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৪৯তম বিসিএস ⎯ তথ্য ও যোগাযোগ প্রযুক্তি (EEE) [ ৮৯২]

পরীক্ষা৪৯তম বিসিএস ⎯ তথ্য ও যোগাযোগ প্রযুক্তি (EEE) [ ৮৯২]তারিখতারিখ অনির্ধারিতসময়35 minutes৪৯ বৈধ · অসম্পূর্ণ
মোট প্রশ্ন৫০
সিলেবাস
Exam 13 i) Power Amplifiers ii) Pulse and Switching Circuits iii) Voltage Regulators iv) Phase Locked Loops (PLL) [Source: Class–10 and relevant books]
ঘনত্ব
উত্তর
উত্তরিতবর্তমানপুনরায় দেখুনঅসম্পূর্ণ

৪৯তম বিসিএস ⎯ তথ্য ও যোগাযোগ প্রযুক্তি (EEE) [ ৮৯২]

৪৯তম বিসিএস ⎯ তথ্য ও যোগাযোগ প্রযুক্তি (EEE) [ ৮৯২] · তারিখ অনির্ধারিত · ৫০ প্রশ্ন

.
Calculate the effective resistance seen looking into the primary of a 15:1
transformer connected to an 8 ohm load.
  1. 1.2 kilo-ohm
  2. 1.0 kilo-ohm
  3. 1.2 ohm
  4. 1.8 kilo-ohm
সঠিক উত্তর:
1.8 kilo-ohm
উত্তর
সঠিক উত্তর:
1.8 kilo-ohm
ব্যাখ্যা

R'L = a2RL = (15)(8 ) = 1800 ohm = 1.8 k-ohm

.
For the  Figure   , calculate the dc efficiency where component values result in a dc base current of 6 mA, and collector current  at Q point is 40 mA . The ac power delivered to the 8 ohm speaker for the circuit is 0.477 W

  1. 31.1%
  2. 3.1%
  3. 34.1%
  4. 34.8%
অনির্ধারিত
ব্যাখ্যা

প্রশ্নটি হওয়া উচিত ছিল - For the  Figure, calculate the dc efficiency where component values result in a dc base current of 6 mA, and collector current  at Q point is 140 mA . The ac power delivered to the 8 ohm speaker for the circuit is 0.477 W
প্রশ্ন অনুযায়ী সঠিক উত্তর না থাকায় প্রশ্নটি বাতিল করা হলো। 
----------------- 

The input (dc) power obtained from the supply is calculated from the supply dc voltage and the average power drawn from the supply:
Pi(dc) = VCCICQ = (10 V)(140 mA) = 1.4 W
The efficiency of the amplifier is then
% η=Po(ac) / Pi(dc) * 100% = 0.477 W/1.4 W * 100% = 34.1%

.
Calculate the efficiency of a transformer-coupled class A amplifier for a
supply of 12 V and outputs of V(p) = 12 V.
  1. 25%
  2. 50%
  3. 75%
  4. 20%
সঠিক উত্তর:
50%
উত্তর
সঠিক উত্তর:
50%
ব্যাখ্যা


Since VCEQ = VCC = 12 V, the maximum and minimum of the voltage swing are,
respectively,
VCEmax = VCEQ + V(p) = 12 V + 12 V = 24 V
VCEmin = VCEQ - V(p) = 12 V - 12 V = 0 V

% η = 50{(24 V - 0 V)2 /(24 V + 0 V)2 }% = 50%

.
For a class B amplifier providing a 40-V peak signal to a 20-ohm load (speaker) and a power supply of VCC = 10 V and Idc = 0. 73  determine the  output power.
  1. 50 W
  2. 40 W
  3. 45 W
  4. 100 W
সঠিক উত্তর:
40 W
উত্তর
সঠিক উত্তর:
40 W
ব্যাখ্যা

Po(ac) = V2L(p)/2RL
= (40) 2/2(20 ) = 40 W

.
For a class B amplifier providing a 20-V peak signal to a 16-ohm load (speaker) and a power supply of VCC = 30 V, determine the circuit efficiency.
  1. 52%
  2. 32%
  3. 92%
  4. 72%
সঠিক উত্তর:
52%
উত্তর
সঠিক উত্তর:
52%
ব্যাখ্যা

.
For a class B amplifier using a supply of Vcc=30V and driving a load of 16 ohm, determine transistor dissipation.
  1. 5.7 W
  2. 7 W
  3. 3.5 W
  4. 7.1 W
সঠিক উত্তর:
5.7 W
উত্তর
সঠিক উত্তর:
5.7 W
ব্যাখ্যা

.
For the circuit of Figure, calculate the power handled by each output transistor  for an input of 12 V rms.

  1. 12.8 W
  2. 13.6 W
  3. 15.8 W
  4. 18.6 W
সঠিক উত্তর:
15.8 W
উত্তর
সঠিক উত্তর:
15.8 W
ব্যাখ্যা

.
Calculate the total harmonic distortion for an output signal having fundamental amplitude of 2.5 V, second harmonic amplitude of 0.25 V, third harmonic amplitude of 0.1 V, and fourth harmonic amplitude of 0.05 V.
  1. 10.95%
  2. 12.95%
  3. 20.95%
  4. 6.95%
সঠিক উত্তর:
10.95%
উত্তর
সঠিক উত্তর:
10.95%
ব্যাখ্যা



.
In power amplifier circuit Class …….... operation gives the maximum distortion.
  1. A
  2. B
  3. C
  4. D
সঠিক উত্তর:
C
উত্তর
সঠিক উত্তর:
C
ব্যাখ্যা

Power Amplifier Classes Overview:
Power amplifiers are classified into several categories based on their conduction angle, efficiency, and the amount of signal distortion they generate. The primary classes of power amplifiers are Class A, B, AB, and C, each having distinct characteristics.

Class A Power Amplifier:
In a Class A amplifier, the output transistor conducts for the entire input signal cycle (360°). This results in the highest linearity and lowest distortion because the output signal is an accurate replica of the input signal. However, the efficiency is low (around 25-30%), as the transistor remains on throughout the signal cycle, consuming significant power even when there is no input signal.

Class B Power Amplifier:
Class B amplifiers only conduct for half of the signal cycle (180°). This improves efficiency compared to Class A (around 78%), but it introduces a significant amount of distortion at the crossover point where the two halves of the waveform meet. This is known as crossover distortion, which makes Class B less ideal for applications requiring high linearity.

Class AB Power Amplifier:
Class AB amplifiers combine features of both Class A and Class B. The output transistor conducts for more than half the input cycle (but less than the full cycle). This reduces crossover distortion while maintaining better efficiency than Class A. Class AB amplifiers are commonly used in audio systems, providing a good balance between linearity and efficiency.

Class C Power Amplifier:
Class C amplifiers only conduct for a fraction of the input signal cycle (less than 180°). This results in the highest efficiency (around 90%) but also the most distortion. Because of the limited conduction period, the output signal is significantly distorted, making Class C suitable only for applications where distortion is not a concern, such as radio frequency (RF) transmission.

১০.
The size of a power transistor is made considerably large to ..
  1. provide easy handling
  2. dissipate heat
  3. facilitate connections
  4. none of the above
সঠিক উত্তর:
dissipate heat
উত্তর
সঠিক উত্তর:
dissipate heat
ব্যাখ্যা

Power transistors are used in high-power applications where significant amounts of heat are generated. When a transistor operates, it consumes electrical energy and converts part of it into heat. If this heat is not effectively managed, it can damage the transistor and reduce its efficiency.

To prevent this, the size of power transistors is increased to dissipate heat more effectively. A larger surface area allows for better heat dissipation, which keeps the transistor cool during operation, ensuring it functions properly and extends its lifespan.

Here’s a brief overview of why the other options are not correct:

(i) Provide easy handling: While a larger transistor might be easier to handle in terms of physical manipulation, this is not the primary reason for making it large. Handling ease is secondary compared to thermal management in power transistors.
(iii) Facilitate connections: The size of the transistor may make connections easier, but this is not the main reason for increasing the size. The key factor in the design of power transistors is heat dissipation.
(iv) None of the above: This is incorrect because, as explained, the size of the transistor is primarily increased to dissipate heat.

১১.
Low efficiency of a power amplifier results in ....
  1. low forward bias
  2. less battery consumption
  3. more battery consumption
  4. none of the above
সঠিক উত্তর:
more battery consumption
উত্তর
সঠিক উত্তর:
more battery consumption
ব্যাখ্যা

Power amplifiers are used to amplify weak signals, and their efficiency is a key factor in determining how much power they consume during operation. Efficiency refers to the ratio of output power (useful power) to the total power supplied to the amplifier.

Low efficiency in a power amplifier means that a significant amount of the input power is wasted, usually in the form of heat, rather than being converted into useful output power.
As a result, more energy needs to be supplied to achieve the same output, leading to more battery consumption or higher power consumption from the power source.
Here’s why the other options are incorrect:

(ক) Low forward bias: Forward bias refers to the condition in which a transistor is operating (whether it allows current to flow easily or not). Low forward bias does not directly relate to amplifier efficiency or power consumption.
(খ) Less battery consumption: This is the opposite of the correct answer. Low efficiency means more power is wasted, so it would not lead to less battery consumption.
(ঘ) None of the above: This is incorrect because the correct answer is (ঘ) more battery consumption.

১২.
If the voltage gain versus frequency curve of a transistor amplifier is not flat, then there is ........ distortion.
  1. amplitude
  2. frequency
  3. intermodulation
  4. none of the above
সঠিক উত্তর:
intermodulation
উত্তর
সঠিক উত্তর:
intermodulation
ব্যাখ্যা

When the voltage gain versus frequency curve of a transistor amplifier is not flat, it indicates that the amplifier's performance varies with frequency, leading to a distortion in the output signal. This type of distortion is commonly referred to as intermodulation distortion.

Intermodulation distortion (IMD) occurs when two or more signals at different frequencies mix within the amplifier, resulting in additional signals at the sum and difference frequencies (which are not part of the original input). This happens when the amplifier behaves non-linearly, often at higher frequencies or when driven beyond its linear operating range. As a result, unwanted harmonics and mixing products appear, distorting the signal.

১৩.
The most costly coupling is………. coupling.
  1. RC
  2. impedance
  3. direct
  4. transformer
সঠিক উত্তর:
transformer
উত্তর
সঠিক উত্তর:
transformer
ব্যাখ্যা

The most costly coupling is transformer coupling.

Transformer coupling involves using a transformer to connect different stages of a circuit, providing isolation and impedance matching. This type of coupling is often more expensive due to the cost of the transformer, which is more complex and requires additional components compared to other coupling methods like RC, impedance, or direct coupling.

১৪.
If the stages have only Rand C components, they must operate in class ............ operation.
  1. A
  2. B
  3. C
  4. AB
সঠিক উত্তর:
A
উত্তর
সঠিক উত্তর:
A
ব্যাখ্যা

If the stages have only R (resistor) and C (capacitor) components, they must operate in Class A operation.

Class A operation is typically used in circuits with resistive and capacitive components, where the transistor or active device conducts throughout the entire input cycle, ensuring a continuous flow of current. This type of operation is common in amplifiers with simple resistive and capacitive coupling.

১৫.
Amplitude distortion is also called........ distortion.
  1. intermodulation
  2. phase
  3. harmonic
  4. resonant
সঠিক উত্তর:
harmonic
উত্তর
সঠিক উত্তর:
harmonic
ব্যাখ্যা

Amplitude distortion is also called harmonic distortion.

Harmonic distortion occurs when the output signal contains additional frequencies that are integer multiples of the original signal's frequency. These additional frequencies, known as harmonics, distort the amplitude of the signal, resulting in a change in its waveform.

১৬.
Which is not considered as a linear voltage regulator?
  1. Fixed output voltage regulator
  2. Adjustable output voltage regulator
  3. Switching regulator
  4.  Special regulator
সঠিক উত্তর:
Switching regulator
উত্তর
সঠিক উত্তর:
Switching regulator
ব্যাখ্যা

 In linear regulator’s the impedance of active element may be continuously varied to supply a desired current to the load. But in the switching regulator, a switch is turned on and off.

১৭.
For the given circuit, let VEB(ON)=1v, ß= 15 and IO=2mA. Calculate the load current

  1. 23.45A
  2. 46.32A
  3. 56.87A
  4. 30.75A
সঠিক উত্তর:
30.75A
উত্তর
সঠিক উত্তর:
30.75A
ব্যাখ্যা

The equation for load current,
IL = [(ß+1)IO] - [ß×(VEB(ON)/R1)]
= 32 - 1.25
= 30.75A.

১৮.
Find the difference between output current having a load of 100Ω and 120Ω for 7805 IC regulator. Consider the following specification: Voltage across the load = 5v; Voltage across the internal resistor = 350mv.
  1.  8.4mA
  2. 7mA
  3. 9mA
  4. 3.4mA
সঠিক উত্তর:
 8.4mA
উত্তর
সঠিক উত্তর:
 8.4mA
ব্যাখ্যা

Given the voltage across the internal resistor to be 350mv, which is less than 0.7v. Hence the transistor in 7805 is off.
When load = 100Ω, IL= IO = Ii = 5v/100 Ω = 50mA
When load = 120Ω, IO = 5v/120 Ω = 41.6mA.
So, the difference between the output voltage = 50-41.6mA = 8.4mA.

১৯.
The change in output voltage for the corresponding change in load current in a 7805 IC regulator is defined as
  1.  Line regulation
  2.  Load regulation
  3.  Input regulation
  4. All of the mentioned
সঠিক উত্তর:
 Load regulation
উত্তর
সঠিক উত্তর:
 Load regulation
ব্যাখ্যা

Load regulation is defined as the change in output voltage for a change in load current and is also expressed in millivolts or as a percentage of output voltage.

২০.
Which of the following statements is true regarding the internal feedback mechanism in a linear voltage regulator under varying load conditions?
  1. The feedback mechanism in a linear regulator is primarily designed to improve the load regulation by adjusting the reference voltage according to the load current.
  2. In a linear voltage regulator, the internal feedback loop compensates for the ripple voltage at the output by adjusting the duty cycle of the switching transistor.
  3. The feedback mechanism ensures that the regulator maintains a constant output voltage, irrespective of the changes in load, by adjusting the pass element's conduction state based on the sensed output voltage.
  4. The feedback loop in a linear voltage regulator operates at a fixed reference voltage and does not adjust for load changes, which is why it is less effective at regulating output voltage under heavy load conditions.
সঠিক উত্তর:
The feedback mechanism ensures that the regulator maintains a constant output voltage, irrespective of the changes in load, by adjusting the pass element's conduction state based on the sensed output voltage.
উত্তর
সঠিক উত্তর:
The feedback mechanism ensures that the regulator maintains a constant output voltage, irrespective of the changes in load, by adjusting the pass element's conduction state based on the sensed output voltage.
ব্যাখ্যা

(ক) The feedback mechanism in a linear regulator is primarily designed to improve the load regulation by adjusting the reference voltage according to the load current.
Explanation: This statement is incorrect. In a linear voltage regulator, the feedback mechanism adjusts the pass element (such as a transistor) to maintain a constant output voltage, not the reference voltage itself. The reference voltage typically remains constant and is compared to the output to regulate the pass element's conduction. The load current does influence the feedback loop, but it doesn't directly adjust the reference voltage.

(খ) In a linear voltage regulator, the internal feedback loop compensates for the ripple voltage at the output by adjusting the duty cycle of the switching transistor.
Explanation: This statement is incorrect. Linear voltage regulators do not use a switching transistor or modulate duty cycles, which is typical of switching regulators (like buck or boost converters). Linear regulators use a pass element, such as a transistor in series with the load, and adjust its conduction to maintain constant output voltage. The feedback loop adjusts for variations in the output voltage but does not directly handle ripple through duty cycle adjustments.

(গ) The feedback mechanism ensures that the regulator maintains a constant output voltage, irrespective of the changes in load, by adjusting the pass element's conduction state based on the sensed output voltage.
Explanation: This statement is correct. In a linear voltage regulator, the feedback mechanism continuously senses the output voltage and adjusts the pass element (typically a transistor) to ensure that the output voltage remains constant. When the load changes, the feedback system adjusts the pass element’s conduction to counteract the voltage drop or rise, ensuring a stable output voltage.

(ঘ) The feedback loop in a linear voltage regulator operates at a fixed reference voltage and does not adjust for load changes, which is why it is less effective at regulating output voltage under heavy load conditions.
Explanation: This statement is incorrect. The feedback loop in a linear voltage regulator does adjust to load changes. When the load current increases or decreases, the feedback loop compensates by adjusting the pass element's conduction to maintain a constant output voltage. The ability to handle load changes is one of the key features of a linear regulator, although it may not be as efficient under heavy load conditions as a switching regulator.

২১.
In a voltage regulator, what happens when the loop gain becomes too high?
  1. The output voltage becomes unstable and oscillates.
  2. The regulation accuracy improves, and the system becomes more reliable.
  3. The regulator switches from linear to switching mode.
  4. The input voltage requirement decreases due to improved feedback.
সঠিক উত্তর:
The output voltage becomes unstable and oscillates.
উত্তর
সঠিক উত্তর:
The output voltage becomes unstable and oscillates.
ব্যাখ্যা

In a voltage regulator, the loop gain refers to the amplification factor of the feedback system that maintains a stable output voltage by continuously adjusting the regulator’s output based on the input voltage and load conditions. When the loop gain becomes too high, it can cause instability in the regulator's behavior. The feedback loop works to correct any discrepancies between the actual and desired output voltage, but if the gain is excessive, the regulator may overcompensate for these discrepancies. This can result in oscillations, where the output voltage continuously swings between high and low values, unable to settle at a stable level.

The reason for this instability lies in the phase shift and delay introduced by the high loop gain. When the gain is too high, the feedback loop may reinforce errors instead of correcting them, leading to oscillatory behavior. This is commonly referred to as a "ringing" or "oscillation" phenomenon in the system. To prevent this, voltage regulators are often designed with a limited loop gain to ensure that the feedback remains effective without causing instability. Thus, excessive loop gain results in poor performance, making the output voltage unstable and oscillating rather than remaining steady.

২২.
Why do switching regulators generally offer higher efficiency than linear voltage regulators, especially when the difference between input and output voltage is large?
  1. Switching regulators regulate the output voltage by adjusting the duty cycle of the pass transistor, which reduces power loss.
  2. Linear regulators use less complex feedback systems, allowing for greater efficiency.
  3. Switching regulators eliminate the need for feedback loops.
  4. Switching regulators use capacitors that store energy, reducing the need for a power supply altogether.
সঠিক উত্তর:
Switching regulators regulate the output voltage by adjusting the duty cycle of the pass transistor, which reduces power loss.
উত্তর
সঠিক উত্তর:
Switching regulators regulate the output voltage by adjusting the duty cycle of the pass transistor, which reduces power loss.
ব্যাখ্যা

Switching regulators offer higher efficiency than linear voltage regulators, especially when there is a significant difference between input and output voltages, due to their method of regulating output voltage. Unlike linear regulators, which drop excess voltage as heat through a pass transistor, switching regulators use a different approach.

In switching regulators, the output voltage is regulated by adjusting the duty cycle of a pass transistor, which is a switching component. This switching action rapidly turns the transistor on and off, and the average voltage delivered to the load is controlled by adjusting the proportion of time the switch is on versus off (the duty cycle). By doing this, the regulator avoids dissipating the excess voltage as heat, as the switch is either fully on or fully off during each cycle, which minimizes energy loss. This process is more efficient because less power is wasted compared to the linear method, where the pass transistor must constantly dissipate excess voltage as heat.

In contrast, linear regulators maintain a constant output by converting the excess voltage into heat, leading to substantial energy loss when there is a significant difference between the input and output voltages. Therefore, switching regulators are particularly advantageous in scenarios with large voltage differences, providing much higher efficiency by reducing the power lost as heat.

২৩.
Which of the following characteristics is most important for ensuring stability in a voltage regulator’s feedback loop?
  1. A high loop gain across all frequencies.
  2. A low pass element resistance to reduce power dissipation.
  3. Proper compensation of phase margins and loop bandwidth.
  4. A high capacitor value at the output to filter noise.
সঠিক উত্তর:
Proper compensation of phase margins and loop bandwidth.
উত্তর
সঠিক উত্তর:
Proper compensation of phase margins and loop bandwidth.
ব্যাখ্যা

The most important characteristic for ensuring stability in a voltage regulator's feedback loop is proper compensation of phase margins and loop bandwidth.

In a voltage regulator, the feedback loop is designed to maintain the output voltage at a stable level despite variations in input voltage or load conditions. The stability of this feedback loop is crucial for avoiding oscillations or instability in the output voltage.

Phase margin refers to the amount of additional phase shift that the system can tolerate before it becomes unstable. A proper phase margin ensures that the regulator can handle transient conditions without overshooting or oscillating.

Loop bandwidth determines how quickly the feedback loop can respond to changes in the output voltage. If the bandwidth is too wide, the system may respond too aggressively, potentially leading to instability. If the bandwidth is too narrow, the system may not correct errors promptly, leading to poor regulation.

To ensure stable operation, both the phase margin and loop bandwidth must be properly compensated. This compensation ensures that the system can handle changes in load or input voltage without becoming unstable, providing steady and reliable voltage regulation.

২৪.
An IC 7840 regulator has an output current =180mA and internal resistor =10Ω. Find the collector current in the output using the transistor specification: ß=15 and VEB(ON) =1.5v.
  1. 270mA
  2. 450mA
  3. 100mA
  4. 50mA
সঠিক উত্তর:
450mA
উত্তর
সঠিক উত্তর:
450mA
ব্যাখ্যা

The collector current from transistor, IC= ßIB
Where, IB = IO-(VEB(ON)/R1) = 180mA-(1.5v-10Ω) = 0.03A.
Therefore, IC = 15×0.03 = 0.45A = 450mA.

২৫.
Calculate the output voltage for LM314 regulator. The current IADJ is very small in the order of 100µA. (Assume VREF=1.25v)

  1. 17.17v
  2. 34.25v
  3. 89.34v
  4. 23.12v
সঠিক উত্তর:
17.17v
উত্তর
সঠিক উত্তর:
17.17v
ব্যাখ্যা

 The output voltage,
VO = VREF[1+(R2/R1)]+(IADJ×R2)
=1.25Vin× [1+(3kΩ/240Ω)] +( 100µA×3kΩ )
= 16.875 + 0.3.
=> VO = 17.17v.

২৬.
Determine the period of the output waveform for the circuit of Fig. when triggered by a negative pulse.

  1. 0.825 ms
  2. 0.625 ms
  3. 0.075 ms
  4. 0.085 ms
সঠিক উত্তর:
0.825 ms
উত্তর
সঠিক উত্তর:
0.825 ms
ব্যাখ্যা

Thigh = 1.1RAC
= 1.1(7.5 × 103)(0.1 × 10-6)
= 0.825 ms.

২৭.
_____ is the chip substance used in 555 timer during fabrication process?
  1. Silicon
  2.   Platinum
  3. Diamond
  4. All the above
সঠিক উত্তর:
Silicon
উত্তর
সঠিক উত্তর:
Silicon
ব্যাখ্যা

The 555 timer IC is typically fabricated using silicon as the semiconductor material. Silicon is the most commonly used material in the production of integrated circuits due to its excellent electrical properties and abundance.

২৮.
What is the typical period of the output signal in an astable multivibrator 555 timer circuit, given the following resistor and capacitor values: RA = 7.5kΩ, RB = 7.5kΩ, C = 0.1µF?
  1. 1.05 ms
  2. 1.575 ms
  3. 635 Hz
  4. 2.1 ms
সঠিক উত্তর:
1.575 ms
উত্তর
সঠিক উত্তর:
1.575 ms
ব্যাখ্যা

To calculate the period of the output signal in an astable multivibrator 555 timer circuit, we use the following formula:

T=0.693×(RA+2×RB)×CT = 0.693
Where:

RA=7.5kΩ
RB=7.5kΩ
C=0.1μF=0.1×10-6 F
Now, substitute the values into the formula:

T=0.693×(7.5kΩ+2×7.5kΩ)×0.1μFT = 0.693 µF
T=0.693×(7.5kΩ+2×7.5kΩ)×0.1μF T=0.693×(7.5kΩ+15kΩ)×0.1μFT = 0.693 µF
 T=0.693×(7.5kΩ+15kΩ)×0.1μF T=0.693×22.5kΩ×0.1μFT = 0.693 µF
T=0.693×22.5kΩ×0.1μF T=0.693×22500×0.1×10−6 = 0.693*10-6F
T=0.693×22500×0.1×10−6 T=0.693×2.25×10−3 
T≈1.575 ms

২৯.
What is the frequency of the output signal in an astable multivibrator circuit if the time constant is 1.575 ms?
  1. 635 Hz
  2. 1.44 kHz
  3. 1.0 kHz
  4. 5.5 kHz
সঠিক উত্তর:
635 Hz
উত্তর
সঠিক উত্তর:
635 Hz
ব্যাখ্যা

To calculate the frequency of the output signal in an astable multivibrator circuit, we use the following formula:

f=1/T

T is the period of the output signal.
Given that the time constant T is 1.575 ms (or 1.575×10−31.575 s), the frequency is:

f=11.575×10−3
f = ≈635Hz

৩০.
In monostable operation of the 555 timer, what happens when the trigger input goes low?
  1. The output at pin 3 goes low.
  2. The output at pin 3 goes high for a specified time period.
  3. The output at pin 3 switches between high and low states.
  4. The capacitor discharges immediately, causing the output to reset.
সঠিক উত্তর:
The output at pin 3 goes high for a specified time period.
উত্তর
সঠিক উত্তর:
The output at pin 3 goes high for a specified time period.
ব্যাখ্যা

In monostable operation of the 555 timer, when the trigger input goes low, it causes the output at pin 3 to go high for a specified time period (based on the external resistor and capacitor values). After this period, the output returns to low and remains in the low state until the next trigger.

৩১.
For a monostable 555 timer circuit, what is the time period of the output signal when R = 7.5kΩR = 7.5kΩR = 7.5kΩ and C = 0.1µFC = 0.1µFC = 0.1µF?
  1. 1.1 ms
  2. 0.825 ms
  3. 2.5 ms
  4. 0.7 ms
সঠিক উত্তর:
0.825 ms
উত্তর
সঠিক উত্তর:
0.825 ms
ব্যাখ্যা

For a monostable 555 timer circuit, the time period TTT (also known as the pulse width) is calculated using the formula:

T = 1.1×R×CT 
Where:

R = 7.5kΩ
C = 0.1μF = 0.1×10−6 F
Now, substituting the given values into the formula:

T = 1.1 × 7.5 × 103 × 0.1 × 10−6
T = 1.1 × 7.5 × 10−3
T = 0.825ms.

৩২.
Find monostable vibrator circuit using 555 timer.
  1.  None of the mentioned
সঠিক উত্তর:
উত্তর
সঠিক উত্তর:
ব্যাখ্যা

৩৩.
Which among the following can be used to detect the missing heart beat?
  1. Monostable multivibrator
  2.  Astable multivibrator
  3. Schmitt trigger
  4. None of the mentioned
সঠিক উত্তর:
Monostable multivibrator
উত্তর
সঠিক উত্তর:
Monostable multivibrator
ব্যাখ্যা

A monostable multivibrator can be used as a missing pulse detector by connecting a transistor between trigger inputs. If a pulse misses, the discharge trigger input goes high & transistor become cut-off and the output goes low. So, this type of circuit can be used to detect missing heart beat.

৩৪.
Determine the frequency and duty cycle of a rectangular wave generator.

  1. Frequency = 63.7kHz; Duty cycle = 50%
  2.  Frequency = 53.7kHz; Duty cycle = 55%
  3. Frequency = 43.7kHz; Duty cycle = 50%
  4. Frequency = 60kHz; Duty cycle = 55%
সঠিক উত্তর:
 Frequency = 53.7kHz; Duty cycle = 55%
উত্তর
সঠিক উত্তর:
 Frequency = 53.7kHz; Duty cycle = 55%
ব্যাখ্যা

Frequency = 1.45/(RA+RB)C .
Where RA = 100Ω + 50Ω = 150Ω,
RB = 100Ω + 20Ω = 120Ω.
=>∴f = 1.45/((150+120) x 0.1µF) = 53703Hz = 53.7kHz.

Duty cycle, D% = [RB/(RA+RB)] x 100%
= 120Ω/(150Ω +120Ω) x 100%
= 0.55×100%
= 55%.

৩৫.
How to obtain symmetrical waveform in Astable multivibrator?
  1. Use clocked RS flip-flop
  2.  Use clocked JK flip-flop
  3. Use clocked D-flip-flop
  4.  Use clocked T-flip-flop
সঠিক উত্তর:
 Use clocked JK flip-flop
উত্তর
সঠিক উত্তর:
 Use clocked JK flip-flop
ব্যাখ্যা

Symmetrical square wave can be obtained by adding a clocked JK flip-flop to the output of Astable multivibrator. The clocked flip-flop acts as a binary divider to the times output and produces 50% duty cycle without any restriction on the choice of resistors.

৩৬.
Determine the upper and lower threshold voltage

  1. VUT = +14.63v, VLT = +14.63v
  2. VUT = -14.63v, VLT = -14.63v
  3. VUT = VLT = ±14.63v
  4.  None of the mentioned
সঠিক উত্তর:
VUT = -14.63v, VLT = -14.63v
উত্তর
সঠিক উত্তর:
VUT = -14.63v, VLT = -14.63v
ব্যাখ্যা

Explanation: Upper threshold voltage,
VUT = [R1/(R1+ R2)] × (+Vsat)
= [10kΩ/(10kΩ +250Ω)] × (+15v)
= +14.63v.

Lower threshold voltage
VLT = [R1/(R1+ R2)] × ( -Vsat)
= [10kΩ /(10kΩ+250Ω)] × (-15v)
= -14.63v.

৩৭.
What happens if the threshold voltages are made longer than the noise voltages in schmitt trigger?
  1.  Enhance the output signal
  2. Reduce the transition effect
  3. Eliminate false output transition
  4. All the mentioned
সঠিক উত্তর:
Eliminate false output transition
উত্তর
সঠিক উত্তর:
Eliminate false output transition
ব্যাখ্যা

In schmitt trigger, if the threshold voltage VUT and VLT are made larger than the input noise voltage. The positive feedback will eliminate the false output transition.

৩৮.
Calculate the hysteresis voltage for the schmitt trigger from the given specification:
R2 =56kΩ , R1 = 100Ω ,Vref = 0v & Vsat = ±14v.

  1.  0 mv
  2.  25 mv
  3. 50 mv
  4. -25 mv
সঠিক উত্তর:
50 mv
উত্তর
সঠিক উত্তর:
50 mv
ব্যাখ্যা

Upper threshold voltage, VUT =[R1/(R1+R2)]×( +Vsat) = [100kΩ/(56kΩ +100 Ω)]×(+14v)= +25mv.
Lower threshold voltage VLT = [R1/(R1+ R2)]×(-Vsat) = [100kΩ /(56kΩ+100Ω)]×(-14v)= -25 mv.
∴ Hysteresis voltage = VUT-VLT = 25-(-25) = 50mv.

৩৯.
What is the primary function of a phase-locked loop (PLL)?
  1. To generate multiple frequencies
  2. To lock the frequency of a voltage-controlled oscillator (VCO) to the input signal frequency
  3. To amplify the input signal
  4. To demodulate AM signals
সঠিক উত্তর:
To lock the frequency of a voltage-controlled oscillator (VCO) to the input signal frequency
উত্তর
সঠিক উত্তর:
To lock the frequency of a voltage-controlled oscillator (VCO) to the input signal frequency
ব্যাখ্যা

The primary function of a Phase-Locked Loop (PLL) is to lock the frequency of a voltage-controlled oscillator (VCO) to the frequency of an input signal. This is achieved through a feedback loop that constantly adjusts the VCO to match the input signal's frequency. The PLL works by comparing the phase difference between the input signal and the VCO output, generating an error signal that is used to correct any discrepancies. The goal is to maintain synchronization between the two signals, hence the term "phase-locked."

A PLL typically consists of three main components:

Phase Detector (PD): The phase detector compares the phase of the input signal with the phase of the VCO output. It produces an error signal based on the phase difference.

Low-Pass Filter (LPF):
The error signal from the phase detector is often noisy, so it is passed through a low-pass filter to smooth out high-frequency components, leaving only the relevant information to adjust the VCO.

Voltage-Controlled Oscillator (VCO): T
he VCO produces a frequency that is adjustable based on the voltage supplied to it. The error signal from the filter is used to adjust the VCO's frequency so that it matches the input signal.
By constantly adjusting the VCO's frequency to match that of the input signal, the PLL ensures that the output signal is phase-locked to the input signal. This synchronization is crucial in many applications, such as frequency synthesis, FM demodulation, and data transmission. The PLL can lock onto a wide range of frequencies, making it a versatile tool in electronics.

৪০.
What are the components of a basic PLL circuit?
  1. Phase detector, low-pass filter, voltage-controlled oscillator
  2. Amplifier, phase comparator, frequency divider
  3. Voltage-controlled oscillator, frequency multiplier, modulator
  4. Low-pass filter, demodulator, mixer
সঠিক উত্তর:
Phase detector, low-pass filter, voltage-controlled oscillator
উত্তর
সঠিক উত্তর:
Phase detector, low-pass filter, voltage-controlled oscillator
ব্যাখ্যা

A Phase-Locked Loop (PLL) is a feedback control system used for synchronizing the frequency of an oscillator (usually a voltage-controlled oscillator, or VCO) with an input signal. The PLL consists of three main components:

Phase Detector (PD):   The phase detector compares the phase of the input signal with the phase of the VCO's output signal. It detects the phase difference between the two signals and generates an error signal that is proportional to the phase difference. The phase detector is responsible for providing the information necessary to adjust the VCO to match the input signal's frequency and phase.

Low-Pass Filter (LPF):
The error signal generated by the phase detector is often noisy and fluctuating. The low-pass filter is used to smooth out high-frequency components from the error signal, leaving only the relevant, low-frequency information. This filtered signal is then used to control the VCO. The LPF helps to stabilize the PLL by ensuring that only
the desired components of the error signal affect the VCO.

Voltage-Controlled Oscillator (VCO):
The VCO generates a periodic signal (oscillates) whose frequency is controlled by an input voltage. The VCO is the key component that is adjusted based on the error signal from the low-pass filter. The goal is for the VCO's frequency to "lock" onto the frequency of the input signal, meaning the VCO adjusts its output to match the phase and frequency of the incoming signal, thus maintaining synchronization.
These three components—phase detector, low-pass filter, and voltage-controlled oscillator—work together in a feedback loop to achieve phase and frequency synchronization, which is the core function of a PLL.

৪১.
What does the phase comparator in a PLL do?
  1. Amplifies the input signal
  2. Compares the phase difference between the input and VCO signals
  3. Modulates the frequency of the VCO
  4. Filters the output signal
সঠিক উত্তর:
Compares the phase difference between the input and VCO signals
উত্তর
সঠিক উত্তর:
Compares the phase difference between the input and VCO signals
ব্যাখ্যা

In a Phase-Locked Loop (PLL), the phase comparator (also called a phase detector) plays a critical role in maintaining synchronization between the input signal and the output signal of the voltage-controlled oscillator (VCO). The primary function of the phase comparator is to compare the phase difference between the input signal and the VCO output signal.

Here's a breakdown of how it works:

Input and VCO Signals:
The input signal is the reference frequency that we want the VCO to match. The VCO, on the other hand, generates a frequency that is adjusted by the PLL to match the input signal.

Phase Detection:
The phase comparator detects the phase difference between the two signals — the input signal and the VCO output. It does not compare their amplitudes or frequencies directly; instead, it focuses on the difference in their phases (the time difference between corresponding points in the signals, such as the peaks or zero-crossings).

Error Signal:
The phase comparator produces an output signal (often called an error signal) that is proportional to the phase difference. If the input signal and the VCO output signal are not in phase, the error signal will drive the PLL system to make adjustments.

Feedback Loop:
This error signal is then sent to the low-pass filter, where it is filtered to remove high-frequency noise. The filtered error signal is used to control the VCO, causing its frequency to adjust until the phase difference is minimized, at which point the PLL "locks" the VCO to the input signal's frequency and phase.

৪২.
In a PLL, what happens when the loop is in lock?
  1. The VCO frequency differs from the input frequency
  2. The phase comparator output voltage is zero
  3. The VCO frequency matches the input signal frequency
  4. The low-pass filter stops working
সঠিক উত্তর:
The VCO frequency matches the input signal frequency
উত্তর
সঠিক উত্তর:
The VCO frequency matches the input signal frequency
ব্যাখ্যা

When a Phase-Locked Loop (PLL) is in lock, it means that the PLL has successfully synchronized the output frequency of the voltage-controlled oscillator (VCO) with the frequency of the input signal. Let's explore what happens during this process:

Phase Detection: The phase comparator continuously monitors the phase difference between the input signal and the VCO output. If there is any discrepancy, the phase comparator generates an error signal proportional to that phase difference.
Error Signal Adjustment: This error signal is sent to a low-pass filter, which smooths the signal and removes any high-frequency noise. The filtered error signal is then used to adjust the VCO.
Frequency Adjustment: The VCO uses this error signal to modify its frequency. If the VCO frequency is too high or too low compared to the input signal, the PLL adjusts the VCO until both frequencies match.
Lock Condition: When the PLL is in lock, the error signal becomes constant (or zero), meaning there is no more phase difference between the input signal and the VCO output. This results in the VCO frequency matching the input frequency. The PLL has successfully "locked" the VCO to the input signal.

৪৩.
What is the capture range of a PLL?
  1. The frequency range over which the PLL can maintain lock
  2. The frequency range over which the PLL can acquire lock with the input signal
  3. The maximum frequency the PLL can handle
  4. The difference between the input signal and the VCO output
সঠিক উত্তর:
The frequency range over which the PLL can acquire lock with the input signal
উত্তর
সঠিক উত্তর:
The frequency range over which the PLL can acquire lock with the input signal
ব্যাখ্যা

The capture range of a Phase-Locked Loop (PLL) refers to the frequency range within which the PLL can acquire lock with the input signal. In other words, it is the range of frequencies over which the PLL can successfully lock its output frequency (VCO frequency) to the frequency of the input signal, even if the two frequencies initially start out of sync.


Initial Frequency Difference: When a PLL is first powered on or when it is trying to lock onto an input signal, there may be a significant difference between the frequency of the input signal and the frequency of the VCO.
Acquiring Lock: The PLL must adjust the VCO frequency to match the input signal’s frequency. The capture range determines how far apart the two frequencies can be and still allow the PLL to acquire lock. If the frequency difference between the input signal and the VCO is within this range, the PLL can eventually lock onto the input signal and synchronize the VCO to its frequency.
Beyond Capture Range: If the difference between the input frequency and the VCO frequency exceeds the capture range, the PLL may fail to acquire lock, and additional techniques or adjustments may be needed to bring the PLL into lock (such as using a wider frequency range PLL or employing a different method to tune the system).

৪৪.
What is the role of the low-pass filter in a PLL?
  1. To pass high-frequency components
  2. To amplify the VCO output
  3. To remove high-frequency noise and pass the lower frequency component
  4. To compare the phase difference of the signals
সঠিক উত্তর:
To remove high-frequency noise and pass the lower frequency component
উত্তর
সঠিক উত্তর:
To remove high-frequency noise and pass the lower frequency component
ব্যাখ্যা

In a Phase-Locked Loop (PLL), the low-pass filter plays a crucial role in ensuring the stability and smooth operation of the system. After the phase comparator detects the phase difference between the input signal and the VCO (Voltage-Controlled Oscillator) output, it generates an error signal. This error signal represents the phase difference and needs to be processed before it can be used to adjust the VCO. The low-pass filter is responsible for cleaning up this error signal, ensuring that the PLL system operates efficiently and effectively.

Role of the Low-Pass Filter in a PLL:
Filtering High-Frequency Noise: The error signal generated by the phase comparator can be noisy and contain high-frequency components that are not useful for controlling the VCO. These high-frequency components could cause instability or unwanted oscillations in the system. The low-pass filter removes these high-frequency components and smooths the error signal, making it suitable for controlling the VCO.
Passing the Low-Frequency Component: After removing the high-frequency noise, the low-pass filter allows the relevant, low-frequency components of the error signal to pass through. This low-frequency error signal is what the PLL uses to adjust the frequency of the VCO. By filtering out noise and passing only the necessary low-frequency information, the low-pass filter ensures that the VCO adjusts in a stable and controlled manner.
Ensuring Stability: The low-pass filter ensures that the VCO frequency change is gradual and smooth, preventing rapid or erratic changes in frequency that could disrupt the phase-locking process. This smooth adjustment helps the PLL to lock the VCO to the input signal efficiently.

৪৫.
Which application does a PLL NOT directly support?
  1. Frequency demodulation
  2. Frequency synthesis
  3. Frequency-shift keying (FSK) decoding
  4. Audio amplification
সঠিক উত্তর:
Audio amplification
উত্তর
সঠিক উত্তর:
Audio amplification
ব্যাখ্যা

A Phase-Locked Loop (PLL) is a versatile electronic circuit used in many applications involving frequency synchronization, modulation, and signal processing. However, it does not directly support audio amplification.
Audio amplification is the process of increasing the amplitude (strength) of an audio signal to a level that can drive speakers or other output devices. This process primarily involves increasing the signal's power without altering its frequency content or phase. Audio amplifiers work by boosting the input signal using transistors or operational amplifiers, and this task does not require phase comparison or frequency synchronization, which are the main functions of a PLL. A PLL is used for applications where the focus is on maintaining or locking a signal's frequency to a reference signal. It is not designed to amplify audio signals or provide power amplification for audio systems.

৪৬.
In an FM demodulator PLL, what is the demodulated output voltage related to?
  1. The phase difference between the input and VCO signals
  2. The frequency variation of the input signal
  3. The amplitude of the input signal
  4. The fixed frequency of the VCO
সঠিক উত্তর:
The frequency variation of the input signal
উত্তর
সঠিক উত্তর:
The frequency variation of the input signal
ব্যাখ্যা

In a Frequency Modulation (FM) demodulator using a Phase-Locked Loop (PLL), the demodulated output voltage is directly related to the frequency variation of the input signal. This process involves using the PLL to extract the variations in frequency from the modulated input signal and convert them into an appropriate output, such as an audio signal or data stream.

How FM Demodulation Works Using a PLL:
FM Signal Characteristics: In FM, the information is encoded by varying the frequency of a carrier signal. The input signal’s frequency shifts up and down around a central carrier frequency, with the rate and extent of the frequency variation carrying the information (e.g., an audio signal in FM radio).

Phase-Locked Loop in FM Demodulation: 
The PLL in an FM demodulator works by locking the output of a voltage-controlled oscillator (VCO) to the frequency of the input signal. As the frequency of the input FM signal varies, the PLL adjusts the frequency of the VCO to stay in sync with the input signal's frequency.

Frequency Detection:
 The PLL tracks the instantaneous changes in the frequency of the input signal. When the frequency of the input signal increases or decreases, the VCO also adjusts its frequency to match the input. The phase comparator in the PLL detects these changes and produces an error signal proportional to the frequency deviation of the input signal.

Demodulated Output:
 The error signal generated by the phase comparator, which reflects the frequency deviation of the input FM signal, is then passed through a low-pass filter. The result is a demodulated output voltage that is directly related to the frequency variation of the input FM signal. This output can then be used for further processing, such as converting it into an audio signal.

৪৭.
How does a frequency synthesizer work with a PLL?
  1. By using a phase comparator to lock the VCO output to a fixed reference signal
  2. By introducing a frequency divider between the VCO output and the phase comparator
  3. By using a low-pass filter to stabilize the frequency
  4. By directly applying the input signal to the VCO
সঠিক উত্তর:
By introducing a frequency divider between the VCO output and the phase comparator
উত্তর
সঠিক উত্তর:
By introducing a frequency divider between the VCO output and the phase comparator
ব্যাখ্যা

A frequency synthesizer is a system that generates a range of output frequencies from a fixed reference signal, typically by using a Phase-Locked Loop (PLL). The PLL allows for precise control over the output frequency, which is a multiple or fraction of the reference frequency. One of the key methods that a frequency synthesizer uses to achieve this is by incorporating a frequency divider between the Voltage-Controlled Oscillator (VCO) output and the phase comparator.

Phase-Locked Loop Components:
The basic components of the PLL in a frequency synthesizer are the phase comparator, low-pass filter, and VCO. The PLL adjusts the frequency of the VCO to match the input signal (or reference signal) in terms of phase and frequency.

Frequency Divider:
In a frequency synthesizer, a frequency divider is inserted between the VCO output and the phase comparator. This is a crucial step for generating output frequencies that are different from the reference frequency.

The VCO generates a frequency that is typically a multiple of the reference signal's frequency.
The frequency divider divides the VCO output by a fixed integer (or sometimes by a variable factor), reducing the VCO frequency to the desired frequency.

Phase Comparator Operation:
 The phase comparator compares the phase of the reference signal and the divided VCO output. If there is a phase difference, the PLL adjusts the VCO to correct this difference and lock the output to the reference signal. The frequency divider ensures that the VCO generates the necessary higher-frequency signal, which is then divided down to the desired output frequency.

Frequency Control:
By changing the division factor of the frequency divider or by adjusting the reference signal, the PLL can generate a wide range of output frequencies. This is the basis for frequency synthesis, where multiple, precise frequencies are produced from a single reference source.

৪৮.
In the 565 PLL, how is the free-running frequency(f0)   set?
  1. By adjusting the phase detector
  2. By selecting the appropriate low-pass filter
  3. Using external resistor R1 and capacitor C1
  4. By applying a fixed voltage to the VCO
সঠিক উত্তর:
Using external resistor R1 and capacitor C1
উত্তর
সঠিক উত্তর:
Using external resistor R1 and capacitor C1
ব্যাখ্যা

The 565 Phase-Locked Loop (PLL) is a specific type of PLL, often used in signal processing, where the VCO (Voltage-Controlled Oscillator) frequency is adjustable. The free-running frequency (fₒ) refers to the frequency at which the PLL oscillates when no input signal is present or when it is not locked to any reference signal. The free-running frequency is determined by the characteristics of the VCO in the PLL circuit.

The Free-Running Frequency is Set in the 565 PLL: In the 565 PLL, the free-running frequency (fₒ) is set by selecting the appropriate external components, specifically an external resistor R1 and capacitor C1. These components control the charging and discharging rates of the internal timing capacitor in the VCO, which directly affects the oscillation frequency.

1. Resistor (R1) and Capacitor (C1): The external resistor (R1) and capacitor (C1) are connected to the timing circuit inside the PLL. These components form an RC network that determines the frequency of oscillation for the VCO.
By selecting different values for R1 and C1, the free-running frequency of the PLL can be adjusted. The larger the value of the resistor and capacitor, the lower the free-running frequency, and vice versa.

2. VCO Frequency Control: 
The frequency of the VCO is inversely related to the product of R1 and C1. The combination of these components controls how quickly the VCO's internal capacitor charges and discharges, which sets the oscillation frequency of the PLL when it's not locked to an external signal.

৪৯.
What is the role of a frequency divider in a PLL-based frequency synthesizer?
  1. To divide the input frequency by a fixed amount
  2. To generate a higher frequency signal than the VCO output
  3. To stabilize the VCO frequency
  4. To lock the PLL loop
সঠিক উত্তর:
To divide the input frequency by a fixed amount
উত্তর
সঠিক উত্তর:
To divide the input frequency by a fixed amount
ব্যাখ্যা

In a PLL-based frequency synthesizer, the frequency divider plays a key role in generating different output frequencies from a fixed reference signal. The frequency divider is used to scale the frequency of the signal coming from the Voltage-Controlled Oscillator (VCO) to achieve the desired output frequency.

Input Signal and VCO Frequency: The input signal (or reference signal) is fed into the phase comparator, where it is compared with the VCO output. The VCO's frequency is adjusted based on the phase difference detected by the phase comparator to lock it to the input signal's frequency.
The VCO, however, generally produces a frequency that is higher than the desired output frequency. This is where the frequency divider comes into play.

Frequency Divider's Role: 
The frequency divider divides the VCO output frequency by a fixed integer or ratio. This is crucial because the VCO typically generates a frequency that is higher than needed.
The division factor of the frequency divider determines the final output frequency. For example, if the VCO output is 1 MHz and the divider ratio is 10, the output frequency will be 100 kHz.

Generating Precise Frequencies:
By adjusting the division factor, a PLL-based frequency synthesizer can generate a wide range of output frequencies that are integer multiples or fractions of the reference signal's frequency. The frequency divider allows the synthesizer to produce these precise frequencies.

৫০.
In FSK decoding using a PLL, what happens when the input signal switches between two carrier frequencies?
  1. The output remains constant
  2. The output voltage shifts between two corresponding levels
  3. The PLL loses lock
  4. The VCO frequency becomes constant
সঠিক উত্তর:
The output voltage shifts between two corresponding levels
উত্তর
সঠিক উত্তর:
The output voltage shifts between two corresponding levels
ব্যাখ্যা

In Frequency Shift Keying (FSK) decoding, the goal is to recover the transmitted data, where the data is encoded by shifting the frequency of the carrier signal between two distinct frequencies. A Phase-Locked Loop (PLL) is commonly used in FSK decoding to track and demodulate these frequency shifts.

FSK Signal Characteristics: 
In FSK, the carrier signal switches between two different frequencies (often referred to as f₁ and f₂) to represent binary data (e.g., '0' and '1'). Each of these frequencies corresponds to a different symbol in the transmitted data.

Role of the PLL: 
The PLL tracks the frequency of the input signal and adjusts the output frequency of the Voltage-Controlled Oscillator (VCO) to match the frequency of the input signal. When the input signal switches between the two frequencies (f₁ and f₂), the PLL adjusts the VCO accordingly to lock onto the new frequency.

What Happens When the Input Signal Switches Between Frequencies:
 As the input signal switches between the two carrier frequencies (f₁ and f₂), the PLL follows these changes.
The output voltage of the phase comparator (which compares the phase of the VCO with the input signal) shifts between two corresponding levels. These levels correspond to the frequency shifts of the input signal, and this variation in the output voltage is a direct reflection of the frequency shifts in the FSK signal.
The phase comparator output voltage essentially indicates whether the signal is at f₁ or f₂, and the voltage shifts between these levels as the signal changes frequencies.