CHAPTER 3 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 18. A signal has passed through three cascaded amplifiers, each with 5dB gain? What is the total gain? How much is the signal amplified? Since dB is a logrithmic measure, total gain in dB is the sum of the component dB values: Total gain = 5dB + 5dB + 5dB = 15dB For power: dB=10log(P2/P1) For power gain of three stages: 3.1623 x 3.1623 x 3.1623 = 31.6 5dB + 5dB + 5dB = 15dB 15dB=10log(P2/P1) (P2/P1)= 10^(15/10) = 31.6 20. What is the total delay ( latency ) for a frame of size 10 million bits that is being sent on a link with 15 routers each having a queuing time of 2 μs and a processing time of 1μs. The length of the link is 3000 Km. The speed of the light inside the link is 2 × 108 m/s. The link has a bandwidth of 6 Mbps. Which component of the total delay is dominant ? Which one is negligible ? Given message size = 10 million bits = 10 × 106 bits = 10000000 bits queuing time = 2 μs = 2 × 10-6 s processing time =1 μs = 1 × 10-6 s length of link, distance d = 3000 Km = 3000× 103 m The speed of light (Propagation speed) = 2 × 108 m/s Bandwidth , B = 6 Mbps = 6 × 106 bps For delay(latency), use following formula Delay (latency) = propagation time + transmission time + queuing time + processing time ………………………… (1)
Propagation time = =
= 0.015 s
Transmission Time = =
= 1.67 s
Now, from equation (1),we get
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delay(latency) = 0.015 + 1.67 + 2 × 10-6 + 1 × 10-6 Delay (latency) = 1.685 s In this case, because the massage size (frame size) is very long and bandwidth is not very high, the dominant factor is transmission time, not the propagation time and hence transmission time = 1.67 s propagation time is negligible (ignored) because it is very less time than transmission time. propagation time = 0.015 s Both these components are the factors of delay (latency). 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
22. A signal has a wavelength of 2 Micrometer in air. How far can the front of the wave travel during 3000 periods? 2 X 3000 = 6000 micrometers i.e., 6 mm 24. If the bandwidth of the channel is 8 Kbps, How long does it take to send a frame of 200,000 bits out of this device?
200000/8000 = 25 sec. 26. What is the frequency of a signal? (figure given in book) 8 cycles/ 4ms i.e. 8/4 cycles/ms Thus 2000Hz will be the frequency! 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
30. What is the bit rate for each of the following signals? a. A signal in which 1 bit lasts 0.001 s
CHAPTER 3 1000 bps b. A signal in which 5 bits last 4 ms 5/4 bit per ms i.e. 1250 bps c. A signal in which 15 bits last 20 15/20 bits per i.e. 750 kbps 32. Given the following periods calculate the corresponding frequencies. a. 8s 1/8 Hz i.e. 0.125Hz [Remember f = 1 / T] b. 10 f = 1 / T Thus 1/10 100 KHz [1s = 1000000 microseconds] c. 200 ns f = 1/T Thus 1/100ns 5 MHz [1s = 1000000000 ns] 34. If the peak voltage value of a signal is 30 times the peak voltage value of the noise, what is SNR? What is SNRdB? SNR = 30 SNRdB = 10 logSNR = 14.77 36. What is the phase shift for the following? a. A sine wave with maximum amplitude at time zero 90 b. A sine wave with maximum amplitude after ¼ cycle 90 c. A sine wave with zero amplitude after ¾ cycle and increasing 90
CHAPTER 3 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
44. How many bits can fit on a link with a 3ms delay if the bandwidth of the link is a. 2 Mbps 2000000 X 0.003 = 6 Kb b. 15 Mbps 15000000 X 0.003 = 45 Kb c. 150 Mbps 150000000 X 0.03 = 450 Kb 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
