Practice Test: Electronics Engineering (ECE)- 14
65 Questions MCQ Test GATE ECE (Electronics) 2023 Mock Test Series | Practice Test: Electronics Engineering (ECE)- 14
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A invest 1/3 part of the capital for 1/6 of the time, B invest 1/4 part of the capital for 1/2 of the time and C invest rest of the capital for rest of the time. Out of a profit of Rs. 23000, B’s share is?
Pipe A can fill a tank three times as fast as pipe B. If together two pipes can fill the tank in 48 min, the slower pipe alone will be able to fill the tank in:
Ratio of efficiency, A:B = 3:1
Ratio of times, A:B = 1:3
Total capacity = Total efficiency × Total time = 4 × 48 = 192 unit
Time taken by slower pipe
In the following question, out of the four alternatives, select the word opposite in meaning to the given word.
Gratuitous
Gratis means without charge; free. Thus, option B is the correct answer.
Find the area bounded between parabola and the line y2 = x,y = 2.
y = 2, y2 = x
⇒ x = 22 = 4
∴ Parabola and intersect at point (4,2)
The bar graph shows the number of employees working under the six different Departments (A, B, C, D, E, F) of a certain company. Study the diagram and answer the following questions.
If departments F and D are merged to create a new department G, then which department will have the least number of employees?
Employees in department A = 25
Employees in department B = 6
Employees in department C = 10
Employees in department E = 15
Employees in department G = 8
∴ Department B has the least number of employees
In the following question, a sentence is given with a blank to be filled in with an appropriate word. Select the correct alternative out of the four and indicate it by selecting the appropriate option.
Confusion prevails in madrasas in Uttar Pradesh over the distribution of free NCERT textbooks as the academic session ____________ from August.
The answer is ‘has begun’ because we use Present Perfect Tense, if the action is important and not the time of action or an action that has recently finished. Thus, option D is the correct answer.
A sum of Rs.400 amounts to Rs.480 in 4 years. What will it amount to if the rate of interest is increased by 2 % for the same time?
We know that,
Now rate is increased by 2 %
So, new rate is 7%
New Amount = S.I + P = 112 + 400 = Rs.512
The question below consists of a set of labelled sentences. These sentences, when properly sequenced form a coherent paragraph. Select the most logical order of sentences from among the options.
P: The Information and Broadcasting Ministry plans to conduct an independent study to gauge the impact of government advertisements on people.
Q: The advertisements are carried on various platforms, including print and visual media.
R: The Directorate of Advertising and Visual Publicity (DAVP) is the nodal agency of the government for advertising on behalf of the various ministries.
S: The initiative comes ahead of the Lok Sabha election in 2019 for which the government is expected to reach out to the people and highlight the works done by it in the past 4 years.
The paragraph talks about the plans and advertisements of The Information and Broadcasting Ministry, which is given in sentence P.
P is the first statement. The word ‘initiative’ given in the sentence S is talking about the plans.
Hence, S must follow P. Now, the introduction the advertising agency is given in the sentence R, which must be the next statement.
Thus, the sequence after rearrangement is PSRQ and option B is the correct answer.
In the following question, some part of the sentence may have errors. Find out which part of the sentence has an error and select the appropriate option. If the sentence is free from error, select 'No error'.
The gold foil used liberal (1)/ in Thanjavur paintings serves (2)/ many objectives that makes the painting more attractive. (3)/ No error
The error is in part (1) of the sentence. Change ‘liberal’ to ‘liberally’ because in this sentence it is in adjective form while the proper usage of liberal is in its adverb form i.e. ‘liberally’ as it qualifies the gold foil here.
A, B and C can do a job in 6 days, 12 days and 15 days respectively. C works till 1/8 of the work is completed and then leaves. Rest of the work is done by A and B together. Time taken to finish the remaining work by A & B together is how much?
Remaining work
∴ Time taken in doing 7/8 part of work
Given a network with values of components depicted in the figure.
Find the sum of current through 5Ω and 4Ω resistor.
Then, we can apply mesh analysis to solve it.
Mesh 1: 15I1 - 10I2 = 5
Mesh 2: -10I1 + 20I2 = 10
On solving them, we have
I1 = 1A
I2 = 1A
Thus, sum = 2A
The number of possible distinct Boolean expression of 3 variables will be ___________.
Given the following system
T{X[n]} = X[n] + 3u[n+1]
Where u[x] represents unit step functions-
Which of the following is a correct representation of the system?
T{X2[n] + X1[n]} = X1[n] + X2[n] +3u[n+1]
And
T{X1[n]} = X1[n] + 3u[n+1]
T{X2[n]} = X2[n] + 3u[n+1]
Since,
T{X2[n] + X1[n]} ≠ T{X1[n]} + T{X2[n]}
Thus, system is non linear.
T{X[n-no]} = X[n-no] + u[n+1]
≠ y[n-no]
Thus, system is Time Variant.
Consider the following sentences regarding a constant signal.
1) A constant signal is a periodic signal because it has no fundamental period.
2) A constant signal is an anti periodic signal because it never repeats itself.
List-1 (pole location) with list-2(shown constant amplitude with impulse response).
Select the correct answer using the codes given below.
If we plot A then it is similar to the 4 which is followed by equation K1+K2e-at+K3eat
For B
If we plot B then it is similar to the 1 which is followed by equation (sinat + sinbt) u(t)
For C
If we plot c then it is similar to the 3 which is followed by equation eatsinbt u(t)
For D
If we plot D then it is similar to the 2 which is followed by equation sinat u(t)
In the root locus for open loop transfer function is 
the ‘break away/in points are located at
s2 + 8s + 15 + K(s+6) = 0
In a PCM system, if the code word length is increased from 6 to 8 bits, the signal to quantization noise ratio improves by the factor
When word length is 8
Thus it improves by a factor of 16.
The op-amp configuration shown below has following transfer function 
The feedback resistance used has 1.5 MΩ , the value of capacitance will be ___________ µf.
A semiconductor sample at room temperature has intrinsic concentration of 2.5 X 1017 /m3. After doping what will be the minority carrier concentration if the majority carrier concentration is given as 5.5 X 1021 /m3.
n1p1 = n2p2 = ni2For Intrinsic Semiconductor, n = p = ni2 and as per questions before doping n1p1 = ni2
Therefore,
An LED is connected as shown in figure. It should glow when V1 is at logic ‘0’ state (0.2V). Calculate R( in Ω). (Assume in active state current as 15 mA)
By KVL
5 - 1.7 - 0.2 - (15 + R)15mA = 0
3.1/ (15 × 10-3) = 15 + R
R = 191.67 Ω
For an ideal p-channel MOSFET, μp = 300cm2/v-s, W = 15μm, L = 1.5μm, tox = 300A, Vt = -0.7V. If the transistor is non-saturation region at VSD = 0.5V, then what is the Transconductance gm?
where
Cox = (3.9 x 85 x 1014)/(300 x 1010).
Cox = 1.15 x 10-7 F/m2. Also, gm = ∂(ID)/∂(VSG).
On substituting and solving, gm = 0.172mS.
In the circuit shown below express the current I0 in terms of Vi
As per virtual short,
From the circuit given below, find out the operating region of the transistors T1 and T2
(VTH = -0.4)
VSD = VS – VD = 1.5 – 0 = 1.5 V
VSD(sat) = VSG + VTH = (1.5 − 0.5) − 0.4
= 1 – 0.4 = 0.6V
Here, VSG > (VTH) & VSD > VSD(sat)
So, T1 is in Saturation region
Similarly for T2,
VSD = VS − V0 = 0.9 − 0.9 = 0
VSD(sat) = VSG + VTH = (0.9 − 0) − 0.4 = 0.5V
Here VSD < VSD(sat) & VSG > (VTH)
∴ T2 in linear region
In the circuit shown below, assume that the voltage drop across a forward biased diode is 0.7 V. The thermal voltage V1 = kT/q = 25 mV. The small signal input Vi = Vp cos(ωt) where Vp = 100 mV.
The bias current IDC through the diodes is
Hence option A is correct
A communication channel having AWGN characteristics is operating in such a way that SNR >> 1. The bandwidth of signal being transmitted is B and capacity C1. Determine the capacity of channel if a signal with half the bandwidth is transmitted through the same channel with same quality.
Now, B2 = 0.5 × B1. SNR is same, thus C2 would be,
A three stage amplifier with identical stages with lower cut-off frequency per stage 'f1' is given overall negative feedback. Depending on the overall gain, the system may oscillate at a low frequency fc given by
Consider the differential equation 3yn(x) + 27y(x) = 0 with initial conditions y(0) = 0 and y'(0) = 2000. The value of y at x = 1 is
Auxiliary equation, 3m2 + 27 = 0 ⇒ m2 + 9 = 0
⇒ m = 0 + 3i
yc = c1 cos 3x + c2 sin3x and yp = 0
∴ yc = c1 cos 3x + c2 sin 3x
y(0) = 0 ⇒ c1 + 0 = 0 ⇒ c1 = 0
∴ y = c2 sin3x
y = 3c2 cos3x
y′(0) = 2000 ⇒ 2000 = 3c2 ⇒ c2 = 2000/3
Here in sin 3, 3 is in radian as it is a value.
Let z be a complex variable. For a counter-clockwise integration around a unit circle C, centred at origin, 
the value of A is
z = 4/5 lies circle
In the given circuit, the equivalent impedance of the circuit between terminals A-B is
KCL at node A
The ROCs of different impulse responses are shown below. Which of the following impulse responses are stable?
The current mirror circuit acts as a _________.
The current being mirrored can be a constant current, or it can be a varying signal dependent upon the requirement and hence the circuit.
Hence it acts as a current regulator circuit.
The given signal is the sum of two-real exponentials:
What will be the region of convergence (ROC) in this case?
The z-transform will be,
Now, for convergence of X(z), both sums must converge. This requires,
Thus, the region of convergence is |z| > 1/2
In the circuit given below, if current through the resistance R is given by I = 1 + 5 cos 2t , then find the value of R.
We have, VA = IR = R(1 + 5 cos 2t)
Separating the AC and DC parts, we get, Vddc = R,Vdac = 5 R cos 2t
Idc = 1,Iac = 5 cos 2t
Taking into account the DC part and eliminating the AC part, we get,
VAde = supply dc = 1V
Comparting it with the previous equation we have R = 1Ω
A ‘dual slope integrating type’ A to D converter, converts analog voltage of 15 volt to (11001010)2. If the clock frequency is 2 MHz, the total conversion time is _______ μsec.
= 2n+1 × TCLK
= (28+1) × ½ usec
= 28 usec
= 256 usec
The open loop transfer function of a unity negative feedback control system is given by 
Find the value of K for the phase margin of the system to be 450 is
If ωg is the gain crossover frequency
Consider a binary channel shown below, the value of P(Y1) and P(Y2), if P(X1) and P(X2) are 0.3 and 0.7 respectively are
Channel matrix
After solving, we get P(Y) = [0.135, 0.865]
The random process Z(t) is defined as Z(t) = X +Y; where X and Y are independent random variables. X is uniformly distributed on (-1, 1) and Y is uniformly distributed on (0, 1)
The auto correlation function of Z(t) is Rz (0) is
∵ X and Y are independent
Rz(0) = E[X2] + E[Y2]
Consider the following program for an INTEL 8085 microprocessor
1) MVI A, 92 H
2) ORA A
3) JP PORT
4) XRI 66 H
5) PORT OUT F2 H
6) HLT
The content of output port F2 H after execution of the program is
1) A = 92H
2) A = 92H Parity flag = 0
3) Checks for parity
4) A = 92 ⊕ 66 = F4H
5) Send the content of accumulator to port F2H
⇒ Content of output port F2 H = F4
Consider a GaAs sample at T = 300 K. Let the Hall effect device is fabricated with the following geometry: d = 0.01cm, W = 0.05 cm and L = 0.5 cm. The electrical parameters are : Ix = 2.5 mA, Vx = 2.2 V and Bz = 2.5 × 10-2 tesla. The hall voltage is VH = -4.5 mV. What will be the resistivity of the sample?
Majority carrier concentration, n is given by
In Hall Effect, mobility μn is given by
= 8182cm2/V−sec
Hence, we obtain the conductance as
σ = eμnn
= 1.6×10−19× 8.68 × 1014 × 8182
= 0.88Ω - cm
Consider an abrupt PN junction (at T = 300 K) shown in the figure. The depletion region width Xn on the N-side of the junction is 0.2 μm and the permittivity of silicon (εsi) is 1.044 x 10-12 F/cm At the junction, the approximate value of the peak electric field (in kV/cm) is _________.
Given an arrangement of memories
1) 8 kB ROM
2) 8 kB RAM – 1
3) 4 kB RAM -2
4) Free space of memory of 4 KB after ROM.
Then the value of chip select (active low) for RAM-2 is ________?.
ROM: 8 kB ⇒ P = 13, N = 8
Free space: 4 kB ⇒ P = 12, N = 8
RAM -1: 8 kB ⇒ P = 13, N = 8
RAM -2: 4 kB ⇒ P = 12, N = 8
Rom:
Free space:
RAM 1:
RAM -2:
Same as free space
0000 to oFFF.
The arrangement:
For an abrupt p-n junction diode, the doping concentrations on p-side is 9 × 1016 cm3 and an n-side is 5 × 1016/cm3. The diode is in reverse bias operating mode with total depletion width of 4 μm. Then the depletion width on n-side is __________ μm.
Hence here maximum frequency terms have frequency
The input impedance of lossless λ/8 transmission line with other end short circuited is given by ______. (Assume characteristic impedance Zo = 50Ω)
For a certain number system (51/4)b = (13)b is true
Then b is:
(51/4)b = (13)b
i.e., (51)b = (4)b x (13)b
∴ 5b + 1 = (4) · (b + 3) = 4b + 12
b = 12 - 1 = 11
Hence b = 11
Following system of equation has
4x + 5y + 9z = 4
2x + y + 3z = 2
X + 2y + 3z = 1
Writing the Augmented matrix [A : B.
Rank (A) = Rank(A : B) = 2 < 3="" (number="" of="" />
Hence the given system of equations have infinite solution.
A 400W, 220V bulb is supplied with 110V; then the power consumed by the bulb will be____?
Hence option C is correct.
For the PNP transistor circuit, given RB = 500kΩ, VCC = 12 volt, ICBO = 18 μA, β = 60. The value of ICQ = ____ mA?
(Assume |VBE| = 0.7 V)
By KVL to Base-emitter loop 0 + 0.7 + (500k) IBQ - 12 = 0
Now we have
ICQ = βIBQ + (1 + β) ICBO
∴ ICQ = (60)(22.6) + (1 + 60) (18)
= 2.454 mA
A waveform of second derivative of a certain signal is shown below.
The Fourier transform of given signal f(t) is.
We have
taking laplace transform,
Hence option (C)
For the signal 
The ratio of power content in 1st harmonics and 8th harmonics is
the period of signal is = T1 = 8T2
i.e., T = 1
∴ ωm = 2πrad/sec - ... Fundamental frequency,
Now
Comparing with
Power in 8th harmonic
Let Q√γ be the BER of a BPSK system over an AWGN channel with two-sided noise power spectral density N0/2. The parameter γ is a function of bit energy and noise power spectral density.
A system with two independent and identical AWGN channels with noise power spectral density N0/2 is shown in the figure. The BPSK demodulator receives the sum of outputs of both the channels.
If the BER of this system is Q b√γ, then the value of b is _____________.
Function of bit energy and noise
Counterllation diagram of BPSK
Channel is AWGN which implies noise sample as independent
Let 2x + n1 + n2 = x' + n'
where x' = 2x
n′ = n1 + n2
Now Bit error rate
E1 is energy in x2
is PSD of h1
E1 = 4E[as amplitudes are getting doubled]
= N0[ independent and identical channel
An LTI system with transfer function H(f) is to be obtained in such a manner that it can generate a random signal x(t) having autocorrelation function Rx(τ) = 3ɳe-3|τ|, by passing white noise n(t) having PSD Sn(f) = ɳ/2 watts/Hz. What would be the suitable transfer function H(f)?
We have autocorrelation function
Then spectral density at output is given by
This will give
We have a relation while filtering through an LTI system given by Sx(f) = |H(f)|2Sn(f)
This relation would give impulse response the LTI system as
A biliary symmetric channel (BSC) has a transition probability of 1/8. If the binary transmit symbol X is such that P(X = 0) = 9/10, then the probability of error for an optimum receiver will be
Py = 1)/(x = 0 P(x = 0)/Py = 1)/(x = 1P(x = 1)
P(x = 0) = 9/10
P(x = 0)+P(x = 1) = 1 So P(x = 1) = 1/10
Transition Probability is nothing but Probability of changing the value from 0 ta 1 and 1 to 0
i.e.
P(0/1) = P(1/0) = 1/8
Pe = P(x = 0)Py = 1)/(x = 0 + P(x = 1)Py = 0)/(x = 1
= (9/10) × (1/8) + (1/10) × (1/8)
= 1/8
But this nat the Pe for optimum Receiver. So,Pe = 1 − Pc
where Pc is Probability of correct detection Use MAP detection Rule:
The received symbal m = m1 if P(y/x1)P(x1)/P(y/xj)P(xj) ≥ 1
This is obtained from following:
Pc(m = mi) = P(mi/y)⋅P(y)=P(xi/y)⋅P(y)
M is the estimated output signal Using Baye's theorem
P(xi/y)⋅P(y)=P(y/xi)⋅P(xi)
when o/py = 0 then Py = 0)/(x = 0P(x = 0)/Py = 0)/(x = 1P(x = 1)
= (7/8⋅9/10)/(1/8⋅1/10) = 63 ≥ 1
So when y = 0, m = 0
Similarly when y = 1 then =(1/8⋅9/10)/(7/8⋅1/10) = 9/7 ≥ 1
So when y = 0 then alsa
So if we choose the rule such that the message signal is 0 for y =0 (or) 1 then the Probability of error is nothing but P(x = 1) = (1/10)
If an electric field is given as:
Determine the work (in nJ) that is to be done in moving a 3μC charge along this path if the path is located at P(-0.3, -4, 0.6).
(Differential because charge is moved through ΔL).
Therefore,
= -Q[-8y2z+40xyz -24xy2] x 10-6 (in μm)
Now it is given that Q = 3 x 10-6
So, we get: dW= -3 x 10-6 [-8y2z +40xyz -24xy2] x 10-6
Now, we have point P = (-0.3, -4, 0.6)
x = -0.3, y = -4 and z = 0.6
Therefore, dW= -3 x 10-6[ (-8 x (-4)2 x 0.6) +40(-0.3) (-4)(0.6) - 24(-0.3)(-4)2] x 10-6
= 4.89 x 10-10 J.
A 15 A source is operating at 100 MHz and feeds a Hertzian dipole of length 6mm situated at origin of space, determine electric field at P (4, 300, 900)
since it is a current source driving a hertzian dipole, We need to get Magnetic field intensity and then Electric field intensity as,
Here, I0 = 15A,r = 4m,η = 377Ω, dl = 6mm,θ = 30∘ and β can be determined,
In SSB modulation system, if signal (cos 30t) is modulated with a carrier signal of frequency 850 rad /sec, then the expression for the lower side band is
m(t) = cos 30t
m(t) = cos(30t − 90∘) = sin30t
For lower side band
s(t) = cos30t cos 850t + sin 30t sin 850t
=cos(850−30)t
= cos 820t
The divide by N counter is shown below. If initially Q0 = 0, Q1 = 1, Q2 = 0 the value of N is _______.
State is repeating after 5 clock pulses.
N = 5
The logic function implemented by the circuit given below is
Consider following connection to the memory
The accessible range of address from memory is
From the given data A00 − A11 are 12 address lines while remaining 4 address lines are decided by OR gate If A12A13A14A15 all are zero then only chip select will active.
Hence A12 − A15 will remain fixed at zero A15A14A13A12A11A10A9A8A7A6A5A4A3A2A1A0
Start address 0000000000000000
End address 0000111111111111
Accessible range is 0000− OFFF H
A box contains 2 identical bags A and B . Bag A contains 2 Red and 5 Green balls. Bag B contains 2 Red and 6 Green Balls. A person draws a ball at random. If the drawn ball was Red what is the probability that it was from bag A?
An intrinsic semiconductor bar of Si is doped with donor type impurity to the extent of 1 atom per 108 silicon atoms, then the resistivity of silicon crystal will be (atomic density of Si crystals =5 × 1022 atoms /cm3 and μn = 1300cm2N − sec)
∵ It is an n -type semiconductor 
The system of equations x - 4y + 7z = 12 , 3x + 8y - 2z = 10, 26 z - 8y = 6
Here, Rank [A : B] = Rank[A] = 3 = n
And n = 3
Therefore, the system has a unique solution.
(D2 − 5D + 6)y = ex cos x is a Partial differential equation whose solution is given by
y = C1e2x + C2e3x − Kex(3sin2x + cos2x)
The value of K is
Evaluate 
using Simpson’s ⅓ rd rule. Using h = 0.5
Simpson's ⅓ rd rule
In a certain animal husbandry weight of animals had a distribution as shown in table below:
Mean weight of animals in the husbandry is______.
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