Television - the art of image transmission and reception CAMERA TELEVISION 1 st question: How to capture an image and convert it into an electrical signal? Early days, camera adopted line scanning

No scanning required now, standard remains the same CCD - solid state image recording TELEVISION Electronic signals in PAL, NTSC standards To understand scanning, we need to look at early video camera design

Figure 1 J.S. Zarach and Noel M. Morris, Television principles & practice Click Me!!

What you have seen only converted the brightness of a single spot to a voltage or current flow. What about converting the entire image plane? A scanning mechanism is required to convert every point on the image to its corresponding voltage or current value. Click Me again !! What does the scanned waveform looks like?

Line 1 Line 2Line 3 Line 2 Line 3 Suppose the image looks like the following

Line 1 Line 2 Line 3 Frame rate: Number of pictures scanned per second A single picture One second Pic 1Pic 2Pic 3Pic M M Frames/Pictures per second

Color images are made up of three primary colors = ++ In early days, few people had color TV. Transmitting three images takes up 3 times the bandwidth. Solution: Split image into Luminance and Chrominance. Luminance contains B/W information, very sensitive to our eyes Chrominance supplements color information, less sensitive to us.

= + U + V Y B/W TELEVISION Color TELEVISION Wide bandwidth Narrow bandwidth

Luminance (Black and White) signal Color (Chrominance) signal + = + Sound = TELEVISION How to combine them together?

f VSB Chroma at 4.43 MHz Sound at 6 MHz The components are modulated and grouped into different parts of the frequency spectrum MHz

Figure 2 Amplitude Modulator Amplitude Modulator Frequency Modulator Frequency Modulator Syn. Pulses Generator Syn. Pulses Generator RF Modulator RF Modulator Audio Signal Video Signal MYMY Y (I.F.)

How are video signals generated and recovered on the Television in practice? The basic concept has been covered. A major consideration: bandwidth Bandwidth determined by: Resolution and Pictures/second. Objective: To send as much information as possible with less amount of bandwidth

Video camera in the early days Synchronization Interlaced scanning Existing TV standards are consequences of early development. Image Acquisition Scanning Image reconstruction Synchroniza -tion Bandwidth restriction Interlaced Scanning Color information Frequency Interleaving Future Bread and Butter More pictures/sec. Less Bandwidth Add color without extra Bandwidth

Figure 3 Resolution: The smallest distance that can exist between two points Maximum number of points that can exist in an area x

Bandwidth determined by Resolution Finest observable checker board pattern Marginal Not Resolvable Pictures per second Movie standard: 25 pictures/second Scanning method Non-interlaced/Interlaced scanning Resolvable

pixels 1 cycle 1 cycle = 2 pixels cpl: cycles per line lpp: lines per picture pps: pictures per second

pixels 1 cycle 1 cycle = 2 pixels Frames per second = 25 BW = 6MHz Lines per picture (frame) = 625 Line freq. = kHz 64us Pixels per line = 768 Cycles per line = 384

cpl: cycles per line lpf d : lines per field f d ps: fields per second Fields per second = 50 BW = 6MHz Lines per picture (field) = Line freq. = kHz Pixels per line = 768 Cycles per line= 384

1. What kind of TV system is adopted in Hong Kong? PAL 2. What kind of scanning is employed in PAL system? Interlaced 3. What is the resolution in PAL system? 768X625(V) 4. How many Fields per second? 50 (25frames) 5. What is the time taken to scan a single line? 64uS 6. What is the baseband bandwidth for PAL signal? 6MHz

Adjacent video lines may be different Usually very similar locally Can be assumed as periodic Discrete spectrum

A sequence of video lines of a homogeneous picture. A t 0 x/2-x/2 T= 64us Waveform of the sequence of video lines

A t 0 x/2-x/2 T= 64us It can be proved easily that b n =0 A sequence of video lines

anan n=0 n=1 n=2 n=3 Noted that the spectrum is “discrete” with lots of empty space

anan n=0 n=1 n=2 n=3 Noted that the spectrum is “discrete” with lots of empty space An alternative representation, noted that f L = 1/T is the line frequency

Y B/W R G B + Y Weighted sum Figure 4a

Y Camera Negative AM MYMY f Y - I.F. Carrier Figure 4b fYfY D.C. USBLSB f Y + 5.5MHz Y Click Me!

fYfY D.C. Signal USBLSB f Y + 5.5MHz Figure 5a fYfY D.C. SSB Signal f Y + 5.5MHz fYfY Filter D.C. Passband

Figure 5b fYfY D.C. Signal USBLSB f Y + 5.5MHz fYfY Filter D.C. Passband fYfY D.C. VSB Signal USBVSB f Y + 5.5MHz

What about Color? Straightforward approach fRfR D.C. f R + 5.5MHz fGfG f G + 5.5MHz fBfB f B + 5.5MHz But this will require THREE TIMES the bandwidth Figure 6

Figure 7 A freq 1 T2T2 T1T1 S 1 (t) T2T2 S 2 (t) TimeSpectrum A freq 1 T1T1

a. Frequency spectrum is not continuous b. Frequency components can only occur in regular spaced slots c. The positions of the slots are determined by the smallest repetitive frequency of the signal THREE important findings for repetitive signals

e. For a continuous sine wave (i.e. infinite duration), the frequency spectrum is a single impulse. f. The position of the impulse of a continuous sine wave is dependent only on the frequency TWO important findings for continuous sine wave

S1S1 T1T1 f2f2 S2S2 1stcycle 2ndcycle 3rdcycle 4thcycle f1f1 A freq 1 T1T1 f2f2 Figure 8a (Enlarged Frequency Scale)

1stcycle 2ndcycle 3rdcycle 4thcycle S1S1 f1f1 A freq 1 T1T1 f2f2 Figure 8b T1T1 f2f2 S3S3 (Carrier) Click Me

Can RGB components be interleaved? Answer: No Figure 8c freq fRfR fGfG fBfB

The 3 color components are only roughly periodic Figure 8d Result: Partial Overlapping between components Distortion is very prominent in smooth region freq R G B

B/W Luminance Y R G B RGB TO YUV Chrominance U, V

. Y = 0.3R G B U = B - Y V = R - Y RGB to YUV transform R = V + Y G = (Y - 0.3R B)/0.59 B = U + Y YUV to RGB transform

Y - Luminance (intensity information) U and V - Chrominance (color information) Y - Wide band (5.5 MHz) U and V - Narrow band (about 2MHz) The Eye is not sensitive to Lumninance at high frequency (e.g. texture) Chrominance, as compare with Luminance

f VSB Chroma at 4.43 MHz MHz Sound at 6 MHz Note: Y and UV are separated by interleaving, what about U and V?

Quadrature Modulation (QM) U V cos  c t cos (  c t + 90 o ) AM Y + CUCU CVCV S Figure 9 Note: Y and UV are separated by interleaving

Phasor representation of Quadrature Modulation CUCU CVCV C U +C V  Color (hue) defined by 

freq Line frequency = 1/T = 15.6kHz Line duration = T = 64  s Color Subcarrier frequency f sc = 283.5/T = 4.43MHz Y U V f sc 284/T 1/T 1/2T Figure 26

Demodulation S-Y cos  c t (LO) cos (  c t + 90 o ) (LO) X X U V Figure 10 LPF Assuming that somehow Y can be eliminated from the video signal, leaving the chrominance components (U,V) behind.

C U cos  c t = U cos 2  c t = U (cos 2  c t + cos (0)) = U after LPF C U cos(  c t + 90 o ) = U cos  c t cos(  c t + 90 o ) = U (cos (2  c t+90) + cos (90 o )) = 0 after LPF

C V cos(  c t+90) = V cos 2 (  c t+90 o ) = V (cos (2  c t+180 o ) + cos (0)) = V after LPF C V cos  c t = V cos  c t cos(  c t + 90 o ) = V (cos (2  c t+90) + cos (90 o ) ) = 0 after LPF

1. The two quadrature carrier signals are not sent to the receiver 2. Phase error in demodulation CUCU CVCV C U +C V   

The two quadrature carrier signals are regenerated in the receiver with a short burst of sine wave The regenerated carrier signals may contain error Consider an error ‘  ’ in the regenerated carrier The carrier (LO) changes from: cos(  c t) to cos(  c t+  ), and cos(  c t+90 o ) to cos(  c t+ 90 o +  )

C U cos(  c t+  )= U cos  c t cos(  c t+  ) = U (cos (2  c t+  ) + cos (  )) = U cos (  ) after LPF C U cos(  c t + 90 o +  ) = U cos  c t cos(  c t + 90 o +  ) = U (cos (2  c t+90 +  ) + cos (90 o +  )) = Ucos (90 o +  ) after LPF

C V cos(  c t+90+  ) = V (cos (2  c t+180 o +  ) + cos (  )) = V cos (  ) after LPF C V cos(  c t+  ) = V (cos (2  c t+90 o +  ) + cos (90 o +  )) = V cos (90 o +  ) after LPF

Error Free phasor diagram CUCU CVCV Correct  C U + C V

CUCU CVCV Correct  Error Error phasor diagram Note: distortion is similar between adjacent lines Every line is subject to distortion

Color Distortion Figure 11 OriginalDistorted (clockwise) Distorted (anticlockwise

1. Odd lines: U modulated by cos(  c t) V modulated by cos(  c t+90 o ) 2. Even lines: U modulated by cos(  c t) V modulated by cos(  c t-90 o ) Under Error Free condition U and V are fully recovered with quadrature demodulation

Odd Lines CUCU CVCV Correct  C U + C V Even Lines CUCU CVCV Correct  C U + C V Error Free Signal

CVCV CUCU Line 1 CVCV CUCU Line 2 CVCV CUCU CVCV CUCU Line 17 Line 18

Implications in video signal Line 1Line 2Line 3Line 4 Y fHfH Line 1Line 2Line 3Line 4 U fHfH Line 1Line 2Line 3Line 4 V f H /2 f H =15.625kHz is the line frequency

LO with Error ‘  ’ Odd Lines CUCU CVCV Error  C U + C V  Even Lines CUCU CVCV Correct  C U + C V 

Odd Lines CUCU CVCV Error  C U + C V LO with Error ‘  ’  CUCU CVCV Error  C U + C V  Even Lines (inverted)

Odd Lines (delayed) CUCU CVCV Error  C U + C V LO with Error ‘  ’  CUCU CVCV Error  C U + C V  + Even Lines (inverted)

LO with Error ‘  ’ Error Free Resultant CUCU CVCV Correct  C U + C V = Phasor addition CUCU CVCV Error (odd)  C U + C V  Error (even)

PAL Color Compensation - Graphical illustration Figure 12a OriginalDistorted

PAL Color Compensation Figure 12b OriginalDistortedCompensated

Line n-1 Line n Line n+1 Line n-1 Line n

PAL Compensation by averaging consecutive lines: 1 st case Subscript “D” denotes delay by 64 us

PAL Compensation by averaging consecutive lines: 1 st case PAL Compensation by averaging consecutive lines: 2 nd case Subscript “D” denotes delay by 64 us