Introduction to Video Signals

Contents
 Picture transmission
 TV transmitter
 TV receiver
 Synchronization
 Receiver controls
 Geometric form and aspect ratio
 Image continuity
 No. of scanning lines
 Interlaced scanning

Contents
 Resolution
 Brightness
 Contrast
 Video signal dimensions
 Horizontal sync composition
 Vertical sync details
 Function of vertical pulse train
 Scanning sequence details
 Perception of brightness and colour
 Additive and subtractive colour mixing

Contents
 Video signals for colour transmission
 Luminance signal (Y)
 Compatibility
 Colour-difference signals
 Encoding of colour difference signals
 Formation of chrominance signal

Picture transmission
 Picture information is optical in nature and may be
thought of as 2-D arrangement of pixels
 Thus it is needed to pick up each pixels simultaneously
for transmission
 It is practically not possible
 Thus ‘Scanning’ is done for each row of pixels
 After scanning, the optical information is converted to
electrical information
 This is done at a very fast rate and hence the eye
cannot detect a change

Synchronization
 It is essential that the same co-ordinates be scanned at any
instant both at the camera tube target plate and at the
raster of picture tube otherwise, the picture details would
split and get distorted
 To ensure perfect synchronization between the scene being
televised and the picture produced on the raster,
synchronizing pulses are transmitted during the retrace
 i.e., fly-back intervals of horizontal and vertical motions
of the camera scanning beam
 Thus, in addition to carrying picture details, the radiated
signal at the transmitter also contains synchronizing pulses

Synchronization
 These pulses which are distinct for horizontal and
vertical motion control, are processed at the receiver
and fed to the picture tube sweep circuitry
 Thus ensuring that the receiver picture tube beam is in
step with the transmitter camera tube beam
 In a colour television system additional sync pulses
called colour burst are transmitted along with
horizontal sync pulses
 These are separated at the input of chroma section and
used to synchronize the colour demodulator carrier
generator

Receiver controls
 Most black and white receivers have on their front
panel
 (i ) channel selector
 ( ii ) fine tuning
 ( iii ) brightness
 ( iv ) contrast
 ( v) horizontal hold
 (vi ) volume controls besides an ON-OFF switch
 Some receivers also provide a tone control

Receiver controls
 The channel selector switch is used for selecting the
desired channel
 The fine tuning control is provided for obtaining best
picture details in the selected channel
 The brightness control varies the beam intensity of the
picture tube and is set for optimum average brightness
of the picture
 The contrast control has actually gained control of the
video amplifier
 This can be varied to obtain desired contrast between
white and black contents of the reproduced picture

Receiver controls
 The hold control is used to get a steady picture in case
it rolls up or down
 The volume and tone controls form part of the audio
amplifier in sound section, and are used for setting
volume and tonal quality of the sound output from the
loudspeaker
 In colour receivers there is an additional control called
‘colour’ or ‘saturation’ control
 It is used to vary the intensity or amount of colours in
the reproduced picture

Receiver controls
 In modern colour receivers that employ integrated
circuits in most sections of the receiver, the hold
control is not necessary and hence usually not
provided

Aspect ratio
 The frame adopted in all television systems is
rectangular with width/height ratio
 In human affairs most of the motion occurs in the
horizontal plane and so a larger width is desirable
 It is not necessary that the size of the picture produced
on the receiver screen be same as that being televised
but it is essential that the aspect ratio of the two be
same
 This is achieved by setting the magnitudes of the
current in the deflection coils to correct values to both
at the TV camera and receiving picture tube

Image continuity
 While televising picture elements of the frame by
means of the scanning process, it is necessary to
present the picture to the eye in such a way that an
illusion of continuity is created
 To achieve this, advantage is taken of ‘persistence of
vision’
 The sensation produced when nerves of the eye’s
retina are stimulated by incident light does not cease
immediately after the light is removed but persists for
about 1/16th of a second

Image continuity
 So when the picture elements are scanned rapidly
enough, they appear to the eye as a complete picture
unit
 In present day motion pictures 24/25 still pictures of
the scene are taken per second and later projected on
the screen at the same rate
 Image continuity can be achieved through scanning

Horizontal scanning
 The linear rise of current in the horizontal deflection
coils deflects the beam across the screen with a
continuous, uniform motion for the trace from left to
right
 At the peak of the rise, the sawtooth wave reverses
direction and decreases rapidly to its initial value
 This fast reversal produces the retrace or flyback

Vertical scanning
 The sawtooth current in the vertical deflection coils
moves the electron beam from top to bottom of the
raster at a uniform speed while the electron beam is
being deflected horizontally
 Both during horizontal retrace and vertical retrace
intervals the scanning beams at the camera tube and
picture tube are blanked and no picture information is
either picked up or reproduced

No. of scanning lines
 The ability of the scanning beam to allow reproduction of
electrical signals and the capability of the human eye to
resolve the scene distinctly depends on the total number of
lines employed for scanning
 More number of scanning lines gives space to cover more
optical information to be converted into digital
 625 line are selected as standard for India(CCIR standard)
 The number of scanning line is limited by below factors
 Large bandwidth requirement
 Small possible thickness of scanning beam
 Capability of the human eye to resolve the scene

Interlaced scanning
 Although the rate of 24 pictures per second in motion
pictures is enough to cause an illusion of continuity,
they are not rapid enough to allow the brightness of
one picture or frame to blend smoothly into the next
through the time when the screen is blanked between
successive frames
 This results in a definite flicker of light that is very
annoying to the observer when the screen is made
alternately bright and dark

Interlaced scanning
 This problem is solved in motion pictures by showing
each picture twice, so that 48 views of the scene are
shown per second although there are still the same 24
picture frames per second
 As a result of the increased blanking rate, flicker is
eliminated
 In television pictures an effective rate of 50 vertical
scans per second is utilized to reduce flicker

Interlaced scanning
 This is accomplished by increasing the downward rate
of travel of the scanning electron beam, so that every
alternate line gets scanned instead of every successive
line
 Then, when the beam reaches the bottom of the
picture frame, it quickly returns to the top to scan
those lines that were missed in the previous scanning
 Thus the total number of lines are divided into two
groups called ‘fields’
 Each field is scanned alternately

Interlaced scanning
 In the 625 lime monochrome system, for successful
interlaced scanning, the 625 lines of each frame or
picture are divided into sets of 312.5 lines
 Each set is scanned alternately to cover the entire
picture area
 To achieve this the horizontal sweep oscillator is made
to work at a frequency of 15625 Hz (312.5 × 50 = 15625)
 To scan the same number of lines per frame (15625/25
= 625 lines)

Interlaced scanning
 Horizontal scanning periods

Interlaced scanning
 Vertical scanning periods

Interlaced scanning
 Scanning sequence

Picture Resolution
 The ability of the image reproducing system to
represent the fine structure of an object is known as its
resolving power or resolution
 Vertical resolution: The extent to which the scanning
system is capable of resolving picture details in the
vertical direction is referred to as its vertical resolution
 Vertical resolution Vr can be represented as Vr = Na x k
 Here Vr = vertical resolution as number of line
 Na = Active number of lines in the system
 k = KELL factor or resolution factor

Picture Resolution
 Assuming a reasonable value of k = 0.69,
Vr = 585 × 0.69 = 400 lines
 Horizontal resolution: The capability of the system to
resolve maximum number of picture elements along
the scanning lines determines horizontal resolution
 This can be evaluated by N = Na × aspect ratio × k
= 585 × 4/3 × 0.69 = 533

Picture Resolution
 A method to determine horizontal resolution

Picture Resolution
 Since along one line there are 533/2 = 267 complete
cyclic changes, 267 complete square wave cycles get
generated during the time the beam takes to travel
along the width of the pattern
 Thus the time duration th of one square wave cycle is
equal to
 The frequency of the square wave will be

Picture Resolution
 A method to study both horizontal and vertical
resolution

Brightness
 Brightness is the overall or average intensity of
illumination
 It determines background light level in the reproduced
picture
 The brightness of a picture, can be varied to obtain
optimum average illumination of the scene

Contrast
 Contrast is the difference in light intensity between
black and white parts of picture over and above the
average brightness level
 A picture with more contrast has bright white and dark
black intensity levels
 Too much contrast makes the picture difficult and
painful to see for a long time
 Too little contrast in picture makes it looking washed
away

Viewing distance
 The viewing distance from the screen of the TV
receiver should not be so large that the eye cannot
resolve details of the picture
 The distance should also not be so small that picture
elements become separately visible
 The above conditions are met when the vertical picture
size subtends an angle of approximately 15° at the eye
 While viewing TV, a small light should be kept ON in
the room to reduce overall contrast
 This prevents strain to the eyes and there is less fatigue

Luminance
 This is the amount of light perceived by the eye
regardless of the colour
 In monochrome television, more lighted parts of
images have more luminance
 In colour TV, colours also have their own intensities
which can be resolved by luminance on a monochrome
TV

HUE
 This is the predominant spectral colour of the received
light
 The colour of any object is distinguished by its hue
 Different hues result from different wavelengths of
spectral radiation

Saturation
 This is the spectral purity of the colour light
 Single hue colours alone occur rarely in nature
 Thus saturation may be taken as an indication of how
little the colour is diluted by white
 A fully saturated colour has no white
 Hue and saturation, when put togather, known as
chrominance

Improvement of saturation
Original image Image with 50 % increase
of saturation

Composite Video Signal
 A composite video signal consists of:
 Camera signal - corresponding to the desired picture
information
 Blanking pulses – to make the retrace invisible
 Synchronizing pulses – to synchronize the transmitter
and receiver scanning
 horizontal sync pulse
 vertical sync pulse
 their amplitudes are kept same
 but their duration are different
 needed consecutively and not simultaneously with the
picture signal – so sent on a time division basis

Composite Video Signal
 Video signal varies between certain limits
 Peak white level: 10 to 12.5%
 Black level : 72%
 Blanking level : Sync pulses added - 75% level
 Pedestal : difference between black level and
blanking level – tend to merge
 Pedestal height : distance between the pedestal
level and the dc level – indicates the average
brightness
 Picture information : 10% - 75%

DC component of the video
signal
 Average value or dc component corresponding to
the average brightness of the scene
 Average brightness can change only from frame
to frame and not from line to line
 Low pedestal height – scene darker
 Larger pedestal height – higher average
brightness

Pedestal height
 The pedestal height is the distance between the pedestal
level and the average value (dc level) axis of the video signal
 This indicates average brightness since it measures how
much the average value differs from the black level
 Even when the signal loses its dc value when passed
through a capacitor-coupled circuit the distance between
the pedestal and the dc level stays the same and thus it is
convenient to use the pedestal level as the reference level to
indicate average brightness of the scene

Blanking pulses
 Make the retrace lines invisible by raising the
signal amplitude slightly above the black level
(75%)
 Repetition rate of horizontal blanking pulse =
scanning freq = 15625Hz
 Freq of vertical blanking pulse = field scanning
freq. = 50 Hz

Interlaced scanning : Revisited

 Horizontal scanning periods

 Vertical scanning periods

 Scanning sequence

Horizontal Sync Composition
 Front porch: This is a brief cushioning period of 1.5 μs
inserted between the end of the picture detail for that
line and the leading edge of the line sync pulse
 This interval allows the receiver video circuit to settle
down from whatever picture voltage level exists at the
end of the picture line to the blanking level before the
sync pulse occurs
 Thus sync circuits at the receiver are isolated from the
influence of end of the line picture details

 Line sync pulse: After the front porch of blanking,
horizontal retrace is produced when the sync pulse
starts
 The flyback is definitely blanked out because the sync
level is blacker than black
 Line sync pulses are separated at the receiver and
utilized to keep the receiver line time base in precise
synchronism with the distant transmitter
 The nominal time duration for the line sync pulses is
4.7 μs. During this period the beam on the raster
almost completes its back stroke (retrace) and arrives
at the extreme left end of the raster

 Back porch: This period of 5.8 μs at the blanking level
allows plenty of time for line flyback to be completed
 It also permits time for the horizontal time-base
circuit to reverse direction of current for the initiation
of the scanning of next line
 The back porch also provides the necessary amplitude
equal to the blanking level and enables to preserve the
dc content of the picture information at the
transmitter

Vertical Sync Details
 The basic vertical sync added at the end of both even
add odd fields is shown in Fig.
 Its width has to be kept much larger than the horizontal
sync pulse, in order to derive a suitable field sync pulse
at the receiver to trigger the field sweep oscillator
 In the 625 line system 2.5 line period (2.5 × 64 = 160 µs)
has been allotted for the vertical sync pulses
 Thus a vertical sync pulse commences at the end of 1st
half of 313th line (end of first field) and terminates at
the end for 315th line

 Similarly after an exact interval of 20 ms (one field
period) the next sync pulse occupies line numbers—
1st, 2nd and 1st half of third, just after the second field
is over
 The horizontal sync information is extracted from the
sync pulse train by differentiation, i.e., by passing the
pulse train through a high-pass filter

 Indeed pulses corresponding to the differentiated
leading edges of sync pulses are used to synchronizes
the horizontal scanning oscillator
 But there is a problem of skipping of horizontal pulses
 The horizontal sync pulses are available both during
the active and blanked line periods but there are no
sync pulses (leading edges) available during the 2.5
line vertical sync period
 This will lead the horizontal oscillator out of sync

 Therefore, it becomes necessary to cut slots in the
vertical sync pulse at half-line-intervals to provide
horizontal sync pulses at the correct instances both
after even and odd fields
 The technique is to take the video signal amplitude
back to the blanking level 4.7 µs before the line pulses
are needed
 The waveform is then returned back to the maximum
level at the moment the line sweep circuit needs
synchronization

 Thus five narrow slots of 4.7 µs width get formed in each
vertical sync pulse at intervals of 32 µs
 The trailing but rising edges of these pulses are actually
used to trigger the horizontal oscillator
 However, the pulses actually utilized are the ones that
occur sequentially at 64 intervals
 Such pulses are marked with line numbers for both the
fields
 During the intervals of serrated vertical pulse trains,
alternate vertical spikes are utilized
 The pulses not used in one field are the ones utilized
during the second field.

 Synchronization of the vertical sweep oscillator in the
receiver is obtained from vertical sync pulses by
integration
 For scanning of 1st field, voltage builds up because
capacitor has more time to charge and 4.7 us time to
discharge
 When the second field comes, the vertical pulse comes
at half line period of horizontal pulse
 Due to this the voltage built up due to charging of
horizontal pulse does not comes to zeros when the
vertical pulse occurs
 This introduces voltage difference between both
vertical pulses

Vertical Sync problem solution

Functions of vertical pulse train
 A suitable field sync pulse is derived for triggering the field
oscillator
 The line oscillator continues to receive triggering pulses at
correct intervals while the process of initiation and
completion of the field time-base stroke is going on
 It becomes possible to insert vertical sync pulses at the end
of a line after the 2nd field and at the middle of a line at the
end of the 1st field without causing any interlace error
 The vertical sync build up at the receiver has precisely the
same shape and timing on odd and even fields

Perception of brightness and
colour
 All objects that we observe are focused sharply by the
lens system of the eye on its retina
 The retina which is located at the back side of the eye
has light sensitive organs which measure the visual
sensations
 The retina is connected with the optic nerve which
conducts the light stimuli as sensed by the organs to
the optical centre of the brain

Perception of brightness and
colour
 The light sensitive organs are of two types—rods and
cones
 The rods provide brightness sensation and thus
perceive objects only in various shades of grey from
black to white
 The cones that are sensitive to colour are broadly in
three different group
 One set of cones detects the presence of blue colour in
the object focused on the retina, the second set
perceives red colour and the third is sensitive to the
green range

Additive colour mixing
 In additive mixing which forms the basis of colour
television, light from two or more colours obtained
either from independent sources or through filters can
create a combined sensation of a different colour
 Thus different colours are created by mixing pure
colours and not by subtracting parts from white
 The impression of white light can also be created by
choosing suitable intensities of these colours
 Red, green and blue are called primary colours. These
are used as basic colours in television

 By pairwise additive mixing of the primary colours the
following complementary colours are produced:
 Red + Green = Yellow
 Red + Blue = Magenta (purplish red shade)
 Blue + Green = Cyan (greenish blue shade

Perception of eye for different
colours

 The brightness (luminance) impression created by the
combined light source is numerically equal to the sum
of the brightnesses (luminances) of the three
primaries
 This property of the eye of producing a response which
depends on the algebraic sum of the red, green and
blue inputs is known as Grassman’s Law
100% White = 30% Red + 59% Green + 11% Blue

Video signals
for colours
 Colour
voltage
amplitudes:

Video signals for colours
 Desaturated colours:
 Any colour is said to be desaturated when mixed with
white
 In a colour camera output signal, Red colour is
desaturated to a small amount, then the Vg and Vb have
lower values
 But as desaturation of red increases, Vg and Vb values
are increased
 For 100% desaturation Vr = Vg =Vb

Video signals for colours
 Colour video Frequencies:
 When the scene is not dominated by one or few colours
the information to be transmitted occupies more
frequency spectrum
 It is discovered that colour frequencies need 1.5 Mhz
band in order to transmit finest details of a scene
 The luminance signal frequency range is up to 5 Mhz

Luminance signal Y
 Luminance refers to the brightness of scene
 It is formed by adding the three camera outputs in the
ratio, Y = 0.3 R + 0.59 G + 0.11 B
 These percentages correspond to the relative
brightness of the three primary colours
 Therefore a scene reproduced in black and white by
the ‘Y ’ signal looks the same as when it is televised in
monochrome

Generation of luminance signal Y

Compatibility
 It is necessary that a colour TV should produce black
and white picture and a black and white TV should be
able to process colour signal to extract the black white
scene information
 This feature is known as compatibility of video signal
 Here we can not transmit Vr, Vg, Vb separately because
of limitation of 5.5 bandwidth
 To solve this problem colour difference signals are
used, which can be accommodated in 5.5 Mhz band

Colour difference signal
 Colour difference voltages are derived by subtracting
the luminance voltage from the colour voltages
 Only (R – Y) and (B – Y) are produced
 It is only necessary to transmit two of the three colour
difference signals since the third may be derived from
the other two
 The circuit for getting colour difference signals is as
follows

 Here by definition we have
 Y = 0.3R + 0.59G + 0.11B
 Therefore,
 (R – Y) = 0.7R – 0.59G – 0.11B
 (B – Y) = 0.89B – 0.59G – 0.3R.
 The colour difference signals equal zero when white or grey
shades are being transmitted
 On peak whites let R = G = B = 1 volt
 Then Y = 0.59G + 0.3R + 0.11B = 0.59 + 0.3 + 0.11 = 1 (volt)
 (R – Y) = 1 – 1 = 0, volt and (B – Y) = 1 – 1 = 0 volt

 On any grey shade let R = G = B = v volts (v < 1)
 Then Y = 0.59v + 0.3v + 0.11v = v
 (R – Y) = v – v = 0 volt and (B – Y) = v – v = 0 volt
 Thus it is seen that colour difference signals during the
white or grey content of a colour scene of during the
monochrome transmission completely disappear and
this is an aid to compatibility in colour TV systems

 Consider we have a desaturated magenta(Purple)
colour to transmit
 Suppose R = 0.7, G = 0.2 and B = 0.6 volts
 The white content is represented by equal quantities of
the three primaries and the actual amount must be
indicated by the smallest voltage of the three, that is,
by the magnitude of G
 Thus white is due to 0.2 R, 0.2 G and 0.2 B. The
remaining, 0.5 R and 0.4 B together represent the
magenta hue

 (i) The luminance signal Y = 0.3 R + 0.59 G + 0.11 B
 Substituting the values of R, G, and B we get Y = 0.3
(0.7) + 0.59 (0.2) + 0.11(0.6) = 0.394 (volts)
 (ii) The colour difference signals are:
 (R – Y) = 0.7 – 0.394 = + 0.306 (volts)
 (B – Y) = 0.6 – 0.394 = + 0.206 (volts)
 (iii) Reception at the colour receiver—At the
receiver after demodulation, the signals, Y, (B – Y)
and (R – Y), become available

 Then by a process of matrixing the voltages B and
R are obtained as:
 R = (R – Y) + Y = 0.306 + 0.394 = 0.7 V
 B = (B – Y) + Y = 0.206 + 0.394 = 0.6 V
 (G – Y) matrix—The missing signal (G – Y) that is
not transmitted can be recovered by using a
suitable matrix based on the explanation given
below:
 Y = 0.3 R + 0.59G + 0.11B
 also (0.3 + 0.59 + 0.11)Y = 0.3R + 0.59G + 0.11B

 Rearranging the above expression we get:
 0.59(G – Y) = – 0.3 (R – Y) – 0.11 (B – Y)
 Substituting the values of (R – Y) and (B – Y)
 (G – Y) = – (0.51 × 0.306) – 0.186(0.206) = – 0.15606 –
0.038216 = – 0.194
 G = (G – Y) + Y = – 0.194 + 0.394 = 0.2

 Unsuitability of (G – Y) Signal for Transmission:
 The proportion of G in Y is relatively large(59%) in
most cases, the amplitude of (G – Y) is small
 The smaller amplitude together with the need for gain
in the matrix would make S/N ratio problems more
difficult then when (R – Y) and (B – Y) are chosen for
transmission

Encoding of colour difference
signals
 The problem of transmitting (B-Y) and (R-Y) video
signals simultaneously with one carrier frequency is
solved by creating two carrier frequencies from the
same colour subcarrier without any change in its
numerical value
 Two separate modulators are used, one for the (B-Y)
and the other for the (R-Y) signal
 However, the carrier frequency fed to one modulator is
given a relative phase shift of 90° with respect to the
other before applying it to the modulator

signals

 The horizontal scanning frequency of camera beam is
15625 Hz
 Therefore, the video frequencies generated on scanning
any scene are multiples of this frequency
 So video information can be shown as below
signals

signals
 Here the chrominance signal is fitted into the gaps of Y
signal frequencies

signals
 The transmitted signal does not contain the subcarrier
frequency but it is necessary to generate it in the
receiver with correct frequency and phase relationship
for proper detection of the colour sidebands
 To ensure this, a short sample of the colour subcarrier
oscillator (8 to 11 cycles) called the ‘colour burst’ is sent
to the receiver along with sync signals
 This is located in the back porch of the horizontal
blanking pedestal

Formation of chrominance signal
 The chroma signal has magnitude and phase angle as
shown below

 Consider that we need to transmit red colour
 For a pure red, R = 1v, G = 0v, B = 0v
 We know that
 (R – Y) = 0.7R – 0.59G – 0.11B
 (B – Y) = 0.89B – 0.59G – 0.3R.
 Putting values for Red colour we have,
 (R – Y) = 0.7(1) – 0.59(0) – 0.11(0) = 0.7v
 (B – Y) = 0.89(0) – 0.59(0) – 0.3(1) = -0.3v

)Y)-(RY)-((B 22
 76.0)(0.7)((-0.3) 22

)(
)(
tan 1
YB
YR


  
104
)3.0(
)7.0(
tan 1


 
 Magnitude of chroma signal can be found as below:
 And the phase of chroma signal with respect to (B-Y)
can found as below
 Thus for pure red colour , the chroma signal falls in
second quadrant

 For cyan(Blue +Green) R = 0v, G = 1v, B = 1v
 Putting values for cyan we have,
 (R – Y) = 0.7(0) – 0.59(1) – 0.11(1) = -0.7v
 (B – Y) = 0.89(1) – 0.59(1) – 0.3(0) = 0.3v
 Magnitude of chroma signal can be found as below:
 And the phase of chroma signal with respect to (B-Y)
can found as below
)Y)-(RY)-((B 22
 76.0)(-0.7)((0.3) 22


104180284
)3.0(
)7.0(
tan 1


 
)(
)(
tan 1
YB
YR


 

 From previous analysis we can say that as cyan is a
complementary colour of red it has the same
magnitude but exactly opposite angle
 In a natural scene we have many combination of
colours in a single horizontal line of an image or video
 Therefore, for a natural scene the chroma signal has
different magnitude and phase angle for each
horizontal line
 The chroma signal decides the hue and saturation of a
colour picture

Introduction to Video Signals

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Introduction to Video Signals