# Band pass Filter MCQ [Free PDF] – Objective Question Answer for Band pass Filter MCQ Quiz

1. A band pass filter is one which

1. Attenuates frequencies between two designated cut off frequencies and passes all other frequencies
2. Passes all frequencies
3. Attenuates all frequencies
4. Passes frequencies between two designated cut off frequencies

A band pass 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

2. 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

1. a highly attenuated square wave at 10 kHz
2. nearly zero
3. a nearly perfect cosine wave at 30 kHz
4. a nearly perfect sine wave at 30 kHz

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.

1. 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

A band-pass filter has a passband between two cut-off frequencies fH and fL. So, any frequency outside this passband is attenuated.

2. Narrow band-pass filters are defined as

A. Q < 10
B. Q = 10
C. Q > 10
D. None of the mentioned

Quality factor (Q) is the measure of selectivity, meaning the higher the value of Q, the narrower its bandwidth.

3. 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

Quality factor of band-pass filter, Q =fc/bandwidth= 566/250=3.46.

4. 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)

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).

5. 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]}

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.

6. When a second-order high pass filter and second-order low pass sections are cascaded, the resultant filter is a

D. None of the mentioned

The order of the band-pass filter depends on the order of the high pass and low pass filter sections.

7. 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

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.

8. 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

Quality factor

Q=√(fh×fL)/(fh-fL)

= √(950Hz×250Hz)/(9950Hz-250Hz) =0.696.

9. 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

|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.

10. The quality factor of a wide band-pass filter can be

A. 12.6
B. 9.1
C. 14.2
D. 10.9

A wide band-pass filter has a quality factor of less than 10.

12. If the gain at center frequency is 10, find the quality factor of the narrow band-pass filter

A. 1
B. 2
C. 3
D. None of the mentioned

The gain of the narrow band-pass filter must satisfy the condition,

AF= 2×Q2

When Q=3,

=> 2×Q2 =2×(32) =18.

=> 10<18.

Hence the condition is satisfied when Q=3.

13. The advantage of a narrow band-pass filter is

A. fc can be changed without changing gain
B. fc can be changed without changing bandwidth
C. fc can be changed without changing resistors
D. All of the mentioned