How Does Linear Feedback Shift Register Work?

1. Voltage to current converter is also called as
a) Current series positive feedback amplifier
b) Voltage series negative feedback amplifier
c) Current series negative feedback amplifier
d) Voltage series positive feedback amplifier
Answer: c
Explanation: Voltage to current converter is also called as current series negative feedback amplifier because the feedback voltage across internal resistor applied to the inverting terminal depends on the output current and is in series with the input difference voltage.

2. Given voltage to current converter with floating load. Determine the output current?

a) 3mA
b) 6mA
c) 4mA
d) 2mA
Answer: d
Explanation: Output current, Io = Vin /R1 = 10/5kΩ =2mA.

3. Which of the following application uses voltage to current converter?
a) Low voltage dc and ac voltmeter
b) Diode match finding
c) Light emitting diode
d) All of the mentioned
Answer: d
Explanation: In all the applications mentioned above, the input voltage Vin is converted into an output current of Vin/R1 or the input voltage appear across resistor.

4. The op-amp in low voltage DC voltmeter cannot be nullified due to
a) D'Arsonaval meter movement
b) Offset voltage compensating network
c) Selection of switch
d) Gain of amplifier
Answer: a
Explanation: The op-amp sometimes cannot be nullified because the output is very sensitive to even slight variation in wiper position of D'Arsonaval meter movement (ammeter with a full scale deflection of 1mA).

5. What is the maximum input voltage that has to be selected to calibrate a dc voltmeter with a full scale voltage range of 1-13v.
a) ≤ ±14v
b) ≥ ±13v
c) ≤ ±15v
d) = ±14v
Answer: a
Explanation: The maximum input voltage has to be ≤ ±14v, to obtain the maximum full scale input voltage of 13v.

6. Higher input voltage can be measured in low voltage DC voltmeter using
a) Smaller resistance value
b) Higher resistance value
c) Random resistance value
d) All of the mentioned
Answer: b
Explanation: Higher resistance values are required to measure relatively higher input voltage. For example, if the range of switch is at x10 position in the low voltage dc voltmeter then, the corresponding resistance value would be 10kΩ. So, it requires a 10v input to get a full scale deflection (if 1v cause full scale deflection in the ammeter with a full scale deflection of 1mA).

7. In the diagram given below, determine the deflection of the ammeter with a full scale deflection of 1mA when the switch is at X2kΩ. Consider resistance of the offset voltage compensating network to be 10Ω.

a) Full scale deflection in the ammeter
b) Half scale deflection in the ammeter
c) Quarter scale deflection in the ammeter
d) No deflection occurs in the ammeter
Answer: b
Explanation: Given Vin=1v ,R1=10+2kΩ ≅2kΩ
Io = Vin/R1= 1v/2kΩ =0.5mA. This means that 2v causes half scale deflection of the ammeter.

8. How to modify a low voltage DC voltmeter to low voltage ac voltmeter
a) Add a full wave rectifier in the feedback loop
b) Add a half wave rectifier in the feedback loop
b) Add a square wave rectifier in the feedback loop
b) Add a sine wave rectifier in the feedback loop
Answer: a
Explanation: A combination of an ammeter and a full wave rectifier can be employed in the feedback loop to form an ac voltmeter.

9. What makes the output voltage equals to zero in practical op-amp?
a) Input offset voltage
b) Output offset voltage
c) Offset minimizing voltage
d) Error voltage
Answer: a
Explanation: Input offset voltage is the differential input voltage that exists between two input terminals of an op-amp without any external input and force the output voltage to zero.

10. What happens due to mismatch between two input terminals in an op-amp?
a) Input offset voltage
b) Output offset voltage
c) Bothe the input and output offset voltage
d) None of the mentioned
Answer: b
Explanation: The input offset voltage in op-amp force the output voltage to zero due to the mismatch between two input terminal, there will be voltage produced at the output and this voltage is called output offset voltage.

11. Define polarity of the output offset voltage in a practical op-amp?
a) Positive polarity
b) Negative polarity
c) Positive or negative polarity
d) None of the mentioned
Answer: c
Explanation: The output offset voltage is a DC voltage, it may be positive or negative in polarity depending on whether the potential difference between two input terminal is positive or negative.

12. The input offset voltage of 741 op-amp has an absolute maximum value of 6mv, which means
a) Minimum difference between input terminals in 741 op-amp can be large as 6mv DC
b) Minimum difference between input terminals in 741 op-amp can be large as 6mv AC
c) Maximum difference between input terminals in 741 op-amp can be large as 6mv DC
d) Maximum difference between input terminals in 741 op-amp can be large as 6mv AC
Answer: c
Explanation: Given, the absolute maximum value for a 741 is Vio= 6mv. Therefore, voltage at the non-inverting input terminal may differ from that at the inverting input terminal by as much as 6mv dc. Also the output offset voltage is a DC voltage and it cannot be AC voltage.

13. If three different 741 op-amps are taken and the corresponding output offset voltage for each of them is measured. The output voltage in these three op-amps have
a) Same amplitude and polarity
b) Different amplitude and polarity
c) Same amplitude and different polarity
d) Different amplitude and same polarity
Answer: b
Explanation: Even though the op-amps are of the same type, the output voltage in these three op-amps are not of the same amplitude and polarity, because of mass production.

14. To reduce the output offset voltage VooT to zero
a) Input offset voltage compensating network is added at the inverting input terminal
b) Input offset voltage compensating network is added at the non-inverting input terminal
c) Input offset voltage compensating network is added at the output terminal
d) None of the mentioned
Answer: d
Explanation: To reduce the VooT to zero, the external circuit is added at the input terminal of the op-amp that will give the flexibility of obtaining input offset voltage of proper amplitude and polarity. The input terminal can be inverting or non-inverting.

15. Which of the following op-amp does not need compensating network?
a) 777
b) 741
c) 748
d) All of the mentioned
Answer: d
Explanation: The compensating network is not needed for these op-amps because, they have offset null pins.

16. Find out the voltage offset null circuit for the 741 op-amp?

Answer: b
Explanation: For 741-type op-amp, the manufacturer recommend a 10kΩ potentiometer be placed across offset null pin1 and 5 and a wiper be connected to the negative supply pin 4. Null output is obtained by adjusting the pot.

17. What will the condition of op-amp, before applying any external input
a) Compensated
b) Biased
c) Balanced
d) Zero
Answer: c
Explanation: Before applying external input to the op-amp, the output offset voltage should be reduced to zero with the help of an offset voltage compensating network. At this condition, the op-amp is said to be balanced or nulled.

18. Choose the compensating network design for non-inverting amplitude

Answer: a
Explanation: If an op-amp is used as an non-inverting amplifier, the compensating network should be connected to the inverting input terminal of the op-amp.

19. Find the thevenin's equivalent for resistance and voltage?

a) 1-iii, 2-ii, 3-1
b) 1-ii, 2-I, 3-iii
c) 1-I, 2-ii, 3-iii
d) 1-ii, 2-iii, 3-i
Answer: b
Explanation: The maximum thevenin equivalent resistance Rmax occurs when the wiper is at the center of the potentiometer and the maximum thevenin equivalent voltage Vmax is equal to +Vcc or –Vee, when wiper is uppermost or lowest in the potentiometer.

20. What is done to compensate the voltage, when V1 > V2?

a) Move the wiper towards +Vcc
b) Move the wiper towards –Vee
c) Keep the wiper at the center of potentiometer
d) None of the mentioned
Answer: a
Explanation: V1 > V2 implies that output offset voltage is positive. This means that V2 should be increased until it is equal to V1. The wiper can be moved towards +Vcc until output offset voltage is reduced to zero.

21. Calculate the maximum thevenin equivalent resistance, if a 10kΩ potentiometer is used?
a) 0.4kΩ
b) 5 kΩ
c) 2.5kΩ
d) 4kΩ
Answer: c
Explanation: Rmax= Ra/2 || Ra/2 = Ra/4.
Given potentiometer, Ra = 10kΩ
=> Therefore, Rmax = 10kΩ/4 =2.5kΩ.

22. Find the input offset voltage for the circuit shown

a) Vio = (Rb*Vmax)/( Rmax+ Rb+ Rc)
b) Vio = Rmax/( Rmax+ Rb+ Rc)
c) Vio = (Rc*Vmax)/( Rmax+ Rb+ Rc)
d) Vio = Vmax/( Rmax+ Rb+ Rc)
Answer: c
Explanation: Compensating network using maximum Thevenin's equivalent for resistance and voltage circuit is shown. Since |V1-V2|- Vio, the maximum value of V2 can be equal to Vio.

23. Find the value of Ra and Rb from the circuit shown?

a) Ra =4.6kΩ ; Rb= 9kΩ
b) Ra =7.3kΩ ; Rb= 3.4kΩ
c) Ra =2.5kΩ ; Rb= 5.1kΩ
d) Ra =4kΩ ; Rb= 10kΩ
Answer: d
Explanation: We know that input offset voltage, Vio =(Rc*Vmax)/ Rb
=> Rb = Vmax*(Rc / Rb ) = (10v/10mv)*10Ω (∵ Vio specified on the datasheet is 10mv for LM307 op-amp).
=> Rb =10000 = 10kΩ.
Since Rb > Rmax let us choose Rb = 10*Rmax. (Where Rmax = Ra/4).
∴ Rb = (10*Rb)/4 and Ra = Rb/2.5 = 10kΩ/2.5=4kΩ.

24. Why does an op-amp without feedback is not used in linear circuit application?
a) Due to high current gain
b) Due to high voltage gain
c) Due to high output signal
d) All of the mentioned
Answer: b
Explanation: In an op-amp without feedback, the voltage gain is extremely high (ideally infinite). Because of the high risk of distortion and clipping of the output signal, an op-amp in open loop configuration is not used in linear circuit application.

25. When the input voltage is reduced to zero in a closed-loop configuration the circuit acts as
a) Inverting amplifier
b) Non-inverting amplifier
c) Inverting and non-inverting amplifier
d) None of the mentioned
Answer: c
Explanation: Since the input signal voltage is reduced to zero, the internal resistance is negligibly small. The output offset voltage is expressed in terms of external resistance and the specified input offset voltage for a given op-amp.
If the non-inverting input terminal is connected to the ground, it acts as inverting op-amp and vice versa.

26. How the value of output offset voltage is reduced in closed-loop op-amp?
a) By increasing gain
b) By reducing gain
c) By decreasing bandwidth
d) By reducing bandwidth
Answer: b
Explanation: The output offset voltage is a product of gain and specified input offset voltage for a given op-amp. Voo= Aoo*Vio. So, the value of output offset voltage can be reduced by reducing the gain value.

27. What happens if R1>>RF in the circuit

a) Some amount of output offset voltage is present
b) Some amount of input offset voltage is present
c) Some amount of gain voltage is present
d) All of the mentioned
Answer: a
Explanation: If R1 >>RF, the gain Aoo≅1, which makes Voo ≅ Vio. Thus, all op-amp circuit has some output offset voltage.

28. Determine the voltage gain for the circuit.

a) 1.1
b) 1.6
c) 1.2
d) 2.2
Answer: d
Explanation: The voltage gain, AF={1+[RF/( R1+ Rc)]} = 1+[15kΩ/(2.5kΩ+10kΩ)] = 2.2.

29. Where does the compensating network connected in an inverting amplifier.
a) Non-inverting input terminal
b) Inverting input terminal
c) Between non-inverting and output terminal
d) Between inverting and output terminal
Answer: a
Explanation: The offset voltage compensating network is connected in the non-inverting terminal for the inverting amplifier and vice versa.

30. Why closed-loop differential amplifiers are difficult to null?
a) Due to compensating network
b) Due to feedback loop
c) Due to input offset voltage
d) None of the mentioned
Answer: a
Explanation: The closed loop differential amplifiers are more difficult to null because the use of compensating network can change the common mode rejection mode.
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31. How to achieve maximum CMRR in the given circuit?

a) R1 = RF
b) RF = R3|| RC+ RB and R1= R2
c) R1= R2 and RF= R3+ RC
d) None of the mentioned
Answer: c
Explanation: To achieve maximum CMRR in the circuit the value of R1= R2 and RF= R3+ RC.

32. What is the advantage of compensated differential amplifier?
a) All of the mentioned
b) Slightly complex circuit
c) Does not affect CMRR
d) Balanced op-amp
Answer: d
Explanation: Since the compensated differential amplifier uses the op-amp with offset voltage null pins. The offset null circuit does not affect the CMRR.

33. The offset voltage in the voltage follower is balanced using
a) Voltage drop across the load resistor
b) Voltage drop across feedback resistor
c) Compensating network connected to inverting input terminal
d) Compensating network connected to non- inverting input terminal
Answer: b
Explanation: Voltage drop across the feedback resistor connect to inverting input terminal is used to cancel the offset voltage in voltage follower.

34. Find the maximum possible output offset voltage, which is caused by the input offset voltage Vio=15mv?

a) 0.075v
b) 0.75v
c) 0.75v
d) 7.5v
Answer: a
Explanation: Aoo=[1+(RF/R1)] =1+(10kΩ/2.5kΩ) = 5.
Voo=5*15mv = 75mv.

35. Compute the output voltage for voltage follower with offset voltage compensating network?

a) 3.6v
b) 10.8v
c) 26v
d) 33v
Answer: b
Explanation: The output voltage is given as Vo= {1+[ RC/( Rb+ (Rmax/4))]}*Vin.
Rmax=Ra/4 = 20kΩ/4 = 5kΩ.
Vo=[1+(39kΩ/(10kΩ+5kΩ))]*3v = 10.8v.

36. Input bias current is defined as
a) Average of two input bias current
b) Summing of two input bias current
c) Difference of two input bias current
d) Product of two input bias current
Answer: a
Explanation: Input bias current is the average of two input bias current flowing into the non-inverting and inverting input of an op-amp.

37. Although the value of input bias current is very small, it causes
a) Output voltage
b) Input offset voltage
c) Output offset voltage
d) All of the mentioned
Answer: c
Explanation: Even a very small value of input bias current can cause a significant output offset voltage in circuits using relatively large feedback resistors.

38. The formula for output offset voltage of an op-amp due to input bias current
a) VOIB= RF*IB
b) VOIB= (RF+R1)/IB
c) VOIB= (1+RF)*IB
d) VOIB= [1+(RF/R1)]*IB
Answer: a
Explanation: The output offset voltage due to input bias current is VOIB = RF*IB.

39. Find the input bias current for the circuit given below

a) 10mA
b) 2mA
c) 5mA
d) None of the mentioned
Answer: c
Explanation: Input bias current, IB=(IB1+ IB2)/2
=> IB =(4mA+6mA)/2 = 5mA.

40. Mention a step to reduce the output offset voltage caused due to input bias current?
a) Use small feedback resistor and resistance at the input terminal
b) Use small feedback resistors
c) Reduce the value of load resistors
d) None of the mentioned
Answer: b
Explanation: Since the output offset voltage is proportional to feedback resistor and input bias current. The amount of VOIB can be reduced by reducing the value of feedback resistor.

41. Given below is a differential amplifier in which V1=V2. What happens to VOIB at this condition?

a) VOIB= 0
b) VOIB= VOIB×10-10
c) VOIB= VOIB/2
d) VOIB= -1
Answer: a
Explanation: The voltage V1 and V2 are caused by the current IB1 and IB2. Although this bias current are very small, if they are made equal, then there will be no output voltage VOIB.

42. Name the resistor that is connected in the non-inverting terminal of op-amp which is in parallel combination of resistor connected in inverting terminal and feedback resistor.
a) Random minimizing resistor
b) Offset minimizing resistor
c) Offset reducing resistors
d) Output minimizing resistors
Answer: b
Explanation: The voltage is product of resistors and input bias current. Therefore, the value of the resistors are adjusted such that the resistors are connected at the inverting input terminal is made equal to resistor connected in non-inverting input terminal. The use of this resistors minimize the amount of output offset voltage and therefore, they are referred to as offset minimizing resistors.

43. Calculate ROM, if the value of IB1 = IB2 in the given circuit.

a) 1173.11Ω
b) 171.31Ω
c) 1171.43Ω
d) 1071.43Ω
Answer: d
Explanation: Offset minimizing resistor, ROM =(R1* RF)/( R1+RF).
=> ROM = (1.2kΩ*10kΩ)/(1.2kΩ+10Ω) = 1071.43Ω.

44. The circuit in which the output voltage waveform is the integral of the input voltage waveform is called
a) Integrator
b) Differentiator
c) Phase shift oscillator
d) Square wave generator
Answer: a
Explanation: Integrator circuit produces the output voltage waveform as the integral of the input voltage waveform.

45. Find the output voltage of the integrator
a) Vo = (1/R×CF)×t∫0 Vindt+C
b) Vo = (R/CF)×t∫0 Vindt+C
c) Vo = (CF/R)×t∫0 Vindt+C
d) Vo = (R×CF)×t∫0 Vindt+C
Answer: a
Explanation: The output voltage is directly proportional to the negative integral of the input voltage and inversely proportional to the time constant RCF.
Vo = (1/R×CF)×t∫0 Vindt+C
Where C-> Integration constant and CF-> Feedback capacitor.

46. Why an integrator cannot be made using low pass RC circuit?
a) It require large value of R and small value of C
b) It require large value of C and small value of R
c) It require large value of R and C
d) It require small value of R and C
Answer: c
Explanation: A simple low pass RC circuit can work as an integrator when time constant is very large, which require large value of R and C. Due to practical limitations , the R and C cannot be made infinitely large.

47. How a perfect integration is achieved in op-amp?
a) Infinite gain
b) Low input impedance
c) Low output impedance
d) High CMRR
Answer: a
Explanation: In an op-amp integrator the effective input capacitance becomes CF×(1-Av). Where Av is the gain of op-amp. The gain is infinite for ideal op-amp. So, effective time constant of the op-amp integrator becomes very large which results in perfect integration.

48. The op-amp operating in open loop result in output of the amplifier to saturate at a voltage
a) Close to op-amp positive power supply
b) Close to op-amp negative power supply
c) Close to op-amp positive or negative power supply
d) None of the mentioned
Answer: c
Explanation: In practice, the output of op-amp never becomes infinite rather the output of the op-amp saturate at a voltage close to op-amp positive or negative power supply depending on the polarity of the input dc signal.

49. The frequency at which gain is 0db for integrator is
a) f=1/(2πRFCF)
b) f=1/(2πR1CF)
c) f=1/(2πR1R1)
d) f=(1/2π)×(RF/R1)
Answer: b
Explanation: The frequency at which the gain of the integrator becomes zero is f=1/(2πR1CF).

50. Why practical integrator is called as lossy integrator?
a) Dissipation power
b) Provide stabilization
c) Changes input
d) None of the mentioned
Answer: d
Explanation: To avoid saturation problems, the feedback capacitor is shunted by a feedback resistance(RF). The parallel combination of RF and CF behave like a practical capacitor which dissipates power. For this reason, practical integrator is called as a lossy integrator.

51. Determine the lower frequency limit of integration for the circuit given below.

a) 43.43kHz
b) 4.82kHz
c) 429.9kHz
d) 4.6MHz
Answer: b
Explanation: The lower frequency limit of integration, f= 1/(2πRFCF) = 1/(2π×1kΩ×33nF) = 4.82kHz.

52. How is the higher order filters formed?
a) By increasing resistors and capacitors in low pass filter
b) By decreasing resistors and capacitors in low pass filter
c) By inter changing resistors and capacitors in low pass filter
d) All of the mentioned
Answer: c
Explanation: High pass filter are often formed by interchanging frequency determining resistors and capacitors in low pass filters. For example, a first order high pass filter is formed from a first order low pass filter by inter changing components Rand C.

53. In a first order high pass filter, frequencies higher than low cut-off frequencies are called
a) Stop band frequency
b) Pass band frequency
c) Centre band frequency
d) None of the mentioned
Answer: b
Explanation: Low cut-off frequency, fL is 0.707 times the pass band gain voltage. Therefore, frequencies above fL are pass band frequencies.

54. Compute the voltage gain for the following circuit with input frequency 1.5kHz.

a) 4dB
b) 15dB
c) 6dB
d) 12dB
Answer: d
Explanation: |VO/Vin|= [AF×(f/fL)]/ [√1+(f/fL)2] = [4×(1.5kHz/225.86)] / √[1+(1.5kHz/225.86)2] =26.56/6.716=3.955 =20log(3.955)=11.9.
|VO/Vin|≅12 dB
AF= 1+(RF /R1)= 1+(12kΩ/4kΩ) =4.
fL= 1/(2πRC) = 1/2π×15kΩ×0.047µF= 1/4.427×10-3 =225.86Hz.

55. Determine the expression for output voltage of first order high pass filter?
a) VO = [1+(RF /R1)]× [(j2πfRC/(1+j2πfRC)] × Vin
b) VO = [-(RF /R1)]× [(j2πfRC/(1+j2πfRC)] × Vin
c) VO = {[1+(RF /R1)]× /[1+j2πfRC] }× Vin
d) None of the mentioned
Answer: a
Explanation: The first order high pass filter uses non-inverting amplifier. So, AF= 1+(RF /R1) and the output voltage, VO = [1+(RF /R1)]× [(j2πfRC/(1+j2πfRC)]× Vin.

56. The internal resistor of the second order high pass filter is equal to 10kΩ. Find the value of feedback resistor?
a) 6.9kΩ
b) 5.86kΩ
c) 10kΩ
d) 12.56kΩ
Answer: b
Explanation: 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.
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57. 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
Explanation: The low cut-off frequency for the given filter is fL =1/√[2π√(R2×R3×C2×C3)]=178.7Hz.

58. 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
Explanation: 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.

59. From the given specifications, determine the value of voltage gain magnitude of 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
Explanation: 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.

60. How is the higher order filters formed?
a) Using first order filter
b) Using second order filter
c) Connecting first and second order filter in series
d) Connecting first and second order filter in parallel
Answer: c
Explanation: 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 filter.

61. State the disadvantage of using higher order filters?
a) Complexity
b) Requires more space
c) Expensive
d) All of the mentioned
Answer: d
Explanation: Although 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.

62. The overall gain of higher order filter is
a) Varying
b) Fixed
c) Random
d) None of the mentioned
Answer: b
Explanation: The overall gain of higher order filter is fixed because all the frequency determining resistor and capacitors are equal.

63. Find the roll-off rate for 8th order filter
a) -160dB/decade
b) -320dB/decade
c) -480dB/decade
d) -200dB/decade
Answer: a
Explanation: For nth order filter the roll-off rate will be -n×20dB/decade.
=>∴ for 8th order filter= 8×20=160dB/decade.

64. Which filter attenuates any frequency outside the pass band?
a) Band-pass filter
b) Band-reject filter
c) Band-stop filter
d) All of the mentioned
Answer: a
Explanation: A band- pass filter has a pass band between two cut-off frequencies fH and fL. So, any frequency outside this pass band is attenuated.

65. Narrow band-pass filters are defined as
a) Q < 10
b) Q = 10
c) Q > 10
d) None of the mentioned
Answer: c
Explanation: Quality factor (Q) is the measure of selectivity, meaning higher the value of Q, the narrower its bandwidth.

66. A band-pass filter has a bandwidth of 250Hz and 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
Explanation: Quality factor of band-pass filter, Q =fc/bandwidth= 566/250=3.46.

67. 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
Explanation: In a wide band-pass filter, the product of high and low cut-off frequency is equal to the square of center frequency
i.e. ( fc)2 =fH×fL
=> fc= √(fh×fL).

68. 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
Explanation: The voltage gain magnitude of the band-pass filters equal to the product of the voltage gain magnitudes of high pass and low pass filter.
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69. 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
Explanation: The order of the band-pass filter depends on the order of the high pass and low pass filter sections.

70. Find the voltage gain magnitude of the wide band-pass filter?
Where total pass band 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
Explanation: 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.

71. 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
Explanation: Quality factor Q=√(fh×fL)/(fh-fL) = √(950Hz×250Hz)/(9950Hz-250Hz) =0.696.

72. The details of low pass filter sections are given as fh =10kHz, AF= 2 and f=1.2kHz. Find the voltage gain magnitude of first order wide band-pass filter, if the voltage gain magnitude of high pass filter section is 8.32dB.
a) 48.13dB
b) 10.02dB
c) 14.28dB
d) 65.99dB
Answer: c
Explanation: |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.

73. 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
Explanation: A wide band-pass filter has quality factor less than 10.

74. Design a narrow band-pass filter, with fc=1kHz, Q= 13 and AF=10 (Take C=0.1µF)

Answer: b
Explanation: Given C =0.1µF.
Therefore, C1=C2 =0.1µF.
R1 =Q/(2π×fc×CAF) =13/( 2π×1kHz×0.1µF×10) =13/6.28 = 2.07 ≅ 2Ω.
R2 =Q/{2π×fc×C ×[(2Q2)- AF]} =13/{(2π×1kHz×0.1µF×[2×(132)-10]} = 13/0.2059=63.11 ≅ 63Ω.
R3 =Q/(π×fc×C) = 13/(π×1kHz×0.1µF) = 13/3.14×10-4 =41.40kΩ ≅41kΩ.

75. If the gain at center frequency is 10, find the quality factor of narrow band-pass filter
a) 1
b) 2
c) 3
d) None of the mentioned
Answer: c
Explanation: 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 condition is satisfied when Q=3.

76. The advantage of 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
Answer: d
Explanation: As the narrow band-pass filter has multiple filters. The center frequency can be changed to a new frequency without changing the gain or bandwidth and is accomplished by changing the resistor to a new value which is given as
R'=R×(fL/fc)2.