Oscillators and Multivibrators

1. What is Barkhausen criterion for oscillation?

a) Aß > 1
b) Aß < 1
c) Aß = 1
d) Aß ≠ 1

Answer: c
Explanation: The Barkhausen criterion for oscillation is Aß = 1.
Where, A-> gain of amplifier and ß-> transfer ratio.
2. At what condition the output signal can be continuously obtained from input signal?
a) When the product of input voltage and feedback voltage is equal to 1
b) When the product of amplifier gain and transfer ratio is equal to 1
c) When the product of feedback voltage and transfer ratio is equal to 1
d) When the product of amplifier gain and input voltage is equal to 1

Answer: b
Explanation: When Aß=1, the feedback signal will be equal to the input signal. At this
condition, the circuit will continue to provide output, even if the external signal is
disconnected. This is because the amplifier cannot distinguish between external signal
and signal from the feedback circuit. Thus, output signal is continuously obtained.
3. An oscillator is a type of
a) Feedforward amplifier
b) Feedback amplifier
c) Waveform amplifier
d) RC amplifier

Answer: b
Explanation: An oscillation is a type of feedback amplifier in which a part of output is fed
back to the input via a feedback circuit.
4. Find the basic structure of feedback oscillator.

a)

b)
c)
d) None of the mentioned

Answer: c
Explanation: The above mentioned diagram is the basic structure of feedback oscillator.
It consists of an amplifier, to the external input (v i) is applied and it have a feedback
network from which the feedback signal (vf) is obtained.
5. What is the condition to achieve oscillations?
a) |Aß|=1
b) ∠Aß=0o
c) ∠Aß=multiples of 2π
d) All the mentioned

Answer: d
Explanation: All the conditions should be simultaneously satisfied to achieve oscillations.
6. What happens if |Aß|<1
a) Oscillation will die down
b) Oscillation will keep on increasing
c) Oscillation remains constant
d) Oscillation fluctuates

Answer: a
Explanation: If |Aß| becomes less than unity, the feedback signal goes on reducing in
each feedback cycle and oscillation will die down eventually.
7. How sustained oscillation can be achieved?
a) Maintaining |Aß| slightly greater than unity
b) Maintaining |Aß| equal to unity
c) Due to non-linearity of transistor
d) Due to use of feedback network

Answer: c
Explanation: When |Aß| is kept slightly greater than unity the signal, however, cannot go
on increasing and get limited due to non-linearity of the device (that is transistor enters
into saturation). Thus, it is the non-linearity of the transistor because of which the
sustained oscillation can be achieved.
8. Why it is difficult to maintain Barkhausen condition for oscillation?
a) Due to variation in temperature
b) Due to variation in supply voltage
c) Due to variation in components life time
d) All of the mentioned

Answer: d
Explanation: The Barkhausen condition |Aß|=1 is usually difficult to maintain in the circuit
as the value of A and ß vary due to temperature variations, aging of components,
change of supply voltage etc.
9. Name the type of noise signal present in the oscillation?
a) Schmitt noise
b) Schottky noise
c) Saturation noise
d) None of the mentioned

Answer: b
Explanation: Schottky noise is the noise signal always present at the input of the
transistor due to variation in the carrier concentration.
10. A basic feedback oscillator is satisfying the Barkhausen criterion. If the ß value is
given as 0.7072, find the gain of basic amplifier?
a) 2.1216
b) 0.7072
c) 1
d) 1.414

Answer: d
Explanation: Barkhausen criterion for oscillation is given as Aß=1
=> A=1/ ß = 1/0.7072 = 1.414.
11. The feedback signal of basic sine wave oscillator is given as
a) Vf = Aß ×Vo
b) Vf = Aß ×Vi
c) Vf = Aß × (Vo/ Vi)
d) Vf = Aß × (Vi/ Vo)

Answer: b
Explanation: The feedback signal of an oscillator is given as the product of external
applied signal & the loop gain of the system.
=> Vf= Aß ×Vi.
12. Express the requirement for oscillation in polar form
a) Aß =1∠360o
b) Aß =1∠90o
c) Aß =1∠πo
d) Aß =1∠270o

Answer: a
Explanation: There are two requirements for oscillation
1. The magnitude of Aß=1
2. The total phase shift of Aß=0 o or 360o.
1. What will be the phase shift of feedback circuit in RC phase shift oscillator?
a) 360o phase shift
b) 180o phase shift
c) 90o phase shift
d) 60o phase shift
Answer: b
Explanation: The RC feedback network provide 180 o phase shift and amplifier used in
RC phase shift oscillator provide 180 o phase shift (op-amp is used in the inverting mode)
to obtain a total phase shift of 360 o.
2. How many RC stages are used in the RC phase shift oscillator?
a) Six
b) Two
c) Four
d) Three

Answer: d
Explanation: The RC stage forms the feedback network of oscillator. It consist of three
identical RC stages, each of 60 o phase shift so as to provide a total phase shift of 180 o.
3. Calculate the frequency of oscillation for RC phase shift oscillator having the value of
R and C as 35Ω and 3.7µF respectively.
a) 1230 Hz
b) 204 Hz
c) 502Hz
d) 673 Hz

Answer: c
Explanation: The frequency of oscillation of RC phase shift oscillator is,
fo=1/(2πRC√6) = 1/(2×3.14×√6×3.7µF×35Ω)
=> fo= 1/ 1.9921×10-3 = 502Hz.
4. What must be done to ensure that oscillation will not die out in RC phase shift
oscillator?
a) Gain of amplifier is kept greater than 29
b) Gain of amplifier is kept greater than 1
c) Gain of amplifier is kept less than 29
d) Gain of amplifier is kept less than 1

Answer: a
Explanation: For a sustained oscillation in RC phase shift oscillator the gain of the
inverting op-amp should be at least 29. Therefore, gain is kept greater than 29 to ensure
the variation in circuit parameter will not make |Aß|<1, otherwise oscillation will die out.
5. Calculate the feedback voltage from phase shit network.

a) Vf = ( VoR3S3C3) / (1+ 6SRC+5S2C2R2+S3R3C3)

b) Vf = ( VoR3S3C3) / (1+ 5SRC+6S2C2R2+S3R3C3)
c) Vf = ( VoR3S3C3) / (1+ 5SRC+5S2C2R2+S3R3C3)
d) Vf = ( VoR2S3C3) / (1+ 6SRC+5S2C2R2+S3R3C3)

Answer: b
Explanation: Applying KVL equation to the circuit, we get
=> I1(R+(1/SC))-I2=Vo -> Equ1
=> -I1R+ I2(2R+(1/SC))- I3R=0 -> Equ2
=> 0- I2R+ I3(2R+(1/SC)=6 -> Equ3
WKT, Vf = I3× 2R,
Solving Equ 1, 2, and 3 for I 3
=> I3= ( VoR2S3C3)/ (1+ 5SRC+6S2C2R2+S3R3C3)
=> Vf = I3× 2R
=( VoR3S3C3) / (1+ 5SRC+6S2C2R2+S3R3C3).
6. Which type of op-amp is avoided for high frequencies?
a) LM318
b) Op-amp 741
c) LF 351
d) None of the mentioned

Answer: b
Explanation: Op-amp741 is generally used for low frequencies < 1 kHz.
7. Find out the constant values of α and ß in phase shift oscillator.
a) α = √6, ß = -1/29
b) α = 6, ß = -1/29
c) α = √6, ß = 1/29
d) α = 6, ß = -1/29

Answer: a
Explanation: From phase shift network, we obtain
ß= 1/(1-5 α2)+j α(6- α2) -> Equ1
For Aß=1, ß should be real and the imaginary terms must be zero
α(6- α2) =0
=> α=√6
Now substituting α2=6 in Equ 1, we get
ß=-1/29 (Negative sign indicates that the feedback network produces a phase shift of
180o).
8. A phase shift oscillator is designed to oscillate at 155Hz. Determine the value of R f.
(Take C=0.30µF)
a) 399Ω
b) 3.98MΩ
c) 13.9kΩ
d) 403kΩ

Answer: d
Explanation: R = 1/(2πC√6×fo)
=> R= 1/7.153×10-4= 1398=13.9kΩ.
9. The value of feedback resistor in phase shift oscillator is 180kΩ. Find its input
resistance?
a) 52kΩ
b) 151kΩ
c) 209kΩ
d) 6.2kΩ

10. Determine the frequency of oscillation (f o) in phase shift oscillator?

a) fo = √6/ωRC
b) fo = 0.56/ωRC
c) fo = 0.065/ωRC
d) fo = 6/ωRC

Answer: c
Explanation: The frequency of oscillation of phase shift oscillator is given as
fo = 1/(2π×RC×√6) = 1/15.38×RC
=> fo = 0.065/RC.
11. The condition for zero phase shift in wein bridge oscillator is achieved by
a) Connecting feedback to non-inverting input terminal of op-amp
b) Balancing the bridge
c) Applying parallel combination of RC to the feedback network
d) All of the mentioned

Answer: b
Explanation: In wein bridge oscillator, the feedback signal in the circuit is connected to
the non-inverting input of op-amp. So, feedback network does not provide any feedback
and the condition of zero phase shift around the circuit is achieved by balancing the
bridge.
1. What is the frequency of oscillation of wein bridge oscillator?
a) fo = 1/(2πRC)
b) fo = 2π/RC
c) fo = RC/2π
d) fo = 2πRC

Answer: a
Explanation: The frequency of oscillation of wein bridge oscillator is f o=2πRC.
2. Sustained oscillation in wein bridge oscillator is possible when the value of ß is
a) 3
b) 1/3
c) 1
d) None of the mentioned

Answer: b
Explanation: The gain |A|≥3, for oscillation to keep growing ( Since, |Aß|≥1 for sustained
oscillation).
3. Determine the value of fo, ß and Rf from the following circuit diagram.

a) fo = 80Hz, ß = 0.162 and Rf = 7kΩ

b) fo = 100Hz, ß = 1.62 and Rf = 7kΩ
c) fo = 60Hz, ß = 0.0162 and Rf = 7kΩ
d) fo = 120Hz, ß = 16.2 and Rf = 7kΩ

Answer: c
Explanation: The frequency of oscillation for the circuit is given as,
fo = 1/(2π×√(R1R2C1C2))
= 1/(2π×√(2.7kΩ×5kΩ×0.1µF×6µF) = 1/(2π×2.85×10 -3)
=> fo = 55.8 = 60Hz.
The value of ß = (R2C1) / (R1C1 + R2C2 + R2C1)
= (5kΩ×0.1µF)/(2.7kΩ×0.1µF+5kΩ×6µF+5kΩ×0.1µF)= 0.0162.
Rf =2 R3
=> Rf = 2×3.5kΩ=7kΩ.
4. What is the problem faced by the wein bridge oscillator?
a) Output sinewave get clipped
b) Output sinewave remain constant without growing
c) Output sinewave keep on decreasing and die out
d) All of the mentioned

Answer: a
Explanation: The gain of wein bridge oscillator is greater than 3, sometimes this may
keep the oscillations growing and it may clip the output sinewave.
5. Find the type of oscillator shown in the diagram

a) Quadrature oscillator
b) Biphasic oscillator
c) RC phase shit oscillator
d) None of the mentioned

Answer: d
Explanation: The circuit shown is the practical wein bridge oscillator with adaptive
negative feedback.
6. Calculate the value of capacitance in wein bridge oscillator, such that f o =1755Hz and
R=3.3kΩ.
a) 2.7µF
b) 0.91µF
c) 0.03µF
d) 0.05µF

Answer: c
Explanation: The frequency of oscillation is given as fo = 0.159/RC
=> C = 0.159/R×fo = 0.159/3.3kΩ×1755Hz
=> C = 0.027µF = 0.03µF.
7. Quadrature oscillators have signals with
a) Different frequency
b) Same frequency
c) Opposite frequency
d) Parallel frequency

Answer: b
Explanation: In Quadrature oscillators, signals have same frequency but have phase
shift with respective to each other.
8. Which of the following component is not used for audio frequency?
a) RC oscillator
b) Wein bridge oscillator
c) LC oscillator
d) None of the mentioned

Answer: c
Explanation: RC and wein bridge oscillator are suitable for audio frequency range
because of size of R and C components becomes very large for generating low
frequencies.
9. Find the signal waveform for Quadrature oscillators?

a)
b)

c)
d) All of the mentioned

Answer: a
Explanation: Quadrature signals are the signals that are of same frequency but have a
phase shift of 90o with respect to each other. The mentioned waveform have a phase
shift of π/2- π=90o phase shift between sine waveform and cosine waveform.
10. If the resistor and capacitor values are same in Quadrature oscillator. Find its
frequency of oscillation for R=50kΩ and C=0.01µF.
a) 112Hz
b) 275Hz
c) 159Hz
d) 318Hz

Answer: d
Explanation: Frequency of Quadrature oscillator, fo = 1/(2πRC)
=> fo= 1/(2π×50kΩ×0.01µF)= 318Hz.
11. What is the possible method used in Quadrature oscillator to remove distortion from
the output waveform?
a) Replace the resistor at the input of non-inverting type amplifier with a potentiometer
b) Replace the resistor at the output of non-inverting type amplifier with a potentiometer
c) Replace the resistor at the input of inverting type amplifier with a potentiometer
d) None of the mentioned

Answer: a
Explanation: The resistor at the input of non-inverting type amplifier is replaced with a
potentiometer in order to eliminate any possible distortion in the output waveform.

1. How are the square wave output generated in op-amp?

a) Op-amp is forced to operate in the positive saturation region
b) Op-amp is forced to operate in the negative saturation region
c) Op-amp is forced to operate between positive and negative saturation region
d) None of the mentioned

Answer: c
Explanation: Square wave outputs are generated where the op-amp is forced to operate
in saturated region, that is, the output of the op-amp is forced to swing repetitively
between positive saturation, +V sat and negative saturation, -Vsat.
2. The following circuit represents a square wave generator. Determine its output voltage

a) -13 v
b) +13 v
c) ± 13 v
d) None of the mentioned

Answer: a
Explanation: The differential output voltage V id = Vin1 – Vin2= 3-7v = -4v.
The output of the op-amp in this circuit depends on polarity of differential voltage V 0= -
Vsat ≅ -Vee = -13 v.
3. Determine the expression for time period of a square wave generator
a) T= 2RC ln×[( R1+ R2) / ( R2)].
b) T= 2RC ln×[( 2R1+ R2) / ( R2)].
c) T= 2RC ln×[( R1+ 2R2) / ( R2)].
d) T= 2RC ln×[( R1+ R2) / (2 R2)].

Answer: b
Explanation: The time period of the output waveform for a square wave generator is T=
2RC ln×[(2R1+ R2)/( R2)].
4. Determine capacitor voltage waveform for the circuit

a)
b)

c)

d)

Answer: c
Explanation: When the op-amp output voltage is at negative saturation, V 1 = [(R1) / (R1+
R2 )] × (-Vsat) = [10kΩ / ( 10 kΩ +11.6 kΩ)] × (-15v) = -7v.
Similarly, when the op-amp’s output voltage is at positive saturation, V 1 = [(R1) / (R1+
R2 )] × (+Vsat) = [10kΩ/ ( 10 kΩ +11.6 kΩ)] × (+15v) = +7v
The time period of the output waveform,T= 2RC ln ×[( 2R1+ R2) / ( R2)] = 2× 10kΩ × 0.05
µF× ln (2×10kΩ + 11.6kΩ) / 11.6kΩ] = 1×10 -3 × ln2.724 = 1ms.
The voltage across the capacitor will be a triangular wave form.
5. What will be the frequency of output waveform of a square wave generator if R 2 = 1.16
R1?
a) fo = (1/2RC)
b) fo = (ln/2RC)
c) fo = (ln /2 ×√RC)
d) fo = (ln/√(2 RC))

Answer: b
Explanation: When R2= 1.16 R1, then fo = 1/2RC× ln[ (2R1+ R2) / R2] = 1/2RC ×ln [(2R1 +
1.161R1 )/ (1.161R1)] = 1/( 2RC×ln2.700)= 1/2RC.
6. What could be the possible output waveform for a free running multivibrator whose
op-amp has a supply voltage of ±5v operating at 5khz?
a)

b)

c)

d) None of the mentioned

Answer: c
Explanation: In a free running multivibrator, the output is forced to swing repetitively
between positive and negative saturation to produce square wave output. Therefore,
+Vsat ≅ +Vcc =+5v and -Vsat ≅ -Vcc =-5v.
=> Frequency= 5khz , f =1/t = 0.2ms.
7. Determine the output frequency for the circuit given below

a) 28.77 Hz
b) 31.97 Hz
c) 35.52 Hz
d) 39.47 Hz

Answer: d
Explanation: The output frequency f o = 1/2RC×ln [ (2R1+ R2)/ R2] = 1 / {(2×33kΩ
×0.33µF)×ln[(2×33kΩ +30kΩ)/30kΩ]} = 1/ (0.02175×ln 32) = 39.47 Hz.
8. The value of series resistance in the square wave generator should be 100kΩ or
higher in order to
a) Prevent excessive differential current flow
b) Increase resistivity of the circuit
c) Reduce output offset voltage
d) All of the mentioned

Answer: a
Explanation: In practice, each inverting and non-inverting terminal needs a series
resistance to prevent excessive differential current flow because the inputs of the op-
amp are subjected to large differential voltages.
9. Why zener diode is used at the output terminal of square wave generator?
a) To reduce both output and capacitor voltage swing
b) To reduce output voltage swing
c) To reduce input voltage swing
d) To reduce capacitor voltage swing

Answer: b
Explanation: A reduced peak-peak output voltage swing can be obtained in the square
wave generator by using back to back zener diodes at the output terminal.
10. A square wave oscillator has f o =1khz. Assume the resistor value to be 10kΩ and find
the capacitor value?
a) 3.9 µF
b) 0.3 µF
c) 2 µF
d) 0.05µF

Answer: d
Explanation: Let’s take R2 = 1.16 R2, therefore the output frequency fo = 1/2RC
=> C = 1/2Rfo = 1/ (2×10kΩ×1khz) = 0.05µF.
1. How a triangular wave generator is derived from square wave generator?
a) Connect oscillator at the output
b) Connect Voltage follower at the output
c) Connect differential at the output
d) Connect integrator at the output

Answer: d
Explanation: The output waveform of the integrator is triangular, if its input is square
wave. Therefore, a triangular wave generator can be obtained by connecting an
integrator at the output of the square wave generator.
2. The increase in the frequency of triangular wave generator.
a) Ramp the amplitude of triangular wave
b) Increase the amplitude of triangular wave
c) Decrease the amplitude of triangular wave
d) None of the mentioned

Answer: a
Explanation: As the resistor value increase or decrease, the frequency of triangular
wave will decrease or increase, respectively. Therefore, the amplitude of the triangular
wave decreases with an increase in it frequency and vice verse.
3. Which among the following op-amp is chosen for generating triangular wave of
relatively higher frequency?
a) LM741 op-amp
b) LM301 op-amp
c) LM1458 op-amp
d) LM3530 op-amp

Answer: b
Explanation: The frequency of the triangular wave generator is limited by the slew rate of
the op-amp. LM301 op-amp has a high slew rate.
4. What is the peak to peak (PP) output amplitude of the triangular wave?
a) VO(pp) = + VRamp + (- VRamp)
b) VO(pp) = – VRamp + (+ VRamp)
c) VO(pp) = + VRamp – (- VRamp)
d) VO(pp) = – VRamp – (+ VRamp)

Answer: c
Explanation: The peak to peak output waveform, V O(pp) = + VRamp-(-VRamp)
Where, – VRamp –> Negative going ramp ;
+ VRamp–> positive going ramp.
5. Determine the output triangular waveform for the circuit.

a)
b)

c)

d)

Answer: b
Explanation: The voltage at which A1 switch from +Vsat to -Vsat
=> -Vramp =(-R2 / R3) × (+Vsat)
= (-10kΩ/40kΩ) ×15v =-3.75v
Similarly, the voltage at which A 1 switch from -Vsat to +Vsat
=> +Vramp = (-R2 / R3) × (-Vsat)
= 10kΩ/40kΩ ×15v =3.75v
∴ Time period, T = (4R1C1R2) / R3
= (4×10kΩ×0.05µF×10kΩ) /40kΩ = 0.5 ms.
6. Find the capacitor value for a the output frequency, f o = 2kHz & VO(pp) = 7v, in a
triangular wave generator. The op-amp is 1458/741 and supply voltage = ±15v. (Take
internal resistor=10kΩ)
a) 0.03nF
b) 30nF
c) 0.3nF
d) 3nF

Answer: d
Explanation: Given, Vsat =15v
∴ VO(pp) = (2R2/R3) × Vsat
=> R2 =(VO(pp) ×R3) / (Vsat×2) = [7/(2×15)]×R3 = 0.233R3
∵ Internal resistor, R2 = R1= 10kΩ
=> R3 = 0.233×10kΩ = 2.33kΩ.
So, the output frequency fO = R3 / ( 4×R1 ×C1× R2)
=> 2khz = 2.33khz/ (4×10kΩ ×10kΩ×C1)
=> C1 = 2.33kΩ / (8×10-11) = 2.9 ×10-9 ≅3nF.
7. Triangular wave form has
a) Rise time < fall time
b) Rise time = fall time
c) Rise time ≥ fall time
d) None of the mentioned

Answer: b
Explanation: The triangular wave form has rise time of the triangular wave always equal
to its fall time, that is, the same amount of time is required for the triangular wave to
swing from -VRamp to +VRamp as from +VRamp to -VRamp.
8. Output of an integrator producing waveforms of unequal rise and fall time are called
a) Triangular waveform
b) Sawtooth waveform
c) Pulsating waveform
d) Spiked waveform

Answer: b
Explanation: Sawtooth waveform has unequal rise and fall times. It may rise positively
many times faster than it falls negatively or vice versa.
9. Find out the sawtooth wave generator from the following circuits.
a)

b)
c)

d) None of the mentioned

Answer: c
Explanation: The triangular wave generator can be converted into a sawtooth wave
generator by inserting a variable dc voltage into the non-inverting terminal of the
integrator.
10. Consider the integrator used for generating sawtooth wave form. Match the list I with
the list II depending on the movement of wiper.

List-I List-II

Rise time
=fall time
(Triangular
wave)

Longer fall
time and
short rise
time
(Sawtooth
wave)

Longer rise
time and
short fall
time
(Sawtooth
wave)
a) 1-iii, 2-ii, 3-i
b) 1-i, 2-ii, 3-iii
c) 1-i, 2-iii, 3-ii
d) 1-ii, 2-iii, 3-i

Answer: c
Explanation: Depending on the duty cycle (movement of the wiper) the type of waveform
is determined.

Active Filters
1. An electrical filter is a
a) Phase-selective circuit
b) Frequency-selective circuit
c) Filter-selective circuit
d) None of the mentioned

Answer: b
Explanation: An electric filter is often a frequency selective circuit that passes a specified
band of frequencies and blocks or alternates signal of frequencies outside this band.
2. Filters are classified as
a) Analog or digital
b) Passive or active
c) Audio or radio frequency
d) All of the mentioned

Answer: d
Explanation: Filters are classified based on the design technique (analog or digital),
elements used for construction (active or passive) and operating range (audio or radio
frequency).
3. Why inductors are not preferred for audio frequency?
a) Large and heavy
b) High power dissipation
c) High input impedance
d) None of the mentioned

Answer: a
Explanation: At audio frequencies, inductor becomes problematic, as the inductors
become large, heavy and expensive.

4. The problem of passive filters is overcome by using

a) Analog filter
b) Active filter
c) LC filter
d) A combination of analog and digital filters

Answer: b
Explanation: The active filters enclose as a capacitor in the feedback loop and avoid
using inductors, this way inductorless active filter are obtained.
5. What happens if inductors are used in low frequency applications?
a) Enhance inductor usage
b) No losses occurs
c) Degrades inductor performance
d) Low power dissipation

Answer: c
Explanation: For low frequency applications more number of turns of wire must be used,
which in turn adds to the series resistance degrading inductor’s performance.

6. Find out the incorrect statement about active and passive filters.
a) Gain is not attenuated in active filter
b) Passive filters are less expensive
c) Active filter does not cause loading of source
d) Passive filters are difficult to tune or adjust

Answer: b
Explanation: Typically active filters are more economical than passive filters. This is
because of the variety of cheaper op-amp and the absence of inductor’s.
7. What are the most commonly used active filters?
a) All of the mentioned
b) Low pass and High pass filters
c) Band pass and Band reject filters
d) All-pass filters

Answer: a
Explanation: All the mentioned filters use op-amp as active element and capacitors &
resistors as passive elements.
8. Choose the op-amp that improves the filter performance.
a) µA741
b) LM318
c) LM101A
d) MC34001

Answer: b
Explanation: LM318 is a high speed op-amp that improves the filter’s performance
through increased slew rate and higher unity gain-bandwidth.
9. Ideal response of filter takes place in
a) Pass band and stop band frequency
b) Stop band frequency
c) Pass band frequency
d) None of the mentioned

Answer: c
Explanation: The ideal response indicates the practical filter response and it lies within
the pass band frequencies.
10. Find out the low pass filter from the given frequency response characteristics.
a)
b)

c)

d)

Answer: a
Explanation: A low pass filter has a constant gain from 0Hz to high cut-off frequency fH.
1. Which filter type is called a flat-flat filter?
a) Cauer filter
b) Butterworth filter
c) Chebyshev filter
d) Band-reject filter

Answer: b
Explanation: The key characteristic of the butterworth filter is that it has a flat pass band
as well as stop band. So, it is sometimes called a flat-flat filter.
2. Which filter performs exactly the opposite to the band-pass filter?
a) Band-reject filter
b) Band-stop filter
c) Band-elimination filter
d) All of the mentioned

Answer: d
Explanation: A band reject is also called as band-stop and band-elimination filter. It
performs exactly the opposite to band-pass because it has two pass bands: 0 < f < f L and
f > f H.
3. 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
Explanation: Bandwidth of a band-pass filter is Bandwidth= fH– fL=10kHz-
2.5kHz=7.5kHz=7500Hz.
4. 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
Explanation: In 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 function of frequency.
5. 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
Explanation: The rate at which the gain of the filter changes in the stop band is
determined by the order of filter. So, for a low pass filter the gain decreases at the rate of
20dB/decade.
6. 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
Explanation: The cauer filter has a ripple pass band and a ripple stop band. So,
generally cauer filter gives the best stop band response among the three.
7. Determine the order of filter used, when the gain increases at the rate of 60dB/decade
on the stop band.
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
Explanation: The gain increases for high pass filter. So, for a third order high pass filter
the gain increases at the rate of 60dB/decade in the stop band until f=f L.
8. Name the filter that has two stop bands?
a) Band-pass filter
b) Low pass filter
c) High pass filter
d) Band-reject filter

Answer: a
Explanation: A band-pass filter has two stop bands: 1) 0 < f < f L and 2) f > fH.
9. The frequency response of the filter in the stop band.
i. Decreases with increase in frequency
ii. Increase with increase in frequency
iii. Decreases with decrease in frequency
iv. Increases with decrease in frequency
a) i and iv
b) ii and iii
c) i and ii
d) ii and iv

Answer: c
Explanation: The order of frequency of the filter in the stop band determines either
steady decreases or increases or both with increase in frequency.
1. Find the voltage across the capacitor in the given circuit

a) VO= Vin/(1+0.0314jf)
b) VO= Vin×(1+0.0314jf)
c) VO= Vin+0.0314jf/(1+jf)
d) None of the mentioned

Answer: a
Explanation: The voltage across the capacitor, V O= Vin/(1+j2πfRC)
=> VO= Vin/(1+j2π×5k×1µF×f)
=> VO= Vin/(1+0.0314jf).
2. Find the complex equation for the gain of the first order low pass butterworth filter as a
function of frequency.
a) AF/[1+j(f/fH)].
b) AF/√ [1+j(f/fH)2].
c) AF×[1+j(f/fH)].
d) None of the mentioned

Answer: a
Explanation: Gain of the filter, as a function of frequency is given as V O/ Vin=A F/(1+j(f/fH)).
3. Compute the pass band gain and high cut-off frequency for the first order high pass
filter.

a) AF=11, fH=796.18Hz
b) AF=10, fH=796.18Hz
c) AF=2, fH=796.18Hz
d) AF=3, fH=796.18Hz

Answer: c
Explanation: The pass band gain of the filter, A F =1+(RF/R1)
=>AF=1+(10kΩ/10kΩ)=2. The high cut-off frequency of the filter, fH=1/2πRC
=1/(2π×20kΩ×0.01µF) =1/1.256×10 -3 =796.18Hz.
4. Match the gain of the filter with the frequencies in the low pass filter

Frequency Gain of the filter

1. f < fH i. VO/Vin ≅ AF/√2

2. f=fH ii. VO/Vin ≤ AF

3. f>fH iii. VO/Vin ≅ AF

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
Explanation: The mentioned answer can be obtained, if the value of frequencies are
substituted in the gain magnitude equation |(V o/Vin)|=AF/√(1+(f/fH)2).
5. Determine the gain of the first order low pass filter if the phase angle is 59.77 o and the
pass band gain is 7.
a) 3.5
b) 7
c) 12
d) 1.71
Answer: a
Explanation: Given the phase angle, φ =-tan-1(f/fH)
=> f/fH=- φtan(φ) = -tan(59.77o)
=> f/fH= -1.716.
Substituting the above value in gain of the filter, |(V O/Vin)| = AF/√ (1+(f/fH)2) =7/√[1+(-
1.716)2)] =7/1.986
=>|(VO/Vin)|=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
Explanation: 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. Cut-off frequency is also called as break
frequency, corner frequency or 3dB frequency.
7. Find the High cut-off frequency if the pass band gain of a filter is 10.
a) 70.7Hz
b) 7.07kHz
c) 7.07Hz
d) 707Hz

Answer: c
Explanation: High cut-off frequency of a filter, fH=0.707×AF =0.707×10
=>fH=7.07Hz.
8. To change the high cutoff frequency of a filter. It is multiplied by R or C by a ratio of
original cut-off frequency known as
a) Gain scaling
b) Frequency scaling
c) Magnitude scaling
d) Phase scaling

Answer: b
Explanation: Once a filter is designed, it may sometimes be a need to change it’s cut-off
frequency. The procedure used to convert an original cut-off frequency fH to a new cut-off
frequency is called frequency scaling.
9. Using the frequency scaling technique, convert 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
Explanation: To change a cut-off frequency from 10kHz to 16kHz,multiply 15.9kΩ
resistor.
[Original cut-off frequency/New cut-off frequency] =10kHz/16kHz =0.625.
∴ R =0.625×15.9kΩ =9.94kΩ. 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 f1=200Hz and f2=3kHz. Specification: AF=2 and fH=1kHz.
a) 4.28 dB
b) 5.85 dB
c) 1.56 dB
d) None of the mentioned

Answer: c
Explanation: When f1=200Hz, VO(1)/Vin =AF/√ [1+(f/fH)2] =2/√ [1+(200/1kHz) 2] =2/1.0198.
=> VO(1)/Vin =1.96
=>20log|(VO/Vin)|=5.85dB.
When f=700Hz, VO(2)/Vin= 2/√ [1+(700/1kHz) 2] =2/1.22=1.638.
=> VO(2)/Vin =20log|(VO/Vin|=20log(1.638) = 4.28.
Therefore, the difference in the gain magnitude is given as V O(1)/Vin-VO(2)/Vin =5.85-4.28
=1.56 dB.
11. Design a low pass filter at a cut-off frequency 1.6Hz with a pass band gain of 2.
a)

b)

c)

d)
Answer: a
Explanation: From the answer, it is clear that all the C values are the same . Therefore,
c= 0.01µF
Given, fH = 1kHz,
=> R= 1/(2πCfm) = 1/2π×0.01µF×1kHz
R= 9.9kΩ ≅ 10kΩ. Since the pass band gain is 2.
=> 2=1+ (RF/R1). Therefore, RF and R1 must be equal.
1. How can a first order low pass filter can be converted into second order low pass filter
a) By adding LC network
b) By adding RC network
c) By adding RC || LC network
d) None of the mentioned

Answer: b
Explanation: The addition of RC network makes the stop band response having a
40dB/decade.
2. Consider the following specifications and calculate the high cut-off frequency for the
circuit given?

a) 95Hz
b) 48Hz
c) 14Hz
d) 33Hz

Answer: d
Explanation: The high cut-off frequency, fH= 1/[2 π√(R2 ×R3× C2× C3)] =
1/[2π√(33kΩ×15kΩ×0.47µF×0.1µF)]= 1/[2π× 4.82×(10 -3)]= 33Hz.
3. Find the gain and phase angle of the second order low pass filter?
Where pass band gain of the filter is 5, frequency and the high cut-off frequency of the
filter are 3000Hz and 1kHz.
a) None of the mentioned
b) Gain magnitude = -1.03dB , φ =63.32o
c) Gain magnitude = -5.19dB , φ =71.56o
d) Gain magnitude = -4.94dB , φ =90o

Answer: c
Explanation: The gain of the second order low pass filter, [V O /Vin] =AF/ √ [1+(f/fh)2] =5/
√[1+(3000/1000)4] =5/9.055 =0.55.
=> [VO /Vin] = 20log(0.55) =-.519dB.
Phase angle of second order low pass filter is given as φ= tan -1(f/fH)
=> φ =71.56o.

4. Design a second order low pass butterworth filter at a high cut-off frequency of
2.2kHz. Given RF=20kΩ and capacitor 0.047µF.
a)

b)

c)

d)
Answer: c
Explanation: Given fH=21.2kHz, C2=C3 =0.047µF
R3 = 1/(2πfHC3 = 1/(2π×2.2kHz×0.047µF) =1/5.9032×10-4 =1.69kΩ.
=>R3 =R2=1.67kΩ
Since, RF=0.586R1
=> R1=RF/0.586 =20kΩ/0.586
R1 = 34.13kΩ.
5. A second order low pass filter is given an input frequency of 30kHz and produce a
output having phase angle of 79 o. Determine the pass band gain of the filter?
a) 11 dB
b) 89.11 dB
c) 46.78 dB
d) None of the mentioned

Answer: c
Explanation: Phase angle of the filter, φ = tan -1(f/fH)
=> fh =f×tan(φ) =30kHz × tan(79<sup<o< sup="">)= 154.34kHz.
Therefore, the pass band gain A F = fH/0.707 = 154.34kHz/0.707
AF= 218.3 =20log(218.3)= 46.78dB.</sup<o<>
6. The pass band voltage gain of a second order low pass butterworth filter is
a) 1.586
b) 8.32
c) 0.586
d) 0.707

Answer: a
Explanation: Second order low pass filter has a pass band voltage gain equal to 1.586
because of equal resistor and capacitor values. This gain is necessary to guarantee
butterworth response.
7. Arrange the series of step involved in designing a filter for first order low pass filter
Step 1: Select a value of C less than or equal to 1µF
Step 2: Choose a value of high cut-off frequency fH
Step 3: Select a value of R1C and RF depending on the desired pass band gain
Step 4: Calculate the value of R
a) Steps- 2->4->3->1
b) Steps- 4->1->3->2
c) Steps- 2->1->4->3
d) Steps- 1->3->4->2
Answer: b
Explanation: The mentioned option is the sequence of steps followed for designing a low
pass filter.
8. Frequency scaling is done using
a) Standard capacitor
b) Varying capacitor
c) Standard resistance
d) None of the mentioned

Answer: a
Explanation: In frequency scaling standard capacitors are chosen, because for non
standard value of resistor, a potentiometer is used.
1. 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.
2. 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.
3. 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.
4. 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.
5. 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, A F =1.586.
=> AF= [1+(RF/R1)] => RF= (AF-1)×R1 =(1.586-1)×10kΩ =5860 =5.86kΩ.
6. 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.
7. 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.
8. 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.
9. 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.
10. 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.
11. 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.
12. 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.
1. 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.
2. 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.
3. 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.
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)

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).
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]}

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.
6. 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.
7. 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
d) 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.
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

Answer: c
Explanation: Quality factor Q=√(fh×fL)/(fh-fL) = √(950Hz×250Hz)/(9950Hz-250Hz) =0.696.
9. The details of low pass filter sections are given as f h =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.
10. 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.
11. Design a narrow band-pass filter, with fc=1kHz, Q= 13 and AF=10 (Take C=0.1µF)
a)

b)

c)

d)
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.
12. 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, A F= 2×Q2
When Q=3,
=> 2×Q2 =2×(32) =18.
=> 10<18. Hence condition is satisfied when Q=3.
13. 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.
1. How many types of band elimination filters are present
a) Three
b) Two
c) Four
d) None of the mentioned

Answer: b
Explanation: Band-reject filters are also called as band elimination filters. They are
classified into two types.
i) Wide band-reject filter and
ii) Narrow band-reject filter.
2. Find the wide band-reject filter.
a)
b)

c)

d) None of the mentioned

Answer: b
Explanation: A wide band-reject filter is made using a low pass filter, a high pass filter
and a summing amplifier.
3. A narrow band-reject filter is commonly called as
a) Notch filter
b) Band step filter
c) Delay filter
d) All of the mentioned

Answer: a
Explanation: A narrow band-reject filter is also called as notch filter because of its higher
quality factor, Q (>10).
4. Find the expression for notch-out frequency?
a) fN = 2πRC
b) fN = 2π/RC
c) fN = 1/2π×√(R/C)
d) fN = 1/2πRC

Answer: d
Explanation: The notch-out frequency is the frequency at which maximum attenuation
occurs: it is given by fN =1/2πRC.
5. The quality factor of passive twin T-network is increased by using
a) Inverting amplifier
b) Non-inverting amplifier
c) Voltage follower
d) Differential amplifier

Answer: c
Explanation: The passive twin T-network has a selectively low figure of merit. The Q of
the network can be increased significantly, if it is used with the voltage follower.
6. Find out the application in which narrow band-reject filter can be used?
a) Embedded system
b) Biomedical instrument
c) Digital computer
d) None of the mentioned

Answer: b
Explanation: Notch filters or narrow band-reject filters are used in biomedical instruments
for eliminating undesired frequencies.
7. Design 120Hzactive notch filter?
a)

b)
c)

d)

Answer: a
Explanation: Since C value < 0.1µF, assume C=0.68µF Notch Frequency, f n = 1/2πRC.
R= 1/(2πfnC)= 1/(2π×120Hz×0.68×10-6) = 1.95kΩ ≅2kΩ.
For R/2, parallel of two 2kΩ resistor
=>R/2 = 2kΩ||2kΩ =(2×2)/(2+2)=1kΩ.
For C/2 , parallel of two 0.68µF capacitor
C/2=> 0.68µF + 0.68µF= 1.36µF.
8. Find the application of area where all-pass filters are used?
a) Cathode ray oscilloscope
b) Television
c) Telephone wire
d) None of the mentioned

Answer: c
Explanation: When signals are transmitted in transmission lines like telephone wire, they
undergo change in phase, all-pass filters are used to compensate these phase changes.
9. Determine the output voltage for all the all-pass filter and express it in complex form?
a) VO =Vin/ [(1-j2πfRC) /(1+ j2πfRC)].
b) VO =Vin× [(1+j2πfRC) /(1- j2πfRC)].
c) VO =Vin ×[(1- j2πfRC) /(1+ j2πfRC)].
d) None of the mentioned

Answer: c
Explanation: The output voltage of all-pass filter is given as VO =Vin× [(1-j2πfRC)
/(1+j2πfRC)] .
10. Determine the input frequency for all-pass filter with phase angle as 62 o. Consider
the value of resistor and capacitor are 3.3kΩ and 4.7µF.
a) Input frequency= -7.65Hz
b) Input frequency= -6.77Hz
c) Input frequency= -3.89Hz
d) Input frequency= -9.65Hz

Answer: d
Explanation: The phase angle is given as Φ = -2tan-1×(2πfRC)
=> f=-tanΦ/4πRC =-tan(62o)/(4π×3.3kΩ×4.7µF)= -1.88/0.1948 =-9.65Hz.
11. Determine the angle for given circuit diagram, if the frequency of input signal is 1khz

a) -45o
b) -180o
c) -270o
d) -90o

Answer: d
Explanation: Phase angle Φ=-2tan-1×(2πfRC/1) = -2tan-1×(2π×1kHz×16kΩ×0.01µF)
= -2tan-1×(1.0048)=-90o.
12. The voltage gain magnitude of all-pass filter is
a) Zero
b) One
c) Infinity
d) None of the mentioned

Answer: b
Explanation: The magnitude of voltage gain of all-pass filter |VO /Vin| = √(1+(2π/RC)2) /
√(1+(2 π/RC)2) =1.
13. What happens if the position of R and C are interchanged in the below circuit
diagram?

a) Vin leads VO
b) Vin lags VO
c) VO leads Vin
d) VO leads Vin
Answer: c
Explanation: For the circuit given, the phase angle changes from 0 to 180 o as frequency
is varied from 0 to ∞. If the positions of R and C are interchanged, the phase shift and
band width input and output becomes positive. That is the output (V O) leads input (Vin).
14. Choose the incorrect statement “In wide band-reject filter” .
a) Low cut-off frequency of low pass filter must be larger than the high cut-off frequency
of the high pass filter.
b) Low cut-off frequency of high pass filter must be equal than the high cut-off frequency
of the high pass filter.
c) Low cut-off frequency of high pass filter must be smaller than the high cut-off
frequency of the low pass filter.
d) None of the mentioned

Answer: d
Explanation: In wide band-reject filter, low cut-off frequency of high pass filter must be
larger than the high cut-off frequency of the low pass filter.