Digital Electronics - Digital Concepts - Discussion
Discussion Forum : Digital Concepts - General Questions (Q.No. 42)
42.
A periodic digital waveform has a pulse width (tw) of 6 ms and a period (T) of 18 ms. The duty cycle is ________.
Discussion:
12 comments Page 2 of 2.
Raj said:
1 decade ago
D = T/P*100 %.
Where D is the duty cycle, T is the time the signal is active, and P is the total period of the signal.
D = 6ms/18ms*100 {where milli = 10^-3}.
So, D = (6x10^-3)/(18x10^-3)*100.
= (6/18)*(10^-3/10^-3)*100.
= 1/3*100.
= 33.3% HP.
Where D is the duty cycle, T is the time the signal is active, and P is the total period of the signal.
D = 6ms/18ms*100 {where milli = 10^-3}.
So, D = (6x10^-3)/(18x10^-3)*100.
= (6/18)*(10^-3/10^-3)*100.
= 1/3*100.
= 33.3% HP.
(1)
Raviprakash G Mishra said:
7 years ago
The Duty Cycle =Ton/(Ton+Toff) that is;
6ms/18ms,
= 33.3%.
6ms/18ms,
= 33.3%.
(1)
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers