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

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.

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)

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

Table: SI photometry quantities [Wikipedia] explanation: see for example in http://home.tir.com/~ms/lightingdefinitions/lightingdefinitions.html

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 [-]

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) 173.2 lx, b) 1558,8 lm, c) 1325 lm, d) 46,86 cd/m2, e) 147,22 lx, f) In = 422 cd

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).

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 = 2.109 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 80 000 lx. Calculate the illuminance of this surface at the 25° elevation of Sun Solution: E2 = 47 800 lx.