bits/pixel

Suppose that a digitized TV picture is to be transmitted from a source that uses a matrix of 480 X 500 picture element (pixels),where each pixel can take on
one of 32 intensity value.Assume that 30 pictures are sent persecond.(This digital sorce is roughly equivalent to broadcast TV standarts that have been adopted.)Find the source rate R (bps). *

         

Each TV picture is 480x500 pixels = 240,000 pixels/picture.

Each pixel can have 32 different intensities, which means log2(32) = 5 bpp (bits/pixel).

So, 

R = (5 bits/pixel) * (240,000 pixels/picture) *  (30 pictures/second)

R = 36,000,000 bits/second or 36,000,000 bps or 36 Mbps

b. Assume that a TV picture to be transmitted over a channel with 4.5–MHz bandwidth, what termal  noise may we expect at its output ?

N = 10 log k + 10 log T + 10 log B

  = 10 log (1.38 * 10^-23) + 10 log 290 + 10 log (4.5 * 10⁷)

  = -228.60120913598763488623456089769 +  

      24.623979978989560873328467629693 +  

      76.532125137753436793763169117857

  = -127.44510401924463721914292415006 dBW

c. Discuss how the parameters given in part (a) could be modified to allow transmission of color TV signal without increasing the required value for R.

The number of possible intensity values allowed for each pixel would increase with the ability of the TV to transmit color.  For example, instead of an intensity of 32, there could be an intensity of 256 represented by 8 bits; log2(256) = 8 bpp (bits/pixel).  With an R of 36 Mbps from part (a), and the number of pictures/second equal to 30, the new size of the picture would be: 

 R = 36,000,000 = (8 bits/pixel) * P *  (30 pictures/second)

      36,000,000 = 240*P

      P = 150,000 pixels/picture

The picture size from part (a) was 480x500, which is a 1/1.042 ratio.  So, the new screen size would be:  

      (y)(1.042y) = 150,000

          1.042y2 = 150,000

              y2  = 143953.935

    y = 380 and  (1.042y) = (1.042 * 380) 395

So, the picture size would be approximately: 380 x 395.

3.19.

Given a channel with intended capacity of 20 Mbps, the bandwidth of the channel   is 3 MHz. Assuming white thermal noise,what signal to noise ratio is required to achieve this capacity? *

       C = W log₂ (1 + S/N)

       20 * 10⁶ = 3 * 10⁶ log₂ (1 + S/N)

       20/3 =  log₂ (1 + S/N)

       S/N = 2^20/3  - 1

          Jadi, signal-to-noise ratio is 2^20/3  - 1.

4.1 

Suppose that data are stored on 1.4 – Mbyte floppy discettes that weight 30 g      each. Suppose that airliner carries 10⁴ kg of this floppies at the speed of 1000 km/h over a distances of 5000 km. What is data transmission rate in bit per second   of this system ?  

Transmission rate   = (1.4 * 10⁶ * 8 * (10⁴ *  10³ / 30) ) / ((5000/1000) * 3600)

                   =  (1.4 * 8 * 10⁶  * 10⁴ ) / (5 * 36 * 3)

                   =  2.074074 *  10⁸  (bps)