A-level Physics (Advancing Physics)/Digital Storage/Worked Solutions

1. An image transmitted down a SVGA video cable is 800 pixels wide, and 600 pixels high. How many pixels are there in the image?

$800\times 600=480000\text{ pixels}$

2. A grayscale image is encoded using 3 bits. How many possible values can each pixel have?

2³ = 8

3. The characters in a text document are numbered from 0 - 255. How many bits should each character be encoded with?

There are 256 possible values.

$b=\frac{\log 256}{\log 2}=8$

4. A page contains 30 lines of text, with an average of 15 characters on each line. Each character is represented by 4 bits. How many megabytes of uncompressed storage will a book consisting of 650 pages like this fill on a computer's hard disk?

Total number of characters = 650 x 30 x 15 = 292 500

Total amount of information = 292500 x 4 = 1 170 000 bits

$1170000\text{ bits}=\frac{1170000}{8}\text{ bytes}=146250\text{ bytes}=\frac{146250}{1024^2}\text{ Mbytes}\approx 0.14\text{ Mbytes}$

5. A 10 cm wide square image is scanned into a computer. Each pixel is encoded using 3 channels (red, green and blue), and each channel can take on 256 possible values. One pixel is 0.01 mm wide. How much information does the scanned image contain? Express your answer using an appropriate unit.

$10\text{cm}=\frac{0.1}{0.01\times 10^{-3}}\text{ pixels}=10000\text{ pixels}$

Total number of pixels = 10 000² = 100 000 000 pixels

$b=\frac{\log 256}{\log 2}=8\text{ bits per. channel per. pixel}=8\times 3\text{ bits per. pixel}=24\text{ bits per. pixel}$

Total information = 24 x 100 000 000 = 2 400 000 000 bits = 300 000 000 bytes

$\frac{300000000}{1024^2}\approx 286\text{ Mbytes}$ (This is why we usually compress images before storage, or at least use fewer bits per. pixel.)