11. The internal resistor of the second-order high pass filter is equal to 10kΩ. Find the value of the feedback resistor?
A. 6.9kΩ
B. 5.86kΩ
C. 10kΩ
D. 12.56kΩ
Answer: B
Pass band gain for second order butterworth response,
AF =1.586.
=> AF= [1+(RF/R1)]
=> RF= (AF-1)×R1
=(1.586-1)×10kΩ
=5860 =5.86kΩ.
12. Consider the following circuit and calculate the low cut-off frequency value?
A. 178.7Hz
B. 89.3Hz
C. 127.65Hz
D. 255.38Hz
Answer: A
The low cut-off frequency for the given filter is
fL =1/√[2π√(R2×R3×C2×C3)] =178.7Hz.
13. Determine voltage gain of second-order high pass Butterworth filter.
Specifications R3 =R2=33Ω, f=250hz and fL=1khz.
A. -11.78dB
B. -26.51dB
C. -44.19dB
D. None of the mentioned
Answer: C
Since R3 =R2
=> C2 = 1/(2π ×fL×R2)
= 1/(2π ×1kHz×33Ω)
=> C3 =C2= 4.82µF.
Voltage gain of filter |VO/Vin|
=AF / [√ 1+(fL/f)4]
= 1.586/[1+(1kHz/250kz)4]
=1.586/252=6.17×10-3
=20log(6.17×10-3)= -44.19dB.
14. From the given specifications, determine the value of voltage gain magnitude of the first order and second-order high pass Butterworth filter?
Pass band voltage gain=2;
Low cut-off frequency= 1kHz;
Input frequency=500Hz.
A. First order high pass filter =-4.22dB , Second order high pass filter=-0.011dB
B. First order high pass filter =-0.9688dB , Second order high pass filter=-6.28dB
C. First order high pass filter =-11.3194dB , Second order high pass filter=-9.3257dB
D. First order high pass filter =-7.511dB , Second order high pass filter=-5.8999dB
Answer: B
For first order high pass filter,
|VO/Vin|=AF ×(f/fL) / [ √1+(f/fL)2]
=(2×(500Hz/1kHz)) /√[1+(500Hz/1kHz)2]
=> |VO/Vin| = 1/1.118= 0.8944
=20log(0.8944) =-0.9686dB.
For second order high pass filter,
|VO/Vin|=AF / [ √ 1 +(fL/f)4]
=2/√[1+ (1kHz/500Hz)2]
=>|VO/Vin|=2/4.123
=0.4851 = 20log(0.4851) = -6.28dB.
15. How are the higher-order filters formed?
A. Using the first-order filter
B. Using second-order filter
C. Connecting first and second-order filters in series
D. Connecting first and second-order filters in parallel
Answer: C
Higher filters are formed by using the first and second-order filters. For example, a third-order low pass filter is formed by cascading first and second-order low pass filters.
16. State the disadvantage of using higher-order filters?
A. Complexity
B. Requires more space
C. Expensive
D. All of the mentioned
Answer: D
Although a higher-order filter than necessary gives a better stop band response, the higher-order type is more complex, occupies more space, and is more expensive.
17. The overall gain of a higher-order filter is
A. Varying
B. Fixed
C. Random
D. None of the mentioned
Answer: B
The overall gain of a higher-order filter is fixed because all the frequency-determining resistors and capacitors are equal.
18. Find the roll-off rate for 8th order filter
A. -160dB/decade
B. -320dB/decade
C. -480dB/decade
D. -200dB/decade
Answer: A
For the nth order filter the roll-off rate will be –
n×20dB/decade.
=>∴ for 8th order filter= 8×20=160dB/decade.
19. 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
20. 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.
21. 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.
22. 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.
23. 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.
