Progress in Block 3 Part 1

I've reached section 2.5 of Block 3 Part 1. Much of the material regarding analog/digital comparisons and binary numbers are duplicated in M150, my previous module. It's good to be refreshed though. Here's my answer for activity 15:

Activity 15

A common bitmap format for displaying reasonably high-quality computer images, such as those generated by digital cameras, is to use 24 bits for each pixel.

(a)  How many pixels are required for 1024 x·768 screen resolution?
How many bits of data does this require for each image?

For a resolution of 1024 x 768:

1024 x 768, or

786,432 pixels are required.

Bits of data required if the picture uses 24 bits per pixel:

1024 x 768 x 24, or

18,874,368 bits.

(b)  At the time of writing this, I am still using a ‘56 kbps’ dial-up modem that connects (if I’m lucky) at 44 kbps. Roughly how long would such an image take to download at 44 kbps?

44 kbps = 44,000 bits per second.

Time required to send 18,874,368 bits at 44,000 bits per second is

(18,874,368 / 44,000) seconds, or

c. 429 seconds, or

7 minutes 9 seconds.

(c)  By the time you read this my broadband connection will be installed, operating at 512 kbps, if not more. How long will a single image take to download over a 512 kbps connection?

Time required to send the same image at 512 kbps (512,000 bits per second) is

(18,874,368 / 512,000) seconds, or

c. 37 seconds.

(d)  My DVD drive sends data to a computer processor at about 3 Mbps. Roughly how long would it take to transfer one image? Try to answer this question without using a calculator, by comparing the DVD transfer rate with the broadband rate

The DVD drive is faster than the broadband connection by a factor of 3,000,000 (3 Mbps) to 512,000 (512 kbps), or about 6.

The approximate time taken for the DVD drive to transfer the picture is therefore about (37 / 6) seconds, or

About 6 seconds.