Chapter 3 Problems

Pataasin ang iyong marka sa homework at exams ngayon gamit ang Quizwiz!

Which signal has a wider bandwidth, a sine wave with a frequency of 100 Hz or a sine wave with a frequency of 200 Hz?

Single waves will always have the same bandwidth

The attenuation of a signal is −10 dB. What is the final signal power if it was originally 5 W?

-10 = 10logBASE(10)(P2 / 5) => P2 = .5 W

What is the bit rate for the signal in Figure 3.35?

1 / bit duration 1 / 16 x 10^-9 / 8 = 500 Mbps

What is the bit rate for each of the following signals? a. A signal in which 1 bit lasts 0.001 s b. A signal in which 1 bit lasts 2 ms c. A signal in which 10 bits last 20 μs

1 / bit duration a.) 1 / .001 = 1000 bps b.) 1 / 2 x 10^-3 = 500 bps c.) 1 / (20 x 10^-6 / 10) = 500000 bps = 500kbps

A TV channel has a bandwidth of 6 MHz. If we send a digital signal using one channel, what are the data rates if we use one harmonic, three harmonics, and five harmonics?

1 harmonic: N = 2xB = 12 MHz 3 harmonics: N = 12 MHz / 3 = 4 MHz 5 harmonics: N = 12 MHz / 5 = 2.4 MHz

A signal has a wavelength of 1 μm in air. How far can the front of the wave travel during 1000 periods?

1000 ( 1 x 10^-6) = .001 meter = 1 mm

If the bandwidth of the channel is 5 Kbps, how long does it take to send a frame of 100,000 bits out of this device?

100000 / 5000 = 20 seconds

A computer monitor has a resolution of 1200 by 1000 pixels. If each pixel uses 1024 colors, how many bits are needed to send the complete contents of a screen?

1200 x 1000 x 10 = 12000000 bits logBASE(2)(1024) = 10

What is the bandwidth of a signal that can be decomposed into five sine waves with frequencies at 0, 20, 50, 100, and 200 Hz? All peak amplitudes are the same. Draw the bandwidth.

200 Hz

If the peak voltage value of a signal is 20 times the peak voltage value of the noise, what is the SNR? What is the SNRdB?

20^2 = 400 W = SNR SNRdB = 10logBASE(10)(400) = 26.02 dB

What is the bandwidth of the composite signal shown in Figure 3.37?

25 Hz

What is the transmission time of a packet sent by a station if the length of the packet is 1 million bytes and the bandwidth of the channel is 200 Kbps?

8 x 1 million / 200 x 10^3 = 40 seconds

What is the theoretical capacity of a channel in each of the following cases? a. Bandwidth: 20 KHz SNRdB = 40 b. Bandwidth: 200 KHz SNRdB = 4 c. Bandwidth: 1 MHz SNRdB = 20

C = B(SNRdB / 3) a.) C = (20x10^3)(40/3) = 267 Kbps b.) C = (200x10^3)(4/3) = 267 Kbps c.) C = (1 x 10^6)(20/3) = 6.67 Mbps

A nonperiodic composite signal contains frequencies from 10 to 30 KHz. The peak amplitude is 10 V for the lowest and the highest signals and is 30 V for the 20-KHz signal. Assuming that the amplitudes change gradually from the minimum to the maximum, draw the frequency spectrum.

In notebook

A periodic composite signal contains frequencies from 10 to 30 KHz, each with an amplitude of 10 V. Draw the frequency spectrum.

In notebook

A periodic composite signal with a bandwidth of 2000 Hz is composed of two sine waves. The first one has a frequency of 100 Hz with a maximum amplitude of 20 V; the second one has a maximum amplitude of 5 V. Draw the bandwidth.

In notebook

The light of the sun takes approximately eight minutes to reach the earth. What is the distance between the sun and the earth?

In notebook

We measure the performance of a telephone line (4 KHz of bandwidth). When the signal is 10 V, the noise is 5 mV. What is the maximum data rate supported by this telephone line?

NA

What is the total delay (latency) for a frame of size 5 million bits that is being sent on a link with 10 routers each having a queuing time of 2 μs and a processing time of 1 μs. The length of the link is 2000 Km. The speed of light inside the link is 2 × 10^8 m/s. The link has a bandwidth of 5 Mbps. Which component of the total delay is dominant? Which one is negligible?

Prop. Time = Distance / Prop. Speed = (2000 x 10^3) / 2x10^8 = .01 s Trans. Time = Frame Size / Bandwidth = 5 mil / 5 x 10^6 = 1 s Queueing Time = 10 x 2 x 10^-6 = .00002 s Processing Time = 10 x 1 x 10^-6 = .00001 s Total = .01 + 1 + .00002 + .00001 = 1.01003 s Transmission Time is the dominant term

A line has a signal-to-noise ratio of 1000 and a bandwidth of 4000 KHz. What is the maximum data rate supported by this line?

SNR = 1000 B = 4000 KHz N = B x logBASE(2)(1 + SNR) => (4000 x 10^3)logBASE(2)(1001) = 40 MHz

A signal with 200 milliwatts power passes through 10 devices, each with an average noise of 2 microwatts. What is the SNR? What is the SNRdB?

SNR = 200 x 10^-3 / 10 x 2 x 10^-6 = 10000 W SNRdB = 10logBASE(10)(10000) = 40 dB

We have a channel with 4 KHz bandwidth. If we want to send data at 100 Kbps, what is the minimum SNRdB? What is the SNR?

SNRdB = 3 x C / B = 3(100x10^3)/4x10^3 = 75 dB SNR = 10^(75/10) = 31622776.6

A signal has passed through three cascaded amplifiers, each with a 4 dB gain. What is the total gain? How much is the signal amplified?

Total Gain = 12 dB 12 = 10logBASE(10)(P) => P = 15.85

How many bits can fit on a link with a 2 ms delay if the bandwidth of the link is a. 1 Mbps? b. 10 Mbps? c. 100 Mbps?

a.) (1x10^6)(2x10^-3) = 2000 bits b.) (10x10^6)(2x10^-3) = 20000 bits c.) (100x10^6)(2x10^-3) = 200000 bits

A device is sending out data at the rate of 1000 bps. a. How long does it take to send out 10 bits? b. How long does it take to send out a single character (8 bits)? c. How long does it take to send a file of 100,000 characters?

a.) 10 / 1000 = .01 s = 10 ms b.) 8 / 1000 = .008 s = 8 ms c.) 100000/1000 = 100 s

A file contains 2 million bytes. How long does it take to download this file using a 56-Kbps channel? 1-Mbps channel?

a.) 8 x 2 mil / 56 x 10^3 = 285.714 seconds b.) 8 x 2 mil / 1 x 10^6 = 16 seconds

What is the phase shift for the following? a. A sine wave with the maximum amplitude at time zero b. A sine wave with maximum amplitude after 1/4 cycle c. A sine wave with zero amplitude after 3/4 cycle and increasing

a.) 90 degrees, pi/2 b.) 0 degrees c.) 90 degrees, pi/2

We need to upgrade a channel to a higher bandwidth. Answer the following questions: a. How is the rate improved if we double the bandwidth? b. How is the rate improved if we double the SNR?

a.) Rate is doubled (2B x logBASE(2)(1 + SNR)) b.) Rate is increased slightly (B x logBASE(2)(1 + 2SNR))

Given the frequencies listed below, calculate the corresponding periods. a. 24 Hz b. 8 MHz c. 140 KHz

a.) T = 1 / 24 = .04166 b.) T = 1 / 8 x 10^6 c.) T = 1 / 140 x 10^3

Given the following periods, calculate the corresponding frequencies. a. 5 s b. 12 μs c. 220 ns

a.) f = 1 / 5 b.) f = 1 / 12 x 10^-6 c.) f = 1 / 220 x 10^-9

A signal travels from point A to point B. At point A, the signal power is 100 W. At point B, the power is 90 W. What is the attenuation in decibels?

dB = 10logBASE(10)(90/100) = -.45757 dB

What is the frequency of the signal in Figure 3.36?

f = 1 / (4 x 10^-3) / 8 = 2000 Hz = 2 kHz

What is the length of a bit in a channel with a propagation speed of 2 × 10^8 m/s if the channel bandwidth is a. 1 Mbps? b. 10 Mbps? c. 100 Mbps?

prop. speed / B a.) 2x10^8 / 1 x 10^6 = 200 m b.) 2x10^8 / 10 x 10^6 = 20 m c.) 2x10^8 / 100 x 10^6 = 2 m


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