Op Amp Active Filter MCQ [Free PDF] - Objective Question Answer for Op Amp Active Filter Quiz

11. Which filter performs exactly the opposite of the band-pass filter?

A. Band-reject filter
B. Band-stop filter
C. Band-elimination filter
D. All of the mentioned

Answer: D

A band rejects is also called a band-stop and band-elimination filter. It performs exactly the opposite of band-pass because it has two passbands: 0 L and f > f_H.

12. Given the lower and higher cut-off frequency of a band-pass filter are 2.5kHz and 10kHz. Determine its bandwidth.

A. 750 Hz
B. 7500 Hz
C. 75000 Hz
D. None of the mentioned

Answer: B

Bandwidth of a band-pass filter is Bandwidth

= f_H– f_L=10kHz-2.5kHz

=7.5kHz=7500Hz.

13. In which filter the output and input voltages are equal in amplitude for all frequencies?

A. All-pass filter
B. High pass filter
C. Low pass filter
D. All of the mentioned

Answer: A

In an all-pass filter, the output and input voltages are equal in amplitude for all frequencies. This filter passes all frequencies equally well and with phase shift and between the two functions of frequency.

14. The gain of the first order low pass filter

A. Increases at the rate 20dB/decade
B. Increases at the rate 40dB/decade
C. Decreases at the rate 20dB/decade
D. Decreases at the rate 40dB/decade

Answer: C

The rate at which the gain of the filter changes in the stopband is determined by the order of the filter. So, for a low pass filter, the gain decreases at the rate of 20dB/decade.

15. Which among the following has the best stop band response?

A. Butterworth filter
B. Chebyshev filter
C. Cauer filter
D. All of the mentioned

Answer: C

The cauer filter has a ripple passband and a ripple stopband. So, generally, the cauer filter gives the best stop band response among the three.

16. Determine the order of filter used, when the gain increases at the rate of 60dB/decade on the stopband.

A. Second-order low pass filter
B. Third-order High pass filter
C. First-order low pass filter
D. None of the mentioned

Answer: B

The gain increases for the high pass filter. So, for a third-order high pass filter, the gain increases at the rate of 60dB/decade in the stopband until f=f_L.

17. Name the filter that has two stopbands?

A. Band-pass filter
B. Low pass filter
C. High pass filter
D. Band-reject filter

Answer: A

A band-pass filter has two stop bands: 1) 0 < f < f_L and 2) f > f_H.

18. The frequency response of the filter in the stopband.

A. Decreases with increase in frequency
B. Increase with an increase in frequency
C. Decreases with a decrease in frequency
D. All of the above

Answer: C

The order of frequency of the filter in the stopband determines either steady decreases or increases or both with an increase in frequency.

19. Find the voltage across the capacitor in the given circuit

A. V_O= V_in/(1+0.0314jf)
B. V_O= V_in×(1+0.0314jf)
C. V_O= V_in+0.0314jf/(1+jf)
D. None of the mentioned

Answer: A

The voltage across the capacitor,

V_O= V_in/(1+j2πfR_C)

=> V_O= V_in/(1+j2π×5k×1µF×f)

=> V_O= V_in/(1+0.0314jf).

20. Find the complex equation for the gain of the first order low pass Butterworth filter as a function of frequency.

A. A_F/[1+j(f/f_H)].
B. A_F/√ [1+j(f/f_H)²].
C. A_F×[1+j(f/f_H)].
D. None of the mentioned

Answer: A

The gain of the filter, as a function of frequency, is given as

V_O/ V_in=A _F/(1+j(f/f_H)).

21. Compute the passband gain and high cut-off frequency for the first order high pass filter.

A. A_F=11, f_H=796.18Hz
B. A_F=10, f_H=796.18Hz
C. A_F=2, f_H=796.18Hz
D. A_F=3, f_H=796.18Hz

Answer: C

The pass band gain of the filter,

A_F =1+(R_F/R₁)

=>A_F=1+(10kΩ/10kΩ)=2.

The high cut-off frequency of the filter

f_H=1/2πRC

=1/(2π×20kΩ×0.01µF)

=1/1.256×10^-3 =796.18Hz.

22. Match the gain of the filter with the frequencies in the low pass filter

Frequency	Gain of the filter
1. f H	i. V_O/V_in ≅ A_F/√2
2. f=f_H	ii. V_O/V_in ≤ A_F
3. f>f_H	iii. V_O/V_in ≅ A_F

A.1-i,2-ii,3-iii
B.1-ii,2-iii,3-i
C.1-iii,2-ii,3-i
D.1-iii,2-i,3-ii

Answer: D

The mentioned answer can be obtained if the value of frequencies is substituted in the gain magnitude equation

|(V_o/V_in)|=A_F/√(1+(f/f_H)²).

24. Determine the gain of the first order low pass filter if the phase angle is 59.77^o and the passband gain is 7.

A. 3.5
B. 7
C. 12
D. 1.71

Answer: A

Given the phase angle

φ =-tan^-1(f/f_H)

=> f/f_H=- φtan(φ)

= -tan(59.77^o)

=> f/f_H= -1.716.

Substituting the above value in gain of the filter

|(V_O/V_in)|

= A_F/√ (1+(f/f_H)²)

=7/√[1+(-1.716)²)] =7/1.986

=>|(V_O/V_in)|=3.5.

25. In a low pass Butterworth filter, the condition at which f=f_H is called

A. Cut-off frequency
B. Break frequency
C. Corner frequency
D. All of the mentioned

Answer: D

The frequency, f=f_H is called cut-off frequency, because the gain of the filter at this frequency is down by 3dB from 0Hz. The cut-off frequency is also called break frequency, corner frequency, or 3dB frequency.

26. Find the High cut-off frequency if the passband gain of a filter is 10.

A. 70.7Hz
B. 7.07kHz
C. 7.07Hz
D. 707Hz

Answer: C

High cut-off frequency of a filter

f_H= 0.707×A_F

= 0.707×10

=>f_H=7.07Hz.

27. To change the high cutoff frequency of a filter. It is multiplied by R or C by a ratio of the original cut-off frequency known as

A. Gain scaling
B. Frequency scaling
C. Magnitude scaling
D. Phase scaling

Answer: B

Once a filter is designed, it may sometimes be a need to change its cut-off frequency. The procedure used to convert an original cut-off frequency f_H to a new cut-off frequency is called frequency scaling.

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