Comunicaciones

CHAPTER 3

Data and Signals
Solutions to Odd-Numbered Review Questions and Exercises

Review Questions
1. Frequency and period are the inverse of each other. T = 1/ f and f = 1/T. 3. UsingFourier analysis. Fourier series gives the frequency domain of a periodic signal; Fourier analysis gives the frequency domain of a nonperiodic signal. 5. Baseband transmission means sending a digital oran analog signal without modulation using a low-pass channel. Broadband transmission means modulating a digital or an analog signal using a band-pass channel. 7. The Nyquist theorem defines the maximumbit rate of a noiseless channel. 9. Optical signals have very high frequencies. A high frequency means a short wave length because the wave length is inversely proportional to the frequency (λ =v/f), where v is the propagation speed in the media. 11. The frequency domain of a voice signal is normally continuous because voice is a nonperiodic signal. 13. This is baseband transmission because nomodulation is involved. 15. This is broadband transmission because it involves modulation.

Exercises
17. a. f = 1 / T = 1 / (5 s) = 0.2 Hz b. f = 1 / T = 1 / (12 μs) =83333 Hz = 83.333 × 103 Hz =83.333 KHz c. f = 1 / T = 1 / (220 ns) = 4550000 Hz = 4.55× 106 Hz = 4.55 MHz 19. See Figure 3.1 21. Each signal is a simple signal in this case. The bandwidth of a simple signal is zero. So thebandwidth of both signals are the same. 23. a. (10 / 1000) s = 0.01 s b. (8 / 1000) s = 0. 008 s = 8 ms

1

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Figure 3.1 Solution to Exercise 19
Frequency domain

0

20

50

100 Bandwidth =200 − 0 = 200

200

c. ((100,000 × 8) / 1000) s = 800 s 25. The signal makes 8 cycles in 4 ms. The frequency is 8 /(4 ms) = 2 KHz 27. The signal is periodic, so the frequency domain is made ofdiscrete frequencies. as shown in Figure 3.2. Figure 3.2 Solution to Exercise 27
Amplitude

10 volts

...
10 KHz 30 KHz

Frequency

29.

31. 33. 35. 37.

Using the first harmonic, data...