Attempt the ISRO Electronics & Communication (ECE) 2015 previous year paper in real CBT exam mode or practice mode with verified answers and explanations.
All ISRO Electronics & Communication (ECE) papers
80 questions. Answers are shown only where the official answer key is available.
Topic: Semiconductor Devices
Electron mobility of the following intrinsic elements in descending order is:
Answer: A. GaAs, Ge, Si
Explanation: The value of mobility of electrons in GaAs, Si, Ge are 8500, 3800, 1300 cm²/V-sec respectively. Hence, GaAs > Ge > Si.
Topic: Semiconductor Devices
A sample of Si is doped with 10^17 donor atoms/cm³. Considering electron mobility in the doped Si 700 cm²/V-sec, the approximate resistivity of the doped Si is:
Answer: C. 0.1 Ω-cm
Explanation: N_d = 10^17 atoms/cm³, μ_e = 700 cm²/V-sec. Conductivity σ = N_d * q * μ_e = 10^17 * 1.6 * 10^-19 * 700 = 11.2 Ω^-1-cm^-1. Thus, resistivity ρ = 1 / σ ≈ 0.1 Ω-cm.
Topic: Transistor Circuits
Common-base current gain of a p-n-p bipolar transistor is 0.99. The common emitter current gain of the transistor is:
Answer: C. 99
Explanation: Common base current gain (α) = 0.99. Common emitter current gain (β) = α / (1 - α) = 0.99 / (1 - 0.99) = 99.
Topic: Optoelectronic Devices
The electrical conductivity of a semiconductor increases when a radiation of wavelength shorter than 1000 nm is incident on it. The band gap of the semiconductor is:
Answer: B. 1.2 eV
Explanation: Wavelength λ = 1000 nm = 1 μm. Bandgap energy E_g ≈ 1.24 / λ(μm) eV = 1.24 / 1 ≈ 1.2 eV.
Topic: Network Theory
The voltage across the resistor R is:
Answer: D. 1 / √2 ∠45° V
Explanation: Z_1 = (1 + j1) Ω = √2 ∠45° Ω and Z_2 = (1 - j1) Ω = √2 ∠-45° Ω. Using current division, I_R = [Z_1 / (Z_1 + Z_2)] * I = [√2 ∠45° / (√2 ∠45° + √2 ∠-45°)] * 1 = (1 / √2) ∠45° A. The voltage across R = 1 Ω is V_R = I_R * R = (1 / √2) ∠45° V.
Topic: Network Theory
The effective resistance faced by the voltage source is:
Answer: C. 3 Ω
Explanation: Since the parallel branches have the same voltage, (3i/4) * 4 = (i/4) * R, which gives R = 12 Ω. The equivalent resistance faced by the source is R_eq = 12 || 4 = 3 Ω.
Topic: Analog Circuits
Common emitter DC current gain of the transistor is 100. The current through the 10 V Zener diode (assuming V_BE of the transistor is 0.7 V) is:
Answer: A. 10.3 mA
Explanation: With β = 100 and V_BE = 0.7 V, applying KVL in loop 1 gives -10 + V_BE + I_E R_E = 0, so I_E = (10 - 0.7) / 10 = 0.93 A = 930 mA. Base current I_B = I_E / (1 + β) = 930 / 100 = 9.3 mA. Total current from supply I_1 = (20 - 10) / 0.5 kΩ = 20 mA. Therefore, Zener diode current I_Z = I_1 - I_B = 20 - 9.3 = 10.7 mA (closest given option is 10.3 mA).
Topic: Analog Circuits
The circuit is with an ideal operational amplifier with ±10 V supply. The output voltage is:
Answer: B. -400 mV
Explanation: Applying virtual ground V_A = V_B = 0 V. Nodal analysis at the inverting node B: (V_B - 40 mV)/2 kΩ + (V_B - 20 mV)/1 kΩ + (V_B - V_out)/10 kΩ = 0. Substituting V_B = 0 V yields V_out = -20 mV * 10 - 20 mV * 10 = -400 mV.
Topic: Network Theory
A step voltage of 1 V applied to the circuit at t = 0. The current through the resistor R just after t = 0 and at steady state are:
Answer: A. 100 mA, 50 mA
Explanation: At t = 0+, the uncharged capacitor acts as a short circuit (v_c = 0 V), making the equivalent resistance 10 Ω, so I_R(0+) = 1 V / 10 Ω = 100 mA. At steady state (t -> ∞), the capacitor acts as an open circuit, giving an equivalent resistance of 20 Ω, so I_R(steady state) = 1 V / 20 Ω = 50 mA.
Topic: Microprocessors & Microcontrollers
Which of the following addressing mode is not usable in XCH instruction of 8051 microcontroller?
Answer: D. Immediate Addressing
Explanation: Explanation not provided in source.
Topic: Information Theory & Coding
In a binary source, 0s occur three times as often as 1s. What is the information contained in the 1s?
Answer: D. 2 bit
Explanation: P(0) = 3 * P(1) and P(0) + P(1) = 1, which gives P(1) = 1/4. Information content contained in 1s is I = log_2(1 / P(1)) = log_2(4) = 2 bits.
Topic: Signals & Systems
Spectrum of a signal x(t), sampled at period T, is given by:
Answer: A. X_s(f) = ∑_{n=-∞}^{∞} X(f - n/T)
Explanation: When x(t) is sampled uniformly with period T, the sampled signal in frequency domain is given by X_s(f) = ∑_{n=-∞}^{∞} X(f - n/T).
Topic: Communications
A bandpass signal occupies the bandwidth 390 kHz to 410 kHz. What minimum sampling frequency would you use from the options given below, so as to avoid aliasing?
Answer: C. 41 kHz
Explanation: F_H = 410 kHz, F_L = 390 kHz. K = floor(F_H / (F_H - F_L)) = floor(410 / 20) = 20. Minimum sampling frequency BW = 2 * F_H / K = (2 * 410) / 20 = 41 kHz.
Topic: Digital Communications
A 12-bit ADC has input signal range of ± 1 V. The signal to quantization noise ratio if a sine wave signal with 0.25 V peak voltage is given as input is:
Answer: A. 61.8 dB
Explanation: For n = 12 bits, range is -1 V to 1 V (4096 levels). With a peak voltage of 0.25 V, only 1024 levels are utilized, corresponding to effective bits n' = 10. Signal to quantization noise ratio = 1.8 + 6 * n' = 1.8 + 60 = 61.8 dB.
Topic: Digital Communications
Identify the modulation schemes for the shown signal constellation diagram
Answer: B. MSK
Explanation: The constellation diagram showing points on a circle corresponds to MSK modulation scheme.
Topic: Communications
Which of the following modulation scheme requires minimum power for transmission?
Answer: A. QPSK
Explanation: Explanation not provided in source.
Topic: Communications
Which of the following channel coding scheme helps in correcting burst errors?
Answer: D. Reed-Solomon
Explanation: Reed-Solomon codes are based on univariate polynomials over finite fields. They are used to detect and correct multiple errors i.e. burst errors.
Topic: Communications
A 70 MHz carrier is QPSK modulated by a 1.544 Mbps T1 data stream. The transmitter employs a raised-cosine filter with α = 0.2. What is the transmitted bandwidth of the signal?
Answer: D. 926.4 KHz
Explanation: F_c = 70 MHz, R_b = 1.544 Mbps, α = 0.2. BW_min = (R_b / 2) * (1 + α) = (1.544 / 2) * (1 + 0.2) = 926.4 kHz.
Topic: Communications
If two tones f_1 and f_2 are amplified by a non-linear amplifier, which frequency components would be present in output?
Answer: D. nf_1 ± mf_2, where n and m are integers
Explanation: As input frequencies f_1 and f_2 are passed through non-linear filter so output signal is of the form V_o = (A_1 cos 2π f_1 t + A_2 cos 2π f_2 t). So, there will be multiple harmonics of frequencies depending on value of β. ∴ Output frequency = nf_1 ± mf_2 where n and m are integers.
Topic: Communications
A phase lock loop with a first order loop filter can track:
Answer: B. frequency offset
Explanation: Phase lock loop is used to track the change of frequency in frequency of input signal i.e. frequency offset.
Topic: Communications
The noise figure of a device is 2. If input SNR is 37 dB, what would be output SNR?
Answer: B. 34 dB
Explanation: Noise figure = 2. Noise figure (in dB) = 10 log_10 (2) = 3 dB. Noise figure (NF) = (S/N)_input / (S/N)_output. ∴ NF (in dB) = (S/N)_input,dB - (S/N)_output,dB. So, (S/N)_output,dB = 37 - 3 = 34 dB.
Topic: Communications
What is the limit of E_b/N_0, below which reliable communication is not possible?
Answer: B. -1.6 dB
Explanation: For reliable communication, limit for E_b/N_0 is E_b/N_0 = ln(2) = 0.693. E_b/N_0 (dB) = -1.6 dB. Thus, limit for reliable communication is -1.6 dB. If it falls below it errors will be very high.
Topic: Satellite Communication
A satellite communication link has uplink C/N_0 of 50 dB-Hz and downlink C/N_0 of 47 dB-Hz. What would be overall link C/N_0?
Answer: B. 45 dB-Hz
Explanation: (C/N)_uplink = 50 dB-Hz = 10^5; (C/N)_downlink = 47 dB-Hz = 10^4.7. 1/CNR_overall = 1/CNR_uplink + 1/CNR_downlink. 1/CNR_overall = 1/10^5 + 1/10^4.7. So, CNR_overall = 33386.05. CNR (dB) = 45.23 dB-Hz.
Topic: Satellite Communication
A signal transmitted from an Earth station bounces back from a geostationary satellite, with an altitude of 35864 km from the Earth surface. The range from the Earth station is 38200 km. The round trip time is about: (excluding processing delays and assuming c = 3 x 10^8m/s)
Answer: A. 255 ms
Explanation: Altitude (h) = 35864 km, C = 3 x 10^8 m/s, Range = 38200 km. As, T = Round-trip distance / Speed = (2 x 38200 km) / (3 x 10^8 m/s) = 255 ms.
Topic: Communications
1 mW in dBm is:
Answer: A. 0 dBm
Explanation: P(dBm) = 10 log_10 (P / 10^-3 W) = 10 log_10 (10^-3 / 10^-3) = 0 dBm.
Topic: Communications
Which of the following is not true?
Answer: C. Source coding/decoding is done to remove errors
Explanation: Source coding/decoding is used to efficiently use the number of message bits while channel coding is used to remove the errors in transmission.
Topic: Digital Circuits
The following Finite State Machine (FSM) is used to detect a particular pattern in input data stream. Whenever the pattern is matched at input, output is set to '1' or else output is cleared to '0'. For which of the following data stream, output goes to '1' twice?
Answer: C. 0011011010010101
Explanation: The given FSM corresponds to detector for 1101, as input sequence 0011011010010101 repeats 1101 two times. Hence output goes to '1' twice.
Topic: Digital Signal Processing
A digital signal processing system is described by the expression: y(n) = 2x(n) + x(n - 1) + 2y(n - 1). The system is:
Answer: D. An unstable IIR filter
Explanation: y(n) = 2x(n) + x(n - 1) + 2y(n - 1) => y(n) - 2y(n - 1) = 2x(n) + x(n - 1). Y(z)/X(z) = (2 + z^-1)/(1 - 2z^-1). As the pole lies at z = 2, which is outside unit circle so it is an unstable system. As the output y(n) consist of y(n - 1) i.e. finite pole, so given system is IIR filter. ∴ System is unstable IIR filter.
Topic: Digital Signal Processing
The rectangular window has relative side-lobe level:
Answer: A. -13 dB
Explanation: Explanation not provided in source.
Topic: Digital Signal Processing
Consider the discrete time signal x(n) = {1, 1, 1, 1, 0.5, 0.5}. y(n) = conv (δ(n - 1), x(n)) is:
Answer: C. x(n - 1)
Explanation: x(n) = {1, 1, 1, 1, 0.5, 0.5}. y(n) = conv(δ(n - 1), x(n)) = x(n - 1) (Using property of impulse function).
Topic: Digital Signal Processing
Consider a sequence x(n) = [2, 4, 6, 8, 0, 1, 3, 5, 7, 9]. Down-sample the sequence by 3 and then up-sample by 2. The resulting sequence is:
Answer: C. [2,0,8,0,3,0,9,0]
Explanation: Given x(n) = {2, 4, 6, 8, 0, 1, 3, 5, 7, 9}. Down-sampling by 3 gives x(3n) = {2, 8, 3, 9}. Up-sampling by 2 gives x(3n/2) = {2, 0, 8, 0, 3, 0, 9}.
Topic: VLSI Design
Power consumed in a CMOS circuit operating at frequency f is proportional to:
Answer: C. V_cc^2 f
Explanation: Power consumed in CMOS = (C * f * V_DD^2) / 2. Therefore, power consumed is proportional to V_DD^2 * f.
Topic: Electromagnetic Theory
The phase velocity of a plane wave given by E_x = E_0 cos(wt-bz) with a frequency of 5.0 GHz and a wavelength in the material medium of 3.0 cm is:
Answer: B. 1.5 x 10^8 m/sec
Explanation: Given f = 5 GHz, wavelength = 3 cm. Phase velocity of plane wave = w/b = f * wavelength = 5 x 10^9 x 3 x 10^-2 = 15 x 10^7 m/sec = 1.5 x 10^8 m/sec.
Topic: Transmission Lines
If the reflected wave at the load of a transmission line is 20 dB below the incident wave, the SWR at the load is:
Answer: B. 1.22
Explanation: Given V_R / V_I = -20 dB = -20 log10 |Gamma|, so |Gamma| = 0.1. SWR = (1 + |Gamma|) / (1 - |Gamma|) = 1.1 / 0.9 = 1.22.
Topic: Transmission Lines
The highest frequency for which a circular coaxial transmission line having outer diameter = 3.1 mm and inner diameter = 1.3 mm can be operated in pure TEM mode (assuming free space medium between the two conductors) should be less than:
Answer: D. 43.4 GHz
Explanation: Explanation not provided in source.
Topic: Antennas and Wave Propagation
In a communication system at 300 MHz, the receiving antenna gain is 8 dBi, the transmitting antenna gain is 10 dBi and the transmitting power level is 25 watts, the distance between the transmitter and the receiver is 1 km, the power received at the receiving antenna port (assuming the propagation medium is loss-less):
Answer: B. 9.97 uW
Explanation: Given f = 300 MHz, G_r = 8 dBi, G_t = 10 dBi, P_T = 25 watts, d = 1 km. Power Received P_r = P_T * G_t * G_r * (lambda / (4 * pi * d))^2 = 25 x 10^0.8 x 10^1 x (1 / (4 * pi * 1000))^2 = 9.97 x 10^-6 W = 9.97 uW.
Topic: Waveguides
A rectangular waveguide with air medium has dimensions a = 22.86 mm and b = 10.6 mm is fed by 3 GHz carrier from a coaxial cable, which of the following is false statement for TE10 mode?
Answer: A. Propagating mode
Explanation: a = 22.86 mm, b = 10.6 mm, f = 3 GHz. As a > b, TE10 is dominant mode. Cut-off frequency for TE10 mode = c / (2a) = (3 x 10^8) / (2 x 22.86 x 10^-3) = 13.12 GHz. As input applied frequency is less than cut-off frequency, the mode is non-propagating.
Topic: Waveguides
TM_01 mode in rectangular perfect metallic waveguide is:
Answer: C. Non-existent mode
Explanation: TM_mn modes for either m = 0 or n = 0 is a non-existing mode for rectangular waveguide.
Topic: Antennas and Wave Propagation
The directivity of a prime-focal parabolic reflector antenna is 30 dBi. If the efficiency of the antenna is 50%, the gain of the antenna is:
Answer: A. 27 dBi
Explanation: Directivity D = 30 dBi, efficiency n = 0.5 (-3 dB). Antenna gain = Directivity + efficiency(dB) = 30 - 3 = 27 dB.
Topic: Electromagnetic Theory
Ratio of skin depths of an Electromagnetic wave inside the conductor for the corresponding frequencies at 4 GHz and 9 GHz, considering the same material properties for both the frequencies is:
Answer: C. 3 : 2
Explanation: Skin depth delta = 1 / $\sqrt{pi * f * mu * sigma}$, so delta is inversely proportional to $\sqrt{f}$. delta(4 GHz) / delta(9 GHz) = $\sqrt{9 / 4}$ = 3 / 2.
Topic: Antennas and Wave Propagation
A Cassegrain reflector antenna has a sub-reflector whose geometrical shape is:
Answer: B. Hyperboloid
Explanation: Sub-reflector of Cassegrain antenna has a hyperboloid shape.
Topic: Antennas and Wave Propagation
The beam-width and the directivity of an antenna are:
Answer: B. Inversely proportional
Explanation: Directivity = 41253 / (theta_1 * theta_2), where theta_1 and theta_2 are beam-width in E, H plane. So directivity is inversely proportional to beam-width.
Topic: Transmission Lines
A lossless transmission line with characteristic impedance Z_0 = 50 ohm is 30 m long and operates at 2 MHz. The line is shorted at the load, if the phase velocity = 0.6 times the velocity of light, the input impedance of the line is:
Explanation: Explanation not provided in source.
Topic: Microwave Engineering
Two microwave sub-systems having respective return loss of 10 dB and 20 dB are cascaded together, the return loss at the input of the integrated system is:
Answer: A. 9.63 dB
Explanation: Explanation not provided in source.
Topic: Antennas and Wave Propagation
A parabolic reflector antenna used for reception of data is 1 meter in diameter and operates at 30 GHz. The far field distance of this antenna for a quadratic phase error of 22.5 degree is:
Answer: A. 200 meter
Explanation: Diameter = 1 m, f = 30 GHz, phase error = 22.5 degree. Far field = 2 * D² / lambda = 2*(1)^2 / 0.01 = 200 m (lambda = c/f = 0.01 m).
Topic: Electromagnetics
A potential field is given by Φ = 2xy^2 - 3y^2 z. If x̂, ŷ, ẑ are the unit vectors along x, y and z directions respectively, the field intensity at (0, 1, 0) is:
Answer: C. -2x̂ + 3ẑ
Explanation: E = -∇Φ = -(2y^2)x̂ - (4xy - 6yz)ŷ - (-3y^2)ẑ. E(0, 1, 0) = -2x̂ + 3ẑ.
Topic: Microwave Engineering
The VSWR of a microwave unit is 1.5, the return loss is:
Answer: A. 14 dB
Explanation: VSWR = 1.5. |Γ| = (VSWR - 1)/(VSWR + 1) = 0.5/2.5 = 1/5. Return loss = -20 log10 |Γ| = 20 log10 5 ≈ 14 dB.
Topic: Microwave Engineering
In two wire transmission line, two consecutive voltage minima are found at 20.6 cm and 25.6 cm. The operating frequency is:
Answer: B. 3 GHz
Explanation: d1 = 20.6 cm, d2 = 25.6 cm. Distance between two successive minima = λ/2 = d2 - d1. So, λ = 2(25.6 - 20.6) = 10 cm. Operating frequency f = c/λ = (3 x 10^10)/10 = 3 GHz.
Topic: Transmission Lines
A TEM mode transmission line is having distributed circuit parameters as R = 1 ohm/m, L = 200 nH/m, G = 300 µS/m, C = 60 pF, the line is:
Answer: C. Distortion-less
Explanation: LG = 200 x 300 x 10^-9 x 10^-6 = 6 x 10^-11 HS/m². RC = 6 x 10^-11 FΩ/m². As LG = RC, so line is distortionless.
Topic: Electromagnetics
A plane wave E = 100 sin (ωt - 10 x) V/m in a loss less medium with μ = 4 μ0, ε = ε0 strikes another medium with 90 degree angle of incidence having μ = 9 μ0, ε = ε0, the reflection coefficient is:
Answer: D. 0.2
Explanation: η1 = √(μ1/ε1) = 2(120π) Ω, η2 = √(μ2/ε2) = 3(120π) Ω. Reflection coefficient Γ = (η2 - η1)/(η2 + η1) = (3(120π) - 2(120π))/(3(120π) + 2(120π)) = 1/5 = 0.2.
Topic: Digital Circuits
What is the functionality of following digital circuit? A is input data, CLK is system clock and Y is output.
Answer: A. Falling edge detection of input A
Explanation: The output Y = A_n' A_{n-1} when the A_n changes from 1 to 0 i.e. falling edge, A_n' = 0, A_{n-1} = 1. Thus, Y = 1.
Topic: Transmission Lines
A lossless transmission line has the distributed circuit parameters of inductance and capacitance per meter as 625 nH/m and 64 pF/m respectively, the phase constant of the line at 100 MHz is:
Answer: A. 3.97 rad/m
Explanation: Vphase = 1/√(LC) = 1/√(625 x 10^-9 x 64 x 10^-12) = 1.58 x 10^8 m/sec. Phase constant β = 2πf/Vphase = (2π x 10^8)/(1.58 x 10^8) = 3.97 rad/m.
Topic: Network Theory
In the circuit if C1 = 2 µF and C2 = 3 µF, the equivalent capacitance between points A and B is:
Answer: C. 1 µF
Explanation: As C1 = 2 µF; C2 = 3 µF. As three capacitance are in series with each other and parallel to C1, the equivalent combination reduces to 3 capacitances of 3 µF in series, so C' = C/n = 1 µF.
Topic: VLSI Design
If n number of MOSFETs with identical W/L are connected in series, then equivalent W/L is given by:
Answer: B. (W/L)/n
Explanation: For n-MOSFETs in series, overall length = nL and width = W. Therefore, (W/L)_overall = W/(nL) = (1/n)(W/L).
Topic: Digital Circuits
Which resource of FPGA implements combinational logic functionality?
Answer: B. Configurable Logic Block (CLB)
Explanation: The building block of FPGA is configurable logic block (CLB).
Topic: Microprocessors and Microcontrollers
What will be the content of Accumulator, Register-B and Overflow Flag (OV) after execution of following 8051 micro-controller assembly code?
MOV A, #65h
MOV 0F0h, #14h
DIV AB
DIV AB
Answer: B. A = 05h, B = 00h, OV = 0
Explanation: Initially A = 65h (101) and B = 14h (20). First DIV AB yields quotient A = 05h and remainder B = 01h. Second DIV AB divides 05h by 01h, resulting in A = 05h, B = 00h, and OV = 0.
Topic: Signals and Systems
Average value of the waveform is:
Answer: D. 4 V
Explanation: Vavg = (14(1) + (-6)(1)) / 2 = 4 V.
Topic: Analog Electronics
Gain and NF of a single stage amplifier is 10 dB and 3 dB respectively. When two such amplifiers cascaded then gain and NF of cascaded amplifier will be:
Answer: A. Gain : 20 dB, NF : 3.2 dB
Explanation: G1 = G2 = 10 dB = 10, F1 = F2 = 3 dB = 2. Overall noise figure = F1 + (F2 - 1)/G1 = 2 + 1/10 = 2.1. Noise figure (dB) = 10 log10 2.1 = 3.2 dB. Overall gain (dB) = G1(dB) + G2(dB) = 20 dB.
Topic: Digital Electronics
Which type of memory has constrain of minimum operating clock frequency?
Answer: C. Dynamic RAM (DRAM)
Explanation: Dynamic RAM (DRAM) needs continuous refreshing for the operation. There is a minimum clock frequency for refreshing to store data property i.e. constrain on minimum operating frequency.
Topic: Microwave Engineering
Main function of helix structure in a Travelling Wave Tube (TWT) is
Answer: A. To reduce axial velocity of RF field.
Explanation: Main purpose of helix structure in TWT is make axial velocity of RF field equals to electron beam velocity so as to transfer the energy over entire length.
Topic: Analog Electronics
Equivalent circuit of FET is shown in the figure below, where r_g = 100 Ή, C_gs = 0.1 pF and g_m = 0.1 mho. For load resistance r_L = 500 Ή, the voltage gain at 1 GHz is:
Answer: A. 50
Explanation: Voltage gain (A_v) = g_m * r_L = 0.1 * 500 = 50.
Topic: Analog Electronics
Which of the following is true for an NMOS transistor operating in linear region?
Answer: A. V_DS < (V_GS - V_T)
Explanation: For NMOS transistor, V_DS < V_GS - V_T corresponds to MOSFET operating in the linear region.
Topic: Electromagnetics
Magnitude of input impedance of a Η/8 lossless 50 Ή transmission line terminated with 25 Ή is:
Answer: C. 50 Ή
Explanation: For l = Η/8, Βl = ΐ/4. Z_in = Z_o * [(Z_L + j Z_o tan(Βl)) / (Z_o + j Z_L tan(Βl))]. Substituting values gives |Z_in| = 50 Ή.
Topic: Microprocessors
Pipelining technique is used in microprocessor to improve which of the following parameter?
Answer: D. Maximum clock frequency
Explanation: Pipelining approach divides an entire process into small and roughly equal time consumable sub-processes, allowing overall throughput to increase significantly as the clock is applied simultaneously to all blocks, thereby increasing maximum clock frequency.
Topic: Microwave Engineering
In case of an ideal Class-F microwave power amplifier, time domain voltage and current waveform of the device have:
Answer: C. No overlap
Explanation: In class F power amplifiers, voltage is a square wave while current is a half sinusoid, which shows zero correlation i.e. no overlap.
Topic: Analog Electronics
Reason for superior high frequency performance of metal-semiconductor diode compared to p-n junction diode is:
Answer: B. No minority carrier storage effect in metal-semiconductor diode
Explanation: Metal semiconductor diodes do not have minority carriers, so reverse recovery time is less. Thus they can be used for switching at high frequencies.
Topic: Microwave Engineering
Which of the following diodes is most suitable for detection of microwave signal?
Answer: B. Schottky barrier diode
Explanation: Schottky barrier diodes (due to metal-semiconductor junction) are most suitable for measurement/detection at microwave frequencies.
Topic: Microwave Engineering
Which device is suitable for higher order RF frequency multiplier?
Answer: D. Step recovery diode
Explanation: Step recovery diodes are used for higher order RF frequency multipliers.
Topic: Microwave Engineering
Input reflection coefficient 'S_11' of the 2-port network for 50 Ή system is:
Answer: C. 1/3 ∠180
Explanation: Normalized impedance Z = 1 / Z_o * Z_parameter. Computing S = (Z - U)(Z + U)^-1 gives S_11 = -1/3 = 1/3 ∠180 .
Topic: Microwave Engineering
The graph below shows operating load line and I-V characteristic of a Schottky diode at two different RF power levels. Under this bias condition RF resistance of the diode:
Answer: C. Increases with the increase of RF power level
Explanation: From the graph, as change of voltage for change in current from maximum to minimum value is large for large power. Hence resistance increases with increase of RF power level.
Topic: Analog Electronics
A BJT is having common emitter current gain 100. Considering 10 V supply and V_BE = 0.7 V what will be the value of R_C and R_B to set the quiescent point at I_C = 10 mA and V_CE = 8 V?
Answer: A. R_C = 200 Ή, R_B = 93 kΉ
Explanation: R_C = (V_CC - V_CE) / I_C = (10 - 8) / 10 mA = 0.2 kΉ = 200 Ή. I_B = I_C / Β = 10 mA / 100 = 0.1 mA. R_B = (V_CC - V_BE) / I_B = (10 - 0.7) / 0.1 mA = 93 kΉ.
Topic: Electromagnetics
Which statement is false about microstripline over stripline?
Answer: A. Less radiative
Explanation: Explanation not provided in source.
Topic: Microwave Engineering
A high gain MESFET packaged device of case-to-channel thermal resistance of the device is 4.5 C/Watt provides 8 W RF output power taking 20 W DC power. What will be the channel temperature of the device if the case temperature of the device maintained at 55 C?
Answer: A. 109 C
Explanation: Θ_{case to channel} = (T_channel - T_case) / (P_in - P_out). 4.5 = (T_channel - 55) / (20 - 8) => T_channel = 55 + 54 = 109 C.
Topic: Analog Electronics
For a class-A FET power amplifier with 10 V drain supply and 2 A drain current bias providing RF load current of 1 A amplitude. What is the DC to RF efficiency for load resistance of 5 Ή?
Answer: D. 12.5%
Explanation: %Η = ( (1 / √2)^2 * 5 / (10 * 2) ) * 100 = 12.5%.
Topic: Analog Electronics
DC to RF efficiency of an ideal class-F amplifier is:
Answer: A. 100%
Explanation: DC to RF efficiency of class F amplifier is 100% which is achieved by using an infinite number of harmonics to yield square and half-wave sinusoid waveform shapes at device drain for voltage and current respectively.
Topic: Analog Electronics
A unilateral transistor has an output impedance Z_out = (10 - j10) Ω. Value of the series and shunt components of the matching network for complex conjugate match at the output of the device to 50 Ω load are:
Answer: A. X_1 = -j25, X_2 = +j30
Explanation: Z_out = (10 - j10) Ω, Z_L = 50 Ω. In order to match the network for complex conjugate at the output of device to 50 Ω: (Z_out + X_2) || X_1 = 50 Ω* = 50 Ω. This condition is fulfilled with only X_1 = -j25 and X_2 = j30 among various options.
Topic: Communication Systems
A communication channel is having a bandwidth of 3000 Hz. The transmitted power is such that the received Signal-to-Noise ratio is 1023. The maximum data rate that can be transmitted error-free through the channel is:
Answer: B. 30 kbps
Explanation: BW = 3000 Hz, SNR = 1023. Channel capacity C = B log_2(1 + SNR) = 3000 log_2(1024) = 30 kbps. From Shannon-Hartley law C ≥ R, so maximum data rate for error-free transmission is 30 kbps.
Topic: Communication Systems
A CDMA system requires E_b / I_0 of 10 dB. Each of the transmitters in the network transmits data at the rate of 1 Kbps with a PN chip rate of 1 Mbps. The maximum number of such users that the network can support is:
Answer: D. 100
Explanation: Explanation not provided in source.
Topic: Signals and Systems
Laplace transform of $e^{θt} \sin(ωt)$ is:
Answer: B. ω / ((S - θ)^2 + ω^2)
Explanation: L[sin ωt] = ω / (s² + ω^2). Using frequency shifting property, L[e^(θt) sin ωt] = ω / ((s - θ)^2 + ω^2).
Topic: Network Theory
Under steady state condition, the energy stored in the circuit is:
Answer: C. 2.88 × 10^-4 J
Explanation: At steady state, capacitors act as open circuits. V_1 = V_2 = V_AD = 8 V. Energy stored = 1/2 C_1 V_1^2 + 1/2 C_2 V_2^2. With C_1 = C_2 = 2 μF, energy stored = 2 × (1/2 × 2 × 10^-6 × 8^2) = 2.88 × 10^-4 J.