11. The voltage gain magnitude of the all-pass filter is
A. Zero
B. One
C. Infinity
D. None of the mentioned
Answer: B
The magnitude of voltage gain of all-pass filter
|VO /Vin| = √(1+(2π/RC.2) / √(1+(2 π/RC.2) =1.
12. What happens if the position of R and C are interchanged in the below circuit diagram?
A. Vin leads VO
B. Vin lags VO
C. VO leads Vin
D. VO leads Vin
Answer: C
For the circuit given, the phase angle changes from 0 to 180o as frequency is varied from 0 to ∞. If the positions of R and C are interchanged, the phase shift and bandwidth input and output become positive. That is the output (VO) leads to input (Vin).
13. Choose the incorrect statement “In wide band-reject filter”.
A. Low cut-off frequency of low pass filter must be larger than the high cut-off frequency of the high pass filter.
B. Low cut-off frequency of the high pass filter must be equal to the high cut-off frequency of the high pass filter.
C. Low cut-off frequency of high pass filter must be smaller than the high cut-off frequency of the low pass filter.
D. None of the mentioned
Answer: D
In a wide band-reject filter, the low cut-off frequency of high pass filter must be larger than the high cut-off frequency of the low pass filter.
14. A bandpass filter is one which
Attenuates frequencies between two designated cut off frequencies and passes all other frequencies
Passes all frequencies
Attenuates all frequencies
Passes frequencies between two designated cut off frequencies
Answer.4
A bandpass filter passes a certain band of frequencies and blocks low and high frequencies.
It has two corner frequencies, smaller (fc1) and larger (fc2).
It attenuates signals with frequencies below fc1 and above fc2
15. A 10 kHz even-symmetric square wave is passed through a bandpass filter with the center frequency at 30 kHz and a 3 dB passband of 6 kHz. The filter output is
a highly attenuated square wave at 10 kHz
nearly zero
a nearly perfect cosine wave at 30 kHz
a nearly perfect sine wave at 30 kHz
Answer.3.
The given signal is 10 kHz even with a symmetrical square wave.
Since the bandpass filter is centered at 30 kHz it follows half-wave symmetry. Hence only odd harmonics exist.
The frequency components present in this wave: are 10 kHz, 30 kHz, 50 kHz, and 70 kHz.
Now, 30 kHz component will pass through filter output is nearly perfect cosine wave at 10 kHz cosine as the signal is even signal.
16. Which filter attenuates any frequency outside the passband?
A. Band-pass filter
B. Band-reject filter
C. Band-stop filter
D. All of the mentioned
Answer: A
A band-pass filter has a passband between two cut-off frequencies fH and fL. So, any frequency outside this passband is attenuated.
17. Narrow band-pass filters are defined as
A. Q < 10
B. Q = 10
C. Q > 10
D. None of the mentioned
Answer: C
Quality factor (Q) is the measure of selectivity, meaning the higher the value of Q, the narrower its bandwidth.
18. A band-pass filter has a bandwidth of 250Hz and a center frequency of 866Hz. Find the quality factor of the filter?
A. 3.46
B. 6.42
C. 4.84
D. None of the mentioned
Answer: A
Quality factor of band-pass filter, Q =fc/bandwidth= 566/250=3.46.
19. Find the center frequency of wide band-pass filter
A. fc= √(fh ×fL)
B. fc= √(fh +fL)
C. fc= √(fh -fL)
D. fc= √(fh /fL)
Answer: A
In a wide band-pass filter, the product of high and low cut-off frequency is equal to the square of the center frequency
i.e. ( fc)2 =fH×fL
=> fc= √(fh×fL).
20. Find out the voltage gain magnitude equation for the wide band-pass filter.
A. AFt×( f/fL)/√[(1+(f/fh)2]×[1+(f/fL)2].
B. AFt/ √{[1+(f/fh)2]×[1+(f/fL)2]}
C. AFt/ √{[1+(f/fh)2]/[1+(f/fL)2]}
D. [AFt/(f/fL)]/ √{[1+(f/fh)2]/[1+(f/fL)2]}
Answer: A
The voltage gain magnitude of the band-pass filters is equal to the product of the voltage gain magnitudes of high pass and low pass filters.
21. When a second-order high pass filter and second-order low pass sections are cascaded, the resultant filter is a
A. ±80dB/decade band-pass filter
B. ±40dB/decade band-pass filter
C. ±20dB/ decade band-pass filter
D. None of the mentioned
Answer: B
The order of the band-pass filter depends on the order of the high pass and low pass filter sections.
22. Find the voltage gain magnitude of the wide band-pass filter?
Where total passband gain is=6, input frequency = 750Hz, Low cut-off frequency =200Hz and high cut-off frequency=1khz.
A. 13.36 dB
B. 12.25 dB
C. 11.71 dB
C. 14.837dB
Answer: D
Voltage gain of the filter,
|VO/Vin|=[AFt×(f/fL)]/{√[1+(f/fL)2]×[1+f/fL)2]}
=[6×(750/20)]/√{[1+(750/200)2]×[1+(750/200)2]}
=22.5/√(15.6×1.56) =5.519.
|VO/Vin|= 20log(5.519) =14.837dB.
23. Compute the quality factor of the wide band-pass filter with high and low cut-off frequencies equal to 950Hz and 250Hz.
A. 0.278
B. 0.348
C. 0.696
D. 0.994
Answer: C
Quality factor
Q=√(fh×fL)/(fh-fL)
= √(950Hz×250Hz)/(9950Hz-250Hz) =0.696.
24. The details of low pass filter sections are given as fh =10kHz, AF= 2, and f=1.2kHz. Find the voltage gain magnitude of the first-order wide band-pass filter, if the voltage gain magnitude of the high pass filter section is 8.32dB.
A. 48.13dB
B. 10.02dB
C. 14.28dB
D. 65.99dB
Answer: C
|VO/Vin|(high pass filter)
= 8.32dB=10(8.32/20) =2.606.
Therefore, the voltage gain of wide band-pass filter |VO/Vin|
= AFt×(f/fL)/√[1+(f/fh)2)]×[1+(f/fL)2)]
={Af/√[(1+(f/fh)2]}×{(Af×f/fL)/√[1+(f/fL)2]}
=Aft /√[1+(f/fh)2]×(2.606)
= [2/√(1+(1.2kHz/10kHz)2]×( 2.606)
= 1.986×2.606 =5.17
=20log×(5.17) =14.28dB.
25. The quality factor of a wide band-pass filter can be
A. 12.6
B. 9.1
C. 14.2
D. 10.9
Answer: B
A wide band-pass filter has a quality factor of less than 10.
