Multiple Choices: 1. Before data can be transmitted, they must be transformed to ________. a. Periodic signals b. Electromagnetic signals c. Aperiodic signals d. Low-frequency sine waves 2. A periodic signal completes one cycle in 0.001 s. What is the frequency? a. 1 Hz b. 100 Hz c. 1 KHz d. 1 MHz 3. As frequency increases, the period ________. a. Decreases b. Increases c. Remains the same d. Doubles 4. Given two sine waves A and B, if the frequency of A is twice that of B, then the period of B is ________ that of A. a. One-half b. Twice c. The same as d. Indeterminate from 5. If the maximum amplitude of a sine wave is 2 V, the minimum amplitude is________ V. a. 2 b. 1 c. –2 d. Between –2 and 2 6. A signal is measured at two different points. The power is P1 at the first point and P2 at the second point. The dB is 0. This means ________. a. P2 is zero b. P2 equals P1 c. P2 is much larger than P1 d. P2 is much smaller than P1 7. ________ is a type of transmission impairment in which the signal loses strength due to the resistance of the transmission medium. a. Attenuation b. Distortion c. Noise d. Decibel 8. ________ is a type of transmission impairment in which the signal loses strength due to the different propagation speeds of each frequency that makes up the signal. a. Attenuation b. Distortion c. Noise d. Decibel
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9. ________ is a type of transmission impairment in which an outside source such as crosstalk corrupts a signal. a. Attenuation b. Distortion c. Noise d. Decibel 10. Using the Shannon formula to calculate the data rate for a given channel, if C = B, then ________. a. The signal is less than the noise b. The signal is greater than the noise c. The signal is equal to the noise d. Not enough information is given to answer the question 11. Data can be ________. a. analog b. digital c. a or b d. none of the above 12. ________is the rate of change with respect to time.
a. Amplitude b. Time c. Frequency d. Voltage 13. _______ data are continuous and take continuous values.
a. analog b. digital c. (a) or (b) d. none of the above
Exercises (From the text book -Data communication and networking-) 16- We measure the performance of a telephone line (4 KHz of bandwidth). When the signal is 20 V, the noise is 6 mV. What is the maximum data rate supported by this telephone line? Capacity=B*log2(1+SNR)= 4000 * (log(1 + (20 / 0.006)) / log(2)) = 46 812.7305 bps 21- Given the frequencies listed below, calculate the corresponding periods. a. 20Hz T = 1 / f = 1 / (20 Hz)=0.0500 s=50 ms. b. 10 MHz T = 1 / f = 1 / (10 MHz) = 0.0000001 = 0.1 × 10–6 s = 0.125 μs c. 150 KHz T = 1 / f = 1 / (150 KHz)=0.0.00000666 s=6.66 × 10–6 s=6.66 ms 26- What is the frequency of the signal in Figure ?
The signal makes 8 cycles in 4 ms. The frequency is 8 /(4 ms) = 2 KHz
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27- A signal travels from point A to point B. At point A, the signal power is 200 W. At point B, the power is 170 W. What is the attenuation in decibels? dB = 10 log10 (170 / 200) = –0.705 dB 32- Given the following periods, calculate the corresponding frequencies. a. 8 s = f = 1 / T =1/8=0.125 Hz b. 10 µs f = 1 / T =1/10 µs =100000 Hz=100 KHz c. 200 ns 6 f = 1 / T = 1 / (200 ns) = 5000000 Hz = 5× 10 Hz = 5 MHz 37- A line has a signal-to-noise ratio of 2000 and a bandwidth of 5000 KHz. What is the maximum data rate supported by this line? 5 000 * (log(2001) / log(2)) = 54832.5273 kbps 40- What is the theoretical capacity of a channel in each of the following cases: a. Bandwidth: 20 KHz SNRdB =40 c= B*((log(1+10^(SNRdb/10)))/log(2))= 20*((log(1+10^(40/10)))/log(2))= 265 kbps b. Bandwidth: 200 KHz SNRdB =6 c=200 * (log(1 + (10^(6 / 10))) / log(2)) = 463.291236 kbps c. Bandwidth: 1 MHz SNRdB =20 c=1 * (log(1 + (10^(20 / 10))) / log(2)) = 6.658 Mbps 41- The attenuation of a signal is -12 dB. What is the final signal power if it was originally 4 W? -12=10*log(p2/4) --> -12/10= log(p2/4) --> -1.2= log(p2/4) --> 10^-1.2=(p2/4) --> p2=0.2523 W 46- A signal with 300 milliwatts power passes through 10 devices, each with an average noise of 3 microwatts. What is the SNR? What is the SNRdB? SNR = (300 mW) / (10 × 3 × μW) =300*10^-3/30*10^-6= 10,000 SNRdB = 10 log10 SNR = 10 log10 10000=40 48- A computer monitor has a resolution of 1300 by 1000 pixels. If each pixel uses 1024 colors, how many bits are needed to send the complete contents of a screen? To represent 1024 colors, we need log2 1024 = 10 bits. The total number of bits are, therefore, 1300 × 1000 × 10 = 13,000,000 bits
