數(shù)字信號處理基礎(chǔ)答案第十二章sol

1、Chapter 12 Solutions12.1Von Neumann architectures rely on one address bus and one data bus, so activities requiring bus usage must wait until the bus is free. Harvard architectures have multiple address and data buses. As a result, less bus contention occurs, and items can be collected from memory i

2、n parallel. Also, Harvard architectures have several memories that can contain program instructions, data, or both; von Neumann architectures feature only one memory that must contain both program instructions and data.12.2Pipelining refers to the overlapping of processor tasks made possible by mult

3、iple buses and multiple memories. A program instruction can be fetched from one memory while a piece of data is collected from another, all while a previous instruction is being executed.12.3(a)4480310 = 1010 1111 0000 0011229910 = 0000 0001 0010 10112(b)The product of the two numbers is 1339609710

4、= 1100 1100 0110 1000 1000 00012. At least twenty-four bits are needed to represent the product.(c)Rounded to its sixteen most significant bits, the product becomes 1100 1100 0110 1001 0000 00002, where bit 8 changes to one because the lower eight bits, 1000 0001 are greater than 1000 0000. The roun

5、ded product equals 1339622410, for an error of 12710.(d)The 32-bit product is 0000 0000 1100 1100 0110 1000 1000 00012. Rounded to the sixteen most significant bits, the product becomes 0000 0000 1100 1100 0000 0000 0000 00002 = 1336934410. An error of 2675310 is committed.12.4PartBinaryDecimal(a)00

6、10 1001 41(b)0101 0010 82(c)0001 0100 20(d)0000 1010 10Note that right shifts halve the number while left shifts double it. When a right shift causes a “one” bit to be lost at the right, precision is lost. For example, half of 41 is 20.5, which cannot be represented. Instead, the result is a truncat

7、ed 20.12.5(a)23745(b)18468(c)1(d)32768(e)3276712.6(a)0000 1100(b)1011 0001(c)1000 0000(d)1111 1111(e)0111 111112.7Before shifting, 0110 11102 = 11010. After shifting, 1101 11002 = 22010.12.8Before shifting, 0110 11102 = 11010. After shifting, 1101 11002 = 3610. The sign of the number changes as a re

8、sult of shifting.12.9(a)An exponent of 4 can be used to normalize all numbers in the block. This choice of exponent ensures one sign bit remains in each number.(b)Before NormalizationAfter Normalization3355360 x 2417862857690414464525840012.10(a)Dynamic Range = 20log(210) = 60.2 dB(b)Dynamic Range =

9、 20log(212) = 72.2 dB(c)Dynamic Range = 20log(214) = 84.3 dB(d)Dynamic Range = 20log(216) = 96.3 dB12.11The weightings1514131211109820212223242526277 6 5 4 3 2 1 02829210211212213214215can be used to find the decimal equivalents:(a)20 + 24 + 26 + 27 + 29 = 0.912109375(b)21 + 22 + 23 + 25 + 213 + 214

10、 + 215 = 0.906463623046875(c)215 = 0.000030517578125(d)20 + 21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 210 + 211 + 212 + 213 + 214 + 215 = 0.000030517578125 (e)20 = 1(f)21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 210 + 211 + 212 + 213 + 214 + 215 = 0.99996948242187512.12(a)The closest 1.15 repres

11、entation is found using truncate(number x 215 + 0.5). This decimal value is converted to binary and hex. For the negative numbers, first re-express as the sum of 1 and a positive number.(i)0111 0000 0000 0000 = 0 x7000(ii)0001 1001 1111 1100 = 0 x19FC(iii)0100 1000 1100 1110 = 0 x48CE(iv)0111 0011 1

12、001 1001 = 0 x7399(v)0.8222 = 1 + 0.1778 = 1000 0000 0000 0000 + 0001 0110 1100 0010 = 1001 0110 1100 0010 = 0 x96C2(vi)0.5194 = 1 + 0.4806 = 1000 0000 0000 0000 + 0011 1101 1000 0100 = 1011 1101 1000 0100 = 0 xBD84(b) The quantized value is the exact decimal value of the 1.15 representation of a nu

13、mber.PartQuantized ValueQuantization Error(i)0.8750.0(ii)0.20300292968750.0000029296875(iii)0.568786621093750.00000337890625(iv)0.00000310546875(v)0.822204589843750.00000458984375(vi)0.000009179687512.13(a)Quantization errors are minimized if the mantissa is maximized, and therefore the exponent is

14、minimized.(i)0.0259 = 0.8288 x 25The mantissa is 0.110 1010 0001; the exponent is 1011, so the floating point number is represented as 1011 0110 1010 0001, or 0 xB6A1. (ii)1.5712 = 0.7856 x 21Mantissa 0.110 0100 1001; exponent 0001. Floating point representation 0001 0110 0100 1001, or 0 x1649.(iii)

15、6.04 = 0.755 x 23Mantissa 0110 0000 1010; exponent 0011. Floating point representation 0011 0110 0000 1010, or 0 x360A.(iv)0.355 = 0.71 x 21 = (1 + 0.29) x 21Mantissa 1010 0101 0010; exponent 1111. Floating point representation 1111 1010 0101 0010, or 0 xFA52.(v)2.111 = 0.52775 x 22 = (1 + 0.47225)

16、x 22Mantissa 1011 1100 0111; exponent 0010. Floating point representation 0010 1011 1100 0111, or 0 x2BC7.(b)Quantization Error = Quantized Value Actual Value, where the quantized value is the exact decimal equivalent of the floating point representation for the number.(i)1.12915 x 107(ii)2.80273 x

17、105(iii)3.90625 x 105(iv) 4.27246 x 106 (v)3.80859 x 10512.14(a)Dynamic Range = = 42.1 dB(b)Dynamic Range = = 186.6 dB12.15(a)Q = R/2N, so 2N = R/Q = 5/0.310 = 16.1 16. The converter appears to use 4 bits.(b)The dynamic range is 20log(2N) = 20log(16) = 24.08 dB.12.16(a)f(x) = x(x(x(x 3) + 2) 1(b)The

18、 original function requires six multiplications. The Horner function requires three multiplications.12.17The exact value of f(0.5) = 2.7182818. (a)Using the first four terms of the power series expansion:(b)12.18(a)To get a 16-point table requires sixteen x values in steps of (/2)/16. xy0.00000.0000

19、0.09820.09800.19630.19510.29450.29030.39270.38270.49090.47140.58900.55560.68720.63440.78540.70710.88360.77300.98170.83151.07990.88191.17810.92391.27630.95691.37440.98081.47260.9952(b)Each waveform can be produced using the sequence of numbers listed in (a).(i)The sine wave sin(x) is produced as foll

20、ows: sequence, reversed sequence, negated sequence, negated reversed sequence.(ii)The cosine wave cos(x) is produced as follows: reversed sequence, negated sequence, negated reversed sequence, sequence.(iii)The sine wave sin(2x) is produced using the same sequences as the sine wave sin(x), but skipping every other sample to increase the frequency.12.19cos(x) The cosine table is constructed from equally-spaced angles between and nearly radians. The last angle is less than radians so that the