Electrical Engineering - Passive Filters - Discussion
Discussion Forum : Passive Filters - General Questions (Q.No. 1)
1.
In a certain parallel resonant band-pass filter, the resonant frequency is 14 kHz. If the bandwidth is 4 kHz, the lower frequency
Discussion:
11 comments Page 1 of 2.
Sanjeev singh said:
1 decade ago
The resonant frequency is 14 kHz.
Band Width is 4 KHz.
High frequency = resonant frequency+B.w/2, So Hf = 14+4/2 = 16.
Lower frequency = resonant frequency-B.w/2, So Lf = 14-4/2 = 12.
Answer C.
Band Width is 4 KHz.
High frequency = resonant frequency+B.w/2, So Hf = 14+4/2 = 16.
Lower frequency = resonant frequency-B.w/2, So Lf = 14-4/2 = 12.
Answer C.
(9)
Alex.jangam said:
1 decade ago
Mid frequency is 14 KHz, Band width =4kHz
So frequesncy range is
14+ 4/2 = 16 Khz on high pass and 14- 4/2 = 12 KHz on low pass.
So frequesncy range is
14+ 4/2 = 16 Khz on high pass and 14- 4/2 = 12 KHz on low pass.
(1)
Karthik.MG said:
7 years ago
14-4/2= 12 how?
Please explain me.
Please explain me.
(1)
Pavan kumar said:
6 years ago
Low pass 14-4/2 = 12kHz.
(1)
Fij said:
2 years ago
Very Clear explanation. Thanks @Sanjeev Sing.
(1)
ASHOK said:
1 decade ago
The resonant frequency is 14 kHz.
Band Width is 4 KHz then multiply the band width with 4.
4KHz * 4= 16KHz.
High frequency is below the 4KHz and Low Frequency is Near the 16KHz then the resultant Low Frequency is 12KHz.
Band Width is 4 KHz then multiply the band width with 4.
4KHz * 4= 16KHz.
High frequency is below the 4KHz and Low Frequency is Near the 16KHz then the resultant Low Frequency is 12KHz.
Chinna said:
1 decade ago
B.W = f2-f1 = 4hz.
fr^2 = f1*f2 = 196.
By solving above eq...f1 = 12hz.
fr^2 = f1*f2 = 196.
By solving above eq...f1 = 12hz.
Rauf said:
9 years ago
It's very good to know and improve the electrical knowledge. Thank you.
Giridhar k said:
9 years ago
Resonant frequency = G.M(geometric mean) of f1 & f2.
fr = root of (f1 & f2).
For High quality factor circuits Q >> 5, ckts approximately A.M of f1 & f2. fr = 0.5(f1 + f2).
fr = root of (f1 & f2).
For High quality factor circuits Q >> 5, ckts approximately A.M of f1 & f2. fr = 0.5(f1 + f2).
Veeresh said:
6 years ago
Thanks for solving this clearly @Sanjeev sing.
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