MPPGCL JE Electronics Mock Test - 5 - Electronics and Communication Engineering (ECE) MCQ

In ASK modulation each symbol is transmitted using a single bit.

Hence, the Bit rate = Baud rate

Baud rate = 300 baud

Note: QPSK transfers 2 bit for 1 symbol

∴ The baud rate for QPSK will be:

Baud rate = 300/2 = 150 baud

The maximum radar range is given by:

Where

P_t is the transmitted power

A₀ is capture area

S = effective area

P_min = minimum power at the receiver which can be recognized

λ = wavelength of EM radiation

Concept:

Y parameter:

I₁ = V₁ Y₁₁ + V₂ Y₁₂

I₂ = V₁ Y₂₁ + V₂ Y₂₂

     ...1)

     ...2)

Calculation:

For the given question Z = 5 Ω

Voltage Controlled Oscillator (VCO):

• For a Voltage controlled oscillator generating a sawtooth waveform, the main component is the capacitor that is charging and discharging and is deciding the formation of the output waveform.
• The voltage is compared with a reference voltage using a comparator.
• When the capacitor voltage exceeds the reference voltage, the comparator generates a high logic output that triggers the transistor, and the capacitor is connected to the ground and starts discharging. Thus the output waveform generated is the representation of the charging and discharging of the capacitor and the frequency is controlled by the Modulating Input.

• A practical example of a voltage-controlled oscillator (VCO) is the LM566.
• The LM566 is a general-purpose VCO that may be used to generate square wave and triangular waveforms as a function input voltage.

Applications:

• Function generator
• Tone generator
• FM modulation
• Frequency shift keying

Concept:

If we pass the input signal to a single T-flip flop, we will get half of the frequency at the output.

Similarly, when we pass the input signal to n-T flip-flops, the output frequency (f_out) will be:

Application:

Given Input frequency f = 200 Hz

f_out = 25 Hz

The frequency spectrum of an Amplitude Modulated wave is given by:

Bandwidth in AM = 2f_m = 6 kHz

f_m = 3 kHz

Upper sideband = f_c + f_m = 600 kHz

∴ The carrier frequency is:

f_c = 600 – f_m = 600 - 3 kHz

f_c = 597 kHz

Concept:

Drain current equation for JFET

I_D = I_DSS (1 -  V_GS/ V_p)²

Where, 

I_D is  the drain current

I_DSS is the saturation current

V_GS is the Gate to Source voltage

V_P is the pinch-off voltage.

Input current to JFET is zero.

Use the KVL equation to obtain the value of V_DS

Calculation:

Given , V_G = -2 V , V_S = 0 V , V_P =  -8 V , I_DSS = 10 mA

V_GS = ( V_G - V_S) = ( -2 - 0) = -2 V

 I_D = I_DSS (1 -  V_GS/ V_p)² = 10 (1 - (-2/-8))² = 5.625 mA

Apply KVL in the path shown by arrow indicated as 1

-16 + 2I_D + V_DS = 0

⇒  V_DS = 16 - 2 (5.625) 

⇒  V_DS = 4.75

Concept 

Zener Diode

• Zener diodes are specially designed to give accurate and stable reverse breakdown voltage.
• When it is forward-biased, it behaves like a standard diode.
• When reverse-biased, a small leakage current flows through it.
• If the reverse voltage across Zener diode is increased, a value of voltage will be attained at which reverse breakdown occurs.
• As the breakdown occurs, a sudden increase in current takes place called as Zener Current.
• Application: Voltage Regulator.

Zener Breakdown:

• Zener diode is a highly doped PN junction diode, due to which the depletion region between P and N becomes very narrow and the electric field is quite high.
• The breakdown easily occurs in reverse biased conditions because of the narrow junction in Zener Diode.
• Even though the electric field is very high, the electrons can tunnel through due narrow depletion region.
• In case, we narrow the space charge region, It is obvious that with the decreased reverse voltage applied reverse breakdown will occur. 
• This causes a high reverse current to flow even in the reverse bias. This can be used as a normal PN junction diode in forward bias because the junction is not destroyed.

Fig: Circuit Diagram or Layer Structure of Zener Diode

Additional Information
Difference between Zener and Avalanche Breakdown:

Calculation:

For the given MUX, the output is given by,

Given I₀ = X, I₁ = Y, I₂ = X and I₃ = 0

Now, 

= 0 + X̅Y + XY̅ + 0

= X̅Y + XY̅

= X ⊕ Y (X-OR gate)

So, option (3)

The correct answer is 40 W

Concept:

Let's take the given below direction of voltage and current as a reference.

If any direction other than the reference was given, then don't forget to take the negative sign in the final answer.

Solution:

The graph shows that the current is zero, which means an open circuit condition. In this case, V is nothing but the Thevenin voltage across that point.

 V_th = - 40 V

But the given direction is opposite to the reference direction. 

 V_th = - (-40 V) = 40 V

From the graph, as we take output voltage zero means short circuit then short circuit current would be I_SC = 4 A

= 10 

According to the maximum power transfer theorem 

= 40 W

Concept:

• When the p-n junction is forward biased the junction potential is lowered which allows the majority carrier holes in the p-type side and majority carrier electron in the n-side to diffuse across the junction because of a high concentration gradient.
• These majority carriers holes and electrons are the minority in n and p sides respectively. Therefore this process of diffusion is also called minority carrier injection.
• This diffusion current dominates the forward bias current and is given by

Where V_a is the forward bias voltage across the p-n junction.

Explanation:

Mechatronics is the synergistic integration of mechanical engineering with electronics and intelligent control algorithms in the design and manufacture of products processes.

Mechatronics has evolved through several stages that are primarily defined in terms of

• Primary level mechatronics
• Secondary level mechatronics
• Tertiary level mechatronics
• Quaternary level mechatronics

Primary Level Mechatronics:

• This level encompasses input/output (I/O) devices such as sensors and actuators that integrate electrical signalling with mechanical action at the basic control level. Electrically controlled fluid valves and relay switches are two examples.

Secondary Level Mechatronics:

• It integrates microelectronics into electrically controlled devices. Sometimes these products are stand-alone. An example is a cassette tape player.

Tertiary Level Mechatronics:

• This level further enhances the quality in terms of sophistication by incorporating advanced feedback functions into the control strategy.
• The mechatronic systems at this level are called smart systems.
• The control strategy uses microelectronics, microprocessors, and other application-specific integrated circuits (ASIC) as bits and pieces for control realization.
• A microprocessor-based electrical motor used for actuation purposes in typical industrial robots can be considered as an example of a tertiary level mechatronic system.

Quaternary Level Mechatronics: Intelligent control is very much a part of mechatronics now.

• Decision-making is done an influence is needed to perform the decided upon action.
• Machine or plant control intelligence is moving into the realm of human-like intelligence.
• This level attempts to improve smartness a step ahead by introducing intelligence and FDI (Fault Detection and Isolation) capability into the systems.
• At this level, mechatronics means many things to many people. Artificial Neural Networks and Fuzzy Logic try to capture some of the intellectual capabilities of intelligence.

Transformer Utilization Factor (TUF) is defined as the ratio of the DC power output of a rectifier to the effective Transformer VA rating used in the same rectifier. The effective VA Rating of the transformer is the average of the primary and secondary VA ratings of the transformer.

This factor indicates the effectiveness of transformer usage by the rectifier.

For half-wave rectifier, it is equal to 0.286

Important Points

• TUF for Half-Wave rectifier is 28.6%
• TUF for Full-Wave centre tapped rectifier is 57.2%
• TUF for Full-Wave bridge rectifier is 81%

Baud Rate is defined as the rate of change of signal on transmission medium after encoding and modulation have occurred.

Baud Rate:

• The baud rate is the rate at which information is transferred to a communication channel.
• The baud rate indicates the number of electrical oscillations per second that occurs within a data transmission.
• Baud rate is commonly used when discussing electronics that use serial communication.
• The higher the band rate, the more bits per second that are transferred.
• In the serial part content, ‘9600 band’ means that the serial port is capable of transferring a maximum of 9600 bits per second.
• At baud rates above 76,800, the cable length will need to be reduced.
• Higher the baud rate, the mere sensitive the able becomes to the quality of installation, due to how much of the wire is untwisted around each device.

​Note:

• Bit rate is the transmission of the number of bits per second whereas baud rate is the number of signal units transmitted per second.

Explanation:

Resolution (R):

Resolution of Digital to Analog Converter(DAC) is a change in analog voltage corresponding to 1 LSB bit increment at the input.

Where

V_r = Reference voltage

n = number of bits

Resolution should be below 0.1% of the maximum value 

(2ⁿ - 1) ≥ 1000

2ⁿ  ≥   1001

n = 10 satisfies the above equation.

Hence n = 10.

Concept:

The maximum range of a radar system is given as

Where:

R_max = Maximum range of the radar

S_min = Minimum Detectable Signal Power

P_t = Peak transmitted power

A_e = Aperture Area

D = Antenna diameter

We observe that 

Rmax ∝ D^1/2 x P_t^1/4  

Calculation:

P(new) = 16 P(old)

D(new) = 2D(old)

R_max ∝ D^1/2 x P_t^1/4  

R(new) = √2D^1/2 x 2P^1/4

R(new) = 2√2R(old)

R(new) = √8R(old)

Hence,  If the peak transmitted power is increased by a factor of 16, and the antenna diameter is increased by a factor of 2, the radar range will be increased by a factor of √8.

IMPATT diode is used as a power generating device in semiconductor microwave oscillators and delivers a relatively high RF power to produce carrier signals for microwave transmission systems, particularly airborne and ground-based low-power radar systems.

IMPATT diode:

IMPATT diode consists of 4 layers according to the construction.

• An IMPATT diode has n+ - p - i - p + structure and is used with reverse bias.
• It exhibits negative resistance and operates on the principle of avalanche breakdown. I
• IMPATT diode circuits are classified as a broadly tunable circuit, low Q circuit, and high Q circuit. The impedance of the IMPATT diode is a few ohms.
• The word IMPATT stands for Impact Avalanche Transit Time diode.

The features of IMPATT diode oscillator are:

1) Frequency 1 to 300 GHz, Power output (0.5 W to 5 W for single diode circuit and up to 40 W for the combination of several diodes),

2) It has an efficiency of about 20%.

3) Its applications include police radar systems for detection of speed, low power microwave transmitters, etc.

Concept:

The conductivity (σ) of a semi-conductor is given by:

   ---(1)

Where,

n₀ and p₀ are the concentration of electrons and holes in doped semiconductor

μ_n is electron mobility

μ_p is hole mobility

Analysis:

For a doped semiconductor, one of the charge concentration will dominate and the conductivity will be:

 (For semiconductors doped with donor impurities only)

 (For semiconductors doped with acceptor impurities only)

Since Boron is a trivalent impurity, the acceptor concentration will be:

N_a = 10¹⁸ atom/cm³

The conductivity of sample A doped with Boron impurity will be:

Similarly, since phosphorus is a pentavalent impurity, the donor concentration will be:

N_d = 10¹⁸ atom/cm³

And the conductivity of sample B doped with Phosphorus impurity will be:

The required ratio will be:

Given N_a = N_d = 10¹⁸ /cm³

Concept:

R – 2R ladder DAC:

3 bit R – 2R DAC  by using Inverting op-amp is shown below:

Output voltage is defined as:

V₀ = - I₁(R_F) ⋯ (i)

Current is calculated as:

     ⋯ (ii)

Current drawn from the source is

I = V / R ⋯ (iii)

Current is calculated as:

The final output voltage is

For n bit DAC output voltage is

Calculation:

Given digital input is 1010 and the reference voltage is 5 V

a₃ = 1, a₂ = 0, a₁ = 1, a₀ = 0

considering the value of the feedback resistor for Op amp is R Ω

Output voltage is

V₀ = -3.125 V

• A mobile identification number (MIN) is a serial number that uniquely identifies a cell phone services subscriber within a mobile carrier network.
• It is used by mobile phone services providers to identify subscribers within its database, specifically when routing calls.
• MIN can also be called a mobile subscriber identification number (MSIN).

Optical fibers are transparent fibers and act as a light pipe to transmit light between its two ends. They are made up of silicon dioxide.

Refractive index: The ratio of the speed of light in a vacuum to the speed of light in a medium is called the refractive index of that medium.

It is also called an absolute refractive index.

Critical angle: When a ray of light is going from a denser medium to a rare medium then the angle of incidence at which the refraction angle is 90° is called the critical angle.

The critical angle is given by:

Θ_C = critical angle = sin^-1(n₂/n₁)

Where n₂ is the refractive index of second medium in which light ray is going and n₁ is the refractive index of first medium from which light is going to second medium.

Advantages and disadvantages of step-index mono-mode optical fiber.

Concept:

Fixed bias circuit or Base bias circuit is represented as:

With I_C = β I_B

V_CE = -I_C R_C + V_CC

Calculation:

Given,

R_B = 470 kΩ, R_C = 2.2 kΩ, and V_CC = 18 V

β = h_fe = 100

The value of the collector current is:

I_C = β Iβ

I_C = 3.68 mA

∴ The required value of V_CE is:

V_CE = V_CC - I_C R_C

V_CE = 18 – 3.68 × 2.2

V_CE = 9.902 V 

Concept:

• Packet filter firewall can investigate up to network layer
• Packet filter firewall cannot investigate in transport layer section hence cannot block user with a specified port number

Explanation:

A packet filter network firewall can investigate and block network traffic based on several fields in a network packet header.
These fields include:

I. the source and destination port numbers.

II. the source and destination IP addresses.

Additional Information
MAC addresses are included in the packet header, they are typically used for local network communication and are not used for filtering traffic by a packet filter firewall.

Note: However, some advanced firewalls may be able to filter traffic based on the MAC address in certain scenarios, such as in virtualized environments or for wireless access points.

Hence, the correct answer is option 2.

Calculation:

Given equation is

y[n] = x[n] - x[n-1]

Applying DTFT,

Y(Ω) = X(Ω) - e-jΩX(Ω)

Y(Ω) = X(Ω)[1 - e-jΩ]

H(Ω) = e-jΩ/2[ejΩ/2 - e-jΩ/2]

H(Ω) = 2 e-jΩ/2 sin[Ω/2]

At Ω = 0 

|H(Ω)| = 0

Hence, it is a high pass filter.

Differential pulse-code modulators:

• The DPCM module implements one of the simplest compression algorithms.
• It combines a spatial prediction loop with a non-uniform quantizer of which the representation levels are Huffman encoded.
• It is a signal encoder that uses PCM but adds some functionalities based on the prediction of the samples of the signal.
• The input to the DPCM can be an analog or digital signal.
• The number of encoded bits is smaller than in the PCM system.
• It needs far fewer bits per error sample than what would have been needed for the original samples themselves.
• This results in fewer bandwidth requirements by the DPCM system and hence has a lower dynamic range than the original message itself.​

Hence, options 1 and 2 are correct.

40% of 900 + x = 30% of 2600
360 + x = 780
x = 420
Hence, Option C is correct. 

The logic followed here is:

The figure in image 2 is water image of the figure in image 1. The figure in image C of answer figure will be the water image of the figure in image 3.

Hence, image C of answer figure is the correct answer.

Checking Statement I:

Since there is no information about Monika's rank. Therefore, Rahul's rank cannot be determined.

Clearly, Statement I alone is not sufficient to answer the question.

Checking Statement II:

Aditi's rank and Monika's rank is not given. Therefore, Rahul's rank cannot be determined.

Clearly, Statement II alone is not sufficient to answer the question.

Checking Both statement I and II:

Ranks are not known for any of them. Thus, Rahul's rank cannot be found.

Clearly, Both statements I and II  is not sufficient to answer the question.

Hence, Option D is correct.