# Conversions

29/01/2017

I know how to convert milliseconds to seconds

20 ms = 20 x 10⁻³ = 0.02 s30 ms = 30 x 10⁻³ = 0.03 s150 ms = 150 x 10⁻³ = 0.15 s1 second = 1000 milliseconds

I also know how to convert bits, bytes, megabytes etc and also know how to convert kbps to Mbps etc

I am using the example from my second TMA

to express 105.96 Mbps in scientific notation105.96 Mbps / 10 / 10= 1.0596 x 10⁻²

to convert 931 kB into bytes I would have to do the following calculation.931 kB in bits is 7,626,752 b- I worked this out below1024 B x 931 kB(this is because there is 1024 B in 1 kB)= 953,344 Bthen I have to multiply that by 8 b (this is because there is 8 bits in 1 Byte)953,344 B x 8b= 7,626,752 b or 7630000 b (to 3 s.f)To convert 105.96 Mbps into bps I have to carry out the following calculation105.96 Mbps = 1000,000 bps x 105.96 Mbps(this is because there is 1000,000 bps in 1 Mbps)= 105,960,000 bps

If I wanted to find out the time it would take to send a file size of 931 kB over a connection speed of 105.96 Mbps I would have to convert 931 kB into bits and convert 105.96 Mbps into bps.

as I have already converts the file size and the connection speed into the correct unit I won’t have to do it again, I will just use the figures above.

the formula would be time = file size / connection speed7,626,752 b / 105,960,000 bps = 0.0720 seconds (to 3 s.f)

having done this calculation, My tutor’s feedback did mention that I should have done this calculation in scientific notation.

Time to download = file size/connection speed= 7.626,752 x 10⁶ bits / 1.0596 x 10⁸ bps= 7.197765194 x 10⁻² seconds (or 0.0720 s) (3 s.f)

I must remember to do them this way in the future. I will of course add to this the more I learn.

# Base eight system

Base eight or the octal system is based on 8 digits 0 to 7 unlike the decimal numbering system, that is based on ten digits 0 to 9.

7x8+3 = 59so the octal number of 73 will be

I will add to this soon

Below shows how to work out how much bandwidth it takes to carry out video conferencing. This was one of the activities on Block 4 Part 4 Activity 26

Question

Suppose that a video conferencing system has a frame rate of 30 frames per second and an image size of 352 pixels by 288 pixels. It requires a bandwidth of 384 kbps. Assuming that all other factors are equal, what bandwidth is required for the following videoconferencing systems?

a. A system with the same image size, but operating at 15 frames per second.

The answer to this is 192 kbps as it is operating at half the speed so

`384 kbps / 2 = 192 kbps`

b. A system operating at 30 frames per second but with an image size of 528 pixels by 384 pixels.

The answer is 768 kbps this is because the file size is double the original size for example:

```
352 x 288 = 101375
528 x 384 = 202752
202752 / 101375 = 2
then 384 kbps x 2 = 768 kbps
```

c. A system operating at 15 frames per second with an image size of 528 pixels by 384 pixels

The answer would be 384 kbps this is because it is the same file size as before but the frame rate is half operating at only 15 frames per second so..

`768 kbps / 384 kbps = 2 which is basically equal to 386 kbps`

This is how you work out the time it takes to download a file the formula is

**time to download = files size / speed**

for example

800 KB x 1024 B = 819,200 B x 8 b = 6,553,600 b 768 kbps x 1000 = 768,000 bps 6,553,600 b / 768,000 bps = 8.50 s (3 s.f)a file size of 800 KB over a connection speed of 768 kbps(you need to convert both numbers in the same format of bps and bits)So the total amount of time it should take to send this file is 8.5 seconds.