st question: How to capture an image and
1
Television
- the art of image transmission
Early days, camera adopted line scanning
convert
it into an electrical signal?
and reception
CAMERA
TELEVISION
To understand
No
scanning required
scanning,
now,
westandard
need to look at
early video camera
remainsdesign
the same
CCD - solid
state image
recording
Electronic signals in PAL, NTSC
standards
TELEVISION
Click Me!!
Figure 1
J.S. Zarach and Noel M. Morris, Television principles & practice
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?
Suppose the image looks like the following
Line 1
Line 2
Line 3
Line 1
Line 2
Line 3
Frame rate: Number of pictures scanned per second
Line 1
Line 2
Line 3
A single picture
One second
Pic 1
Pic 2
Pic 3
M Frames/Pictures per second
Pic M
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.
Narrow bandwidth
Wide bandwidth
Y
=
B/W TELEVISION
+
U
+
Color TELEVISION
V
Luminance (Black and White) signal
+
Color (Chrominance) signal
=
+
Sound
How to combine them together?
=
TELEVISION
The components are modulated and grouped into different parts of
the frequency spectrum.
Chroma at
4.43 MHz
-1.75 MHz
Sound at 6 MHz
VSB
f
Audio
Signal
Frequency
Modulator
Video Y
Signal
Amplitude
Modulator
MY
(I.F.)
Syn. Pulses
Generator
Figure 2
RF
Modulator
The basic concept has been covered.
How are video signals generated and
recovered on the Television in practice?
A major consideration: bandwidth
Bandwidth determined by: Resolution and Pictures/second.
Objective: To send as much information as possible with
less amount of bandwidth
Existing TV standards are consequences of early development.
Bread and Butter
More pictures/sec.
Less Bandwidth
Add color without
extra Bandwidth
Image
Acquisition
Image
reconstruction
Bandwidth
restriction
Color
information
Scanning
Synchroniza
-tion
Interlaced
Scanning
Frequency
Interleaving
Video camera in the early days
Synchronization
Interlaced scanning
Future
Resolution:
• The smallest distance that can exist between
two points
• Maximum number of points that can exist in
an area
x
Figure 3
Bandwidth determined by
Resolution
Finest observable checker board pattern
Pictures per second
Movie standard: 25 pictures/second
Scanning method
Non-interlaced/Interlaced scanning
Resolvable
Marginal
Not Resolvable
pixels
1 cycle
1 cycle = 2 pixels
Bandwidth  cpl  lpp  pps
cpl: cycles per line
lpp: lines per picture
pps: pictures per second
pixels
64us
1 cycle
1 cycle = 2 pixels
Pixels per line = 768
Cycles per line = 384
Lines per picture (frame) = 625
Line freq. = 15.625 kHz
Frames per second = 25
BW = 6MHz
Pixels per line = 768
Cycles per line= 384
Lines per picture (field) = 312.5
Line freq. = 15.625 kHz
Fields per second = 50
BW = 6MHz
Bandwidth  cpl  lpf d  f d ps
cpl: cycles per line
lpfd: lines per field
fdps: fields per second
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.
T=
64us
A
t
0
-x/2
x/2
Waveform of the sequence of video lines
T=
64us
A sequence of video lines
A
t
0
-x/2
x/2

f t   a0   an cos nt  bn sin nt 
1
a0 
T
x
2
n 1
Ax
 x Adt  T
2
It can be proved easily that bn=0

f t   a0   an cos nt  bn sin nt 
n 1
2
an 
T
T
2
2
T f t cos ntdt  T
2
x
2
 A cos ntdt
x
2
  nx  
sin


2 Ax
sin y 
2 


K
an 
 Ksinc  y 
y
T  nx 


2
n=0
n=1
n=2
n=3
an
f
0
1
T
2
T
3
T
4
T
Noted that the spectrum is “discrete” with lots of empty space
An alternative representation, noted that fL= 1/T is the line frequency
n=0
n=1
n=2
n=3
an
f
0
fL
2 fL
3 fL
4 fL
Noted that the spectrum is “discrete” with lots of empty space
B/W
Y
B/W
R
B/W
G
B/W
B
Figure 4a
+
Y
Weighted sum
Camera
Y
Negative
AM
Click Me!
MY
fY - I.F. Carrier
D.C.
fY
5.5MHz
Y
LSB
Figure 4b
USB
fY +
5.5MHz
D.C.
Signal
fY
LSB
D.C.
fY + 5.5MHz
USB
fY
Filter
Passband
D.C.
fY
SSB
Signal
Figure 5a
fY + 5.5MHz
D.C.
Signal
fY
LSB
USB
fY
D.C.
Filter
Passband
D.C.
VSB
Signal
fY + 5.5MHz
fY
VSB
Figure 5b
fY + 5.5MHz
USB
What about Color?
Straightforward approach
D.C.
fR
fR+
fG +
fB+
5.5MHz
5.5MHz
5.5MHz
fG
fB
But this will require THREE TIMES the bandwidth
Figure 6
Time
S1(t)
Spectrum
A
1
T1
T1
S2(t)
T2
freq
A
1
T2
Figure 7
freq
THREE important findings for repetitive signals
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
TWO important findings for continuous sine wave
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
S1
1st
cycle
3rd cycle
2nd
cycle
4th cycle
f1
S2
T1
f2
A
(Enlarged
Frequency
Scale)
f2
1
T1
freq
Figure 8a
1st
S1
cycle
3rd cycle
2nd
cycle
4th cycle
f1
S3
T1
(Carrier)
A
f2
f2
freq
1
T1
Figure 8b
Can RGB components be interleaved?
fR
fG
fB
freq
Answer: No
Figure 8c
The 3 color components are only roughly periodic
R
G
B
freq
Result: Partial Overlapping between components
Distortion is very prominent in smooth region
Figure 8d
B/W
R RGB
B/W
G TO
B/W
B
YUV
Luminance
Y
Chrominance
U, V
.
RGB to YUV transform
Y = 0.3R + 0.59G + 0.11B
U=B-Y
V=R-Y
YUV to RGB transform
R=V+Y
G = (Y - 0.3R - 0.11B)/0.59
B=U+Y
• 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
Chroma at
4.43 MHz
Sound at 6 MHz
VSB
f
-1.75 MHz
Note: Y and UV are separated by
interleaving, what about U and V?
Quadrature Modulation (QM)
cos  c t
U
AM
CU
Y
V
+
AM
cos ( c t + 90o)
Figure 9
S
CV
Note: Y and UV are
separated by interleaving
Phasor representation of Quadrature Modulation
CU
+CV
CV

CU
Color (hue) defined by 
Line duration = T = 64s
Line frequency = 1/T = 15.6kHz
Color Subcarrier frequency fsc = 283.5/T = 4.43MHz
1/T
freq
Y
1/2T
fsc
U
V
Figure 26
284/T
Demodulation
cos  c t (LO)
X
LPF
U
X
LPF
V
S-Y
Assuming that
somehow Y can be
eliminated from the
video signal, leaving
the chrominance
components (U,V)
behind.
cos ( c t + 90o) (LO)
Figure 10
CU cos c t = U cos2 c t
= U (cos 2 c t + cos (0))
= U after LPF
CU cos( c t + 90o)
= U cos c t cos( c t + 90o)
= U (cos (2 c t+90) + cos (90o))
= 0 after LPF
CV cos( c t+90) = V cos2( c t+90o)
= V (cos (2 c t+180o) + cos (0))
= V after LPF
CV cos c t
= V cos c t cos( c t + 90o)
= V (cos (2 c t+90) + cos (90o) )
= 0 after LPF
1. The two quadrature carrier signals are not sent
to the receiver
2. Phase error in demodulation
d
CU
+CV
CV

CU
d
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 ‘d ’ in the regenerated carrier
The carrier (LO) changes from:
cos( c t) to cos( c t+ d ) , and
cos( c t+90o) to cos( c t+ 90o+d )
CU cos( c t+d )= U cos c t cos( c t+d )
= U (cos (2 c t+ d ) + cos (d ))
= U cos (d ) after LPF
CU cos( c t + 90o +d )
= U cos c t cos( c t + 90o +d )
= U (cos (2 c t+90 +d ) + cos (90o +d ))
= Ucos (90o +d ) after LPF
CV cos( c t+d )
= V (cos (2 c t+90o +d ) + cos (90o +d ))
= V cos (90o +d ) after LPF
CV cos( c t+90+d )
= V (cos (2 c t+180o +d ) + cos (d ))
= V cos (d ) after LPF
Error Free phasor diagram
Correct
CU + CV
CV

CU
Error phasor diagram
Error
Correct
CV
Note: distortion
is similar between
adjacent lines
Every line is
subject to
distortion

CU
Color Distortion
Distorted
(anticlockwise
Original
Figure 11
Distorted
(clockwise)
1. Odd lines:
U modulated by cos( c t)
V modulated by cos( c t+90o)
2. Even lines:
U modulated by cos( c t)
V modulated by cos( c t-90o)
Under Error Free condition U and V are fully
recovered with quadrature demodulation
Error Free Signal
Odd Lines
CV
Even Lines

Correct
CU + CV
CU
Correct
CU + CV

CU
CV
Error Free Signal
CV
Line 1
CU
CU
Line 2
CV
CV
Line 17
CU
CU
Line 18
CV
Implications in video signal
fH
Y
Line 1
Line 2
Line 3
Line 4
fH
U
Line 1
Line 2
Line 3
Line 4
fH/2
V
Line 1
Line 2
Line 3
Line 4
fH=15.625kHz is the line frequency
LO with Error ‘d ’
Odd Lines
CV
Even Lines

Error
CU + CV
CU
d
d
Correct
CU + CV

CU
CV
LO with Error ‘d ’
Odd Lines
CV
Even Lines (inverted)
Error
CU + CV
CV
Error
CU + CV
d
d


CU
CU
LO with Error ‘d ’
Odd Lines (delayed)
CV
Even Lines (inverted)
CV
Error
CU + CV
Error
CU + CV
+
d

d

CU
CU
LO with Error ‘d ’
Phasor addition
Error Free Resultant
Error (odd)
CV
CV
Correct
CU + CV
CU + CV
Error
(even)
d

=

CU
CU
PAL Color Compensation - Graphical illustration
Original
Distorted
Figure 12a
PAL Color Compensation
Original
Distorted
Figure 12b
Compensated
Line n-1
Line n-1
Line n
Line n
Line n+1
  line period  64s
C t   C t     C t  2   
 U t  cost   jV t  cost 
 U t    cos t     jV t    cos t   
 U t  2  cos t  2   jV t  2  cos t  2   
PAL Compensation by averaging consecutive lines: 1st case
Ccomp t   0.5CD t   C t   
U t  cost   jV t  cost   U t    cos t     j V t    cos t   
2
jV t  cost   jV t    cos t   

2

Subscript “D” denotes delay by 64 us
PAL Compensation by averaging consecutive lines: 1st case
Ccomp t   0.5C D t   C t   
U D t  cos t     jVD t  cos t     U t    cos t     j V t    cos t   
2
jV t    cos t     jV t    cos t   

 jV t    cos t   
2

PAL Compensation by averaging consecutive lines: 2nd case
Ccomp t   0.5CD t   C t   
U D t  cos t     jVD t  cos t     U t    cos t     V t    cos t   
2
U t    cos t     U t    cos t   

 U t    cos t   
2

Subscript “D” denotes delay by 64 us