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}).

5. 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.

6. 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.

7. 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.

8. 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.

9. Using the frequency scaling technique, convert the 10kHz cut-off frequency of the low pass filter to a cutoff frequency of 16kHz.(Take C=0.01µF and R=15.9kΩ)

A. 6.25kΩ
B. 9.94kΩ
C. 16kΩ
d )1.59kΩ

Answer: B

To change a cut-off frequency from 10kHz to 16kHz, multiply the 15.9kΩ resistor.

However, 9.94kΩ is not a standard value. So, a potentiometer of 10kΩ is taken and adjusted to 9.94kΩ.

10. Find the difference in gain magnitude for a filter, if it is the response obtained for frequencies f_{1}=200Hz and f_{2}=3kHz. Specification: A_{F}=2 and f_{H}=1kHz.

A. 4.28 dB
B. 5.85 dB
C. 1.56 dB
D. None of the mentioned

Answer: C

When f_{1}=200Hz, V_{O}(1)/V_{in} =A_{F}/√ [1+(f/f_{H})^{2}]

=2/√ [1+(200/1kHz) ^{2}] =2/1.0198.

=> V_{O}(1)/V_{in} =1.96

=>20log|(V_{O}/V_{in})|=5.85dB.

When f=700Hz

V_{O}(2)/V_{in}= 2/√ [1+(700/1kHz) ^{2}]

=2/1.22=1.638.

=> V_{O}(2)/V_{in} =20log|(V_{O}/V_{in}|

=20log(1.638) = 4.28.

Therefore, the difference in the gain magnitude is given as