GATE 2021 EC Electronics Engineering Detailed Solutions PDF

<br> # MADE EASY - GATE 2021 EC Electronics Engineering Detailed Solutions ## Exam Held on: 07-02-2021 (Afternoon Session) #### **MADE EASY** India's Best Institute for IES, GATE & PSUS **GATE 2021** **EC** **Electronics Engineering** **Detailed Solutions** **Exam held on:** **07-02-2021** **Aft...

<br> # MADE EASY - GATE 2021 EC Electronics Engineering Detailed Solutions ## Exam Held on: 07-02-2021 (Afternoon Session) #### **MADE EASY** India's Best Institute for IES, GATE & PSUS **GATE 2021** **EC** **Electronics Engineering** **Detailed Solutions** **Exam held on:** **07-02-2021** **Afternoon Session** ## SECTION - A ### GENERAL APTITUDE **Q.1** A diagram of points P and Q is shown with a line connecting them. The least number of squares that must be added so that the line P-Q becomes the line of symmetry is: - 7 - 4 - 6 - 3 **Ans. (b)** A second diagram of points P and Q is shown with a line connecting them. To make P-Q as a symmetric line, the minimum number of squares added = 6 **Q.2** A diagram of an equilateral triangle is shown. Corners are cut out of the triangle to produce a regular convex hexagon. The ratio of the area of the regular convex hexagon to the area of the original equilateral triangle is: - 5:6 - 2:3 - 4:5 - 3:4 **Ans. (c)** Let the side of the two larger equilateral triangle = a Then side of regular hexagon = a/3 Area of regular hexagon = 6x (a/3)^2 Area of triangle = √3/4 (a)^2 Required ratio = [6√3a^2/9] / [√3/4 (a)^2] = 6:9 = 2:3 ## SECTION - B ## TECHNICAL **Q.1** A message signal having peak-to-peak value of 2 V, root mean square value of 0.1V and bandwidth of 5 kHz is sampled and fed to a pulse code modulation (PCM) system that uses a uniform quantizer. The PCM output is transmitted over a channel that can support a maximum transmission rate of 50 kbps. Assuming that the quantization error is uniformly distributed, the maximum signal to quantization noise ratio that can be obtained by the PCM system (rounded off to two decimal places) is _____. **Ans. (30.72)** Vp-p = 2V; RMS[m(t)] = 0.1V, fm = 5kHz; C = 50kbps S/Nmax = ? S = RMS[m(t)]^2= (0.1)^2=0.01 C ≥ R, 50kbps ≥ nfs; fs = NR = 2fm = 10kHz, n≤5 Nmax = 5 No = 4^2 / 12 = 16/3 Amin = Vp-p / 2Nmax = 2V / 2*5 = 1/5 (No)min = Amin^2 / 12 = 1/5^2 / 12 = 1/150 = 3.25x10^-4 S/Nmax = S / (No)min = 0.01 / 3.25x10^-4 = 30.72 **Q.2** For a unit step input u[n], a discrete-time LTI system produces an output signal (2δ(n) + 8δ(n+1) + 8δ(n-1) ). Let y[n] be the output of the system for an input (1/2)^n *u(n). The value of y[0] is _____. **Ans. (0)** Let, h(n) = s(n) - s(n-1) = 2δ(n+1) + δ(n)+ δ(n-1) - 2δ(n) - δ(n-1) - δ(n-2) = 2δ(n+1) - δ(n) - δ(n-2) x1(n) = (1/2)^n * u(n), then output y1(n) = x1(n) * h(n) y1(n) = [1/2]^n * u(n) * [2δ(n+1) - δ(n) - δ(n-2)] y1(n) = [1/2]^n * u(n) * 2δ(n+1) - [1/2]^n * u(n) * δ(n) - [1/2]^n * u(n) * δ(n-2) y1(n)= 2[1/2]^(n+1)u(n+1) - [1/2]^n * u(n) - [1/2]^(n-2) * u(n-2) y1(n) = 2[1/2]^(n+1) u(n+1) - [1/2]^n u(n) - [1/2]^(n-2) u(n-2) y1(0) = 0-1-1 y1(0) = 0-2 y1(0) = -2 **Q.3** The block diagram of a feedback control system is shown in the figure. A block diagram with input X(s), output Y(s) and feedback loop H is shown. It contains two blocks labeled G1 and G2. The transfer function Y(s)/X(s) of the system is: - (G1+G2) / (1+GH) - (G1 + G2 + G1G2H) / (1+G1H) - (G1 + G2) / (1+G+H+G2H) - (G1 + G2) / (1+ G1H + G2H) **Ans. (a)** R = [G1(1-0) + G2(1-0)] / [1 - (-GH)] R = (G1+G2) / (1+GH) **Q.4** Consider two 16-point sequences x[n] and h[n]. Let the linear convolution of x[n] and h[n] be denoted by y[n], while z[n] denotes the 16-point inverse discrete Fourier transform (IDFT) of the product of the 16-point DFTs of x[n] and h[n]. The value(s) of k for which z[k]=y[k] is/are: - k = 0,1,2,15 - k = 0 and k = 15 - k = 15 - k = 0 **Ans. (b)** If two 'N' point signals x(n) and h(n) are convolving with each other linearly and circularly then, y(k) = z(k) at k = N-1 Where, y(n) = Linear convolution of x(n) and h(n) z(n) = Circular convolution of x(n) and h(n) N = 16 (given) Therefore, y(k) = z(k) at k = N-1 = 15 **Q.5** If (1235)x = (3033)y, where x and y indicate the bases of the corresponding numbers, then: - x=7 and y=5 - x=6 and y=4 - x=8 and y=6 - x=9 and y=7 **Ans. (b)** x^3 + 2x^2 + 3x + 5 = 3y^3 + 3y + 3 Option (b) will satisfy the equation. **Q.6** A silicon P-N junction is shown in the figure. The doping in the P region is 5x10^16 cm-3 and doping in the N region is 10×10^16 cm-3 . The parameters given are: - Built-in voltage (Φbi) = 0.8V - Electron charge (q) = 1.6 × 10^−19 C - Vacuum permittivity (∈₀) = 8.85 × 10^−12 F/m - Relative permittivity of silicon (∈si) = 12 A diagram of a silicon P-N junction is shown with P region 1.2 µm wide and N region 0.2µm wide. The magnitude of reverse bias voltage that would completely deplete one the two regions (P or N) prior to the other (rounded off to one decimal place) is ____ V. **Ans. (8.239)** NA = 5 × 10^16 cm^-3, ND = 10 × 10^16 cm^-3, Φbi=0.8V , q = 1.6 x 10^-19 C , ∈₀ = 8.85 × 10^-12 F/m = 8.85 × 10^-14 F/cm , ∈si = 12 Doping on both sides is comparable, so smaller region would deplete first. So, depletion region width on N-side = x = 0.2 µm = 0.2 x 10^-4 cm x = √2∈si(NA/ND)(Φbi + VR) / (q(NA + ND)) 0.2 x 10^-4 = √2×12×8.85×10^-14(5 × 10^16/ 10 × 10^16)(Φbi + VR) / (1.6×10^-19(15 × 10^16)) V = 9.039-0.8 VR = 8.239V **Q.7** An 8-bit unipolar (all analog output values are positive) digital-to-analog converter (DAC) has a full-scale voltage range from 0 V to 7.68 V. If the digital input code is 10010110 (the leftmost bit is MSB), then the analog output voltage of the DAC (rounded off to one decimal place) is ____ V. **Ans. (4.5)** Vps = 7.68V , n = 8 bit Resolution (k) = Vps/2^n-1 = 7.68/2^8-1 = 0.03 VDAC = k * {Decimal equivalent} = 0.03 * {150} = 4.5V **Q.8** For the circuit with an ideal Op-Amp shown in the figure, VREF is fixed. A circuit with an ideal Operational Amplifier is shown. It contains an input voltage Vin, an output voltage Vout, and VREF. Resistors R1 and RF are connected as feedback to the Op-Amp. If Vout = 1 Volt for Vin = 0.1 Volt and Vout = 6 Volt for Vin = 1 Volt, where Vout is measured across R1 connected at the output of this Op-Amp, the value of RF/RIN is _____. **Ans. (b)** Vout = (Rin * Vout + Vin * RF) / (Rin+RF) = VREF * R2 / (R1+R2) For Vin = 0.1V and Vout = 1V, we get, 1 * Rin + 0.1 * RF = VREF * R2 / (R1+R2) For Vin = 1V and Vout = 6V, we get, 6 * Rin + 1 * RF = VREF * R2 / (R1+R2) Therefore, 1 * Rin + 0.1 * RF = 6 * Rin + 1 * RF -5 * Rin = 0.9 * RF RF/Rin = -5/0.9 = -5.55 **Q.9** A speech signal, band limited to 4 kHz, is sampled at 1.25 times the Nyquist rate. The speech samples, assumed to be statistically independent and uniformly distributed in the range -5 V to +5 V, are subsequently quantized in an 8-bit uniform quantizer and then transmitted over a voice-grade AWGN telephone channel. If the ratio of transmitted signal power to channel noise power is 26 dB, the minimum channel bandwidth required to ensure reliable transmission of the signal with arbitrarily small probability of transmission error (rounded off to two decimal places) is ___ kHz. **Ans. (9.25)** fm = 4kHz, fs = 1.25 * NR = 1.25 * (2 * fm) = 10kHz, S/N = 26dB = 10^2.6 n = 8 bits/sample C ≥ R Blog2(1+S/N) ≥ nfs , Blog2(1+10^2.6) ≥ 8 * 10k, B ≥ 9.25kHz Bmin = 9.25kHz **Q.10** Consider a superheterodyne receiver tuned to 600 kHz. If the local oscillator feeds a 1000 kHz signal to the mixer, the image frequency (in integer) is ___ kHz. **Ans. (1400)** fs = 600kHz, fLO = 1000kHz IF = fLO - fs = 400kHz fsi = fs + 2IF = 600k + 800k = 1400kHz **Q.11** If the vectors (1, 0, -1, 2), (7, 0, 3, x) and (2, 0, 3, 1.0) and R^3 are linearly dependent, the value of x is ____. **Ans. (8)** |(1, -1, 2) , (7, 3, x) , (2, 3, 1) | = 0 ⇒ 1(3-3x) + 1(7-2x) + 2(15) = 0, -5x = -40 ⇒ x = 8 **Q.12** Consider a carrier signal which is amplitude modulated by a single-tone sinusoidal message signal with a modulation index of 50%. If the carrier and one of the sidebands are suppressed in the modulated signal, the percentage of power saved (rounded off to one decimal place) is _____. **Ans. (94)** μ = 0.5 % of power saved = (P_c + μ^2 / 4) / (P_c + P_c(1+μ^2/4) / (P_c + μ^2 / 4)) = (4 + μ^2 / 4 ) / (4 + 2μ^2 / 4) = (4 + (0.5)^2 ) / (4 + 2 × (0.5)^2) = 0.944 = 94.4% **Q.13** An symmetrical periodic pulse train vin of 10 V amplitude with on-time TON = 1 ms and off-time TOFF = 1 µs is applied to the circuit shown in the figure. The diode D1 is ideal. A circuit is shown. It contains voltage source Vin, a parallel combination of capacitor with 20nF and resistor 500 kOhms, and a diode D1. The difference between the maximum voltage and minimum voltage of the output waveform vo (in integer) is ____ V. **Ans. (10)** V = 10V, Diode is ON A circuit with Vin = 10V, 20nF capacitor and 500kOhms resistor is shown, with current flowing through the diode. The capacitor charges up to 10V. Vin = 0V: diode is OFF A circuit with Vin = 0V, 20nF capacitor and 500kOhms resistor is shown, with no current flowing through the diode. Discharging time constant = RC = 10 msec (discharging >> TOFF) Capacitor discharges negligibly. In steady state : Vc = 10V, Vout = Vin - Vc = 10V - 10V = 0V When Vin = 10V, Vout = 10-10 = 0V When Vin = 0V, Vout = 0-10 = -10V Vout(max) - Vout(min) = 0-(-10)= 10V **Q.14** For the transistor M1 in the circuit shown in the figure, µnCox=100 µA/V2 and (W/L) = 10, where µn is the mobility of electron, Cox is the oxide capacitance per unit area, W is the width and L is the length. A circuit is shown. It contains a MOSFET labelled M1, with VDD = 3V, RD = 20kOhms and input voltage VGS. The channel length modulation coefficient is ignored. If the gate-to-source voltage VGS is 1V to keep the transistor at the edge if saturation, then the threshold voltage of the transistor (rounded off one decimal place) is ____ V. **Ans. (0.5)** µnCox=100 µA/V^2, W/L = 10, VDD = 3V, RD = 20kOhms, VGS = 1V IDS = µnCox/2 × (W/L) × (VGS-VT)^2 VGS = 1V, IDS = µnCox/2 × (W/L) × (VGS-VT)^2 = 100/2×10×(VGS - VT)^2 VDS = 3 - 20 × IDS = 3-20 × (100/2 × 10 × (VGs-VT)^2) VDS = 3-1000 (1-VT)^2 MOSFET operates in saturation if VDS ≥ VGS - VT VDS = VGS - VT VGS-VT = 3 - 1000(1-VT)^2 We take, 1 = 3-1000(1-VT)^2, 1-VT = x , 3-1000x^2=x 1000x^2 + x - 3 = 0 x = 1±√1+1200/2000, x = 1±11/2000, x = 0.5 and -0.6 VGS > VT , VT < VGS, VT < 1 . Therefore, VT = 0.5V **Q.15** Consider a rectangular coordinate system (x, y, z) with unit vectors âx, â, and âz. A plane wave traveling in the region z≥0 with electric field vector E = 10cos(2×10^8t + βz)ày is incident normally on the plane at z = 0, where β is the phase constant. The region z≥0 is in free space and the region z<0 is filled with a lossless medium (permittivity ∈₂ = ∈₀, permeability μ₂ = 4μ₀, where ∈₀ = 8.85 × 10^−12 F/m and μ₀ = 4π × 10^−7 H/m). The value of the reflection coefficient is ____. **Ans. (1/3)** E = 10cos(2n × 10^8 t + βz)ây for z≥0 (free space), ∈r2 = 1, μr2 = 4 for z<0 (medium) η₂ = 120π √(µ₂/∈₂) = 120π √(4/1) = 120π √4 η₁ = 120π √(µ₁/∈₁) = 120π √(1/1) = 120π Γ = (η₂ - η₁) / (η₂ + η₁) = 120π √4 - 120π / 120π √4 + 120π = 120π (2-1) / 120π (2+1) = 1/3 **Q.16** For an n-channel silicon MOSFET with 10 nm gate oxide thickness, the substrate sensitivity (∂VT/∂VBS) is found to be 50 mV/V at a substrate voltage |VBS| = 2V, where VT is the threshold voltage of the MOSFET. Assume that, |VBS | >> 2ΦB, where qΦB is the separation between the Fermi energy level EF and the intrinsic level Ei in the bulk. Parameters given are: - Electron charge (q) = 1.6 × 10^−19 C - Vacuum permittivity (∈₀) = 8.85 × 10^−12 F/m - Relative permittivity of silicon (∈si) = 12 - Relative permittivity of oxide (∈ox) = 4 The doping concentration of the substrate is ____ cm^-3. **Ans. (d)** tox = 10 nm = 10 x 10^-7 cm, ∂VT/∂VBS = 50mV/V = 50 x 10^-3 V/V , |VBS| = 2V, q = 1.6 x 10^-19 C, ∈₀ = 8.85 × 10^-12 F/m = 8.85 × 10^-14 F/cm, ∈si = 12, ∈ox = 4, |VBS | >> 2ΦB VT = Φms + √2∈si q NA (2ΦB - VBS) / Cox + 2ΦB |VBS| = |VSB| VT = Φms + √2∈si q NA (2ΦB + VSB) / Cox + 2ΦB ∂VT/∂VBS = √2∈si q NA / Cox * (1/ 2√(2ΦB + VSB)) + 0 50 x 10^-3 = √2x8.85×10^-14×12×1.6×10^-19 NA / (4×8.85×10^-14×(10×10^-7)) *(1/2√(2×2)) NA = 7.375x10^15 cm^-3 **Q.17** A standard air-filled rectangular waveguide with dimension a = 8 cm, b = 4 cm, operates at 3.4 GHz. For the dominant mode of wave propagation, the phase velocity of the signal is vp. The value (rounded off to two decimal places) of vp/c, where c denotes the velocity of light, is ____. **Ans. (1.198)** a = 8cm, b = 4cm, f = 3.4GHz f_c10 = c / 2a = 3×10^8 / 2(8×10^-2) = 1.875GHz Vp = c * √(1 - (f_c10 / f)^2) = c * √(1 - (1.875 / 3.4)^2)) = 1.198c Vp/c = 1.198 **Q.18** A box contains the following three coins. - A fair coin with head on one face and tail on the other face. - A coin with heads on both the faces. - A coin with tails on both the faces. A coin is picked randomly from the box and tossed. Out of the two remaining coins in the box, one coin is then picked randomly and tossed. If the first toss results in a head, the probability of getting a head in the second toss is ____. **Ans. (1/3)** Let P(H2) = Probability of getting head in second toss P(H1) = Probability of getting head in first toss P(H2 / H1) = P(H2∩H1) / P(H1) P(H1) = 1/3×1/2 + 1/3×1 + 1/3×0 = 1/2 To get head in second toss when head came in first toss, following cases can be made 1. Fair coin: Both heads coin = 1/3×1/2×1/2 = 1/12 Both tail coin = 1/3×1/2×0 = 0 2. Both head coin: Fair coin = 1/3×1×1/2 = 1/6 Both tail coin = 0 P(H2∩H1) = 1/12 + 1/6 + 0 = 1/6 P(H2/H1) = (1/6) / (1/2) = 1/3 **Q.19** In a high school having equal number of boy students and girl students, 75% of the students study Science and the remaining 25% students study Commerce. Commerce students are two times more likely to be a boy than are Science students. The amount of information gained in knowing that a randomly selected girl student studies Commerce (rounded off to three decimal places) is ___ bits. **Ans. (3.32)** P(S) = 1/2, P(C) = 1/4, P(B) = 1/2; Let P(B/S) = x, P(B/C) = 2x P(B) = P(S) P(B/S) + P(C) * P(B/C) , 1/2 = 1/2x + 1/4 * 2x x = 2/5 P(B/C) = 2x = 4/5 P(C/G) = P(C) * P(G/C) / P(G) P(C/G) = (1/4) * (4/5) * (5/4) = (1/4) P(C/G) = 1/4 From equation (i), P(G/C) = 1 - 1/4 = 1/4 = 5/20 I(C/G) = log2(P(G/C)) = log2(5/20) = log2(1/4) = -2 = 3.32 **Q.20** A real 2×2 non-singular matrix A with repeated eigenvalue is given as A = [x -3] [3 4] where x is a real positive number. The value of x (rounded off to one decimal place) is ____. **Ans. (10)** |λI - A| = |λ - x 3| = 0 | - 3 λ - 4| λ^2 - (4+x)λ + (4x + 9) = 0 b^2 - 4ac = 0 (4+x)^2 - 4(4x+9) = 0 16 + x^2 + 8x - 16x - 20 = 0 x^2 - 8x + 20 = 0 x = 8±√64+80 / 2 = 8±√144 / 2 = 8±12 / 2 x = 10 x = -2 **Q.21** The electrical system shown in the figure converts input source current is(t) to output voltage v(t). A circuit is shown. It contains an input current is(t), a voltage source with v(t) across a 1F capcitor and a 1H inductor. There is a 1 ohms resistor connected in series with the inductor. Current i1(t) in the inductor and voltage v(t) across the capacitor are taken as the state variables, both assumed to be initially equal to zero, i.e., i1(0) = 0 and v(0) = 0. The system is: - Completely observable but not state controllable - Completely state controllable as well as completely observable - Neither state controllable nor observable - Completely state controllable but not observable **Ans. (c)** is = V + v(t) V = -V + is VL = L*İ1 VL = (is - i1) L*i1 = is - i1 i1 = -İ1 + is V = v(t) x = A*x + B*u x = [ -1 0; 0 -1 ] * x + [ 1 ; 0 ] * u y = [ 0 1 ] * x + [ 0 ] * u A = [-1 0 ; 0 -1] Qc = [BAB] =[1 -1] Qc = 0 Q = [CAC] = [1 1] Q = 0 **Q.22** The complete Nyquist plot of the open-loop transfer function G(s)H(s) of a feedback control system is shown in the figure. The Nyquist plot with ReG(s)H(s) along the x-axis and Im(G(s)H(s) along the y-axis is shown. The point (-1 , j0) lies on the graph. If G(s)H(s) has one zero in the right-half of the s-plane, the number of poles that the closed-loop system will have in the right-half of the s-plane is ____. **Ans. (*)** The given Nyquist plot is not matched according to the data given in the question. **Q.23** Consider the circuit with an ideal Op-Amp shown in the figure. A circuit is shown with an ideal operational amplifier. It contains an input Vin, an output Vout, and VREF. There are three resistors R1, R and RF connected. The negative terminal of the Op-Amp is connected to VREF and the positive terminal is connected to the inverting input. Assuming |VN| << |Vcc| and |VREF| << |Vcc|, the condition at which Vout equals to zero is: - Vin = 0.5 VREF - Vin = 2 + VREF - Vin = 2 VREF - Vin = VREF **Ans. (d)** For ideal op-amp: V- = V+ = 0 KCL at node V: (Vin - 0)/R - (VREF - 0)/R - (Vout - 0)/RF = 0 Vout / RF = (Vin - VREF) / R Vout = RF /R * (VREF - Vin) **Q.24** A sinusoidal message signal having root mean square value of 4 V and frequency of 1 kHz is fed to a phase modulator with phase deviation constant 2 rad/volt. If the carrier signal is c(t) = 2cos(2π106t), the maximum instantaneous frequency of the phase modulated signal (rounded off to one decimal place) is ____ Hz. **Ans. (1011313.7)** Vrms = 4V, fm = 1kHz, K = 2 rad/volt, c(t) = 2 cos(2π× 10^6t) Vpeak = √2 × Vrms = √2 × 4V m(t) = 4√2 sin(2π 10^3t) d/dt(fi)max = fc + K * d/dt(m(t)) d/dt(m(t)) = 4√2 ×(2π×10^3) cos(2n × 10^3t) d/dt(m(t))max = 4√2×2π×10^3 (fi)max = 10^6 + 2π * (4√2×2×10^3) / (2π) = 1011313.7Hz **Q.25** Consider a polar non-return-to-zero (NRZ) waveform using +2V and -2V for representing binary '1' and '0' respectively, is transmitted in the presence of additive zero-mean white Gaussian noise with variance 0.4V^2. If the a priori probability of transmission of a binary '1' is 0.4, the optimum threshold voltage for a maximum a posteriori (MAP) receiver (rounded off to two decimal places) is ____ V. **Ans. (0.0405)** Y = X + N H1: X = +2V H0: X = -2V Var[N] = σ^2= 0.4V E[N] = 0 P(1) = 0.4 P(0)= 0.6 Opt. Vth by using MAP theorem Vin[a1 - a2] / σ^2 = ln(P(0)/P(1)) a1 = E[2+N] = E[2] + E[N] = 2 a2 = -2V = E[-2+N] = E[-2] + E{N] = -2 σ^2 = Var[Y] = Var[X+N] = Var[X] + Var[N] = 0 + 0.4 = 0.4 VTh = (2 + 2)/ (2 × 0.4) * ln(0.6/0.4) VTh = 4/0.8 * ln(1.5)= 0.0405 **Q.26** The propagation delays of the XOR gate, AND gate and multiplexer (MUX) in the circuit shown in the figure are 4 ns, 2 ns and 1 ns, respectively. A circuit is shown. It contains an XOR gate, two MUX gates (one with two inputs and one with four inputs) and five input signals P, Q, R, S and T. If all the inputs P, Q, R, S and T are applied simultaneously and held constant, the maximum propagation delay of the circuit is - 3 ns - 6 ns - 5 ns - 7 ns **Ans. (c)** For T = 0, the propagation delay is 2(XOR) + 2(MUX1) + 1 (MUX2) = 3 + 2 + 1 = 6ns For T=1, the propagation delay is 2 (XOR) + 2(MUX2) + 1 (MUX1) = 3 + 2 + 1 = 6ns **Q.27** Consider the two-port network shown in the figure. A circuit is shown. It contains two voltage sources V1 and V2 connected in series. The current I1 flows from the source V1 through a 1 ohm resistor and a 1 ohm resistor connected to source V2, and then further through another 1 ohm resistor back to source V1. The current I2 flows back from source V2 through a 3V dependant voltage source (3V2), a 1 ohm resistor connected to source V1, and then further through another 1 ohm resistor back to source