TV Broadcast Distribution

• Terrestrialo TV Programming is distributed by terrestrial microwave systems.o The principal advantage is its infrastructure that provided many routing alternatives.o Its primary disadvantage was the poor performance of long repeater cascades (coast-to-coast) as compared

to satellite transmission circuits.• Satellite Program Distribution

o Space stations in the Domestic Satellite Service are being used extensively for distribution of TV programming.

o It has vastly improved performance over terrestrial networks provided that carrier-to-noise limitations can be overcome.

o Allotted bands are 3.7 to 4.2 GHz (downlink) and 5.925 to 6.425 GHz (uplink) for C Band and 11/14 GHz in the Ku band.

• DBS (Direct Broadcast Satellite)o Broadcasting direct to the public (direct home reception)o Allotted bands are 12.2 to 12.7 GHz (DL) and 17.3 to 17.8 GHz (UL).

TV Principles and Concepts

• TV Picture• Scanning• Synchronization• Video Bandwidth

TV Picture• Picture Element

-The smallest area of light or shade in an image is a picture element called PIXEL or PEL.-Picture elements are converted to a electrical signal by a camera tube at the studio. This signal becomes the video signal to be broadcast to receivers. -The picture tube in the receiver converts the video back to visual information.

Picture Qualities• Brightness

It is the overall or average intensity of illumination. It determines the background level in the reproduced picture.• Contrast

It is the difference in intensity between black parts and white parts of the reproduced picture.• Detail

It is described by the resolution of definition of an image. The more picture elements, the higher the resolution, the clearer the picture.

• Color LevelThe color information superimposed on a monochrome picture. It is dependent on the amplitude of the chrominance signal in a color TV.

• Hue or TintThe color of an object. It is dependent on the phase angle of the chrominance signal in a color TV.

• Aspect RatioWidth-to-Height ratio of the picture frame. The standard aspect ratio is 4:3.

Picture Definition

• Number of Active Lines (nv)

nv = N – NS where:N = no. of lines per frameNs = no. of lines suppressed during retrace.

• Width of Line (w)

w=V/ nv V=vertical dimension of the CRT • Aspect Ratio (a)

a=H/V = nh/ nv H=horizontal dimension of CRTnh = no. of active pixel in a line.

• Maximum No. of Pixel per line (NL)

NL = nh/ 0.835

Example• In an NTSC system, if the no. of lines per frame is 525 and the number of suppresed lines is 40 per frame, find the

picture height and width. Also find the number of pixel in a line.

Solving for the height:nv=N – Ns = 525 – 40 = 485 linesSolving for the width:nh= a x nv = (4/3) x 485 = 647 linesSolving for the no. of pixels in a line:NL = nh / 0.835 = 647/0.835 = 775 pixels

Scanning

• Horizontal Scanningo The sawtooth current in the horizontal deflection coils

deflect the beam across the screen with a continuous motion from left to right.

o At the peak of the rise, the sawtooth wave reverses direction and decreases rapidly to its initial value. This fast reversal produces the retrace/flyback.

• Horizontal Line-Scanning Frequency (fh)fh = N x P N = no. of lines per frame

P = no. of frames per second• Horizontal Line-Scanning Time (Th)

Th = 1/fh• Horizontal Flyback or Retrace Time (Trh)

Trh = 0.10 (Th)

• Vertical Scanningo The sawtooth current in the verticall deflection coils

deflect the beam to move from top to bottom of the raster.

o The trace part deflects the beam to the bottom of the raster then the retrace returns the beam to the top.

• Vertical Line-Scanning Frequencyfv = 2 x P P = no. of frames per second

• Vertical Line-Scanning Time (Tv)Tv = 1/fv

• Vertical Flyback or Retrace Time (Trv)Trv = 0.10 (Tv)

Interlaced Scanning

Synchronization

• Synchronizing Pulses

Video Bandwidth

BWv = 0.35 (fhNL)

o Where: fh = horizontal line scanning frequency NL = maximum no. of pixels per line

Example:

• In an NTSC System, (1) find the horizontal and vertical synchronization frequencies, and the time required to scan one line, and (2) if the number of pixels per line is 775, find the video bandwidth.

• Solving for the horizontal sync frequency:

fh = NxP = 525 x 30 = 15,750 Hz

• Solving for the horizontal sync frequency:

fv = 2xP = 2 x 30 = 60 Hz

• Solving for the time required to scan one line

Th = 1/fh = 1/15,750 = 63.5 microsec

• Solving for the video bandwidth

BWv = 0.35 (fhNL) = 0.35 (15,750)(775) = 4.27 MHz