Networking
EX-03 CHAPTER 3- Exercises
Solutions
1. Baseband transmission means sending a digital or an analog signal without modulation using a low-pass channel. Broadband transmission means modulating a digital or an analog signal using a band-pass channel. 2. Optical signals have very high frequencies. A high frequency means a short wavelength because the wavelength is inversely proportional to the frequency ( = c/f), where c is the propagation speed in the media. 3. The frequency domain of a voice signal is normally continuous because voice is a nonperiodic signal.
4. This is broadband transmission because it involves modulation.
5. Each signal is a simple signal in this case. The bandwidth of a simple signal is zero. So the bandwidth of both signals is the same. 6. dB = 10 log10 (90 / 100) = –0.46 dB 7. –10 = 10 log10 (P2 / 5) log10 (P2 / 5) = 1 (P2 / 5) = 10-1P2 = 0.5 W 8. The total gain is 3 4 = 12 dB. The signal is amplified by a factor 10 = 15.85
1.2
9. 100,000 bits / 5 Kbps = 100,000 bits / 5,000 bps = 20 seconds 10. Distance traveled = 1 m 1000 = 1000 m = 1 mm 11. We use Shannon formula: C = BW log2(1+SNR)
C= 4,000 log2 (1 + 1,000) 40 Mbps
12. We use Shannon formula:
C = BW log2(1+SNR)
C = 4,000 log2 (1 + 10 / 0.005) = 43,866 bps
13. To represent 1024 colors, we need log21024 = 10 bits. The total number of bits are, therefore, 1200 1000 10 = 12,000,000 bits
16. Note that Nyquist BitRate= 2 BW log2 L ; Shannon Capacity C = BW log2(1+SNR) a. The data rate is doubled. b. Assume the SNR is doubled to 2 SNR. If C1 = BW log2(1+SNR), then C2 = BW log2(1+2SNR) and the data rate increases slightly. 18. We have
(bit length) = (propagation speed) (bit duration)
The bit duration is the inverse of the bandwidth. 8 a. Bit length = (2 10 m) [(1 / (1 Mbps)] = 200 m. This means a bit occupies 200 meters on a transmission medium. 8 b. Bit length = (2 10 m) [(1 / (10 Mbps)] = 20 m. This means a
Solutions
1. Baseband transmission means sending a digital or an analog signal without modulation using a low-pass channel. Broadband transmission means modulating a digital or an analog signal using a band-pass channel. 2. Optical signals have very high frequencies. A high frequency means a short wavelength because the wavelength is inversely proportional to the frequency ( = c/f), where c is the propagation speed in the media. 3. The frequency domain of a voice signal is normally continuous because voice is a nonperiodic signal.
4. This is broadband transmission because it involves modulation.
5. Each signal is a simple signal in this case. The bandwidth of a simple signal is zero. So the bandwidth of both signals is the same. 6. dB = 10 log10 (90 / 100) = –0.46 dB 7. –10 = 10 log10 (P2 / 5) log10 (P2 / 5) = 1 (P2 / 5) = 10-1P2 = 0.5 W 8. The total gain is 3 4 = 12 dB. The signal is amplified by a factor 10 = 15.85
1.2
9. 100,000 bits / 5 Kbps = 100,000 bits / 5,000 bps = 20 seconds 10. Distance traveled = 1 m 1000 = 1000 m = 1 mm 11. We use Shannon formula: C = BW log2(1+SNR)
C= 4,000 log2 (1 + 1,000) 40 Mbps
12. We use Shannon formula:
C = BW log2(1+SNR)
C = 4,000 log2 (1 + 10 / 0.005) = 43,866 bps
13. To represent 1024 colors, we need log21024 = 10 bits. The total number of bits are, therefore, 1200 1000 10 = 12,000,000 bits
16. Note that Nyquist BitRate= 2 BW log2 L ; Shannon Capacity C = BW log2(1+SNR) a. The data rate is doubled. b. Assume the SNR is doubled to 2 SNR. If C1 = BW log2(1+SNR), then C2 = BW log2(1+2SNR) and the data rate increases slightly. 18. We have
(bit length) = (propagation speed) (bit duration)
The bit duration is the inverse of the bandwidth. 8 a. Bit length = (2 10 m) [(1 / (1 Mbps)] = 200 m. This means a bit occupies 200 meters on a transmission medium. 8 b. Bit length = (2 10 m) [(1 / (10 Mbps)] = 20 m. This means a