples are plotted.n91011121314yn0.7400.44380.73370.44020.73590.4415n1516171819yn0.73510.44110.73540.44120.7353ynn(b)The samples values in one cyc

9、le of the output remain within 1% of the samples in the preceding cycle beginning with sample n = 9. Thus, an approximate steady state is reached after 9 samples.5.9(a)The convolution of the impulse response and the input produces the output samples listed.n01234yn5.00007.50008.75009.37509.6875n5678

10、9yn9.84389.92199.96099.98059.9902(b)Beginning with sample n = 6, the output changes by less than 1% from the previous sample. Steady state begins at this point. The approximate steady state output level is close to one.5.10(a)The period of the input is 2= 2/2/5 = 5.(b)The output is obtained through

11、convolution. xk0.000.9510.5880.5880.9510.000.951hk2y0 = 0.0h1k32y1 = 1.902h2k432y2 = 4.029h3k432y3 = 4.392h4k432y4 = 1.314xk0.5880.9510.000.9510.5880.5880.951h5k432y5 = 5.2043h6k432y6 = 1.902h7k432y7 = 4.028h8k432y8 = 4.392h9k432y9 = 1.314(c)ynnsteady state behaviortransient behavior(d)In steady sta

12、te, the output repeats every five samples, just like the input.5.11(a)xk111111hk000y0 = 0h1k1000y1 = 0h2k11000y2 = 0h3k111000y3 = 1h4k1111000y4 = 2h5k11111000y5 = 3h6k01111100y6 = 4h7k00111110y7 = 5h8k00011111y8 = 5h9k00001111y9 = 5ynn(b)The steady state portion of the output begins with sample n =

13、7. The transient portion comprises samples n = 0 to n = 6.5.12(a)To obtain the impulse response of the cascaded system, an impulse function is applied at the input of the first system. The output is h1n, by definition. This output acts as the input to the second system, and is convolved with h2n to

14、find the overall impulse response hn. This is general result: The impulse response of a two filters cascaded together is the convolution of the filters impulse responses.h1k100.100.2h2k0.36790.60651h0 = 1.0h21k0.22310.36790.60651h1 = 0.6065h22k0.22310.36790.60651h2 = 0.2679h23k0.22310.36790.60651h3

15、= 0.1625h24k0.22310.36790.60651h4 = 0.1632h25k0.22310.36790.60651h5 = 0.0990h26k0.22310.36790.6065h6 = 0.0736h27k0.22310.3679h7 = 0.0446h28k0.2231h8 = 0.0(b)The step response may be found by convolving the impulse response hn with the step function, or by computing cumulative sums of the impulse res

16、ponse samples. The step response samples are listed in the table, and both the impulse and step responses of the system are plotted below.n012345678sn1.01.60651.87442.03692.20012.29912.37272.41732.4173snhnnn5.13The step response is the response of the system to an input step, xn = un.xk111111hk0.135

17、0.3681y0 = 1.0h1k0.1350.3681y1 = 1.368h2k0.1350.3681y2 = 1.503h3k0.1350.3681y3 = 1.503h4k0.1350.3681y4 = 1.503h5k0.1350.3681y5 = 1.503The output approaches a constant value because the input has a constant value.5.14To re-express the difference equation, the impulse response must be found first. The

18、 impulse response can be found using the difference equation, that is,hn = 0.8hn1 + nThe first ten samples are:n01234hn10.80.640.5120.410n56789hn0.3280.2620.2100.1680.134The difference equation becomes yn = xn 0.8xn1 + 0.64xn2 0.512xn3 + 0.410 xn4 0.328xn5 + or, more compactly, .5.15The impulse resp

19、onse samples are given in the table:n01234hn1.00000.74080.54880.40660.0000This is a finite impulse response, so the coefficients bk for the difference equation match the impulse response samples:yn = xn + 0.7408xn1 + 0.5488xn2 + 0.4066xn3From this difference equation, the step response may be obtain

20、ed assn = un + 0.7408un1 + 0.5488un2 + 0.4066un3The first eight samples are listed in the table:n01234567sn1.00001.74082.28962.69622.69622.69622.69622.6962Alternatively, the step response may be found using convolution. The table below shows the calculations. Convolution and the difference equation

21、give the same results.xk111111hk0.74081.000y0 = 1.000h1k0.54880.74081.000y1 = 1.7408h2k0.40660.54880.74081.000y2 = 2.2896h3k0.40660.54880.74081.000y3 = 2.6962h4k0.40660.54880.74081.000y4 = 2.6962h5k0.40660.54880.74081.000y5 = 2.6962h6k0.40660.54880.7408y6 = 2.6962h7k0.40660.5488y7 = 2.69625.16The im

22、pulse response for a nine-term moving average filter ishn = Convolution shows that the step response samples are cumulative sums of the impulse response samples.n01234567sn1/92/93/94/95/96/97/98/9n89101112131415sn11111111Eight samples are affected by boundary effects. Steady state begins with sample

23、 n = 8.5.17Since the impulse response for an 11-term moving average filter has 11 terms, the first 11 1 = 10 samples and the last 10 samples are influenced by boundary effects. Therefore, 20 samples are affected, while 2048 20 = 2028 are not.5.18The impulse response for a 5-term moving average filte

24、r is:hn = Sixteen samples of the output are computed in the convolution table, and are plotted below. The filter has the effect of smoothing out the abrupt transitions from the otherwise steady signal.xk0.50.50.510.50.50.51hk0.20.2y0 = 0.1h1k0.20.20.2y1 = 0.2h2k0.20.20.20.2y2 = 0.3h3k0.20.20.20.20.2

25、y3 = 0.5h4k0.20.20.20.20.2y4 = 0.6h5k0.20.20.20.20.2y5 = 0.6h6k0.20.20.20.20.2y6 = 0.6h7k0.20.20.20.20.2y7 = 0.7xk0.50.50.510.50.50.510.50.5h8k0.20.20.20.20.2y8 = 0.6h9k0.20.20.20.20.2y9 = 0.6h10k0.20.20.20.20.2y10 = 0.6h11k0.20.20.20.20.2y11 = 0.7h12k0.20.20.20.20.2y12 = 0.6h13k0.20.20.20.20.2y13 =

26、 0.6 xk0.510.50.50.510.50.50.51h14k0.20.20.20.20.2y14 = 0.6h15k0.20.20.20.20.2y15 = 0.7yn n5.19(a)The difference equation isThe impulse response is(b)The first few samples are calculated as follows:xk0.0000.5000.8661.000.8660.5hk1/31/31/3y0 = 0.0h1k1/31/31/3y1 = 0.167h2k1/31/31/3y2 = 0.455h3k1/31/31

27、/3y3 = 0.789h4k1/31/31/3y4 = 0.911h5k1/31/31/3y5 = 0.789xk0.8660.50.0000.5000.8661.000.8660.5h6k1/31/31/3y6 = 0.455h7k1/31/31/3y7 = 0.0h8k1/31/31/3y8 = 0.455h9k1/31/31/3y9 = 0.789h10k1/31/31/3y10 = 0.911h11k1/31/31/3y11 = 0.789xk0.8660.50.0000.5000.8661.000.8660.5h12k1/31/31/3y6 = 0.455h13k1/31/31/3

28、y7 = 0.0h14k1/31/31/3y8 = 0.455nnxnynInput SignalOutput Signal(c)As the above table shows, only the samples for n = 0 and n = 1 are influenced by boundary effects. These samples also form the transient portion of the output response. The output begins to show its steady state behavior with the sampl

29、e at n = 2.5.20(a)The easiest way to deduce the impulse response is to identify impulse response samples one at a time from the table below.xk321hk?2y0 = 6h1k?02y1 = 4h2k?102y2 = 1h3k0102y3 = 2h4k0102y4 = 1h5k010y5 = 0The impulse response is hn = 2n n2.(b)The output is calculated in the convolution

30、table below.xk20100.5hk102y0 = 4h1k102y1 = 0h2k102y2 = 4h3k102y3 = 0h4k102y4 = 2h5k102y5 = 0h6k10y6 = 0.5h7k1y7 = 0.05.21The 12 x 12 original image has the gray scale values:255255255255255255000000255255255255255255000000255255255255255255000000255255255255255255000000255255255255255255000000255255255255255255000000000000255255255255255255000000255255255255255255000000255255255255255255000000255255255255255255000000255255255255255255000000255255255255255255The convolution kernel for the smoothing filter isTo avoid boundary effects