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Chapter 1: Problem 29

For the following, as in the previous problem, assume that no data compression is done. Calculate the bandwidth necessary for transmitting in real time: (a) HDTV high-definition video at a resolution of \(1920 \times 1080,24\) bits/pixel, 30 frames/second. (b) POTS (plain old telephone service) voice audio of 8-bit samples at \(8 \mathrm{KHz}\). (c) GSM mobile voice audio of 260 -bit samples at \(50 \mathrm{~Hz}\). (d) HDCD high-definition audio of 24-bit samples at \(88.2 \mathrm{KHz}\).

Short Answer

Expert verified
Meningkatkan bandwidth untuk HDTV: 1.49 Gbps, POTS: 64 Kbps, GSM: 13 Kbps, HDCD: 2.12 Mbps.

Step by step solution

01

- Understanding HDTV Video Bandwidth Requirement

To calculate the bandwidth for HDTV high-definition video, use the formula: Bandwidth (in bits/second) = Resolution (pixels/frame) × Color Depth (bits/pixel) × Frame Rate (frames/second). Given: Resolution = 1920 × 1080 pixels, Color Depth = 24 bits/pixel, Frame Rate = 30 frames/second.
02

- Calculate HDTV Video Bandwidth

Calculate the total bandwidth:dBandwidth = (1920 × 1080) pixels/frame × 24 bits/pixel × 30 frames/second.= 1920 × 1080 × 24 × 30 bits/second= 1,492,992,000 bits/second= 1.49 Gbps
03

- Understanding POTS Voice Audio Bandwidth Requirement

For POTS voice audio, the formula is: Bandwidth (in bits/second) = Sample Size (bits/sample) × Sampling Rate (samples/second). Given: Sample Size = 8 bits/sample, Sampling Rate = 8 KHz.
04

- Calculate POTS Voice Audio Bandwidth

Calculate the total bandwidth:dBandwidth = 8 bits/sample × 8,000 samples/second.= 64,000 bits/second= 64 Kbps
05

- Understanding GSM Voice Audio Bandwidth Requirement

For GSM voice audio, the formula is: Bandwidth (in bits/second) = Sample Size (bits/sample) × Sampling Rate (samples/second). Given: Sample Size = 260 bits/sample, Sampling Rate = 50 Hz.
06

- Calculate GSM Voice Audio Bandwidth

Calculate the total bandwidth:dBandwidth = 260 bits/sample × 50 samples/second.= 13,000 bits/second= 13 Kbps
07

- Understanding HDCD Audio Bandwidth Requirement

For HDCD audio, the formula is: Bandwidth (in bits/second) = Sample Size (bits/sample) × Sampling Rate (samples/second). Given: Sample Size = 24 bits/sample, Sampling Rate = 88.2 KHz.
08

- Calculate HDCD Audio Bandwidth

Calculate the total bandwidth:dBandwidth = 24 bits/sample × 88,200 samples/second.= 2,116,800 bits/second= 2.12 Mbps

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

HDTV video bandwidth
HDTV, or High-Definition Television, requires a substantial amount of data to be transmitted for high-quality video. This data is typically measured in terms of bandwidth. Bandwidth refers to the rate at which data is transmitted, usually in bits per second. To determine the bandwidth for HDTV video, you need to know three key parameters:

  • Resolution: This is the number of pixels in each frame. For an HDTV with a resolution of 1920x1080 pixels, the total number of pixels per frame is 1920 multiplied by 1080.
  • Color Depth: This indicates how many bits are used to represent the color of each pixel. For HDTV, it is commonly 24 bits per pixel.
  • Frame Rate: This indicates how many frames are displayed per second. Commonly, HDTV uses 30 frames per second.

Using these parameters, you can calculate the bandwidth using the formula:

\[ \text{Bandwidth} = \text{Resolution} \times \text{Color Depth} \times \text{Frame Rate} \]

The calculation for a 1920x1080 resolution at 24 bits per pixel and 30 frames per second is as follows:

\[1920 \times 1080 \times 24 \times 30 = 1,492,992,000 \text{ bps} = 1.49 \text{ Gbps} \]

This means you need 1.49 Gbps of bandwidth to transmit HDTV video in real-time. This highlights the substantial data requirements for high-definition video.
POTS voice audio
POTS or Plain Old Telephone Service refers to the traditional telephone system that has been in use for decades. Transmitting voice over POTS requires significantly less bandwidth than HDTV.

To calculate the bandwidth needed for POTS voice audio, two key parameters are used:

  • Sample Size: This is the number of bits used to represent each sample of audio. For POTS, it is typically 8 bits per sample.
  • Sampling Rate: This is the number of audio samples taken per second. For POTS, this is 8,000 samples per second (8 KHz).

The formula to determine the bandwidth is simple:

\[ \text{Bandwidth} = \text{Sample Size} \times \text{Sampling Rate} \]

Using the given numbers:

\[8 \text{ bits/sample} \times 8,000 \text{ samples/second} = 64,000 \text{ bps} = 64 \text{ Kbps} \]

Therefore, the bandwidth required for POTS voice audio is 64 Kbps. This relatively low bandwidth requirement allows POTS to efficiently transmit voice audio with the limited data rates available on traditional telephone lines.
GSM mobile voice audio
GSM, or Global System for Mobile Communications, is a standard developed to describe the protocols for second-generation (2G) digital cellular networks. Voice transmission in GSM is more data-efficient than POTS due to data compression techniques used in GSM.

For GSM mobile voice audio, the necessary parameters for bandwidth calculation are:

  • Sample Size: GSM voice samples are 260 bits each.
  • Sampling Rate: The GSM standard uses a sampling rate of 50 Hz (samples per second).

To calculate the required bandwidth, use the formula:

\[ \text{Bandwidth} = \text{Sample Size} \times \text{Sampling Rate} \]

Applying the values for GSM:

\[260 \text{ bits/sample} \times 50 \text{ samples/second} = 13,000 \text{ bps} = 13 \text{ Kbps} \]

Thus, GSM mobile voice audio requires a bandwidth of just 13 Kbps. This low bandwidth requirement allows mobile networks to handle many simultaneous voice calls efficiently.
HDCD high-definition audio
HDCD, or High Definition Compatible Digital, is an audio format used to deliver higher-quality audio. Like HDTV, HDCD requires more bandwidth than standard formats to transmit its higher quality.

Two main parameters are essential for calculating the bandwidth of HDCD audio:

  • Sample Size: HDCD uses 24 bits per sample.
  • Sampling Rate: The standard sampling rate for HDCD is 88.2 KHz (88,200 samples per second).

To determine the needed bandwidth for HDCD audio, the formula is:

\[ \text{Bandwidth} = \text{Sample Size} \times \text{Sampling Rate} \]

Substituting the given values:

\[24 \text{ bits/sample} \times 88,200 \text{ samples/second} = 2,116,800 \text{ bps} = 2.12 \text{ Mbps} \]

Therefore, 2.12 Mbps is required to transmit HDCD audio in real-time. This high bandwidth facilitates the delivery of superior audio quality, making HDCD an excellent choice for audiophiles.
bandwidth calculation formula
Understanding how to calculate bandwidth is crucial for various data transmission applications. The formula to calculate the bandwidth required for transmitting digital data involves the following parameters:

  • Sample Size: The number of bits used to store each sample of data.
  • Sampling Rate: The number of samples taken per second.

The general formula to compute bandwidth is:

\[ \text{Bandwidth} = \text{Sample Size} \times \text{Sampling Rate} \]

For instance, in the case of audio, imagine you have a sample size of 16 bits per sample and a sampling rate of 44.1 KHz (common for CD-quality audio). The bandwidth calculation would be:

\[16 \text{ bits/sample} \times 44,100 \text{ samples/second} = 705,600 \text{ bps} \]

This example can be applied to various types of data transmission scenarios including video, voice, and high-definition audio. Understanding the bandwidth calculation helps students and professionals design and assess systems for effective data transmission.

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Most popular questions from this chapter

Use a Web search tool to locate useful, general, and noncommercial information about the following topics: MBone, ATM, MPEG, IPv6, and Ethernet.

For the following, assume that no data compression is done; this would in practice almost never be the case. For (a)-(c), calculate the bandwidth necessary for transmitting in real time: (a) Video at a resolution of \(640 \times 480,3\) bytes/pixel, 30 frames/second. (b) \(160 \times 120\) video, 1 byte/pixel, 5 frames/second. (c) CD-ROM music, assuming one CD holds 75 minutes' worth and takes \(650 \mathrm{MB}\). (d) Assume a fax transmits an \(8 \times 10\)-inch black-and-white image at a resolution of 72 pixels per inch. How long would this take over a 14.4-Kbps modem?

Consider a simple protocol for transferring files over a link. After some initial negotiation, A sends data packets of size \(1 \mathrm{~KB}\) to B; B then replies with an acknowledgment. A always waits for each ACK before sending the next data packet; this is known as stop-and-wait. Packets that are overdue are presumed lost and are retransmitted. (a) In the absence of any packet losses or duplications, explain why it is not necessary to include any "sequence number" data in the packet headers. (b) Suppose that the link can lose occasional packets, but that packets that do arrive always arrive in the order sent. Is a 2-bit sequence number (that is, \(N\) mod 4) enough for \(\mathrm{A}\) and \(\mathrm{B}\) to detect and resend any lost packets? Is a 1-bit sequence number enough? (c) Now suppose that the link can deliver out of order, and that sometimes a packet can be delivered as much as 1 minute after subsequent packets. How does this change the sequence number requirements?

Calculate the total time required to transfer a \(1000-\mathrm{KB}\) file in the following cases, assuming an RTT of \(100 \mathrm{~ms}\), a packet size of \(1 \mathrm{~KB}\) and an initial \(2 \times\) RTT of "handshaking" before data is sent. (a) The bandwidth is \(1.5 \mathrm{Mbps}\), and data packets can be sent continuously. (b) The bandwidth is \(1.5 \mathrm{Mbps}\), but after we finish sending each data packet we must wait one RTT before sending the next. (c) The bandwidth is "infinite," meaning that we take transmit time to be zero, and up to 20 packets can be sent per RTT. (d) The bandwidth is infinite, and during the first RTT we can send one packet \(\left(2^{1-1}\right)\), during the second RTT we can send two packets \(\left(2^{2-1}\right)\), during the third we can send four \(\left(2^{3-1}\right)\), and so on. (A justification for such an exponential increase will be given in Chapter \(6 .)\)

What differences in traffic patterns account for the fact that STDM is a costeffective form of multiplexing for a voice telephone network and FDM is a costeffective form of multiplexing for television and radio networks, yet we reject both as not being cost-effective for a general-purpose computer network?
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