GATE Electronics practice paper - 1

ANSWER: Photo-lithography

Explanation:
Photo-lithography basically deals with the photo-engraving technique where the silicon wafer is coated with a photo-resist along with an exposure to UV light.

However, the exposed portions of resist are completely removed by keeping the unexposed areas at an intact or an integral level

ANSWER: Common Base Amplification Factor

Explanation:
The point of operation that moves on the characteristic curve for AC operation is basically represented as,
α _AC = Δ I_c / Δ I_E ( V_CB = constant)
where, α _AC is expressed as common base short amplification factor.
In DC mode of operation, collector current and emitter current are specified levels of current on the basis of majority carriers & hence can be relatively expressed as,
α _AC = I_c / I_E

ANSWER: Early Effect

Explanation:
The effective base width decreases as compared to metallurgical base-width because the collector junction is in reverse-biased mode.
This effect is also renowned as base -width modulation since effective base width exhibits variations in accordance to biasing conditions.

ANSWER: A1 & A3

Explanation:
Fermi-Level in P-type semiconductor occurs exactly above the acceptor energy level. However, by increasing the addition of impurity during the existence of Fermi-level effect results in the downward shifting of p-type in fermi-level energy (E_F) .Besides these, it also moves towards the centre with the rise in temperature level.

ANSWER: Only A1 is true

Explanation:
JFET comprises a wedge-shaped channel & gate to source operates only under the revere biased condition. The reverse biasing of gate to source junction increases the input resistance of JFET.
Hence, the channel width cross-sectional area get reduced upon the application of drain to source voltage. This ultimately increases the channel current density.

ANSWER: S1: Correct, S2: Correct, S3:Correct , S4: Correct

Explanation:
SiO₂ performs the major functions associated with an ion implantation process in IC fabrication process. It serves to be a dielectric, isolation as well as passivation layer depending upon the type of application.
It also allows the precise doping level of windows between SiO₂ with greater proficiency at ion implantation process.

ANSWER: Closed-loop Control System

Explanation:
In an open-loop system, the control characteristics are totally independent of the system output & hence are generally stable, while the output relies on the controlling action (input) along with the previous output
in a closed-loop control system. As a result, closed loop systems may sometimes become unstable.

ANSWER: Blocks in Parallel

Explanation:
According to the drawing procedure of block diagram, sequential block reduction priorities are assigned for those connected in series , parallel,minor feedback loops and then summing point shift.
However, the transfer function of the serially connected blocks is evaluated on the basis of multiplication while the transfer function of the blocks connected in parallel combination is evaluated on the basis of an algebraic addition.
Associative law is only applicable when two or more directly connected summing points are interchangeable.

ANSWER: Transmittance

Explanation:
Basically, signal flow graph is a graphical representation of a set associated with simultaneous equations & applicable for linear time invariant systems by reducing the time required for block reduction techniques.
Sine the transmittance is normally represented or written on the branch near the arrow by specifying the input-output relationship, it is also referred as a transfer function.

ANSWER: Summing & Take-off Points should be represented by separate nodes

Explanation:
According to the construction rule of signal flow graph, all summing & take-off points are absolutely represented by nodes.
Therefore, a single node is sufficient to represent the summing and take-off points,if the position of summing point is before the take-off point.
But, if the position of summing point lies after the take-off point, then both of them are represented with the assistance of separate nodes connected by a branch possessing unit value of transmittances.

ANSWER: Path Factor

Explanation:
Mason's Gain Formula is used to evaluate an overall transmittance (gain), which can be expressed as,

T = k=1∑ⁿ P_K Δ_K / Δ
where, P_k = forward path transmittance of kth path
Δ = graph determinant comprising closed loop transmittances & mutual interactions between non-touching loops.
Δ_K = path factor consisting of all isolated closed loops from the forward path in the graph.

ANSWER: It consists of at least one pole or zero at RHS of an imaginary axis

Explanation:
If the transfer function possesses at least one pole or zero at RHS of an imaginary axis, then the system is said to be in ' Non-minimum Phase while options (b) &(d) represent the conditions of all-pass and minimum phase systems/transfer functions respectively

ANSWER: B, A, F, C, E, D

Explanation:
Root locus is a graphical mechanism which is concerned to the plotting of roots of the characteristic equation in s-plane for several parameter values. It is mandatory to determine the root locus properly in proper sequential steps specified in option (c) so as to estimate the presence or location of any point on it by verifying the roots of characteristic equation with the assistance of angle and magnitude criteria.

ANSWER: zero & infinity

Explanation:
Ideally, the diode operates in forward-biased condition .This is so because the current flows in forward-biased direction only when the resistance across the voltage is zero. Without any form of obstruction in the flow of current, diodes operates in forward-biasing.
On the contrary, the diode has to pass through an infinite resistance in reverse-biasing mode, which ultimately prevent it from conduction.
However, these both operating conditions of an ideal diode enable it to act as a perfect switch.

ANSWER: Collector Current

Explanation:
Collector current is independent of collector voltage and depends only on the emitter current in the active region of BJT.
Therefore, the magnitude of collector current is slightly less than that of emitter current because the value of α_DC is less than one but, in fact, almost equal to unity.
These reasons enable the collector current to act as a constant form of current source in an active region of BJT

ANSWER: sine wave

Explanation:
In accordance to the configuration arrangement of push-pull amplifier, the centre-tapped secondary winding of transformer provides equal and opposite voltage to base circuits of the two transistors.
Thus , the supply voltage is connected between bases and centre tap. Transistor T₁ provides amplification to every half cycle of the signal by leading it to appear in the upper half of the primary winding of output transformer by the ultimate result of combining two collector currents.
This eventually generates the output in the form of sine wave.

ANSWER: A is true and B is false

Explanation:
Clipper circuits are applicable for selecting that portion of input wave which is located above or below the reference levels.
For the clipper circuit that clips the input voltage above the reference voltage , the necessary operational conditions are

1) If V_i < V_R, then diode is reverse-biased & does not operate or conduct Thus, V₀= V_i &
2) If V_i>V_R , then diode operates in forward biased condition & hence V_o = V_R

ANSWER: Increase in collector voltage increases the space charge width & decreases base width

Explanation:
Due to the phenomenon of early effect, the collector current increases slightly with voltage. This results in the reduction of base current.
However, increase in collector voltage also increases the space charge but reduces effective base-width. Therefore, there is less possibility for recombination within the base region.

ANSWER: Widlar Current Sources

Explanation:
Generally, it is mandatory for large resistors to maintain small currents of few micro ampere. Therefore, large areas are occupied by these large resistors on the IC chip. However, Widler current sources are preferred for replacing these large resistors.

ANSWER:

Explanation:
J-K Flipflop has an ability to overcome the drawbacks of S-R flipflops .
When both the inputs of JK flipflop are zero, the output is Q_n while when both the inputs are 1, the output is the complement of Q_n i.e (Q
_n).
When the inputs are in the form of '0' , '1' & '1','0' respectively, then the outputs are generated in accordance to the J input value; i.e, outputs are '0' & '1' respectively for both the conditions.

ANSWER: 3 kHz

Explanation:
Given:
Input frequency = 768 kHz
The output frequency of eight flipflops = (1 / 2⁸) x Input frequency
= (1 / 2⁸ ) x 768
= 768 / 256
= 3 kHz

ANSWER: Maximum Power Consumption

Explanation:
Transistors of ECL always operates in an active or cut-off regions & never goes in saturation region thereby eliminating the storage time.
ECL comprises difference amplifiers in addition to emitter followers. These emitter followers are utilized at the output of difference amplifiers for shifting the DC level by consuming maximum power at the requisite level.

ANSWER: B , C , E, F, A, D

Explanation:
When the number of variables increases beyond six or multiple form, it becomes very arduous to form the group and simplify the boolean expression.
However, it is possible to overcome this predicament by using Quine-Mc Clusky method for the simplification purpose of the boolean expression.
Therefore, Option (c) specifies the precise sequential order of the steps that are followed or undertaken in this method.

ANSWER: A & B

Explanation:
From the given statements, C & D are correct while A & B are incorrect ones. This is so because

In MROM, the user specified data is absolutely stored to the manufacturer of the memory, while PROM allows the data pattern to be programmed on electrical basis with an assistance of a special device called as PROM programmer. Once the data is programmed in PROM, it cannot be changed.

ANSWER: 0,1 & flip-flop is set

Explanation:
Set & Reset actions of S-R clocked flipflop are performed only on the basis of a clock pulse. So, when
S=0 & R =0, outputs are 1,1 ( output remains unchanged)
S=0 & R = 1, outputs are 1,0 ( Reset action of flipflop)
S= 1 & R=0, outputs are 0,1 ( Set action of flipflop)
S= 1 & R=1, outputs are 0,0 (output remains undefined since they ought to be 1)

ANSWER: 64

Explanation:
According to the given data,

t_nd = 10 ns & f = 16 MHz
Assume that 'n' number of flipflops are connected . Hence, the frequency can be expressed as,
f = 1 / n. t_d
16 x 10⁶ = 1 / (n x 10 x 10 ^-9 )

n = 1 / 16 x 10⁶ x 10 x 10^-9
= 1 / 16 x 10^-2
n = 6.25 ≈ 6
Thus, number of flipflops = 6
Number of MOD ripple counters = 2ⁿ = (2) ⁶ = 64

ANSWER: Loop

Explanation:
According to the properties of a loop, it ensures the presence of at least two branches in addition to exactly two paths between any pair of nodes in it .Besides these, the number of branches are always equal to the number of nodes and hence, the two branches are exactly incident at each node.

ANSWER: n^n-2

Explanation:
For any kind of network, the number of trees can be always determined on the basis of its branches and nodes. This can be implied as,
No. of trees = n^n-2
where, n represents the nodes & n-2 represents the number of branches in a tree

ANSWER: Gyrator

Explanation:
Gyrator is a four terminal device whose coefficient value is dependent on the Op-amp parameters as well as externally connected R,C values.
It exhibits a unique feature of impedance inversion phenomenon due to which the input impedance is capacitive for an inductive load impedance and vice-versa. In addition to this, simulation of equivalent inductance values is also a major function performed by a gyrator

ANSWER: A₁ & A₂

Explanation:
Only the statements A₁ & A₂ satisfy the properties of a tree. The statements A₃ & A₄ are not precise because a tree contains all the nodes on the graph and does not contain any closed path.

ANSWER: (4/5) RC

Explanation:
By replacing the voltage source in terms of internal impedance, the equivalent circuit is shown below.

From the above equivalent network, it is obvious that

R_eq = (R x 4R) / 5R
= 4R /5
Also, the time constant (τ) = R_eq x C
= (4 /5) RC

ANSWER: It absorbs 512 W

Explanation:
From the given network,
Power = V x I
= (2 Ix) x Ix
= 2 I²x

Since I_x = 16A

P = 2 x (16)²
= 512 W (absorption)
Hence, the total power of about 512W is absorbed in the given circuit.

ANSWER: A & B

Explanation:
No explanation is available for this question!

ANSWER: even signal

Explanation:
A signal can be said to be symmetric on Y-axis only when the symmetrical nature of curve obtained from the signal exhibits two equal portions of the curve exactly similar to each other over the Y-axis.
Since this kind of symmetry is shown by the signal has even (equal) form of distributions in terms of curve portions,amplitudes & time instants, they are generally referred as 'Even Signals'

ANSWER: x(-t) = -x(t) & x[-n] = -x[n]

Explanation:
The necessary condition of an odd signal is that it must be zero at t=o or n=o . Thus, it can be expressed as x(-t) = -x(t) & x[-n] = -x[n].
This implies that negative value of time instants are equal to the negativity of an entire term in an odd signal, while the negative value of time instants are equal to the positive values of an entire term in an even signal.
However, sin t is an odd signal while cos t is an even signal

ANSWER: Single Valued Property

Explanation:
The conditions which provide chances to the periodic function to get expanded in a convergent Fourier series , are renowned as Dirichlet Conditions.
Thus, the very first Dirichlet condition states that, the signal should have only the finite values over any time instant for a finite interval of time so the signal can be considered to be fourier transformable.
This condition implies the single-valued property that does not change or tend to increase over the infinite time intervals.

ANSWER: Y₁ only

Explanation:
The statement Y₁ is only precise because every signal should not be necessarily even or odd. It can also belong to some other category of signals. Also, the signal which is neither odd nor even odd is basically expressed as the addition (sum) of even and odd signals.

ANSWER: C & D

Explanation:
According to the property of energy and power signals, a signal is said to be an energy signal only when the value of normalized energy lies within zero and infinity i.e. normalized energy should be greater than zero but less than infinity.
Similarly, Power signals lie within the same range of zero and infinity. This implies that , a signal is known as a power signal only if the value of normalized average power is greater than zero but less than infinity.
Signals which do not satisfy any of these properties are generally known as neither energy signals nor power signals.

ANSWER: Time Shifting, Time Scaling & Time Reversal

Explanation:
In accordance to the properties of continuous time signals, when the stages of time shifting, scaling and reversal are to be performed,then it is always obligatory to initially perform time shifting, and then scaling and the reversal.
This is so because scaling can only be done after shifting the time signal. Time shifting property does not affect the amplitude of signal but it shows a slightly delayed version of original signal.
Eventually, once the signal undergoes time scaling, it can further undergo the time reversal phase, where the exact reflection of the time-scaled signal is generated about Y-axis.

ANSWER: r sin θ dФ

Explanation:
The locus of points which comprises similar powers corresponding to the two spheres is known as radical plane.
Therefore, the distance between two radical planes of the spheres is equal to the product of radius vector measured from Z-axis, sine angle between Z-axis & radius vector and the angle between X-axis and radius vector projection in XY plane.

ANSWER: 0 ≤ θ ≤ π , 0 ≤ Ф ≤ 2 π & r = constant

Explanation:
A sphere is formed only when the value of an angle vector originated due to Z-axis and radius vector lies between zero and π; and the value of angle vector between X-axis & radius vector projection over XY plane is greater than zero but less than or equal to 2π ,by keeping radius vector as a constant

ANSWER: Laplacian

Explanation:
A function is said to be harmonic only when its laplacian operator value ▼²f is zero. Similarly, a scalar field is also said to be harmonic only when its laplacian operator becomes zero despite the other operators.

ANSWER: By the Volume integral of the charge density

Explanation:
Guass's law states that the total flux over the outer closed surface is always equivalent to the total charge within the surface.
Similarly, Guass's law can also be defined in terms of volume integral which states that the total charge present over the closed surface is equal to the volume integral function of the charge density.

ANSWER: Only B is correct

Explanation:
Conductor develops an equal potential throughout the surface and the static electric field intensity within the inner portion of a conductor is zero. Thus, the properties of a conductor implies that the static electric field intensity over the surface of a conductor is directed equally and normal (perpendicular) to the surface.

ANSWER: Both A & B are true

Explanation:
The conductors in the lossless line are perfect and the dielectric medium that separates them is absolutely lossless.
Besides this, a transmission line where the value of an attenuation constant is independent of frequency & phase constant is linearly dependent on frequency.
Hence, it is obvious that a lossless line can also be a distortionless line but a distortionless line cannot be essentially a lossless line.

ANSWER: Only when time averages are equal to the ensemble averages

Explanation:
Random Processes can be generally and completely specified on the basis of ensemble and the time statistics since only PDFs are insufficient to describe them completely.
So, any random process possessing the time averages exactly equal to averages [Probability Distribution Factor PDF (f_x(t) (x)] which are derived from ensemble, then it is said to be an ergodic process.

ANSWER: A is true & B is false

Explanation:
A random process is said to be Stationary in Strict Sense (SSS) if the joint probability distribution factor remains invariant to the translation of time origin. On the contrary, it is said to be Stationary in Wide Sense if mean value m_x (t) is independent of time and the autocorrelation function Rx (t_k, t_i) depends only on the time difference (t_k - t_i).
Hence, it is obvious that all SSS are definitely WSS but all WSS are not necessarily SSS.

ANSWER: 1900 kHz

Explanation:
Given data :
I_F = 350 kHz
Carrier Frequency = f_c = 1200 kHz

We know that,
Image Frequency = 2 I_F + f_c
= 2x 350 + 1200
= 1900 kHz

ANSWER: 4.4 kHz

Explanation:
Given data :
Frequency Deviation = Δf = 10 x ( 2π 200) rad
Deviation ratio = D = [10 x ( 2π 200)] / [( 2π 200)]
Therefore, D = 10

Bandwidth of Frequency Modulated Signal is given by,
B = 2 (D + 1) f_m
= 2 (10 +1 ) x 200
= 2 (11) x 200
= 22 x 200
= 4400
= 4.4 kHz

ANSWER: R_XY , (t₁, t₂ ) = 0 for every t₁ and t₂

Explanation:
Two random X(t) & Y(t) processes are usually said to be mutually orthogonal only and only if R_XY, (t₁,t₂) becomes exactly equal to zero; where
t₁ & t₂ are the two time instants at which the processes are observed.

ANSWER: A & B

Explanation:
Non-uniform quantizer has has non-linear characteristics while the step-size consistently keeps on varying & depends on the amplitude of input signal.
This process of non-uniform quantization resemble to the mechanisms of passing the baseband signal through compressor and then eventually applying the compressed signal to a uniform quantizer.