MPPGCL JE Electronics Mock Test - 2 - Electronics and Communication Engineering (ECE) MCQ

Concept:

The minimum number of Taps required to realize FIR filter is given as:

Δf = f_STOP - f_pass

Given:

f_s = 200 kHz

f_stop = 15 kHz

f_pass = 10 kHz

≈ 110

Concept:

The inductor does not allow sudden change of current.

Hence, i_L(0^-) = i_L(0⁺)

Current expression for an RL circuit would be;

i(t) = i(∞) + [i(0⁺) - i(∞) ]e^-Rt/L

Calculation:

at t = 0^-

Switch is connected to position 1

inductor will be short circuited.

i(0^-) = -20/10 = -2 A

at t = 0⁺

The switch will be connected to position 2 and the circuit will become source free.

As inductor does not allow sudden change of current

 i_L(0^-) = i_L(0⁺) = -2 A

As there is no source the current will decay exponentially

So, at t = ∞ 

i(∞ ) = 0 A

Thevenin equivalent resistance after the operation of the switch seen from inductor

R = (10 + 10) = 20 Ω 

Current expression;

i(t) = i(∞) + [i(0⁺) - i(∞) ]e^-Rt/L

⇒  i(t) = 0 + [-2 - 0 ]e^-20t/4

 ∴ i(t) = -2e^-5t

Reference bit – 1

Input sequence – 0 1 0 0 1

Output sequence – 0 0 1 0 0

Output phase sequence - π π 0 π π

Process control system

An automatic control system in which the output is a variable is called a process control system.

An automatic control system is also known as a closed-loop system because the feedback will automatically correct the change in output due to disturbances.

An example of a basic process control system is a thermostat, a heating element, and a cooling element within a room. As the temperature in the room fluctuates beyond set boundaries, the thermostat turns on either the heating or cooling system to keep the room at a specific temperature.

The feedback path in an op-amp differentiator consists of a resistor

At low frequencies input impedance Is high

At high frequency the input impedance is low

Thus, with an increase in frequency circuit sensitivity to noise increases.

The input signal to the differentiator is applied to the capacitor. The capacitor blocks any DC content so there is no current flow to the amplifier summing point, X resulting in zero output voltage.

The capacitor only allows AC type input voltage changes to pass through and whose frequency is dependent on the rate of change of the input signal.

the DSB-SC signal

The envelope

Now in general, the envelope of an expression

is given by

Thus we have, envelope of .

Cathode Ray Oscilloscope (CRO):

• The CRO stands for a Cathode Ray Oscilloscope.
• It is typically divided into four sections which are display, vertical controllers, horizontal controllers, and Triggers.
• Most of the oscilloscopes are used the probes and they are used for the input of any instrument.
• We can analyze the waveform by plotting amplitude along with the x-axis and y-axis.

• The applications of CRO are mainly involved in the radio, TV receivers, also in laboratory work involving research and design.
• In modern electronics, the CRO plays an important role in the electronic circuits.

Basic Controls of CRO: 

The basic controls of CRO mainly include position, brightness, focus, astigmatism, blanking & calibration.

• Position:
  • In the oscilloscope, the position control knob is mainly used for position control of the intense spot from the left side to the right side.
  • By regulating the knob, one can simply control the spot from left side to the right side.
• Brightness:
  • The ray’s brightness mainly depends on the intensity of the electron.
  • The control grids are accountable for the electron intensity within the electron ray.
  • So, the grid voltage can be controlled by adjusting the electron ray brightness.
• Focus:
  • The focus control can be achieved by regulating the applied voltage toward the center anode of the CRO.
  • The middle & other anodes in the region of it can form the electrostatic lens.
  • Therefore, the main length of the lens can be changed by controlling the voltage across the center anode.
• Astigmatism:
  • In CRO, this is an extra focusing control & it is analogous to astigmatism within optical lenses.
  • A ray focused in the middle of the monitor would be defocused on the screen edges as the electron path lengths are dissimilar for the center & the edges.
• Blanking Circuit: The time base generator present in the oscilloscope generated the blanking voltage.
• Calibration Circuit:
  • An oscillator is necessary for the purpose of calibration within an oscilloscope.
  • However, the oscillator which is used should generate a square waveform for preset

Concept:

Min Terms: 

• A minterm is a boolean expression written for all those terms whose value is 1 in the K-map.
• It is denoted by: F(x, y, z) = ∑(minterms)

​MaxTerms: 

• A maxterm is a boolean expression written for all those terms whose value is 0 in the K-map.
• It is denoted by: F(x, y, z) = π(max terms)
• Max terms are the compliments of minterms.

Explanation:

Given,  F(x, y, z) = ∑(1, 3, 6, 7)

Binary representation is:

 F(x, y, z) = ∑(001, 011, 010, 011)

It is minterm representation.

F(x, y, z) = ∑(000, 010, 100, 101)

F(x, y, z) = ∏ (0, 2, 4, 5)

It is max max-term representation.

EXPLANATION:

Faraday’s Law states that a change in magnetic flux induces an emf in a coil.

Also, Lenz’s Law states that this induced emf produces a flux which opposes the flux that generates this emf, i.e.

     ----(1)

EMF is also defined as:

Also, 

Putting the above in Equation (1), we get:

Additional Information

Maxwell's Equations for time-varying fields is as shown:

The given circuit is redrawn as:

DIAGRAM

The output Q will be:

Applying De-Morgan's Law, the above expression becomes:

Q = A ⊕ B, i.e. A EXOR B

The correct answer is Magnetic Tape.

• Magnetic tape is a secondary storage device used for storing files of data; Example: a company’s payroll record.
• Access is sequential and consists of records that can be accessed one after another as the tape moves along a stationary read-write mechanism
• A hard disk is an electromechanical data storage device from which data is accessed in a random-access manner, individual blocks of data can be stored or retrieved in any order
• Both sequential and random access of data is possible in hard disk
• In random-access memory (RAM) the memory cells can be accessed for information transfer from any desired random location
• That is, the process of locating a word in memory is the same and requires an equal amount of time no matter where the cells are located physically in memory; Hence RAM is random access

Concept:

The cutoff frequency for dominant mode is given as:

a and b are the dimensions of the waveguide

Calculation:

The mode of operation is TE₁₀, i.e. m = 1, n = 0.

The cut off frequency for this mode will be:

f_c (1, 0) = 2.5 GHz.

Since 2.5 GHz is the cut-off frequency for TE₁₀ mode to operate, signals with frequencies below 2.5 GHz will not be able to pass.

Figure of merit (FOM):

A figure of merit is used to decide the quality of a communication technique

For Angle modulation:

FOM of FM is given by,

Thus, as transmission bandwidth W is doubled, FOM becomes one fourth.

Important Point

For Amplitude modulation:

The signal to noise ratio at output Is a maximum 1/3 (for μ = 1) of the signal to noise ratio at the input.

FOM is 1 for DSB, SSB, VSB.

​Phase Modulation:

Concept:

A system is linear, if the operations on the input signal are all linear and no signal-independent terms are contained. Basic linear operations are given below-

• Multiplication of input signal with a constant value: y(t) = ax(t)
• Time-shifting the input signal y(t) = x(t−t₁)
• Scaling of the input signal y(t) = x(bt)
• Combinations of scalling,shifting and multiplication , e.g. y(t) = ax(b(t−t₁))
• Summations of terms that are linear, e.g. y(t) = ax(t)+x(t−t₁)
• Convolution of two input signal, eg y(t) = x(t)*h(t)

Some non-linear operations are given below-

• Multiplication of the signal with itself, i.e. y(t) = x(t)⋅x(t)
• Applying any non-linear function to the signal, e.g. y(t) = sin(x(t))
• Adding constant terms, which are independent of the signal, i.e. y(t) = x(t)+a

Analysis:

(I)  d²y(t)/dt² + 9a₁dy(t)/dt + a₂y(t) = u(t) is linear

(ii) y(t) dy(t)/dt + a₁y(t) = a₂u(t) is non-linear due to multiplication of y(t)

(iii).  2d²y(t)/dt² + dy(t)/dt + t²y(t) = 5 is Non Linear because of the contant term 5.

Optical fiber communication:

The communication system that uses lightwave for data transmission .

Optical fiber communication is carried out at a very high frequency 10¹⁴ to 10¹⁵ Hz. 

UltraViolet wave operates at 10¹⁴ to 10¹⁷.

Visible light band operates at 1014 to 1015.

Infrared band operates at 1013 to 1014.

Hence correct option is "4".

In on-off control, thyristor switches connect the load to the AC source for a few cycles of input voltage and then disconnect it for another few cycles. It is also called burst firing, integral control.

 In phase angle control, thyristor switches connect the load to the AC source for some portion of each cycle of the input voltage.

Concept:

In an extrinsic semiconductor, the conductivity (hence resistivity) depends on the number of carriers present and is given by:

n0 = majority carrier electron concentration

p₀ = majority carrier hole concentrationμn and μp are the electron and hole mobilities respectively.

The resistivity is the inverse of conductivity, i.e.

Calculation:

Since the intrinsic carrier concentration of Germanium is of the order of 10¹³, and the given donor density is of the order of 10²⁰, we can write:

n₀ = Nd = 1020 atoms/m3

Since there is no acceptor impurity, p₀ << n₀ and can be easily neglected and the resistivity becomes:

ρ = 0.164 Ωm

The correct answer is All of the above.

Key Points

• The Internet is a huge collection of networks, a networking infrastructure. It encapsulates millions of computers together globally, forming a network in which any computer can communicate with any other computer as long as they are both connected to the Internet. In other words, it is a worldwide system of cross-connected computer networks, connecting millions of devices through which exchange of information such as data, news, and opinions, etc. is possible.
• It utilizes the TCP/IP (Transmission Control Protocol/Internet Protocol) to aid millions of users across the globe. So, TCP/IP can be called the backbone of the Internet. It is considered as a network of networks that consists of thousands of private and public, academic, business, and government interconnections. The Internet is often considered as “The Information Highway”, which implies that there is a straight and clear way of obtaining information. It connects thousands of computer networks. Each device connected to the Internet is known as the host and is independent. Through telephone wires, Fiber optical cable, and satellite links, Internet users can share a variety of information.
• The Internet is defined as an Information superHighway, to access information over the web. However, It can be defined in many ways as follows:
  • The Internet is a worldwide global system of interconnected computer networks.
  • The Internet uses the standard Internet Protocol (TCP/IP).
  • Each device connected to the internet is identified by its unique IP address.

The correct answer is option 4):(the same as that of Y)

Concept:

• The given circuit resembles the wheat stone bridge
• 
• The Wheatstone bridge works on the principle of null deflection, i.e. the ratio of their resistances is equal, and no current flows through the circuit. Under normal conditions, the bridge is in an unbalanced condition where current flows through the galvanometer. The bridge is said to be balanced when no current flows through the galvanometer.
• If the resistance of each resistor is 100 ohm, The bridge is at balance condition no current flows in the centre arm. the potential at X is equal to y

ASK System:

1. For ASK Transmitter on-off keying is used.

2. In Amplitude Shift Keying (ASK) binary 1 is represented with the presence of carrier and binary 0 is represented with the absence of the carrier.

1 : s₁(t) = A_c cos 2πf_ct

0 : s₂ (t) = 0

The given

Narrow bandpass filter

T = 0.25 msec

Now applying Fourier series formula on x(t), we get:

ω₀T = 2π

So, odd components are a₁, a₃, a₅, a₇, a₉…., and so on.

Since the filter passes frequencies only between 20 kHz and 40 kHz, the components lying between this band of frequency will be passed by the filter.

The fundamental frequency given is 4 kHz.

a₁ = 4 kHz

a₃ = 12 kHz

These 3 components of 20 kHz, 28 kHz, and 36 kHz will be passed and the rest of the components will be rejected by the filter.

a₁₁ → 44 kHz

a₁₃ → 52 kHz

       ⋮

And so on

∴ Option 2 is correct.

• The File Transfer Protocol (FTP) is a standard network protocol used for the transfer of computer files between a client and server on a computer network
• Simple Network Management Protocol (SNMP) is an Internet Standard protocol for collecting and organizing information about managed devices on IP networks and for modifying that information to change device behaviour
• The Simple Mail Transfer Protocol (SMTP) is a communication protocol for electronic mail transmission
• Radio Free Ethernet (RFE) is a network audio broadcasting system, It consists of programs and tools that allow packets of audio data to be transmitted around a network

Notes:

• FTP is Transport layer protocol
• SNMP, SMTP is an application layer protocol

PSK(Phase Shift Keying):

In PSK (phase shift keying), binary 1 is represented with a carrier signal and binary 0 is represented with 180° phase shift of a carrier, i.e. the phase of the carrier signal is changed by varying the sine and cosine inputs at a particular time.

For binary ‘1’ → S₁ (A) = Acos 2π f_Ct

For binary ‘0’ → S₂ (t) = A cos (2πf_Ct + 180°) = - A cos 2π f_Ct

The Constellation Diagram Representation is as shown:

Important Point

ASK, PSK and FSK are signaling schemes used to transmit binary sequences through free space. In these schemes, bit-by-bit transmission through free space occurs.

FSK (Frequency Shift Keying):

In FSK (Frequency Shift Keying) binary 1 is represented with a high-frequency carrier signal and binary 0 is represented with a low-frequency carrier, i.e. In FSK, the carrier frequency is switched between 2 extremes.

For binary ‘1’ → S₁ (A) = Acos 2π f_Ht

For binary ‘0’ → S₂ (t) = A cos 2π f_Lt . The constellation diagram is as shown:

ASK(Amplitude Shift Keying):

In ASK (Amplitude shift keying) binary ‘1’ is represented with the presence of a carrier and binary ‘0’ is represented with the absence of a carrier:

For binary ‘1’ → S₁ (t) = Acos 2π f_ct

For binary ‘0’ → S₂ (t) = 0

The Constellation Diagram Representation is as shown:

where ‘I’ is the in-phase Component and ‘Q’ is the Quadrature phase.

From the given question, by using Pythagoras theorem, we get,

We need to find 
⇒AB² + BC² + CD² + DE²  
By putting the value from the equation (1), we get,
⇒ AC² + CD² + DE²
⇒ By putting the value from the equation (2), we get,
⇒ AD2 + DE²
By putting the value from the equation (3), we get,
⇒ AE²
Hence, the correct answer is AE².

The correct answer is Markanda.

Key Points

• Aruna is the ancient name of Markanda river.
  • ​The Markanda river flows in Sirmaur district, Ambala district, and Markanda town in Kurukshetra district.
  • The river flows through Haryana and Himachal Pradesh.
  • The Markanda river originates from the Shivalik hills.
  • The total length of the river is 120 km out of which 90 km is in Haryana.
  • A barrage at Jalbehra in Kurukshetra district has been built on the river.
  • The Begna river is the main tributary of the Markanda river.

Important Points

• The Sahibi River originates in Jaipur Distt. of Rajasthan and it passes through Alwar Distt. 
  • It drains into the Yamuna at Delhi, where its channelled course is also called the Najafgarh drain, which also serves as Najafgarh drain bird sanctuary.
• The Ghaggar River rises in Shivalik hills in the Solan district of Himachal Pradesh.
  • It enters Haryana from the Panchkula district.
  • And at the end, the river leaves Sirsa District of Haryana State and reaches Sri Ganga Nagar district, Rajasthan.
  • Kaushalya river is a left bank tributary of Ghaggar River.
• Yamuna River is the largest tributary of the Ganga River.
  • It originates from Yamunotri Glacier.
  • It meets the Ganges at the Sangam (where Kumbh Mela is held) in Prayagraj, Uttar Pradesh after flowing through Uttarakhand, Himachal Pradesh, Haryana and Delhi.
  • Length: 1376 km
  • Important Dam: Lakhwar-Vyasi Dam (Uttarakhand), Tajewala Barrage Dam (Haryana) etc.
  • Important Tributaries: Chambal, Sindh, Betwa and Ken.

As 'Forest' and 'Vivarium' have the same meaning in the same 'Sea' and 'Aquarium' have the same meaning.

From the common explanation, we get O sits exactly opposite to Q.

Hence, Option B is correct.

Final Arrangement:

Common Explanation:

References:

M is sitting fourth to the right of N.

P is sitting second to the left of N.

Inferences:

From the above reference, we get the following arrangements:

References:

Q is sitting third to the right of S.

Inferences:

From the above reference, we get two different cases:

Case 1:

Case 2:

References:

S is not sitting beside P.

T is sitting second to the left of O.

O is not sitting beside M.

Inferences:

So, from this case 1 will be eliminated as S is not sitting beside P.

Thus, the final arrangement,