Question

A common measure of transmission for digital data is the number of bits transmitted per second. Generally, transmission is accomplished in packets consisting of a start bit, a byte (8 bits) of information, and a stop bit. Using these facts, answer the following:

a. Compute the time required to transmit an image of 1200x800 pixels with 8 bits for gray of each pixel using a 50 M bits/sec. modem?

b.What would the time be at 3 M bits/sec, a representative of download speed of a DSL connection?

(c) Repeat a and b when the image is RGB colored with 8 bits for each primary color. f 20 colored frames per second

Answer 1

Answer:  a) 0,19 seg. b) 3,2 seg. c) 11,5 seg.  d) 192 seg.

Explanation:

a)  For each pixel in the image, we use 8 bits + 1 start bit + 1 stop bit= 10 bits.

• Number of Pixels: 800*1200= 960,000.
• Bits transmitted: 960,000 x 10 = 9.6 * 10⁶ bits = 9.6 Mbits
• if  the modem is able to transmit up to 50 Mbits in a second, we can calculate how much time it will be needed to transmit 9.6 Mbits, using this equality:

• 50 Mbit = 1 sec
• 9.6 Mbit = x  ⇒  t= 9.6 / 50 = 0.19 sec

b)  If the modem speed changes to 3 Mb/s, all we need to do is just use the same equality, as follows:

• 3 Mbit = 1 sec
• 9.6 Mbit = x  ⇒  t= 9.6 / 3 = 3.2 sec

c) Now, if we need to transmit a colored image , at a rate of 20 f/sec, we need to calculate first how many bits we need to transmit, as follows:

• 1 Frame= 800*1200* (24 bits + 3 start bits + 3 stop bits)=  28.8  Mbits
• 1 Second= 20 frames/ sec = 28.8 Mbits *20 = 576 Mbits.
• If the modem speed is 50 Mb/s, we can use the same formula that we used for a) and b), as follows:

• 50 Mbit = 1 sec

• 576 Mbits = x  ⇒  t= 576 / 50 = 11.5 sec

d) Same as c) replacing 50 Mb/s by 3 Mb/s, as follows:

• 3 Mbit = 1 sec
• 576 Mbits = x  ⇒  t= 576 / 3 = 192 sec

Answer 2

The time that's required to transmit an image of 1200x800 pixels with 8 bits for gray is 151.6 million seconds.

How to calculate the time taken?

The time required to transmit an image of 1200x800 pixels with 8 bits for gray will be:

= (1200 × 800 × 8) / 50

= 151.6 million seconds.

The time needed to be at 3 M bits/sec, a representative of download speed of a DSL connection will be:

= 7.68/3

= 2.56 seconds.

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