数字信号处理英文版课后习题.pdf

Chapter 2 Solutions 2.1 The minimum sampling rate is twice the maximum frequency in the signal, or 44.1 kHz. 2.2 (a) Since = 2f = 20 rad/sec, the frequency of the signal f = 3.18 Hz. The Nyquist rate for the signal is 6.37 Hz (b) 3 5000 , so f = 833.3 Hz. The Nyquist rate is 1666.7 Hz. (c) 7 3000 , so f = 214.3 Hz. The Nyquist rate is 428.6 Hz. 2.3 (a) 125 8000 1 f 1 T S S s (b) The maximum recoverable frequency is half the sampling rate, or 4000 kHz. 2.4 = 4000 rad/sec, so f = 4000/(2) = 2000/ Hz, and T = /2000 sec. Therefore, five periods cover 5/2000 = /400 sec. For this signal, the Nyquist sampling rate is 2(2000/) = 4000/ Hz, so the sampling rate in this case is fS = 4(4000/) = 16000/ Hz. Therefore, the total number of samples collected in five periods is (16000/ samples/sec )(/400 sec) = 40. 2.5 = 2500 rad/sec, so f = 2500/(2) = 1250 Hz, and T = 1/1250 sec. Therefore, five periods cover 5/1250 sec. For this signal, the Nyquist sampling rate is 2(1250) = 2500 Hz, so the sampling rate in this case is fS = 7/8(2500) = 2187.5 Hz. The total number of samples collected is (2187.5 samples/sec)(5/1250 sec) = 8.75. This means only eight samples are collected while five periods of the analog signal elapse. In fact, for this signal, an integer number of samples can never be collected in an integer number of analog periods. 2.6 Copies of the spectrum occur symmetrically around every multiple of the sampling frequency. 2.7 Copies of the signal appear at kfS f. In other words, copies of the spectrum appear around every multiple of the sampling rate. (a) The sampling rate meets Nyquist requirements, so no aliasing occurs. (b) The sampling rate is too low for this signal, so aliasing occurs. Spectral images are shown by dashed lines, while the final spectrum is shown as a solid line. 2.8 The cell phone signal is band-limited. The transmission range is 30 kHz wide. The minimum sampling rate must be at least 60 kHz. In this case, a 60 kHz sampling rate is adequate. A baseband copy of the transmission can be recovered by a filter with a 30 kHz cut-off. 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 Frequency, kHz 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 Frequency, kHz -200020040060080010001200 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1Magnitude Frequency 2.9 (a) Copies of the signal at 300 Hz appear at , 3001000, 3000, 3001000, 3002000 Hz, . These translate to copies at 1300, 700, 300, 700, 1300, 1700, 2300 Hz and so on. Only one of these lies in the Nyquist range (the range recovered after sampling, between 0 and 500 Hz in this case), 300 Hz, the true signal frequency. No aliasing occurs in this case. (b) Copies of the signal appear at , 6001000, 6000, 6001000, 6002000, , or 1600, 600, 400, 400, 600, 1400, 1600, 2600 Hz, etc. The only frequency on this list that falls within the Nyquist range is 400 Hz. This is the aliased frequency. (c) Copies of the signal appear at , 13001000, 13000, 13001000, 13002000, , or 2300, 1300, 300, 300, 700, 1300, 2300, 3300 Hz, etc. Only the 300 Hz signal lies in the Nyquist range. This is the aliased frequency. 2.10 (a) The signal aliases to the range 100 to 400 Hz. Spectral inversion does not occur in the baseband. (b) The band-limited signal aliases to the range 50 to 200 Hz. Spectral inversion occurs in the baseband. 0 30 60 90 120 899970 900000 900030 f (kHz) -500050010001500 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 fS -500050010001500 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 fS Magnitude Frequency Magnitude Frequency 2.11 A 8 kHz sampling rate allows a Nyquist range of 0 to 4 kHz. After sampling, spectral copies of the signal appear at 24000 25000, 16000 25000, 8000 25000, 0 25000, 8000 25000, 16000 25000, Hz. The only copy that lies in the Nyquist range occurs at 1000 Hz. This is the aliased frequency. 2.12 The easiest way to see that these two signals have the same samples is to compute them. The sampling instants are given by nTS, where n is the sample number and TS is the sampling interval, 1/150 sec. n t = nTS x1(t) x2(t) 0 0 1.000 1.000 1 1/150 0.809 0.809 2 2/150 0.309 0.309 3 3/150 0.309 0.309 4 4/150 0.809 0.809 5 5/150 1.000 1.000 6 6/150 0.809 0.809 2.13 Since images of the target appear at 0 0.2 MHz, 2 0.2 MHz, 4 0.2 MHz, 900 0.2 MHz, the actual frequency of the target is 900.2 MHz. 2.14 In each revolution of the wheel, the tire covers d = .635 = 1.995 m. The bicycle travels 15 km/h, or 15000/3600 = 4.17 m/sec. Thus, each tire of the bicycle revolves at the rate of cycles/sec 09. 2 m/cycle 1.995 m/sec 17. 4 . According to Nyquist, at least two samples must be taken per cycle, for a sampling rate of 4.18 samples/sec. This can be accomplished by taking 418 snapshots in every 100 seconds. 2.15 When the signal is sampled at 600 Hz, images of the sine waves frequency on either side of every multiple of the sampling frequency. Since the aliased frequency is 150 Hz, copies appear at 0 150, 600 150, 1200 150, etc. Only 150, 450 and 750 Hz lie below 1 kHz. When the sampling rate is 550 Hz, the aliased frequency is 200 Hz, so copies must appear at 0 50, 550 200, 1100 200, etc. Only 200, 350, 750 and 900 Hz lie below 1 kHz. The only sine wave that is consistent with both sampling results has a frequency of 750 Hz. 2.16 The range of analog voltages is 6 V. (a) quantization step = 6/24 = 375 mV (b) quantization step = 6/28 = 23.44 mV (c) quantization step = 6/216 = 91.55 V 2.17 (a) 28 = 256 (b) 210 = 1024 (c) 212 = 4096 2.18 Quantization maps an infinite number of analog signal levels to a finite number of digital signal levels, determined by the number of bits used. For any finite number of bits, non-zero errors must occur because the quantization step size is non-zero. 2.19 The quantization errors are calculated by subtracting the true signal value from the quantized value. n 0 1 2 3 4 5 6 7 8 9 Quantization Error 0.4 0.3 0.1 0.4 0.3 0.1 0.4 0.1 0.2 0.2 2.20 Quantization Step = Range/2N = 4/24 = 0.25 V The table below describes the quantization method. The bottom range is half a step wide, while the top range is one and a half steps wide. All other ranges are one quantization step wide. The quantization diagram is provided as well. Digital Code Quantized Level (V) Range of Analog Input Mapping to this Digital Code (V) 0000 2.0 2.0 x 1.875 0001 1.75 1.875 x 1.625 0010 1.5 1.625 x 1.375 0011 1.25 1.375 x 1.125 0100 1.0 1.125 x 0.875 0101 0.75 0.875 x 0.625 0110 0.5 0.625 x 0.375 0111 0.25 0.375 x 0.125 1000 0 0.125 x 0.125 1001 0.25 0.125 x 0.375 1010 0.5 0.375 x 0.625 1011 0.75 0.625 x 0.875 1100 1.0 0.875 x 1.125 1101 1.25 1.125 x 1.375 1110 1.5 1.375 x 1.625 1111 1.75 1.625 x 2.0 2.21 n Analog Sample (V) Digital Code Quantized Level (V) Quantization Error (V) 0 0.5715 001 0.625 0.0535 1 4.9575 111 4.375 0.5825 2 0.625 001 0.625 0.0000 3 3.6125 110 3.750 0.1375 4 4.0500 110 3.750 0.3000 5 0.9555 010 1.250 0.2945 6 2.7825 100 2.500 0.2825 7 1.5625 011 1.875 0.3125 8 2.7500 100 2.500 0.25 9 2.8755 101 3.125 0.2495 2.22 Dynamic Range = 20 log (2N) (a) 24.1 dB (b) 48.2 dB (c) 96.3 dB 2.23 Dynamic Range = 20 log (2N) = 60.2 dB This gives 2N = 1023.29. The value of N that most closely satisfies this equation is N = 10. 2.24 Range = R, Quantization Step = Q R = 1 V -2-1.5-1-0.500.511.52 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 Analog Sample Value Quantized Value 0111 0110 0101 0100 0011 0010 0001 0000 1111 1110 1101 1100 1011 1010 1001 1000 Digital Code 0.5Q = 0.1 V Thus, Q = 0.2 V R/Q = 2N = 5 Two quantization bits (N = 2) are not sufficient to meet the requirement. Three quantization bits (N = 3) must be used. 2.25 Bit rate = #bits/sample x #samples/sec = 16 x 8000 = 128 kbps 2.26 The maximum mid-range quantization error is half a quantization step Q. The step size is R/2N for a range R and a number of bits N. Therefo