數(shù)字信號(hào)處理基礎(chǔ)答案第十三章sol

1、Chapter 13 Solutions13.1The ADSP-2181 uses 16 bits for both signed and unsigned formats.(a)0000 1001 1011 0011(b)0000 1001 1011 0011For positive integers, there is no difference between the signed and unsigned formats.13.2(a)(i)The 1.15 representation is given by truncate(0.93207 x 215 + 0.5) = 3054

2、210 = 0 x774E = 0111 0111 0100 11102.(ii)The 6.10 representation is given by truncate(0.93207 x 210 + 0.5) = 95410 = 0 x03BA = 0000 0011 1011 10102.(iii) The 12.4 representation is given by truncate(0.93207 x 24 + 0.5) = 1510 = 0 x000F = 0000 0000 0000 11112. (b)(i)The quantized value is 0.932067871

3、09375, for a quantization error of 0.00000212890625.(ii)The quantized value is 0.931640625, for an error of 0.000429375.(iii)The quantized value is 0.9375, giving a quantization error of 0.00543. 13.3(a)The largest value is 0111 11.11 1111 11112, or 31.9990234375, and the smallest value is 1000 00.0

4、0 0000 0000, or 32, so the range is 63.9990234375. The smallest resolvable difference is 210. Thus, the dynamic range is 20log(full scale range/smallest resolvable difference) = 96.33 dB.(b)The largest value is 0111 1111 1111.11112, or 2047.9375, and the smallest value is 1000 0000 0000.0000, or 204

5、8, so the range is 4095.9375. The smallest resolvable difference is 24. Thus, the dynamic range is 20log(full scale range/smallest resolvable difference) = 96.33 dB.(c)The largest value is 0111 1111 1111 11112, or 32767, and the smallest value is 1000 0000 0000 0000, or 32768, so the range is 65535.

6、 The smallest resolvable difference is 1. Thus, the dynamic range is 20log(full scale range/smallest resolvable difference) = 96.33 dB.Note that all three fixed point formats have the same dynamic range.13.4(a)In 1.3 format, 0.61310 = 0.1012. Therefore, the quantized value is 0.625.In 2.2 format, 1.

7、8510 = 01.112, so the quantized value is 1.75.(b)The product of the quantized numbers is (0.625)(1.75) = 1.0937510 = 01.00 0110 0000 0000. A minimum 2.5 format is needed to represent this product exactly. Note that a format 1.3 number multiplied by a 2.2 number will give at worst a number in (1+2).(

8、3+2) = 3.5 format.(c)The true product is (0.613)(1.85) = 1.13405, 0.0403 (or 3.55%) greater than the quantized product.13.5The coefficients for the filter are 5, 1, 5. To use 1.15 format, the coefficients must be scaled down by a factor of 8, or 23, which gives scaled coefficients 0.625, 0.125, 0.62

9、5. The inputs are too large too be accommodated by 1.15 format. Because they lie between 2 V and 2 V, they must be scaled down by a factor of 2. The scaled filter implementation isxSn = 0.5xnySn = 0.625xSn + 0.125xSn1 + 0.625xSn2yn = 2(8ySn) = 16ySnThe difference equation diagram is shown below. At

10、the right, the scaling factors, 8 for the coefficients and 2 for the inputs, are re-introduced to correct the calculation.xnyn+delay0.625delayxn1xn2821/20.6250.12513.6(a)The direct form 2 equations arewn = xn + 2.0651wn1 1.5200wn2 + 0.3861wn3yn = 0.0086wn + 0.0258wn1 + 0.0258wn2 + 0.0086wn3(b)The bk

11、 coefficients do not require scaling down, as they already fall within the range for a 1.15 representation, but they can benefit from scaling up, to improve the precision of the calculations. The bk coefficients can be scaled up by a factor of 32 and still fit into the 1.15 range. The ak coefficient

12、s must be scaled down by a factor of 4. The inputs do not require scaling, though they are divided by 4 in the process of scaling down the ak coefficients. If the inputs become too small as a result, they can be scaled appropriately by scaling factor C. The new equations are:wSn = 0.25xn + 0.516275w

13、n1 0.3800wn2 + 0.096525wn3wn = 4wSnySn = 0.2752wn + 0.8256wn1 + 0.8256wn2 + 0.2752wn3yn = = 0.03125ySn(c)The difference equation diagram is shown below. To accommodate the limited size of the diagram, the coefficient 0.516275 is rounded to 0.5163, and the coefficient 0.096525 is rounded to 0.0965.yn

14、xnwndelaydelaydelay0.27520.82560.09650.51630.380+0.2540.031250.82560.275213.7No scaling is necessary. b0 = b6 = 0.000 0000 1100 0001 = 0 x00C1b1 = b5 = 1.111 1001 1011 1110 = 0 xF9BEb2 = b4 = 0.001 0101 1111 0111 = 0 x15F7b3 = 0.101 1111 0000 1110 = 0 x5F0E13.8Because register l1 is zero, the buffer

15、 is linear. After ten reads, the my1 register contains the value 0 x000a, or 10.13.9Because register l1 is eight, the buffer is circular. After ten reads, the my1 register contains the value 0 x0002, or 2.13.10(a) 0001 0001 0100 1010 + 0010 0000 0000 00100011 0001 0100 1100 The register ar contains

16、the sum 0 x314C.(b)Subtracting is the same as adding the twos complement. The twos complement of 0 x2002 = 0010 0000 0000 0010 is 1101 1111 1111 1110, or 0 xDFFE. Therefore,0001 0001 0100 10101101 1111 1111 11101111 0001 0100 1000 The register ar contains the sum 0 xF148.(b)The round mode assumes tw

17、o signed operands, so the analysis proceeds in the same manner as in (a). The last step, however, is to round the result to the top 16 bits. The result in (a) was 1.110 1101 0101 0101 1100 0000 0000 0000, with the bottom 16 bits highlighted. Rounding is required because the highlighted bits exceed 0

18、 x8000. To round, a one is added at bit position 15, top left-most highlighted bit, which in this case causes a carry into bit 16 to give a result in mr of 1.110 1101 0101 0110 0 xxx xxxx xxxx xxxx, where the x bits are irrelevant. With sign extension into mr2, the 24-bit result in mr2 mr1 is 0 xFF

19、ED56.(c)Both numbers are unsigned. In this format, bit 16 has the weighting 2013.12The binary representation of the number in register ar is 0010 1110 1001 0110.(a)The number is copied into sr0 and shifted to the right by three positions. After shifting, sr contains 0000 0000 0000 0000 0000 0101 110

20、1 0010, or 0 x0000 05D2.(b)The number is copied into sr0 and shifted to the left by one position. After shifting, sr contains 0000 0000 0000 0000 0101 1101 0010 1100, or 0 x0000 5D2C.(c)The number is copied into sr1 and shifted to the right by two positions. After shifting, sr contains 0000 1011 101

21、0 0101 1000 0000 0000 0000, or 0 x0BA5 8000.(d)The number is copied into sr1 and shifted to the left by three positions. After shifting, sr contains 0111 0100 1011 0000 0000 0000 0000 0000, or 0 x74B0 0000.13.13The loop in this factorial program segment multiplies the values 6, 5, 4, 3, 2 and 1 to g

22、ive a product of 720. Integer arithmetic is used in this program, so results of the multiplications appear in mr0. At the end of the loop, my1 = mr0 = 0 x02D0, or 720, and ar = 0. It would be reasonable to guess that the expired counter has the value 0, but at the conclusion of the loop cntr = undef

23、ined.13.14This program segment computes a single filter output for an FIR filter with four terms. A circular buffer “coeff” contains the decimal values 0.25, 0.5, 0.25 and 0.125. A circular buffer “data_in” contains the values 0.5625, 0.25, 0.75 and 0.625. The filter output is (0.25)(0.5625) + (0.5)(0.25) + (0.25)(0.75) + (0.125)(0.625) = 0.53125. The first three partial sums are 0 x00 1200 0000, 0 x00 2200 0000, and 0 x00 3a00 0000. After the final sum and rounding step, the final answer is stored in register mr1 as 0 x4400, which is the 1.15 representation f