Electrical Engineering - Passive Filters - Discussion
Discussion Forum : Passive Filters - General Questions (Q.No. 4)
4.
The maximum output voltage of a certain low-pass filter is 15 V. The output voltage at the critical frequency is
Discussion:
11 comments Page 1 of 2.
Selam said:
3 years ago
Vc = Vs * 0.707,
Vc =15v * 0.707,
Vc = 10.62.
Vc =15v * 0.707,
Vc = 10.62.
(2)
Vin said:
6 years ago
So, which one is the correct formula of all the formulae that has been presented here. I am confused.
Udaypratapsingh said:
6 years ago
As we know, that voltage output is 70.7% of input voltage at critical frequency.
(1)
Shailendra said:
9 years ago
Db to Voltage:
20log(V) = 20.52,
log(V) = 20.52/20 = 1.026,
V = 10^(1.026),
V = 10.62.
20log(V) = 20.52,
log(V) = 20.52/20 = 1.026,
V = 10^(1.026),
V = 10.62.
(1)
Mitrabhanu said:
9 years ago
How to convert dB into the voltage? Please answer this.
Nyce said:
1 decade ago
Critical frequency is the 70.7% of the original value or -3 db in terms of db.
A_jediknight_IB said:
1 decade ago
I think it is more intuitive to think of it in terms of dB.
Max voltage is 15V, convert this to dB.
15V = 20*log(15) = 23.52dB.
Voltage at the critical freq will be 3dB less than this max.
23.52dB - 3dB = 20.52dB.
Then convert back to voltage 20.52dB = 10.62V.
Max voltage is 15V, convert this to dB.
15V = 20*log(15) = 23.52dB.
Voltage at the critical freq will be 3dB less than this max.
23.52dB - 3dB = 20.52dB.
Then convert back to voltage 20.52dB = 10.62V.
Ganga said:
1 decade ago
Critical frequency < 1/sqrt(2) of o/p voltage.
=> critical freq = 15/sqrt(2) = 10.605 V.
=> critical freq = 15/sqrt(2) = 10.605 V.
Swe said:
1 decade ago
How did 70. 7 arise, can you please explain me ?
Arundeep said:
1 decade ago
Critical frequency is less then 70. 7% by, O/P voltage.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers