Question

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?

Short answer

221.184 Mbps, 768 Kbps, 1.16 Mbps, and 28.8 seconds, respectively.

Step-by-step solution

Calculate bandwidth for 640x480 video

First, determine the total number of pixels per frame: \[ 640 \times 480 = 307,200 \text{ pixels/frame} \]Next, find the total bytes per frame: \[ 307,200 \text{ pixels} \times 3 \text{ bytes/pixel} = 921,600 \text{ bytes/frame} \]Then, calculate the total bytes per second: \[ 921,600 \text{ bytes/frame} \times 30 \text{ frames/second} = 27,648,000 \text{ bytes/second} \]Convert the result to bits per second (bps): \[ 27,648,000 \text{ bytes/second} \times 8 = 221,184,000 \text{ bps} \]Thus, the required bandwidth is 221.184 Mbps.

Calculate bandwidth for 160x120 video

Determine the total number of pixels per frame: \[ 160 \times 120 = 19,200 \text{ pixels/frame} \]Calculate the total bytes per frame: \[ 19,200 \text{ pixels} \times 1 \text{ byte/pixel} = 19,200 \text{ bytes/frame} \]Find the total bytes per second: \[ 19,200 \text{ bytes/frame} \times 5 \text{ frames/second} = 96,000 \text{ bytes/second} \]Convert it to bits per second (bps): \[ 96,000 \text{ bytes/second} \times 8 = 768,000 \text{ bps} \]Thus, the required bandwidth is 768 Kbps.

Calculate bandwidth for CD-ROM music

First, convert the minutes to seconds: \[ 75 \text{ minutes} \times 60 = 4,500 \text{ seconds} \]Determine the total bytes for the music: \[ 650 \text{ MB} \times 10^6 \text{ bytes/MB} = 650,000,000 \text{ bytes} \]Calculate bytes per second: \[ 650,000,000 \text{ bytes} \times \frac{1}{4500 \text{ seconds}} \approx 144,444.44 \text{ bytes/second} \]Convert this to bits per second (bps): \[ 144,444.44 \text{ bytes/second} \times 8 \approx 1,155,555.56 \text{ bps} \]Thus, the required bandwidth is approximately 1.16 Mbps.

Calculate fax transmission time

Determine the total number of pixels: \[ 8 \times 10 \text{ inches} \times 72 \text{ pixels/inch} \times 72 \text{ pixels/inch} = 414,720 \text{ pixels} \]As it's black-and-white, each pixel is 1 bit: \[ 414,720 \text{ pixels} = 414,720 \text{ bits} \]Determine the time to transmit at 14.4 Kbps (14,400 bps): \[ \frac{414,720 \text{ bits}}{14,400 \text{ bps}} \approx 28.8 \text{ seconds} \]Thus, the fax transmission time is approximately 28.8 seconds.

Key concepts

Bandwidth Calculation

Bandwidth is essential for determining how much data can be transferred over a network in a given amount of time. It is usually measured in bits per second (bps) or its derivatives like Kbps and Mbps. Calculating bandwidth involves understanding the data rate and the amount of data that needs to be transmitted. For instance, in the given exercise, we calculated bandwidths for different scenarios—video at various resolutions and frame rates, and audio. This provides a sense of how much data transmission capability you need.

Video Transmission

Video transmission involves sending video data from one point to another. The quality of the video depends on its resolution and frame rate, and both factors significantly impact the required bandwidth. For instance, for a video with a resolution of 640x480 pixels, 3 bytes per pixel, and 30 frames per second, we calculated a high bandwidth requirement of 221.184 Mbps. Lower resolution and frame rates, like those in the second scenario (160x120 resolution, 1 byte per pixel, 5 frames per second), resulted in a much lower bandwidth requirement of 768 Kbps.

Data Rate

Data rate is the amount of data transmitted per unit of time, often expressed in bits per second (bps). Understanding data rates is crucial for real-time applications like streaming or live conferencing. To calculate data rate, you need to know the size of the individual data units (e.g., video frames or audio samples) and the rate at which they are sent. For example, in our exercise, converting bytes to bits (since data transmission rates are typically expressed in bits) was key to determining the appropriate data rate for video and audio.

Audio Data Calculation

Audio data calculation is similar to video in terms of data rate and bandwidth needs but often involves fewer data because audio files are typically smaller than video. For example, in our CD-ROM audio calculation, we assumed a CD holds 75 minutes of music totaling 650 MB. We broke this down into bytes per second and then to bits per second, resulting in a required bandwidth of approximately 1.16 Mbps. Through these steps, you comprehend how audio quality and storage duration impact the necessary bandwidth.

Fax Transmission Time

Fax transmission time is the time required to send a digital image of a document via a fax machine. This involves converting the image into pixels and then transmitting them at a defined modem speed. For example, in our exercise, we calculated the time required to transmit an 8x10 inch black-and-white image at a resolution of 72 pixels per inch over a 14.4 Kbps modem. The total transmission time turned out to be approximately 28.8 seconds. This helps you understand the correlation between image resolution, data size, and transmission speed on the total time taken.