1. Digital TV Systems Exercise 1
Ľ. Maceková
Oct. 2017
Technical University of Kosice, Slovakia

2. Plan of exercises:
Introduction. Home assignements allocation. Examples: screen resolution and observing distance, photometric quantities.
Examples: Colorimetry. Decibels calculation.
Test – examples. Cameras. Video cutting and processing.
Short student presentations: handy cameras specifications.
Video streaming.
Presentation of student videos.
Cable TV – energy budget.
Test – example energy budget. „Debts“ and consultations.

3. 1. Homework – home assignment
Processing captured video
shoot your own video (by handy camera or by smartphone) – Content optional (laboratory, university campus, college, Kosice City, nature, technical process or procedure. Video downloaded from Internet is also permissible.
software for video processing: Movie studio Platinum / Sony Vegas, Pinneacle, Windows Movie Maker, Shotcut, Wondershave Filmora, VideoPad,etc.
cut both, video and audio tracks, add more tracks and process/combine/capture them by more methods/tools of software, add text, headlines, music, effects
picture in picture imaging methode
Finish the video until the end of semester (24-th Nov. 2017)

4. 2.
Short homework (see for example Sony FDR-AX53):
Prepare the short presentation - description/explanation of two camera technical parameters:
- Sensor/effective pixels, zoom
max. apperture, focal distance
recording parameters (recording media, recording time,
still image resolution, video formats,
norm/colour standard, audio channel properties and formats

5. Table: SI photometry quantities
[Wikipedia]
explanation: see for example in

6. Lv= ∆𝐼 𝛼 ∆𝑆 𝑐𝑜𝑠 𝛼 [cd/m2] ... or LV= I / S ... LV=E. kR /π
𝐼= ∆Φ ∆ω [𝑐𝑑]
Δω ... space angle
𝐼= 𝛷 4𝛱 [cd]
.... for spot light source
𝐼= Φ Π . cos α [𝑐𝑑]
... Lambert formula; flat light source, α ...angle between the normal and the light direction
E= ∆𝛷 ∆𝑆 [lx]
E= 𝐼 𝑐𝑜𝑠 𝑎 𝑙 2 [lx]
... for spot light source and sloping impact
Lv= ∆𝐼 𝛼 ∆𝑆 𝑐𝑜𝑠 𝛼 [cd/m2] or LV= I / S ... LV=E. kR /π
ΔIα= ΔIn . cos α
contrast ... K = LV max / LV min [-]
reflex coefficient ... kR = Φ0 / Φc [-]
(Lambert’s emission low )
transparency coefficient ... kT = ΦT / Φc [-]

7. Example 1
The white diffuse surface with dimensions 3 x 3 [m2] is illuminated from the distance 2 m by lamp, which has luminous intensity 800 cd in this direction. The lamp illuminates the surface with angle to the normal line α = 30°. The reflection coefficient of this white surface is kR = 0,85.
Calculate:
Illuminance E [lx] E = (I / l 2 ) cos α
Incident Luminous flux Φd [lm] ... Φd = E . S
Reflected Luminous flux Φr [lm] ... Φd = Φd . kR
Luminance of the surface LV [cd] .... LV = E . kR/ π (Lambert law)
Emission (radiation) of the surface ER [lx] ... ER = E . kR (the secondary source of light)
Luminous intensity of the surface In [cd] in the direction of the normal line ... I = LV . S
Results:
a) lx, b) 1558,8 lm, c) lm, d) 46,86 cd/m2, e) 147,22 lx, f) In = 422 cd

8. Example 2 - contrast
The screen reproduces the image with luminance LV max = 200 cd/m2 and LV min = 5 cd/m2 .
Calculate a) contrast C, if we are watching the screen in total dark ambient,
b) contrast CR , if we an another light is illuminating the screen with the LV a = cd/m2
c) contrast CRF in conditions by b), if we are watching the screen via filter with transparency coefficient kT = 0.2.
Results: a) 40, b) KR=(LV max+LV a)/(LV min+ LV a) = 6,57  decreasing of contrast - degradation of the visual sensing, c) we must consider 2 x transitions across the cez filter  CRF =18,73  increasing, amelioration in comparasion with b).

9. Example 4 - Illuminance of the Earth surface
Example 3 - what is 1 cd?
The human eye is able to sense the light at the conditions of minimal illuminance of the retina by E = lx. Calculate the distance of the eye from the flame of a candle with luminous intensity 1 cd, where we can see it in the completely dark ambient.
Result: E = (I / r2 ) cos α, α = 0°  r= (I/E) -1/2 = ... = 22,4 km
Example 4 - Illuminance of the Earth surface
If the elevation of the Sun above the horizon is 45 °, the illuminance of the horizontal surface of the earth is lx.
Calculate the illuminance of this surface at the 25° elevation of Sun
Solution: E2 = lx.