SN LLC

By Sakif Newaj January 05, 2026

How do 64us per second scanning in india television system?

In the Indian television system (PAL, 625 lines, 25 frames/sec), the horizontal scanning time is 64 µs per line , not per second. Let’s break it down clearly: 1. Frame rate: 25 frames per second. 2. Total lines per frame: 625 lines (including blanking lines). 3. Active lines per frame: About 585 visible lines; the rest are blanked for synchronization and retrace. 4. Time for 1 frame: 1 25 25 1 second = 40,000 µs . 5. Time for 1 line (including blanking): 40 , 000 μ s 625 lines = 64 μ s per line . 625 lines 40 , 000 μ s = 64 μ s per line . What happens in 64 µs? ~52 µs → Active trace (beam moves left to right, drawing the picture). ~12 µs → Horizontal blanking & retrace (beam returns invisibly to the left). This 64 µs per line repeats 625 times to complete 1 frame in 40,000 µs, and 25 such frames are scanned per second. So 64 µs is the...

Posts

Latest posts

By Sakif Newaj January 05, 2026

How do 64us per second scanning in india television system?

By Sakif Newaj October 08, 2025

AnaCom (Chapter 2) (Mid)

Answer: 2.1-2(a) We have two signals x ( t ) x ( t ) and y ( t ) y ( t ) in Fig. P2.1-2a (not fully drawn here, but let's assume they are simple pulses). Step 1: Energy of a signal The energy of a continuous-time signal x ( t ) x ( t ) is: E x = ∫ − ∞ ∞ ∣ x ( t ) ∣ 2 d t E x = ∫ − ∞ ∞ ∣ x ( t ) ∣ 2 d t Similarly for y ( t ) y ( t ) : E y = ∫ − ∞ ∞ ∣ y ( t ) ∣ 2 d t E y = ∫ − ∞ ∞ ∣ y ( t ) ∣ 2 d t Step 2: Energy of x ( t ) + y ( t ) x ( t ) + y ( t ) and x ( t ) − y ( t ) x ( t ) − y ( t ) Let’s compute E x + y E x + y : E x + y = ∫ [ x ( t ) + y ( t ) ] 2 d t E x + y = ∫ [ x ( t ) + y ( t ) ] 2 d t = ∫ [ x ( t ) 2 + y ( t ) 2 + 2 x ( t ) y ( t ) ] d t = ∫ [ x ( t ) 2 + y ( t ) 2 + 2 x ( t ) y ( t )] d t = E x + E y + 2 ∫ x ( t ) y ( t ) d t = E x + E y + 2 ∫ x ( t ) y ( t ) d t Similarly: E x − y = E x + E y − 2 ∫ x ( t ) y ( t ) d t E x − y = E x + E y − 2 ∫ x ( t ) y ( t ) d t Step 3: W...