Acommon 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

 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 Mbitsif  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 sec9.6 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 sec9.6 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  Mbits1 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 sec576 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 sec576 Mbits = x  ⇒  t= 576 / 3 = 192 sec

Falsea long text data type can store a variable amount of text or combination of text and numbers where the total number of characters may exceed 255.